When developing power circuits for static converters, measures to protect power transistors from thermal runaway are paramount. Since MOSFET field-effect transistors do not have secondary breakdown, thermal calculations can be based on the maximum temperature and maximum power dissipation values. The total power released by the transistor in its switching mode is determined from the expression:

where R p is the total power dissipation;

R per - power loss during switching;

R pr - losses on the active resistance of the open transistor channel;

Pynp - control losses in the gate circuit;

Pyr - power loss due to leakage in the closed state.

where L L(op) is the resistance of the transistor in the open state (reference parameter).

Conduction losses P pr are the main component of losses in a field-effect transistor. These losses can be calculated by knowing the effective (rms) value of the drain current:

Power losses caused by leakage current (P^) are negligible (if, of course, the transistor is working), so there is no point in taking them into account at all. In addition, since one of the main advantages of a field-effect transistor is extremely low losses in its control circuit (P control), therefore, the value of control losses can be excluded from the calculations. Taking into account the assumptions made, formula (2.1.7) for calculating total losses takes the following convenient form:

Here it is necessary to make some clarifying digression and remind the reader that the calculation of power dissipation is performed in order to ensure the thermal conditions of power transistors. This calculation will be useful when designing cooling radiators for transistors (for details, you can refer to publications and). A very important parameter, without which it will not be possible to design a cooling element, is the so-called “chip-to-case” thermal resistance R thjc of the transistor. Research has shown that this resistance largely depends on the switching frequency of the transistor, as well as on the duty cycle of control pulses, determined by the ratio of the open state time to the full switching period. In the technical specifications for transistors, the so-called normalized transient thermal impedance junction-to-case characteristics are usually given. As can be seen from Fig. 2.1.11, due to the inertia of thermal processes at high switching frequencies and low duty cycle, the thermal resistance of the “crystal-case” is significantly reduced. In any case, the developer needs to estimate this resistance according to the schedule, so as not to design the radiator for cooling the power elements “by eye.” The reader should be aware that those shown in Fig. 2.1.11 graphs are included in the main set of parameters presented by manufacturing companies for the power element base. If, when choosing an element base, the developer is faced with the fact that these graphs are not in the documentation, it is better not to trust such a manufacturing company and not to use its products in your developments.

Taking into account graphs 2.1.11, the thermal resistance of the “crystal body” is determined by the following formula:

where ZjJJ, D) is the transition coefficient of resistance “crystal body”;

R Q (JC) - thermal resistance “crystal-case” in the mode of large duty cycles of control pulses or at direct current.

In Fig. 2.1.11 there is another curve called single pulse. It is removed for a single (non-repeating) current pulse. This mode of operation is typically used for protection and triggering circuits that operate once. In this case, as a rule, the heat generation is small and the power element does not require a radiator.

But let's return to heat losses. The situation with switching losses is much more complicated. If the load of the field effect transistor is purely

Rice. 2.1.11. Graph of the dependence of the normalized thermal resistance on the frequency and duty cycle of the pulses: a - IRFP250; b - IRJL3103D1; in -FB180SA10

active, switching losses are small and can often be simply ignored. However, active load is a rare case in power converter technology. Much more often, transistors of static converters “work” on loads with a strongly pronounced reactive (inductive-capacitive) component, which is characterized by a mismatch of maximum currents and voltages. In addition, in transistors operating in push-pull circuits (this includes half-bridge, bridge and three-phase circuits), specific reverse recovery losses of opposed diodes occur. We will immediately turn to methods for calculating dynamic losses in push-pull circuits, since it is on their basis that powerful converter technology is built.

In a push-pull circuit, it is necessary to consider the influence of inductance L on the remaining elements of the circuit. It should be remembered that in reality the inductance L is the magnetizing inductance of the primary winding of a high-frequency transformer (if the device being designed is a static converter for powering typical loads), or the inductance of the motor winding (if an adjustable frequency electric drive is being developed).

Let's turn to Fig. 2.1.12 and consider the switching processes occurring in the presented typical circuit. Initially (what

opening of the key element. It is clear that the amplitude of the surges cannot become greater than the supply voltage or ground potential, since the opposed diodes will open and “discharge” the surges to the power source. And yet, if the energy of the oscillatory process is high enough, it may not end by the time the key element is next discovered. Switching when current flows through the reverse diode will lead to a so-called “hard switching” situation, when the power transistor will briefly be in the “through current” mode. To “extinguish” these emissions, an RC circuit with a capacitor and resistor connected in series is connected in parallel to the primary winding of the transformer.

We have just examined the so-called “light” mode of operation of the transistor in push-pull circuits, when control pulses arrive at the gates VT1 and VT2 symmetrically, and at the moment of commutation the currents do not pass through the opposed diodes. It is not difficult to calculate the switching loss power in this case. For each transistor operating in a half-bridge or bridge circuit with a standard transformer load, it can be calculated using the formula

where /^max is the maximum drain current.

There is another case when transistors are forced to operate in a “heavy” switching mode. This case is usually considered in frequency control devices for motors with significant winding inductance. Here, the duration of the open state of the “upper” (VT1) and “lower” (VT2) key elements of the half-bridge and bridge may be unequal: in the extreme case, the opening pulses of one of the power switches disappear altogether. In the case of asymmetry of control pulses, the current in the inductive load does not change its direction, which means that, for example, after turning off the transistor VT2, the current i L (Fig. 2.1.12 c) will flow through its opposed diode. Consequently, turning off transistor VT1 will occur in short-term short circuit mode, since diode VD2 will not be able to instantly restore the locked state. The longer the box diode delays restoring the off state, the more heat will be generated in the transistor. Therefore, to calculate switching losses in “heavy” mode, it is necessary to take into account both the dynamic switching losses of the transistor and the reverse recovery losses of the opposed diodes. The following formula will help you calculate switching losses:

where Q rr is the reverse recovery charge of the boxer diode (reference parameter).

You should also know that the reverse recovery charge of a boxed diode (according to Fig. 2.1.14) slightly depends on the forward current flowing through the diode after turning off the transistor, but is largely determined by the magnitude of the change in the forward current over time at the reverse recovery stage, that is, the magnitude derivative of the current. In practice, this means that slowing down the switching process that causes reverse recovery can reduce the charge, and therefore the energy released. Therefore, in “heavy” switching mode, it is necessary to slow down the process of opening the field-effect transistors. The opening speed can be reduced by limiting the gate current by increasing the gate resistor, as well as bypassing the drain-source junctions of the transistors with RC circuits that limit the switching speed. True, at the same time, switching dynamic switching losses increase.

