Composite Darlington transistor operation and device. Standard TTL series How a multivibrator works

The basic logical element of the series is the AND-NOT logical element. In Fig. Figure 2.3 shows diagrams of the three initial NAND TTL elements. All circuits contain three main stages: transistor input VT1, implementing the logical AND function; phase separating transistor VT2 and a push-pull output stage.

Fig 2.3.a. Schematic diagram of the basic element of the K131 series

The operating principle of the logical element of the K131 series (Fig. 2.3.a) is as follows: when a low-level signal (0 - 0.4V) is received at any of the inputs, the base-emitter junction of the multi-emitter transistor VT1 is forward-biased (unlocked), and almost the entire the current flowing through resistor R1 is branched to ground, as a result of which VT2 closes and operates in cutoff mode. The current flowing through resistor R2 saturates the base of transistor VT3. Transistors VT3 and VT4 connected according to the Darlington circuit form a composite transistor, which is an emitter follower. It functions as an output stage to amplify the signal power. A high logic level signal is generated at the output of the circuit.

If a high-level signal is supplied to all inputs, the base-emitter junction of the multi-emitter transistor VT1 is in closed mode. The current flowing through resistor R1 saturates the base of transistor VT1, as a result of which transistor VT5 is unlocked and a logical zero level is set at the output of the circuit.

Since at the moment of switching transistors VT4 and VT5 are open and a large current flows through them, a limiting resistor R5 is introduced into the circuit.

VT2, R2 and R3 form a phase separating cascade. It is necessary to turn on the output n-p-n transistors one by one. The cascade has two outputs: collector and emitter, the signals on which are antiphase.

Diodes VD1 - VD3 are protection against negative impulses.

Fig 2.3.b, c. Schematic diagrams of the basic elements of the K155 and K134 series

In microcircuits of the K155 and K134 series, the output stage is built on a non-composite repeater (only a transistor VT3) and a saturable transistor VT5 with the introduction of a level shift diode VD4(Fig. 2.3, b, c). The last two stages form a complex inverter that implements the logical NOT operation. If you introduce two phase separating stages, then the OR-NOT function is implemented.

In Fig. 2.3, and shows the basic logical element of the K131 series (foreign analogue - 74N). The basic element of the K155 series (foreign analogue - 74) is shown in Fig. 2.3, b, a in Fig. 2.3, c - element of the K134 series (foreign analogue - 74L). Now these series are practically not developed.

TTL microcircuits of the initial development began to be actively replaced by TTLSh microcircuits, which have junctions with a Schottky barrier in their internal structure. The Schottky junction transistor (Schottky transistor) is based on the well-known circuit of an unsaturated transistor switch (Fig. 2.4.a).

Figure 2.4. Explanation of the principle of obtaining a structure with a Schottky transition:
a - unsaturated transistor switch; b - transistor with a Schottky diode; c - symbol of the Schottky transistor.

To prevent the transistor from entering saturation, a diode is connected between the collector and the base. The use of a feedback diode to eliminate transistor saturation was first proposed by B. N. Kononov. However, in this case it can increase to 1 V. The ideal diode is a Schottky barrier diode. It is a contact formed between a metal and a lightly doped n-semiconductor. In a metal, only some of the electrons are free (those outside the valence zone). In a semiconductor, free electrons exist at the conduction boundary created by the addition of impurity atoms. In the absence of bias voltage, the number of electrons crossing the barrier on both sides is the same, i.e., there is no current. When forward biased, electrons have the energy to cross the potential barrier and pass into the metal. As the bias voltage increases, the barrier width decreases and the forward current increases rapidly.

When reverse biased, electrons in a semiconductor require more energy to overcome the potential barrier. For electrons in a metal, the potential barrier does not depend on the bias voltage, so a small reverse current flows, which remains practically constant until an avalanche breakdown occurs.

The current in Schottky diodes is determined by the majority carriers, so it is greater at the same forward bias and, therefore, the forward voltage drop across the Schottky diode is less than at a conventional p-n junction at a given current. Thus, the Schottky diode has a threshold opening voltage of the order of (0.2-0.3) V, in contrast to the threshold voltage of a conventional silicon diode of 0.7 V, and significantly reduces the lifetime of minority carriers in the semiconductor.

