Resistors values for class AB amplifier

Question

I'm designing a class AB amplifier with a vbe multiplier to amplify the output current. I try to choose the resistors values by doing a DC analysis but when I simulate the circuit in AC, the output current is much smaller than the input. I think my mistake has smth to do with the way the amplifier works because the transistors Q2 and Q3 only work half a cycle, but I don't how to take that into account. Here are my calculations and the full circuit. (I can't change the value of Vin, Vcc, C2, CL and RL because I use them to simplify my calculations, they replace others circuits)

For the DC analysis, I remove the input and the output part and I neglect the current Ib of all the transistors.

• I want the two resistors in the vbe multiplier to be equal. I have with KVL voltage law so VR2=Vbe=0.7V=VR3, as the current is the same in R2 and R3.
• I want an output current of 4.5mA so I choose Ic=4.5mA, as the Ic current from the transistors will be the output current. I supposed than Ic is the same for the 3 transistors. I have then R2=R3=VR2/(Ic1+Ic2)=78 Ohm, which is way too small.
• By KVL voltage law, I have VR1+VR2+VR3+VR4=Vcc so R1+R4=(Vcc-2VR2)/Ic1 => R1=800-R4, which will give really small values too.

Any help is welcome!

Answer 1

Your amplifier has many errors.

Answer 2

What do you mean by "output current"?

If you mean the current passing through RL, then to understand why it's so small, then just imagine what current passes through 11kΩ when it has the full 5V across it:

$$ I = \frac{V}{R} = \frac{5}{11000} = 450\mu A $$

On top of that, CL causes the voltage across RL to go between -2.5V and +2.5V, meaning that current is more like ±225μA. And on top of that, CL and RL form a high-pass filter with cut-off frequency:

$$ f_C = \frac{1}{2\pi R_LC_L} = \frac{1}{2\pi \times 11000 \times 300\times10^{-12}} = 48kHz $$

Your signal frequency is 1kHz, well below that, and is attenuated to almost nothing, so you can expect almost no current through RL at all.

The only current parameter that an amplifier like this can control, is maximum output current. As long as that maximum is not exceeded, the actual current drawn by the load is defined by the load itself, and your load is minuscule. Let me say that again:

It's the load (RL and CL) that determines output current, not the amplifier.

This is a voltage amplifier, with gain 1, and with the benefit that it can supply more current than the original source of input signal. That doesn't mean it always outputs more current, only that it can. The actual output current is a function of the load, not the amplifier, and it can be nanoamps or hundreds of milliamps (in your case).

If you want bigger output current, give it a significant load. One whose cut-off frequency is well below the test signal frequency, and draws enough current to stress the amplifier somewhat.

Try CL = 100μF, RL = 22Ω

Those values will have a cut-off frequency of:

$$ f_C = \frac{1}{2\pi \times 22 \times 100\times10^{-6}} = 70Hz $$

With full output voltage swing (0V to +5V), the voltage across RL will be ±2.5V, for an output current of:

$$ I = \pm \frac{2.5}{22} \approx \pm 100mA $$

Now that's a load which will draw significant current, and permit you to test its ability to actually maintain a voltage gain of 1 under much more demanding conditions.

I know you said you can't change CL or RL, but if you don't, output current is naturally going to be tiny, much smaller than any current you inject at the input, and I question the need for a push-pull stage like this anyway. You could drive that load (with less distortion) using a single-transistor common-collector (emitter-follower) stage, and only one resistor. Why so complicated for a load so tiny?

Answer 3

This configuration should work, but ... with some "good" changes.

As outputs BJT are "followers", voltage gain is < 1.

Or this ...

Answer 4

I also made an error on the circuit I fixed causing its idle current too be too high. I improved it today and also made its maximum output level much higher (if you want) with very low distortion. Its voltage gain is a little less than 1.

Answer 5

Your "amplifier" does not amplify. Also, its output current is extremely low. C4 allows the input signal V1 directly drive the base of Q2 like this:

Answer 6

How to understand electronic amplifiers

It is a well-known fact that more complex electronic circuits can be explained and understood through simpler equivalent electrical circuits. Typically, controllable current and voltage sources are used to model the behavior of active devices. But they are abstract models that give only a formal idea and work well in circuit theory.

For the purpose of intuitive (true) understanding of circuits we need to go down to an even lower level and that is the concept of "controllable" and "dynamic" resistors. It is underestimated only for the simple reason that "resistance" usually means "linear" ("ohmic") resistance and transistors have non-linear resistance. For purposes of intuitive understanding, however, this is not essential.

That is why, I have explained below the idea of ​​the complementary (push-pull) amplifier stage using only the concepts of variable resistor (rheostat), voltage divider and potentiometer. They are sufficient to build a functional idea of ​​these circuit solutions. To show the relationship between the resistor and transistor circuit, I have considered them in parallel from the start to finish.

The push-pull stage is bipolar (split) supplied; so there is no need for coupling capacitors. This simplifies things and does not distract us with unnecessary details.