Rice. 2.1.14. Dependence of the reverse recovery charge of the diode on the speed of the switching process

Quite often in the practice of developing static converters, there are cases when it is necessary to switch a current whose value is higher than the limiting current of a single transistor. And if it turns out to be difficult to choose a more powerful device, you can simply connect several devices in parallel, designed for lower currents. Then the total current will be evenly distributed among the individual transistors. To connect them in parallel, you need to have devices with close threshold voltage values. As a rule, transistors of the same type have very close threshold voltage values, so it is extremely undesirable to choose transistors of different types for parallel operation. Or better yet, take transistors from the same production batch, manufactured under the same conditions.

To ensure uniform heating of the line of transistors, they need to be installed on a common radiator and, if possible, closer to each other. It must also be remembered that through two parallel-connected transistors, you can pass twice the current without reducing the load capacity of single devices, but at the same time, the input capacitance, and therefore the charge of the combined gate, doubles. Accordingly, the control circuit for parallel-connected transistors must be able to provide the specified switching time.

But here too there are some peculiarities, some “tricks”. If you connect the gates of field-effect transistors directly, you can get a very unpleasant “ringing” effect when turned off - influencing each other through the gates, the transistors will randomly open and close, not obeying the control signal. To eliminate “ringing”, it is recommended to put small ferrite tubes on the gate terminals to prevent mutual influence of the gates, as shown in Fig. 2.1.15, a.

This method is very rare today (since the technology for producing ferrite tubes is quite complex). A simpler and more accessible circuit design is shown in Fig. 2.1.15, b,

Rice. 2.1.15. Parallel connection of MOSFET: a - with quenching ferrite tubes; b - with gate resistors

which consists in installing identical resistors with a resistance of tens to hundreds of ohms in the circuits of each gate. The value of gate resistors is usually selected from the ratio:

where Q g is the value of the gate charge for one transistor.

After this, it is necessary to determine the amount of current provided by the transistor gate control device. This current is determined from the condition of the action of voltage U g on parallel connected gate resistors. That is, the value of R g obtained from formula (2.1.13) must be reduced during calculations by as many times as the number of transistors connected in parallel.

Rice. 2.1.16. Option for parallel connection of MOSFET transistors

Transistors VTl...VT4 are installed on a common radiator as close to each other as possible, which ensures their uniform heating. Power buses, which can be made of either printed or solid conductors (for example, copper strip or tinned wire), are connected to the drain and source of all transistors. Gate resistors Rg can be placed above the power busbars. The transistors are secured to the radiator using screws and pressure springs. Sometimes to improve thermal contact between

radiator housings use the following technology: transistors are attached with their heat-removing plates to a common strip of copper (or its alloys), and it, in turn, is screwed to the radiator, previously lubricated at the point of contact with heat-conducting paste. And, of course, it is necessary to ensure electrical insulation of individual groups of transistors in order to avoid short circuits in places where they should not exist according to the electrical circuit.

In Fig. 2.1.17 shows the appearance of a variant of the structural unit of a three-phase controlled bridge, composed of parallel-connected MOSFET transistors, and in Fig. 2.1.18 - electrical diagram for connecting transistors. The radiator has through channels through which it is forced to be blown with air flow.

Published 05/15/2014

The design of the power section usually begins with the selection of keys. The most suitable field-effect transistors for this are MOSFETs. The choice of power transistors is made based on data on the maximum possible current and voltage of the motor supply network.

Selection of power transistors

Transistors must withstand the operating current with some margin. Therefore, field-effect transistors with an operating current of 1.2-2 times the maximum motor current are chosen. The characteristics of field-effect transistors may indicate several current values ​​for different modes. Sometimes they indicate the current that the crystal can withstand Id (Silicon Limited)(it is larger) and the current is limited by the capabilities of the transistor body Id (Package Limited)(it's smaller). For example:

In addition, the current for the pulse mode appears ( Pulsed Drain Current), which is significantly greater (several times) than the maximum possible direct current.

It is necessary to select transistors for direct current, and not pay attention to the parameters indicated for the pulse mode. When selecting a transistor, only the DC current value is taken into account. In this case - 195A.

If it is impossible to select a transistor with the required operating current, several transistors are connected in parallel.

In this case, be sure to use the resistors indicated in the diagram. Their nominal value is units of Ohms, but thanks to them, the transistors connected in parallel open simultaneously. If these resistors are not installed, a situation may arise when one of the transistors opens, but the rest do not yet. During this short time, all the power falls on one transistor and disables it. The determination of the value of these resistors is discussed below. Two transistors connected in parallel can withstand twice the current. 3 – 3 times more. But you should not abuse this and build switches from a large number of small transistors.

The selection of field-effect transistors by voltage is also carried out with a margin of at least 1.3 times. This is done in order to avoid failure of transistors due to voltage surges during switching.

In addition to the above parameters, you should ask about the maximum operating temperature of the transistor and whether it will withstand the required current at this temperature. One of the most important characteristics is the resistance of an open transistor. Its values ​​can reach several milliohms. At first glance, it is very small, but at high currents, significant amounts of heat will be generated on it, which will have to be removed. The power that will heat the transistor in the open state is calculated by the formula:

P=Rds*Id^2

Where:
Rds– open transistor resistance;
Ids– the current that flows through the transistor.

Well, it's a transistor irfp4468pbf If the reference is 2.6 mOhm, then under the hour of transmission of 195 A, 98.865 Watts of heat will be seen. In the case of three-phase bridge circuits, only two keys are open at any given time. So, on two closed transistors, the same amount of heat will be seen (98.865 W each, overall – 197.73 W). All the stinks do not work for the whole hour, but after a while - in pairs, then the skin pair of keys works for 1/3 of an hour. It is correct to say that the heat on all keys will be 197.73 W of heat, and on the skin of the keys (98.865 / 3 = 32.955 W). This is to ensure continuous cooling of the transistors.