In the diagram of Fig. 2.4, b transistor VT1 is kept from going into saturation by a Shatky diode with a low opening threshold (0.2...0.3) V, so the voltage will increase slightly compared to a saturated transistor VT1. In Fig. 2.4, c shows a circuit with a “Schottky transistor”. Based on Schottky transistors, microcircuits of two main TTLSh series were produced (Fig. 2.5)

In Fig. 2.5, and shows a diagram of a high-speed logic element used as the basis of microcircuits of the K531 series (foreign analogue - 74S), (S is the initial letter of the surname of the German physicist Schottky). In this element, the emitter circuit of a phase separating cascade made on a transistor VT2, the current generator is turned on - transistor VT6 with resistors R4 And R5. This allows you to increase the performance of the logic element. Otherwise, this logical element is similar to the basic element of the K131 series. However, the introduction of Schottky transistors made it possible to reduce tzd.r doubled.

In Fig. 2.5, b shows a diagram of the basic logical element of the K555 series (foreign analogue - 74LS). In this circuit, instead of a multi-emitter transistor, a matrix of Schottky diodes is used at the input. The introduction of Shatky diodes eliminates the accumulation of excess base charges, which increase the turn-off time of the transistor, and ensures the stability of the switching time over a temperature range.

Resistor R6 of the upper arm of the output stage creates the necessary voltage at the base of the transistor VT3 to open it. To reduce power consumption when the gate is closed (), a resistor R6 connect not to the common bus, but to the output of the element.

Diode VD7, connected in series with R6 and parallel to the collector load resistor of the phase separating cascade R2, allows you to reduce the turn-on delay of the circuit by using part of the energy stored in the load capacitance to increase the transistor collector current VT1 in transition mode.

Transistor VT3 is implemented without Schottky diodes, since it operates in active mode (emitter follower).

In this article we will talk about the multivibrator, how it works, how to connect a load to the multivibrator and the calculation of a transistor symmetrical multivibrator.

Multivibrator is a simple rectangular pulse generator that operates in self-oscillator mode. To operate it, you only need power from a battery or other power source. Let's consider the simplest symmetrical multivibrator using transistors. Its diagram is shown in the figure. The multivibrator can be more complicated depending on the necessary functions performed, but all the elements presented in the figure are mandatory, without them the multivibrator will not work.

The operation of a symmetrical multivibrator is based on the charge-discharge processes of capacitors, which together with resistors form RC circuits.

I wrote earlier about how RC circuits work in my article Capacitor, which you can read on my website. On the Internet, if you find material about a symmetrical multivibrator, it is presented briefly and not intelligibly. This circumstance does not allow novice radio amateurs to understand anything, but only helps experienced electronics engineers remember something. At the request of one of my site visitors, I decided to eliminate this gap.

How does a multivibrator work?

At the initial moment of power supply, capacitors C1 and C2 are discharged, so their current resistance is low. The low resistance of the capacitors leads to the “fast” opening of the transistors caused by the flow of current:

— VT2 along the path (shown in red): “+ power supply > resistor R1 > low resistance of discharged C1 > base-emitter junction VT2 > — power supply”;

— VT1 along the path (shown in blue): “+ power supply > resistor R4 > low resistance of discharged C2 > base-emitter junction VT1 > — power supply.”

This is the “unsteady” mode of operation of the multivibrator. It lasts for a very short time, determined only by the speed of the transistors. And there are no two transistors that are absolutely identical in parameters. Whichever transistor opens faster will remain open—the “winner.” Let's assume that in our diagram it turns out to be VT2. Then, through the low resistance of the discharged capacitor C2 and the low resistance of the collector-emitter junction VT2, the base of the transistor VT1 will be short-circuited to the emitter VT1. As a result, transistor VT1 will be forced to close - “become defeated”.

Since transistor VT1 is closed, a “fast” charge of capacitor C1 occurs along the path: “+ power supply > resistor R1 > low resistance of discharged C1 > base-emitter junction VT2 > — power supply.” This charge occurs almost up to the voltage of the power supply.

At the same time, capacitor C2 is charged with a current of reverse polarity along the path: “+ power source > resistor R3 > low resistance of discharged C2 > collector-emitter junction VT2 > — power source.” The charge duration is determined by the ratings R3 and C2. They determine the time at which VT1 is in the closed state.

When capacitor C2 is charged to a voltage approximately equal to the voltage of 0.7-1.0 volts, its resistance will increase and transistor VT1 will open with the voltage applied along the path: “+ power supply > resistor R3 > base-emitter junction VT1 > - power supply.” In this case, the voltage of the charged capacitor C1, through the open collector-emitter junction VT1, will be applied to the emitter-base junction of transistor VT2 with reverse polarity. As a result, VT2 will close, and the current that previously passed through the open collector-emitter junction VT2 will flow through the circuit: “+ power supply > resistor R4 > low resistance C2 > base-emitter junction VT1 > — power supply.” This circuit will quickly recharge capacitor C2. From this moment, the “steady-state” self-generation mode begins.