So, if the transistor irfp4468pbf has a resistance of 2.6 mOhm, then at a current of 195 A it will generate 98.865 watts of heat. In the case of a three-phase bridge circuit, only two switches are open at any given time. That is, two open transistors will generate the same amount of heat (98.865 W each, for a total of 197.73 W). But they do not work all the time, but in turn - in pairs, that is, each pair of keys works 1/3 of the time. So it is correct to say that in general, 197.73 W of heat will be generated on all keys, and on each of the keys (98.865 / 3 = 32.955 W). Proper cooling of the transistors must be ensured.

But there is one “but”

We have approximately calculated the heat losses that occur during the period when the keys are fully open. However, we must not forget that keys are characterized by such phenomena as transient processes. It is at the moment of switching, when the switch resistance changes from practically zero to almost infinity and vice versa, that the greatest heat generation occurs, which is significantly greater than the losses that occur with open switches.

It is clear that we can load 0.55 Ohm. The live voltage is 100V. When the switches are open, the output voltage is 100/0.55 = 181 A. The transistor turns off and at the current moment its contact drops to 1 Ohm. In one hour after another the flow rate is 100/(1+0.55)=64.5A Do you remember the formula for calculating heat stress? It turns out that this is a very short hour of heat consumption on the transistor (1+0.55)*(64.5^2) = 6448 W. What is significantly lower when the key is unlocked. If the transistor reaches 100 Ohms, it will consume 99.45 W. If the transistor grows up to 1 KOhm, the consumption will be 9.98 W. If the transistor grows up to 10 KOhm, the consumption will be 0.99 W.

Let's imagine that we have a load of 0.55 Ohm. Supply voltage 100V. With the switches fully open, we obtain a current of 100 / 0.55 = 181 A. The transistor closes and at some point its resistance reaches 1 Ohm. At this time, a current of 100 / (1 + 0.55) = 64.5A flows through it. Remember the formula by which thermal power is calculated? It turns out that at this very short moment the heat loss on the transistor is (1 + 0.55) * (64.5 ^ 2) = 6448 W. Which is significantly more than with a public key. When the transistor resistance increases to 100 Ohms, the loss will be 99.45 W. When the transistor resistance increases to 1 kOhm, the loss will be 9.98 W. When the transistor resistance increases to 10 kOhm, the loss will be 0.99 W.

If you create a very powerful cooling system, more heat will be generated in the transistor than it can physically remove from itself (see: Maximum Power Dissipation), it will burn.

So, it is not difficult to understand that the faster the keys switch, the less heat loss, and the lower the temperature of the keys.

The switching speed of the switches is affected by: the gate capacitance of the field-effect transistor, the value of the resistor in the gate circuit, and the power of the switch driver. How effectively the keys will work depends on the correct choice of these elements.

Sometimes people believe that they can increase the power of the regulator only by changing the keys to more powerful ones. This is not entirely true. More powerful transistors have a larger gate capacitance, and this increases the opening time of the transistor, which affects their temperature. This rarely happens, but I had a case where simply replacing transistors with more powerful ones increased their temperature due to the fact that their switching time increased. So, more powerful transistors require more powerful drivers.

MOSFET key drivers

What is a key driver and why is it needed? Why do we need drivers at all? You can turn on field-effect transistors as shown in the diagram:

Yes, in this case bipolar transistors act as drivers. This is also acceptable. There are also circuits where transistors with a P-channel are used as the upper switches, and with an N-channel as the lower ones. That is, two types of transistors are used, which is not always convenient. In addition, high power P-channel transistors are almost impossible to find. Typically, this combination of transistors with different channels is used in low-power controllers to simplify the circuit.

It is much more convenient to use transistors of the same type, usually only N-channel ones, but this requires compliance with some requirements for controlling the upper transistors of the bridge. The gate voltage of the transistors must be applied relative to their sources (Source). In the case of the lower switch, no questions arise, its turn (Source) is connected to the ground and we can safely apply voltage to the gate of the lower transistor relative to the ground. In the case of the upper transistor, everything is somewhat more complicated, since the voltage at its source (Source) changes relative to ground.

Will explain. Let's imagine that the upper transistor is open and current flows through it. In this state, a fairly small voltage drops across the transistor and we can say that the voltage at the Source source of the upper transistor is almost equal to the motor supply voltage. By the way, in order to keep the upper transistor open, you need to apply a voltage to its gate, the voltage at its source (Source) is higher, that is, the motor supply voltage is higher.

If the upper transistor is closed and the lower one is open, then at the source (Source) of the upper transistor the voltage reaches almost zero.

The upper switch driver supplies the gate of the field-effect transistor with the required voltage relative to its sources (Source), and ensures the generation of a voltage greater than the motor supply voltage to control the transistor. This, and not only this, is what MOSFET switch drivers do.

Driver selection and variety

The variety of drivers is quite large. We are interested in drivers that have two inputs for the upper and lower keys (upper and lower key drivers). For example: IR2101, IR2010, IR2106, IR21064, IR2181, IR2110, IR2113 etc. You need to pay attention to the parameter Vgs your transistors. Most drivers are designed for Vgs=20V. If Vgs transistors, the output voltage of the drivers is lower, for example Vgs transistor = 5V, then drivers with an output voltage of 20V will damage such transistors.

Most drivers are powered by a voltage of 10-20V and support input signals of various levels -3.3V, 5V, 15V.

There are drivers for three-phase bridge circuits, for example:
IR3230, IRS2334, IRS2334, IR21363, IR21364, IR21365, IR21368, IRS2336, IRS23364D, IRS2336D, IRS26310DJ, IR2130, IR2131, IR2132, IR2133, IR2135, IR 2136, IRS2330, IRS2330D, IRS2332, IRS2332D, IR2233, IR2235, IR2238Q, IRS26302DJ.
Such key drivers may be the most suitable option. In addition, some three-phase drivers have an additional feature to protect the switches from too much current, etc. Quite an interesting series of drivers IRS233x(D). It provides a wide range of protections, including surge protection, short circuit protection, overload protection, bus undervoltage protection, power undervoltage protection, and crossover protection.

One of the most important indicators of drivers is the maximum output current. Typically from 200mA to 4000mA. It may seem that 4 Amps is too much. But the calculator decides everything. As noted above, the speed of switching keys is a very important thing. The more powerful the driver, the less time is spent switching keys. You can roughly calculate the key switching time using the formula:

ton = Qg*(Rh+R+Rg)/U

Where:
Qg– full charge of the gate of the field-effect transistor;
Rh– internal driver resistance. It is calculated as U/Imax, where U is the driver supply voltage, Imax is the maximum output current. Note that the maximum output current may be different for the upper and lower transistor;
R– resistance of the resistor in the gate circuit;
Rg– internal gate resistance of the transistor;
U– driver supply voltage.

For example, if we use a transistor irfp4468pbf and driver IR2101 with a maximum current of 200mA. And in the gate circuit there is a 20 Ohm resistor, then the switching time of the transistor is:

540*(12/0.2 + 20 + 0.8)/12 = 3636 nS

Replacing the driver with IR2010, with a maximum current of 3A, and a resistor in the gate circuit of 2 ohms, we obtain the following switching time:

540*(12/3+2+0.8)/12 = 306 nS

That is, with the new driver, the switching time has been reduced by more than 10 times. So the thermal losses on the transistors will be significantly reduced.

Calculation of resistors in the gate circuit

I have developed the following rule for myself: the resistance of the resistor in the gate circuit of the field-effect transistor must be no less than the internal resistance of the driver, divided by 3. For example, the driver IR2101 Powered by a voltage of 12V, maximum current – ​​0.25A. Its internal resistance: 12V / 0.25 = 48Ohm. In this case, the resistor in the gate circuit of the field-effect transistor must be greater than 48/3 = 16 Ohm. If the switching time of transistors with the selected resistors is not satisfactory, you should choose a more powerful driver.

I cannot call this technique ideal, but it has been tested in practice. If anyone can clarify this point, I would be grateful.

Sometimes a diode with or without a resistor is added to the gate circuit of a transistor.

Since in many cases power transistors operate with inductive loads, protection diodes must be used. If they are not there, then when the transistor is turned off due to transient processes, an overvoltage will occur on the inductances (motor windings), which in many cases breaks through the transistor and disables it.

Many power transistors already have internal protection diodes and there is no need to use external diodes. But do not forget to check this in the documentation for the transistor.

Dead-Time

Changing the state of the power switches in the controller of a three-phase brushless motor is performed in the following sequence:

• turn off the key that needs to be turned off;
• we wait for some time (Dead-Time) until the transistor closes (we calculated the approximate switching time of the transistor earlier) and the transient processes associated with switching end;
• turn on the key that needs to be turned on.

All upper and lower switch drivers have a delay between output signals to prevent both transistors from opening at the same time (see:). But this delay is too short. Some upper and lower key drivers have real Dead-Time. But in our case, this will not help at all, because if we remember how the keys are switched (see: ), then we will see that there is never a situation when the keys of one arm change states. So, manage Dead-Time must microcontroller. The only exception can be if you use a special three-phase driver that controls all six keys and has a real Dead-Time.

Current sensors

Traditionally, a shunt is used as a current sensor. Knowing its resistance, measure the voltage across it and calculate the current. But for powerful systems, the use of a shunt is not always technically justified due to too large heat losses on it. Hall effect current sensors have virtually zero resistance, so they do not heat up. In addition, as a rule, the power supply and output signal level of such sensors are in the 5V range, which is very convenient for implementing a regulator on microcontrollers. Currently, the company's current sensors are quite popular Allegro MicroSystems, for example series ACS71X, ACS75X.

In addition to the usual measurement of current level with a microcontroller, it is reasonable to create a hardware protection circuit against exceeding a critical current level. The microcontroller takes some time to measure the current level. In addition, the current is measured periodically after some time. Such delays, as well as possible software errors, can create a situation where a critical current manages to damage the device before the next measurement arrives. The circuit must turn off the power switches when the current exceeds a critical value, regardless of the operation of the microcontroller. To implement such a circuit, a comparator is usually used, the input of which is supplied with a signal from the current sensor and a reference signal. When the permissible current is exceeded, the comparator is triggered. The comparator output is used as a discrete signal in logic circuits; the switches are turned off in an emergency. This implementation has the lowest latency.

Introduction

Recently, field-effect transistors, otherwise called channel or unipolar, have become very widespread. The main advantage of a field-effect transistor is its high input resistance, which can be the same as that of electronic tubes, and even more. Currently, bipolar transistors are increasingly being replaced by field-effect transistors.

In this course work, the main electrical parameters of a field-effect transistor were calculated (resistance of a fully open channel, cut-off voltage, gate capacitance) and the transfer characteristic and related parameters were determined (initial drain current, saturation voltage).

Calculation of parameters of a field-effect transistor with a control p-n junction

1) Brief theoretical information about field-effect transistors with a control p-n junction.

A field-effect transistor is a three-electrode semiconductor device in which the current is created by the main charge carriers under the action of a longitudinal electric field, and the current value is controlled by a transverse electric field created by a voltage applied to the control electrode.

A successful model, called a junction transistor, was made in 1950. It consisted of a thin p-type layer sandwiched between two n-type layers, with metal contacts in each layer. This device worked exactly as Shockley predicted. Junction transistors became widely used instead of point-contact types because they were easier to manufacture and performed better. Shockley's early idea, a field-effect transistor, was not realized for a long time because there were no suitable materials available. A working field-effect transistor was built using silicon crystals when methods for growing and purifying crystals were quite advanced.

Like a vacuum tube, transistors allow a small current flowing in one circuit to control a much larger current flowing in another circuit. Transistors quickly replaced radio tubes everywhere except in applications where very high power needs to be controlled, such as in radio broadcasting or in industrial radio frequency heating applications. Bipolar transistors are usually used where high speed is required, as well as in high-frequency applications where there is no urgent need to use vacuum tubes. Field effect transistors are the main type of transistors used in electronic devices. It is easier to manufacture and consumes even less energy than a bipolar transistor.

Although some transistors are still made from germanium, most are made from silicon, which is more resistant to high temperatures. With further development of technology, it has become possible to place up to a million transistors in one piece of silicon, and this number continues to increase. Such silicon blocks serve as the basis for the rapid development of modern computers, communications and control.

It is known that the input resistance of a bipolar transistor depends on the load resistance of the cascade, the resistance of the resistor in the emitter circuit and the base current transfer coefficient. Sometimes it can be relatively small, making it difficult to match the cascade with the input signal source. This problem disappears completely if you use a field-effect transistor - its input resistance reaches tens and even hundreds of megaohms.