Operation of a symmetrical multivibrator in “steady-state” generation mode

The first half-cycle of operation (oscillation) of the multivibrator begins.

When transistor VT1 is open and VT2 is closed, as I just wrote, capacitor C2 is quickly recharged (from a voltage of 0.7...1.0 volts of one polarity, to the voltage of the power source of the opposite polarity) along the circuit: “+ power supply > resistor R4 > low resistance C2 > base-emitter junction VT1 > - power supply.” In addition, capacitor C1 is slowly recharged (from the power source voltage of one polarity, to a voltage of 0.7...1.0 volts of the opposite polarity) along the circuit: “+ power supply > resistor R2 > right plate C1 > left plate C1 > collector- emitter junction of transistor VT1 > - - power source.”

When, as a result of recharging C1, the voltage at the base of VT2 reaches a value of +0.6 volts relative to the emitter of VT2, the transistor will open. Therefore, the voltage of the charged capacitor C2, through the open collector-emitter junction VT2, will be applied to the emitter-base junction of the transistor VT1 with reverse polarity. VT1 will close.

The second half-cycle of operation (oscillation) of the multivibrator begins.

When transistor VT2 is open and VT1 is closed, capacitor C1 is quickly recharged (from a voltage of 0.7...1.0 volts of one polarity, to the voltage of the power source of the opposite polarity) along the circuit: “+ power supply > resistor R1 > low resistance C1 > base emitter junction VT2 > - power supply.” In addition, capacitor C2 is slowly recharged (from the voltage of the power source of one polarity, to a voltage of 0.7...1.0 volts of the opposite polarity) along the circuit: “right plate of C2 > collector-emitter junction of transistor VT2 > - power supply > + source power > resistor R3 > left plate C2". When the voltage at the base of VT1 reaches +0.6 volts relative to the emitter of VT1, the transistor will open. Therefore, the voltage of the charged capacitor C1, through the open collector-emitter junction VT1, will be applied to the emitter-base junction of transistor VT2 with reverse polarity. VT2 will close. At this point, the second half-cycle of the multivibrator oscillation ends, and the first half-cycle begins again.

The process is repeated until the multivibrator is disconnected from the power source.

Methods for connecting a load to a symmetrical multivibrator

Rectangular pulses are removed from two points of a symmetrical multivibrator– transistor collectors. When there is a “high” potential on one collector, then there is a “low” potential on the other collector (it is absent), and vice versa - when there is a “low” potential on one output, then there is a “high” potential on the other. This is clearly shown in the time graph below.

The multivibrator load must be connected in parallel with one of the collector resistors, but in no case in parallel with the collector-emitter transistor junction. You cannot bypass the transistor with a load. If this condition is not met, then at a minimum the duration of the pulses will change, and at a maximum the multivibrator will not work. The figure below shows how to connect the load correctly and how not to do it.

In order for the load not to affect the multivibrator itself, it must have sufficient input resistance. For this purpose, buffer transistor stages are usually used.

The example shows connecting a low-impedance dynamic head to a multivibrator. An additional resistor increases the input resistance of the buffer stage, and thereby eliminates the influence of the buffer stage on the multivibrator transistor. Its value should be no less than 10 times the value of the collector resistor. Connecting two transistors in a “composite transistor” circuit significantly increases the output current. In this case, it is correct to connect the base-emitter circuit of the buffer stage in parallel with the collector resistor of the multivibrator, and not in parallel with the collector-emitter junction of the multivibrator transistor.

For connecting a high-impedance dynamic head to a multivibrator a buffer stage is not needed. The head is connected instead of one of the collector resistors. The only condition that must be met is that the current flowing through the dynamic head must not exceed the maximum collector current of the transistor.

If you want to connect ordinary LEDs to the multivibrator– to make a “flashing light”, then buffer cascades are not required for this. They can be connected in series with collector resistors. This is due to the fact that the LED current is small, and the voltage drop across it during operation is no more than one volt. Therefore, they do not have any effect on the operation of the multivibrator. True, this does not apply to super-bright LEDs, for which the operating current is higher and the voltage drop can be from 3.5 to 10 volts. But in this case, there is a way out - increase the supply voltage and use transistors with high power, providing sufficient collector current.

Please note that oxide (electrolytic) capacitors are connected with their positives to the collectors of the transistors. This is due to the fact that on the bases of bipolar transistors the voltage does not rise above 0.7 volts relative to the emitter, and in our case the emitters are the minus of the power supply. But at the collectors of the transistors, the voltage changes almost from zero to the voltage of the power source. Oxide capacitors are not able to perform their function when connected with reverse polarity. Naturally, if you use transistors of a different structure (not N-P-N, but P-N-P structure), then in addition to changing the polarity of the power source, you need to turn the LEDs with the cathodes “up in the circuit”, and the capacitors with the pluses to the bases of the transistors.

Let's figure it out now What parameters of the multivibrator elements determine the output currents and generation frequency of the multivibrator?

What do the values ​​of collector resistors affect? I have seen in some mediocre Internet articles that the values ​​of collector resistors do not significantly affect the frequency of the multivibrator. This is all complete nonsense! If the multivibrator is correctly calculated, a deviation of the values ​​of these resistors by more than five times from the calculated value will not change the frequency of the multivibrator. The main thing is that their resistance is less than the base resistors, because collector resistors provide fast charging of capacitors. But on the other hand, the values ​​of collector resistors are the main ones for calculating the power consumption from the power source, the value of which should not exceed the power of the transistors. If you look at it, if connected correctly, they do not even have a direct effect on the output power of the multivibrator. But the duration between switchings (multivibrator frequency) is determined by the “slow” recharging of the capacitors. The recharge time is determined by the ratings of the RC circuits - base resistors and capacitors (R2C1 and R3C2).

A multivibrator, although it is called symmetrical, this refers only to the circuitry of its construction, and it can produce both symmetrical and asymmetrical output pulses in duration. The pulse duration (high level) on the VT1 collector is determined by the ratings of R3 and C2, and the pulse duration (high level) on the VT2 collector is determined by the ratings R2 and C1.

The duration of recharging capacitors is determined by a simple formula, where Tau– pulse duration in seconds, R– resistor resistance in Ohms, WITH– capacitance of the capacitor in Farads:

Thus, if you have not already forgotten what was written in this article a couple of paragraphs earlier:

If there is equality R2=R3 And C1=C2, at the outputs of the multivibrator there will be a “meander” - rectangular pulses with a duration equal to the pauses between pulses, which you see in the figure.

The full period of oscillation of the multivibrator is T equal to the sum of the pulse and pause durations:

Oscillation frequency F(Hz) related to period T(sec) through the ratio:

As a rule, if there are any calculations of radio circuits on the Internet, they are meager. That's why Let's calculate the elements of a symmetrical multivibrator using the example .

Like any transistor stages, the calculation must be carried out from the end - the output. And at the output we have a buffer stage, then there are collector resistors. Collector resistors R1 and R4 perform the function of loading the transistors. Collector resistors have no effect on the generation frequency. They are calculated based on the parameters of the selected transistors. Thus, first we calculate the collector resistors, then the base resistors, then the capacitors, and then the buffer stage.

Procedure and example of calculating a transistor symmetrical multivibrator

Initial data:

Supply voltage Ui.p. = 12 V.

Required multivibrator frequency F = 0.2 Hz (T = 5 seconds), and the pulse duration is equal to 1 (one) second.

A car incandescent light bulb is used as a load. 12 volts, 15 watts.

As you guessed, we will calculate a “flashing light” that will blink once every five seconds, and the duration of the glow will be 1 second.

Selecting transistors for the multivibrator. For example, we have the most common transistors in Soviet times KT315G.

For them: Pmax=150 mW; Imax=150 mA; h21>50.

Transistors for the buffer stage are selected based on the load current.

In order not to depict the diagram twice, I have already signed the values ​​of the elements on the diagram. Their calculation is given further in the Decision.

Solution:

1. First of all, you need to understand that operating a transistor at high currents in switching mode is safer for the transistor itself than operating in amplification mode. Therefore, there is no need to calculate the power for the transition state at the moments of passage of an alternating signal through the operating point “B” of the static mode of the transistor - the transition from the open state to the closed state and back. For pulse circuits built on bipolar transistors, the power is usually calculated for the transistors in the open state.

First, we determine the maximum power dissipation of the transistors, which should be a value 20 percent less (factor 0.8) than the maximum power of the transistor indicated in the reference book. But why do we need to drive the multivibrator into the rigid framework of high currents? And even with increased power, energy consumption from the power source will be large, but there will be little benefit. Therefore, having determined the maximum power dissipation of transistors, we will reduce it by 3 times. A further reduction in power dissipation is undesirable because the operation of a multivibrator based on bipolar transistors in low current mode is an “unstable” phenomenon. If the power source is used not only for the multivibrator, or it is not entirely stable, the frequency of the multivibrator will also “float”.