By analogy with bipolar field-effect transistors, there are different “structures”: with a p-channel and an n-channel. Unlike bipolar ones, they can be with a gate in the form of a p-n junction (channel, or unipolar, transistors); and with an insulated gate (MDS or MOS transistors).

Figure 1 shows a schematic representation of the design of a field-effect transistor and its connection diagram.

field effect transistor transfer current

The basis of the field-effect transistor is a silicon wafer (gate), in which there is a thin region called a channel (Fig. 1, a). On one side of the channel there is a drain, on the other there is a source. When the positive terminal of the power battery GB2 (Fig. 1, b) is connected to the source of the transistor and to the drain of the negative terminal, an electric current arises in the channel. The channel in this case has maximum conductivity.

As soon as you connect another power source - GB1 - to the source and gate terminals (plus to the gate), the channel “narrows”, causing an increase in resistance in the drain-source circuit. The current in this circuit immediately decreases. By changing the voltage between the gate and source, the drain current is controlled. Moreover, there is no current in the gate circuit; the drain current is controlled by an electric field (that’s why the transistor is called field effect), created by the voltage applied to the source and gate.

The above applies to a transistor with a p-channel, but if the transistor is with an n-channel, the polarity of the supply and control voltages is reversed (Fig. 1, c). Most often you can find a field-effect transistor in a metal case - then, in addition to the three main terminals, it may also have a housing terminal, which during installation is connected to the common wire of the structure.

Thin layer semiconductor type P ( or R), bounded on both sides by electron-hole transitions is called channel. The inclusion of the channel in the electrical circuit is ensured using two ohmic electrodes, one of which (I) is called source, and the second (C) - drain. The pin connected to the p-type areas is the control electrode and is called - shutter ( 3). Pins I, C and 3 correspond (in order of listing) to the cathode, anode and grid of an electric vacuum triode or the emitter, collector and base of a conventional bipolar transistor.

The amount of current in the channel depends on the voltage Uс, applied between drain and source, load resistance and wafer resistance between drain and source. At U c And R n=const current in the channel I c (drain current) depends only on the effective cross-sectional area of ​​the channel. Source e zi creates a negative voltage at the gate, which leads to an increase in thickness r-p transition and reducing the current-carrying cross-section of the channel.

With a decrease in channel cross-section The resistance between source and drain increases and the current decreases I With. A decrease in gate voltage causes a decrease in channel resistance and an increase in current. Iс. Connected in series with e ZI amplified alternating voltage source U input, you can change the current through the channel according to the law of changes in the input voltage. Drain current passing through load resistance R n , creates a voltage drop across it that varies according to the law U input. With appropriate selection of the value R n it is possible to achieve an increase in the output voltage level compared to the input voltage, i.e. strengthen the signal.

Field-effect transistors with an insulated gate have a metal-dielectric (oxide) - semiconductor structure. Therefore they are often called MDP - or MOS transistors. The operating principle of these devices is based on the field effect in the surface layer of the semiconductor.

Figure 2 schematically shows the design of such a transistor. The basis of the device is a plate (substrate) of monocrystalline silicon p-type. The source and drain regions are regions of silicon heavily doped with impurities P-type. The distance between source and drain is approximately 1 µm. In this area there is a narrow, lightly doped strip of l-type silicon (channel). The gate is a metal plate isolated from the channel by a dielectric layer approximately 0.1 microns thick. A silicon dioxide film grown at high temperatures can be used as a dielectric.

Depending on the polarity of the voltage applied to the gate (relative to the source), the channel may become poor or get rich charge carriers (electrons). When the gate voltage is negative, conduction electrons are pushed out of the channel region into the bulk of the substrate semiconductor. In this case, the channel is depleted of charge carriers, which leads to a decrease in the current in the channel. A positive gate voltage encourages conduction electrons to be drawn from the substrate into the channel. In this mode, called the enrichment mode, the channel current increases.

Thus, unlike a field-effect transistor with r-p transitions, an insulated gate transistor can operate with zero, negative, or positive gate voltage.

In Fig. 3, A shows an approximate view of a family of weekends ( stock) current-voltage characteristics of a field-effect transistor with r-p transitions

Let the voltage between gate and source Uzi = 0. When increasing positive voltage Uс current at drain Iс will increase. In the beginning, addiction will be almost linear (section OA in Fig. 3, A). However, with increasing Iс the voltage drop across the channel increases, the reverse bias increases for r-p transitions (especially near the drain), which leads to a narrowing of the cross-section of the current-carrying channel and slows down the growth of current Iс. Ultimately, at the drain end of the plate, the channel narrows so much that a further increase in voltage no longer leads to an increase Iс(plot AB in Fig. 7.24, a). This mode is called saturation mode, and the voltage Uс, at which saturation occurs is called saturation voltage (U c. us.). If we remove the current dependence Iс from voltage Uс for a number of gate voltages (U zi< 0), то получим семейство выходных характеристик полевого транзистора.

Dependency at U c =const was called the drainage characteristic (Fig. 3, b).

The output characteristics of a field-effect transistor with an insulated gate have the same form as the characteristics of a transistor with r-p transitions. The only difference is that transistors with r-p transitions can only operate in the depletion (narrowing) mode of the channel, and transistors of the MOS (or MOS) type operate both in the depletion mode (with negative gate voltages) and in the enrichment mode (with positive gate voltages). For the same reason, the drain characteristic of an insulated gate transistor can cover the region of positive voltages between the gate and source .

The main parameters of field-effect transistors are:

One of the parameters of a field-effect transistor is the initial drain current (I from start), i.e. current in the drain circuit at zero voltage at the transistor gate (in Fig. 4, and the variable resistor slider is in the lower position in the diagram) and at a given supply voltage.

If you smoothly move the resistor slider up the circuit, then as the voltage at the transistor’s gate increases, the drain current decreases (Fig. 4, b) and at a specific voltage for a given transistor it will drop to almost zero. The voltage corresponding to this moment is called the cut-off voltage (U ZIots).