We determine the maximum power dissipation: Pdis.max = 0.8 * Pmax = 0.8 * 150 mW = 120 mW

We determine the rated dissipated power: Pdis.nom. = 120 / 3 = 40mW

2. Determine the collector current in the open state: Ik0 = Pdis.nom. / Ui.p. = 40mW / 12V = 3.3mA

Let's take it as the maximum collector current.

3. Let’s find the value of the resistance and power of the collector load: Rk.total = Ui.p./Ik0 = 12V/3.3mA = 3.6 kOhm

We select resistors from the existing nominal range that are as close as possible to 3.6 kOhm. The nominal series of resistors has a nominal value of 3.6 kOhm, so we first calculate the value of the collector resistors R1 and R4 of the multivibrator: Rк = R1 = R4 = 3.6 kOhm.

The power of the collector resistors R1 and R4 is equal to the rated power dissipation of the transistors Pras.nom. = 40 mW. We use resistors with a power exceeding the specified Pras.nom. - type MLT-0.125.

4. Let's move on to calculating the basic resistors R2 and R3. Their rating is determined based on the gain of transistors h21. At the same time, for reliable operation of the multivibrator, the resistance value must be within the range: 5 times greater than the resistance of the collector resistors, and less than the product Rк * h21. In our case Rmin = 3.6 * 5 = 18 kOhm, and Rmax = 3.6 * 50 = 180 kOhm

Thus, the values ​​of resistance Rb (R2 and R3) can be in the range of 18...180 kOhm. We first select the average value = 100 kOhm. But it is not final, since we need to provide the required frequency of the multivibrator, and as I wrote earlier, the frequency of the multivibrator directly depends on the base resistors R2 and R3, as well as on the capacitance of the capacitors.

5. Calculate the capacitances of capacitors C1 and C2 and, if necessary, recalculate the values ​​of R2 and R3.

The values ​​of the capacitance of capacitor C1 and the resistance of resistor R2 determine the duration of the output pulse on the collector VT2. It is during this impulse that our light bulb should light up. And in the condition the pulse duration was set to 1 second.

Let's determine the capacitance of the capacitor: C1 = 1 sec / 100 kOhm = 10 µF

A capacitor with a capacity of 10 μF is included in the nominal range, so it suits us.

The values ​​of the capacitance of capacitor C2 and the resistance of resistor R3 determine the duration of the output pulse on the collector VT1. It is during this pulse that there is a “pause” on the VT2 collector and our light bulb should not light up. And in the condition, a full period of 5 seconds with a pulse duration of 1 second was specified. Therefore, the duration of the pause is 5 seconds – 1 second = 4 seconds.

Having transformed the recharge duration formula, we Let's determine the capacitance of the capacitor: C2 = 4 sec / 100 kOhm = 40 µF

A capacitor with a capacity of 40 μF is not included in the nominal range, so it does not suit us, and we will take the capacitor with a capacity of 47 μF that is as close as possible to it. But as you understand, the “pause” time will also change. To prevent this from happening, we Let's recalculate the resistance of resistor R3 based on the duration of the pause and the capacitance of capacitor C2: R3 = 4sec / 47 µF = 85 kOhm

According to the nominal series, the closest value of the resistor resistance is 82 kOhm.

So, we got the values ​​of the multivibrator elements:

R1 = 3.6 kOhm, R2 = 100 kOhm, R3 = 82 kOhm, R4 = 3.6 kOhm, C1 = 10 µF, C2 = 47 µF.

6. Calculate the value of resistor R5 of the buffer stage.

To eliminate the influence on the multivibrator, the resistance of the additional limiting resistor R5 is selected to be at least 2 times greater than the resistance of the collector resistor R4 (and in some cases more). Its resistance, together with the resistance of the emitter-base junctions VT3 and VT4, in this case will not affect the parameters of the multivibrator.

R5 = R4 * 2 = 3.6 * 2 = 7.2 kOhm

According to the nominal series, the nearest resistor is 7.5 kOhm.