The dependence of the drain current on the gate voltage is quite close to a straight line. If we take an arbitrary increment in the drain current and divide it by the corresponding increment in the voltage between the gate and source, we obtain the third parameter - the slope of the characteristic (S). This parameter is easy to determine without removing the characteristics or searching for it in the directory. It is enough to measure the initial drain current, and then connect, say, a galvanic element with a voltage of 1.5 V between the gate and the source. Subtract the resulting drain current from the initial one and divide the remainder by the element voltage - you get the value of the slope of the characteristic in milliamps per volt.

Knowledge of the features of a field-effect transistor will complement familiarity with its stock output characteristics (Fig. 4, c). They are removed when the voltage between drain and source changes for several fixed gate voltages. It is easy to see that up to a certain voltage between the drain and the source, the output characteristic is nonlinear, and then within significant voltage limits it is almost horizontal.

Of course, a separate power supply is not used in real designs to supply bias voltage to the gate. The bias is formed automatically when a constant resistor of the required resistance is connected to the source circuit.

Input resistance R input between gate and source (determined at the maximum permissible voltage between these electrodes)

Output impedance ( determined in saturation mode)

at U zi = const.

The output resistance is characterized by tangent. the angle of inclination of the output characteristics. In the working area, this angle is close to zero and, therefore, the output resistance turns out to be quite large (hundreds of kilo-ohms).

The channel thickness of the field-effect transistor W in operating condition depends on the SCR r-p transition and in accordance with Fig. 5 is equal. In turn, it depends on the voltage W of the earth r-p transition. Using the expression, you can obtain the dependence of the channel thickness on the gate voltage where is the contact potential difference; - dielectric constant of the semiconductor; - dielectric constant, ; - electron charge, ; N- impurity concentration.

To calculate the cutoff voltage, it is necessary to equate the expression to zero, and we get

The drain-source resistance at U ZI =0 is determined by the expression where L,d,Z- length, thickness and width of the open channel, respectively;

Semiconductor resistivity P-type of electrical conductivity, where is the mobility of electrons in the channel.

This equation corresponds to the saturation point “A” in Fig. 3, a.

The gate capacitance, called the barrier capacitance, represents the charge increment to the voltage increment that caused this change.

where is the area r-p transition.

In addition to those mentioned above, field-effect transistors are characterized by a number of other maximum permissible parameters that determine the maximum operating modes of the device.

The most important advantages of field-effect transistors include:

1. High input impedance, reaching in channel transistors with r-p transitions of magnitude 10 6 - 10 a Ohm, and in transistors with an insulated gate 10 13 - 10 13 Ohm. Such a high value of input resistance is explained by the fact that in transistors with r-p transitions, the electron-hole junction between gate and source is switched in the opposite direction, and in insulated gate transistors the input resistance is determined by the very large leakage resistance of the dielectric layer.

2. Low level of own noise, since in field-effect transistors, unlike bipolar ones, charges of only one sign are involved in current transfer, which eliminates the appearance of recombination noise.

3. High - resistance to temperature and radioactive influences.

4. High density of elements when using devices in integrated circuits.

Field-effect transistors can be used in circuits of amplifiers, generators, and switches. They are especially widely used in low-noise amplifiers with high input impedance. Their use (with an insulated gate) in digital and logic circuits is also very promising.

2) Calculation task

1. Structure: field effect transistor with control p - p silicon-based junction with channel P - type of electrical conductivity and two gates (Fig.)

2. Geometric dimensions of the channel: thickness d= 1 µm, width Z= 500 µm, length L= 25 µm.

3. Electrical parameters: donor impurity concentration in the channel N D= 6 10 15 cm -3, concentration of acceptor impurity in the p - regions of the gates N a= 1 10 18 cm -3

DEFINE

1. Basic electrical parameters: fully open channel resistance R CI open, cut-off voltage U ZI open, gate capacitance C ZI, maximum operating frequency

2. Transfer characteristic and related parameters: initial drain current I C start, saturation voltage U CI us, the slope of the transmission characteristic.

3) Calculation procedure

1. Determine the main electrical parameters.

2. Resistance of a completely open channel at U SI = 0 and U SI = 0 We find using the expression resistivity of the source material we find from a given concentration of donor impurity in the channel using the graph Fig. 6

The cutoff voltage is determined by the formula

Intrinsic concentration of charge carriers. In silicon, the intrinsic concentration of charge carriers is equal. - Boltzmann constant 1.38 10 -23 J K -1 = 300 TO- electron charge, = 1.6 10 -19 C. Then it will be equal

Dielectric constant of silicon =12

We calculate the gate capacitance as the barrier capacitance r-p transition at a gate voltage U з = 0 for an abrupt transition, which is true in the case of shallow diffusion, when the impurity concentration gradient in r-p transition is large, or in the case of alloy gate formation technology, we get

The maximum (operating) frequency can be found using the formula

The driver is a power amplifier and is intended to directly control the power switch (sometimes keys) of the converter. It must amplify the control signal in terms of power and voltage and, if necessary, provide its potential shift.

When choosing a driver, it is necessary to match its output parameters with the input parameters of a powerful switch (MOSFET transistor, IGBT).

1. MOS transistors and IGBTs are voltage-controlled devices, however, to increase the input voltage to the optimal level (12-15 V), it is necessary to provide an appropriate charge in the gate circuit.

3. To limit the rate of rise of current and reduce dynamic noise, it is necessary to use series resistances in the gate circuit.

Drivers for controlling complex conversion circuits contain a large number of elements, so they are produced in the form of integrated circuits. These microcircuits, in addition to power amplifiers, also contain level conversion circuits, auxiliary logic, delay circuits for forming “dead” time, as well as a number of protections, for example, against overcurrent and short circuit, undervoltage and a number of others. Many companies produce a large functional range: lower switch bridge circuit drivers, upper bridge circuit drivers, upper and lower switch drivers with independent control of each of them, half bridge drivers, which often have only one control input and can be used for a symmetrical control law, drivers to control all transistors in the bridge circuit.

A typical circuit for connecting the driver of the upper and lower keys from International Rectifier IR2110 with the bootstrap power supply principle is shown in Fig. 3.1, a. Both keys are controlled independently. The difference between this driver and others is that the IR2110 has an additional level conversion circuit in both the lower and upper channels, which allows you to separate the power supply of the microcircuit logic from the driver supply voltage by level. It also contains protection against low voltage supply to the driver and high-voltage “floating” source.