With a resistor value of R5 = 7.5 kOhm, the buffer stage control current will be equal to:

Icontrol = (Ui.p. - Ube) / R5 = (12v - 1.2v) / 7.5 kOhm = 1.44 mA

In addition, as I wrote earlier, the collector load rating of the multivibrator transistors does not affect its frequency, so if you do not have such a resistor, then you can replace it with another “close” rating (5 ... 9 kOhm). It is better if this is in the direction of decrease, so that there is no drop in the control current in the buffer stage. But keep in mind that the additional resistor is an additional load for transistor VT2 of the multivibrator, so the current flowing through this resistor adds up to the current of collector resistor R4 and is a load for transistor VT2: Itotal = Ik + Icontrol. = 3.3mA + 1.44mA = 4.74mA

The total load on the collector of transistor VT2 is within normal limits. If it exceeds the maximum collector current specified in the reference book and multiplied by a factor of 0.8, increase resistance R4 until the load current is sufficiently reduced, or use a more powerful transistor.

7. We need to provide current to the light bulb In = Рн / Ui.p. = 15W / 12V = 1.25 A

But the control current of the buffer stage is 1.44 mA. The multivibrator current must be increased by a value equal to the ratio:

In / Icontrol = 1.25A / 0.00144A = 870 times.

How to do it? For significant output current amplification use transistor cascades built according to the “composite transistor” circuit. The first transistor is usually low-power (we will use KT361G), it has the highest gain, and the second must provide sufficient load current (let’s take the no less common KT814B). Then their transmission coefficients h21 are multiplied. So, for the KT361G transistor h21>50, and for the KT814B transistor h21=40. And the overall transmission coefficient of these transistors connected according to the “composite transistor” circuit: h21 = 50 * 40 = 2000. This figure is greater than 870, so these transistors are quite enough to control a light bulb.

Well, that's all!

If we take, for example, a transistor MJE3055T it has a maximum current of 10A, and the gain is only about 50; accordingly, in order for it to open completely, it needs to pump about two hundred milliamps of current into the base. A regular MK output won’t handle that much, but if you connect a weaker transistor between them (some kind of BC337) capable of pulling this 200mA, then it’s easy. But this is so that he knows. What if you have to make a control system out of improvised rubbish - it will come in handy.

In practice, ready-made transistor assemblies. Externally, it is no different from a conventional transistor. Same body, same three legs. It’s just that it has a lot of power, and the control current is microscopic :) In price lists they usually don’t bother and write simply - a Darlington transistor or a composite transistor.

For example a couple BDW93C(NPN) and BDW94С(PNP) Here is their internal structure from the datasheet.

Moreover, there are Darlington assemblies. When several are packed into one package at once. An indispensable thing when you need to steer some powerful LED display or stepper motor (). An excellent example of such a build - very popular and easily available ULN2003, capable of dragging up to 500 mA for each of its seven assemblies. Outputs are possible include in parallel to increase the current limit. In total, one ULN can carry as much as 3.5A through itself if all its inputs and outputs are parallelized. What makes me happy about it is that the exit is opposite the entrance, it is very convenient to route the board under it. Directly.

The datasheet shows the internal structure of this chip. As you can see, there are also protective diodes here. Despite the fact that they are drawn as if they were operational amplifiers, the output here is an open collector type. That is, he can only short circuit to the ground. What becomes clear from the same datasheet if you look at the structure of one valve.

7.2 Transistor VT1

As transistor VT1 we use transistor KT339A with the same operating point as for transistor VT2:

Let's take Rk = 100 (Ohm).

Let's calculate the parameters of the equivalent circuit for a given transistor using formulas 5.1 - 5.13 and 7.1 - 7.3.

Sk(req)=Sk(pass)*=2×=1.41 (pF), where

Sk(required)-capacitance of the collector junction at a given Uke0,

Sk(pasp) is a reference value of the collector capacity at Uke(pasp).

rb= =17.7 (Ohm); gb==0.057 (Cm), where

rb-base resistance,

Reference value of the feedback loop constant.

rе= ==6.54 (Ohm), where

re-emitter resistance.

gbe===1.51(mS), where

gbe-base-emitter conductivity,

Reference value of the static current transfer coefficient in a common emitter circuit.

Ce===0.803 (pF), where

C is the emitter capacity,

ft-reference value of the transistor cutoff frequency at which =1

Ri= =1000 (Ohm), where

Ri is the output resistance of the transistor,

Uke0(add), Ik0(add) - respectively, the nameplate values ​​of the permissible voltage on the collector and the constant component of the collector current.

– input resistance and input capacitance of the loading stage.

The upper limit frequency is provided that each stage has 0.75 dB of distortion. This value of f satisfies the technical specifications. No correction needed.