Capacitors C D, C C are designed to suppress high-frequency interference in the logic and driver power circuits, respectively. The high-voltage floating source is formed by capacitor C1 and diode VD1 (bootstrap power supply).

The driver outputs are connected to power transistors using gate resistors R G1 and R G2.

Since the driver is built on field elements and the total power spent on control is insignificant, capacitor C1 can be used as a power source for the output stage, recharged from the power supply U PIT through the high-frequency diode VD1. Capacitor C1 and diode VD1 together form a high-voltage “floating” power supply designed to control the upper transistor VT1 of the bridge stand. When the lower transistor VT2 conducts current, the source of the upper transistor VT1 is connected to the common power wire, the diode VD1 opens and the capacitor C1 is charged to voltage U C1 = U PIT - U VD1. On the contrary, when the lower transistor goes into the closed state and the upper transistor VT1 begins to open (Figure 3.1), the diode VD1 is supported by the reverse voltage of the power supply. As a result of this, the driver output stage begins to be powered exclusively by the discharge current of capacitor C1. Thus, capacitor C1 constantly “walks” between the common wire of the circuit and the wire of the power supply (point 1).

When using the IR2110 driver with bootstrap power, special attention should be paid to the selection of elements of the high-voltage “floating” source. Diode VD1 must withstand a high reverse voltage (depending on the power supply of the circuit), a permissible forward current of approximately 1 A, a recovery time t rr = 10-100 ns, i.e. be fast. The literature recommends the SF28 diode (600 V, 2 A, 35 ns), as well as diodes UF 4004...UF 4007, UF 5404...UF 5408, HER 105... HER 108, HER 205...HER 208 and other “ultra - fast” classes .

The driver circuit is designed in such a way that a high logical signal level at any input HIN and LIN corresponds to the same level at its output HO and LO (see Fig. 3.1 b, common-mode driver). The appearance of a high level logic signal at the SD input leads to the blocking of the transistors of the bridge rack.

It is advisable to use this microcircuit to control inverter switches with PWM output voltage regulation. It must be remembered that in the control system it is necessary to provide time delays (“dead” time) in order to prevent through currents when switching bridge rack transistors (VT1, VT2 and VT3,VT4, Fig. 1.1).

Capacity C1 is a bootstrap capacity, the minimum value of which can be calculated using the formula:

Where Q 3– value of the gate charge of a powerful switch (reference value);

I pit– driver consumption current in static mode (reference value, usually I pit≈ I G c t powerful key);

Q 1– cyclic change in driver charge (for 500-600-volt drivers 5 nK);

V p– supply voltage of the driver circuit;

– voltage drop across the bootstrap diode VD1;

T– switching period of powerful keys.

Fig.3.1. Typical circuit diagram for switching on the IR2110 driver (a) and timing diagrams of its signals at the inputs and outputs (b)

V DD – microcircuit logic power supply;

V SS – common point of the logical part of the driver;

HIN, LIN – logical input signals that control the upper and lower transistors, respectively;

SD – logical input to disable the driver;

V CC – driver supply voltage;

COM – negative pole of the power supply V CC;

HO, LO – driver output signals that control the upper and lower transistors, respectively;

V B – supply voltage of the high-voltage “floating” source;

V S is the common point of the negative pole of the high-voltage “floating” source.

The resulting value of the bootstrap capacitance must be increased by 10-15 times (usually C within 0.1-1 μF). This should be a high-frequency capacitance with low leakage current (ideally tantalum).

Resistors RG 1, R G 2 determine the turn-on time of powerful transistors, and diodes VD G 1 and VD G 2, bypassing these resistors, reduce the turn-off time to minimum values. Resistors R1, R2 have a small value (up to 0.5 Ohm) and equalize the spread of ohmic resistance along the common control bus (required if a powerful switch is a parallel connection of less powerful transistors).

When choosing a driver for high-power transistors, you must consider:

1. Law of control of powerful transistors:

For symmetrical law, high and low switch drivers and half bridge drivers are suitable;

Single-ended law requires upper and lower key drivers with independent control of each powerful key. Drivers with transformer galvanic isolation are not suitable for an asymmetrical law.

2. Parameters of a powerful key (I to or I drain).

An approximate approach is usually used:

I out dr max =2 A can control powerful VT with current up to 50 A;

I out dr max =3 A – control a powerful VT with a current of up to 150 A (otherwise the on and off time increases significantly and the power losses for switching increase), i.e. If a high-quality transistor is chosen incorrectly, it loses its main advantages.

3. Accounting for additional functions.

Companies produce drivers with numerous service functions:

Various powerful key protection;

Driver undervoltage protection;

With built-in bootstrap diodes;

With adjustable and non-adjustable delay time for turning on a powerful VT in relation to the moment of turning off the other (combat through currents in the half-bridge);

With or without built-in galvanic isolation. In the latter case, a galvanic isolation microcircuit (most often a high-frequency diode optocoupler) must be connected to the driver input;

In-phase or anti-phase;

Driver power supply (bootstrap power supply or three galvanically isolated power supplies are required).

If several types of drivers are equivalent, preference should be given to those that switch the gate current of powerful transistors using bipolar VTs. If this function is performed by field-effect transistors, then there may be failures in the driver’s operation under certain circumstances (overloads) due to the “latching” trigger effect.

After selecting the type of driver (and its data), measures are necessary to combat through currents in the half-bridge. The standard method is to turn off a powerful key instantly, and turn on a locked one with a delay. For this purpose, diodes VD G 1 and VD G 2 are used, which, when closing VT, bypass the gate resistors, and the shutdown process will be faster than unlocking.

In addition to shunting the gate resistors R G 1 and R G 2 using diodes (VD G 1, VD G 2, Fig. 3.1) to combat through currents in the P-circuit of a powerful cascade, companies produce integrated drivers that are asymmetric in the output switching current VT I other out m akh on on and off I other out m ah off(For example I other out m akh on=2A, I other out m ah off=3A). This sets the asymmetric output resistances of the microcircuit, which are connected in series with gate resistors R G 1 and R G 2.

where all values ​​in the formulas are reference data for a specific driver.

For a symmetrical (current) driver, the following equality is true:

.