7.2.1 Calculation of the thermal stabilization scheme

As was said in paragraph 7.1.1, in this amplifier, emitter thermal stabilization is most acceptable since the KT339A transistor is low-power, and in addition, emitter stabilization is easy to implement. The emitter thermal stabilization circuit is shown in Figure 4.1.

Calculation procedure:

1. Select the emitter voltage, divider current and supply voltage;

2. Then we will calculate.

The divider current is chosen to be equal to, where is the base current of the transistor and is calculated by the formula:

The supply voltage is calculated using the formula: (V)

The resistor values ​​are calculated using the following formulas:

8. Distortion introduced by the input circuit

A schematic diagram of the cascade input circuit is shown in Fig. 8.1.

Figure 8.1 - Schematic diagram of the cascade input circuit

Provided that the input impedance of the cascade is approximated by a parallel RC circuit, the transmission coefficient of the input circuit in the high frequency region is described by the expression:

– input resistance and input capacitance of the cascade.

The value of the input circuit is calculated using formula (5.13), where the value is substituted.

9. Calculation of C f, R f, C r

The amplifier circuit diagram contains four coupling capacitors and three stabilization capacitors. The technical specifications say that the distortion of the flat top of the pulse should be no more than 5%. Therefore, each coupling capacitor should distort the flat top of the pulse by no more than 0.71%.

Flat top distortion is calculated using the formula:

where τ and is the pulse duration.

Let's calculate τ n:

τ n and C p are related by the relation:

where R l, R p - resistance to the left and right of the capacitance.

Let's calculate C r. The input resistance of the first stage is equal to the resistance of parallel-connected resistances: input transistor, Rb1 and Rb2.

R p =R in ||R b1 ||R b2 =628(Ohm)

The output resistance of the first stage is equal to the parallel connection Rк and the output resistance of the transistor Ri.

R l =Rк||Ri=90.3(Ohm)

R p =R in ||R b1 ||R b2 =620(Ohm)

R l =Rк||Ri=444(Ohm)

R p =R in ||R b1 ||R b2 =48(Ohm)

R l =Rк||Ri=71(Ohm)

R p =R n =75(Ohm)

where C p1 is the separating capacitor between Rg and the first stage, C 12 - between the first and second cascade, C 23 - between the second and third, C 3 - between the final stage and the load. By placing all other containers at 479∙10 -9 F, we will ensure a decline that is less than required.

Let's calculate R f and C f (U R Ф =1V):

10. Conclusion

In this course project, a pulse amplifier has been developed using transistors 2T602A, KT339A, and has the following technical characteristics:

Upper limit frequency 14 MHz;

Gain 64 dB;

Generator and load resistance 75 Ohm;

Supply voltage 18 V.

The amplifier circuit is shown in Figure 10.1.

Figure 10.1 - Amplifier circuit

When calculating the characteristics of the amplifier, the following software was used: MathCad, Work Bench.

Literature

1. Semiconductor devices. Medium and high power transistors: Directory / A.A. Zaitsev, A.I. Mirkin, V.V. Mokryakov and others. Edited by A.V. Golomedova.-M.: Radio and Communication, 1989.-640 p.

2. Calculation of high-frequency correction elements of amplifier stages using bipolar transistors. Educational and methodological manual on course design for students of radio engineering specialties / A.A. Titov, Tomsk: Vol. state University of Control Systems and Radioelectronics, 2002. - 45 p.

Working direct. The working line passes through the points Uke=Ek and Ik=Ek÷Rn and intersects the graphs of the output characteristics (base currents). To achieve the greatest amplitude when calculating a pulse amplifier, the operating point was chosen closer to the lowest voltage since the final stage will have a negative pulse. According to the graph of the output characteristics (Fig. 1), the values ​​IKpost = 4.5 mA, ... were found.

Calculation of Sf, Rf, Wed 10. Conclusion Literature TECHNICAL ASSIGNMENT No. 2 for course design in the discipline “Nuclear power plant circuitry” for student gr. 180 Kurmanov B.A. Project topic: Pulse amplifier Generator resistance Rg = 75 Ohm. Gain K = 25 dB. Pulse duration 0.5 μs. The polarity is "positive". Duty ratio 2. Settling time 25 ns. Release...

That in order to match with the load resistance it is necessary to install an emitter follower after the amplification stages, let's draw the amplifier circuit: 2.2 Calculation of the static mode of the amplifier We calculate the first amplification stage. We select the operating point for the first amplifier stage. Its characteristics:...