There are three capacitances in the MOSFET transistor structure: gate-source capacitance (input capacitance), source-drain capacitance (output capacitance), gate-drain capacitance (pass-through For the IGRT transistor, respectively, , . When a voltage of (15-20) V is applied to the gate, it starts the input capacitance will charge exponentially and at a voltage of 8-10 V a current will appear in the transistor... This period of time is given in the form of the turn-on delay parameter (Fig. 3.2) at a certain resistance in the gate circuit

When a drain current appears in the VT structure, the input capacitance will be charged at a different exponential rate, since this process is influenced by the output capacitance, then ultimately the input capacitance will accumulate charge Q (reference value). The output current (voltage reduction at the source-drain electrodes) will mainly depend on the processes in the circuit, without a significant influence of the gate current.

The capacity discharge time is also given in the VT reference parameters in the form of on time.

When the transistor is turned off, the capacitance will first discharge to a value (), then the source current will begin to decrease to 0 (). Thus, the delay for turning on and off VT will depend on the value of the resistor in the gate circuit, and with the use of a driver, the total resistance in the gate circuit will have two components: (with an unbalanced driver and ) - const and an additional gate resistor, which can be changed for adjustments delays Figure 3.2 presents the above arguments in the form of simplified graphs.

Rice. 3.2. Timing diagrams: (a) - when VT is turned on; (b) - when VT is turned off.

The reference data does not provide the parameters of the input and output capacitances of the transistor, but it is known from mathematics that the initial portion of the exponential (up to 0.7) is approximated by a straight line, the angle of inclination of which is directly proportional to RC, which allows estimation calculations in the form of proportions.

So, to prevent the occurrence of through currents, it is necessary to select the total value of resistance in the gate circuit ( , and regulates the charging rate of the gate capacitance VT) to provide a delay in turning on the transistor greater than or equal to the time spent closing VT (see Fig. 3.2).

(3.1)

where is the decay time of the drain current (reference value);

– delay time of the beginning of VT switching off relative to the moment the blocking voltage is applied to the gate. With shunt gate diodes (VD G 1, VD G 2, Fig. 3.1), the discharge rate is uniquely determined by the resistance . Therefore, to determine the following proportion is solved (assuming that it will be shunted by the diode VD G)

MOP (in bourgeois MOSFET) stands for Metal-Oxide-Semiconductor, from this abbreviation the structure of this transistor becomes clear.

If on the fingers, then it has a semiconductor channel that serves as one plate of the capacitor and the second plate is a metal electrode located through a thin layer of silicon oxide, which is a dielectric. When voltage is applied to the gate, this capacitor is charged, and the electric field of the gate pulls charges to the channel, as a result of which mobile charges appear in the channel that can form an electric current and the drain-source resistance drops sharply. The higher the voltage, the more charges and lower the resistance, as a result, the resistance can drop to tiny values ​​- hundredths of an ohm, and if you raise the voltage further, a breakdown of the oxide layer and the Khan transistor will occur.

The advantage of such a transistor, compared to a bipolar one, is obvious - voltage must be applied to the gate, but since it is a dielectric, the current will be zero, which means the required the power to control this transistor will be scanty, in fact, it only consumes at the moment of switching, when the capacitor is charging and discharging.

The disadvantage arises from its capacitive property - the presence of capacitance on the gate requires a large charging current when opening. In theory, equal to infinity on infinitely small periods of time. And if the current is limited by a resistor, then the capacitor will charge slowly - there is no escape from the time constant of the RC circuit.

MOS transistors are P and N duct. They have the same principle, the only difference is the polarity of the current carriers in the channel. Accordingly, in different directions of the control voltage and inclusion in the circuit. Very often transistors are made in the form of complementary pairs. That is, there are two models with exactly the same characteristics, but one of them is N channel, and the other is P channel. Their markings, as a rule, differ by one digit.

My most popular MOP transistors are IRF630(n channel) and IRF9630(p channel) at one time I made about a dozen of them of each type. Possessing a not very large body TO-92 this transistor can famously pull through itself up to 9A. Its open resistance is only 0.35 Ohm.
However, this is a fairly old transistor; now there are cooler things, for example IRF7314, capable of carrying the same 9A, but at the same time it fits into an SO8 case - the size of a notebook square.

One of the docking problems MOSFET transistor and microcontroller (or digital circuit) is that in order to fully open until completely saturated, this transistor needs to drive quite a bit more voltage onto the gate. Usually this is about 10 volts, and the MK can output a maximum of 5.
There are three options:

But in general, it is more correct to install a driver, because in addition to the main functions of generating control signals, it also provides current protection, protection against breakdown, overvoltage, as an additional bauble, optimizes the opening speed to the maximum, in general, it does not consume its current in vain.

Choosing a transistor is also not very difficult, especially if you don’t bother with limiting modes. First of all, you should be concerned about the value of the drain current - I Drain or I D you choose a transistor based on the maximum current for your load, preferably with a margin of 10 percent. The next important parameter for you is VGS- Source-Gate saturation voltage or, more simply, control voltage. Sometimes it is written, but more often you have to look at the charts. Looking for a graph of the output characteristic Dependency I D from VDS at different values VGS. And you figure out what kind of regime you will have.

For example, you need to power the engine at 12 volts, with a current of 8A. You screwed up the driver and only have a 5 volt control signal. The first thing that came to mind after this article was IRF630. The current is suitable with a margin of 9A versus the required 8. But let’s look at the output characteristic:

If you are going to use PWM on this switch, then you need to inquire about the opening and closing times of the transistor, choose the largest one and, relative to the time, calculate the maximum frequency of which it is capable. This quantity is called Switch Delay or t on,t off, in general, something like this. Well, the frequency is 1/t. It’s also a good idea to look at the gate capacity C iss Based on it, as well as the limiting resistor in the gate circuit, you can calculate the charging time constant of the RC gate circuit and estimate the performance. If the time constant is greater than the PWM period, then the transistor will not open/close, but will hang in some intermediate state, since the voltage at its gate will be integrated by this RC circuit into a constant voltage.

When handling these transistors, keep in mind the fact that They are not just afraid of static electricity, but VERY STRONG. It is more than possible to penetrate the shutter with a static charge. So how did I buy it? immediately into foil and don’t take it out until you seal it. First ground yourself to the battery and put on a foil hat :).