The resistance of the input signal source, and therefore changing the optimality condition during irradiation does not lead to an additional increase in noise. Radiation effects in the IOU. Impact of AI on IOU parameters. Integrated operational amplifiers (IOAs) are high-quality precision amplifiers that belong to the class of universal and multifunctional analog...

When designing circuits for radio-electronic devices, it is often desirable to have transistors with parameters better than those models offered by manufacturers of radio-electronic components (or better than what is possible with the available transistor manufacturing technology). This situation is most often encountered in the design of integrated circuits. We usually require higher current gain h 21, higher input resistance value h 11 or less output conductance value h 22 .

Various circuits of composite transistors can improve the parameters of transistors. There are many opportunities to implement a composite transistor from field-effect or bipolar transistors of different conductivities, while improving its parameters. The most widespread is the Darlington scheme. In the simplest case, this is the connection of two transistors of the same polarity. An example of a Darlington circuit using npn transistors is shown in Figure 1.

Figure 1 Darlington circuit using NPN transistors

The above circuit is equivalent to a single NPN transistor. In this circuit, the emitter current of transistor VT1 is the base current of transistor VT2. The collector current of the composite transistor is determined mainly by the current of transistor VT2. The main advantage of the Darlington circuit is the high current gain h 21, which can be approximately defined as the product h 21 transistors included in the circuit:

(1)

However, it should be kept in mind that the coefficient h 21 depends quite strongly on the collector current. Therefore, at low values ​​of the collector current of transistor VT1, its value can decrease significantly. Dependency example h 21 from the collector current for different transistors is shown in Figure 2

Figure 2 Dependence of transistor gain on collector current

As can be seen from these graphs, the coefficient h 21e practically does not change for only two transistors: the domestic KT361V and the foreign BC846A. For other transistors, the current gain depends significantly on the collector current.

In the case when the base current of transistor VT2 is sufficiently small, the collector current of transistor VT1 may be insufficient to provide the required current gain value h 21. In this case, increasing the coefficient h 21 and, accordingly, a decrease in the base current of the composite transistor can be achieved by increasing the collector current of transistor VT1. To do this, an additional resistor is connected between the base and emitter of transistor VT2, as shown in Figure 3.

Figure 3 Composite Darlington transistor with an additional resistor in the emitter circuit of the first transistor

For example, let's define the elements for a Darlington circuit assembled on BC846A transistors. Let the current of transistor VT2 be equal to 1 mA. Then its base current will be equal to:

(2)

At this current, the current gain h 21 drops sharply and the overall current gain may be significantly less than the calculated one. By increasing the collector current of transistor VT1 using a resistor, you can significantly gain in the value of the overall gain h 21. Since the voltage at the base of the transistor is a constant (for a silicon transistor u be = 0.7 V), then we calculate according to Ohm’s law:

(3)

In this case, we can expect a current gain of up to 40,000. This is how many domestic and foreign superbetta transistors are made, such as KT972, KT973 or KT825, TIP41C, TIP42C. The Darlington circuit is widely used in the output stages of low frequency amplifiers (), operational amplifiers and even digital ones, for example.

It should be noted that the Darlington circuit has the disadvantage of increased voltage U ke. If in ordinary transistors U ke is 0.2 V, then in a composite transistor this voltage increases to 0.9 V. This is due to the need to open transistor VT1, and for this a voltage of 0.7 V should be applied to its base (if we are considering silicon transistors).

In order to eliminate this drawback, a compound transistor circuit using complementary transistors was developed. On the Russian Internet it was called the Siklai scheme. This name comes from the book by Tietze and Schenk, although this scheme previously had a different name. For example, in Soviet literature it was called a paradoxical pair. In the book by W.E. Helein and W.H. Holmes, a compound transistor based on complementary transistors is called a White circuit, so we will simply call it a compound transistor. The circuit of a composite pnp transistor using complementary transistors is shown in Figure 4.

Figure 4 Composite pnp transistor based on complementary transistors

An NPN transistor is formed in exactly the same way. The circuit of a composite npn transistor using complementary transistors is shown in Figure 5.

Figure 5 Composite npn transistor based on complementary transistors

In the list of references, the first place is given by the book published in 1974, but there are BOOKS and other publications. There are basics that do not become outdated for a long time and a huge number of authors who simply repeat these basics. You must be able to tell things clearly! During my entire professional career, I have come across less than ten BOOKS. I always recommend learning analog circuit design from this book.

Last file update date: 06/18/2018

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