How to explore amplifier stages

I have used CircuitLab to simulate the circuits as follows. To keep the schematics simple, I have not used potentiometers as variable resistors but constant resistors with an arrow in pale gray. You can change their resistance by opening the parameters window and setting values ​​in the field. If the window covers the instruments or labeled nodes, you can pull them out (along with the connecting wires) temporarily while setting values ​​and then put them back. Also, you can make cloned instruments and labeled nodes.

The initial schematics look a bit oddly drawn because the idea is to complete them step by step. They are bordered in pale gray to appear at the same scale.

The parameters are adjusted to obtain the same values ​​of voltage drops and currents in individual frames. The idea of ​​this is to show that resistor and transistor circuits are equivalent. Resistance values ​​are exemplary; I have chosen 1 k to be simple to calculate.

Building a simple amplifier stage

In fact, there is no amplification in the literal sense of the word; there is only regulation (reduction) of the power supply voltage.

Producing a positive voltage...

The first thing that comes to our mind to reduce the voltage is to connect a variable resistor R1 in series with the power supply and to change its resistance.

... by unloaded rheostat... But if there is no load connected, current does not flow, there is no drop and nothing changes - the output voltage remains equal to the supply voltage (I have chosen 10 V for simplicity). This goes against human intuition and it is good to experiment - open the R1 parameters and set different values ​​of the resistance.

^{simulate this circuit – Schematic created using CircuitLab}

... and NPN emitter follower unloaded. In a real amplifier (emitter follower here), an NPN transistor plays the role of R1. So open Vin parameters and set various input voltages. As you can see, the emitter output voltage does not follow Vin and stays equal to 10 V as above. So, we conclude, a transistor alone is not enough to make a follower... something else is needed...

... by loaded rheostat... That thing is the load RL. Let's load the rheostat R1 first. Now a load current IL begins flowing, a voltage drop VR1 = IL.R1 appears across R1... and we can change the output voltage Vout across the load within wide limits by varying R1. We conclude that R1 and RL form a voltage divider; so to change the voltage, we need two resistors forming a voltage divider.

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... and NPN emitter follower loaded. Let's now load the transistor and begin changing the input voltage. Since there is a negative feedback, the transistor adjusts its "static" collector-emitter resistance Rce (corresponding to R1 above) so that to make Vout = Vin; hence the name "follower". As you can see, I have adjusted Vin to 5.701 V to obtain exactly 5 V at the output. Thus the Q1's collector-emitter "static resistance" is exactly Vce/IL = 5/5 = 1 k as the R1 resistance above. The transistor can only source current ("push" the load).

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Producing a negative voltage...

All the configurations above are supplied by a positive voltage source V+; so the output voltage is also positive. If we need a negative output voltage, we should supply the voltage divider by a negative voltage V-.

... by unloaded rheostat... Vout is constantly -10 V.

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... and PNP emitter follower unloaded. Also here.

... by loaded rheostat... Now we can change Vout. R2 and RL form another voltage divider.

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... and PNP emitter follower loaded. Here the transistor follows the negative input voltage. It can only sink current ("pull" the load).

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Building a "push-pull" amplifier stage

With the two types of configurations we assembled above, we can amplify either a positive or negative input voltage. In order to be able to amplify both, we obviously need to "stick" two complementary configurations together.

Differential voltage divider

Thus we get another (third) voltage divider composed of two complementary resistors R1 and R2 with oppositely varying resistances. Here again we need a bit more dexterity to change R1 and R2 "simultaneously" and oppositely while looking at the output voltage across the load.

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In these differentially changing R1 and R2 we can see the ubiquitous potentiometer.

Potentiometer

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It is interesting to see how the load impacts the potentiometer. Here is the result of a DC sweep simulation with three load resistances - 100 ohm, 1 k and 100 k. As you can see, the 100 ohm load is too heavy for the 1 k potentiometer. Figuratively speaking, there are three resistors "pulling" the common output point - R1 "pulls" it up, R2 "pulls" it down and RL "pulls" it to the middle point... and the first two must be stronger than the third.

Push-pull stage

Now it remains only to replace the resistors with transistors and we get the famous push-pull stage. The NPN follower will act during the positive input half wave and the PNP follower will act during the negative half wave.

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But the problem with the unwanted Vbe voltage of the transistor appears.

0.7 V > Vin > -0.7 V. Actually, when the input voltage "wiggles" between -0.7 V and 0.7 V, both transistors are off... and we can remove them from the schematic.

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Vin > 0.7 V. When Vin increases enough, the output voltage begins following the positive input voltage less by 0.7 V.

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Vin < -0.7 V. Also, when Vin decreases enough, the output voltage begins following the negative input voltage more by 0.7 V.

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Graphically presented.

Improving the "push-pull" amplifier stage

Biasing

The remedy is obvious - we have to apply a tiny initial voltage (about 0.7 V) across each base-emitter junction. This means to add this voltage to the input voltage and apply the sum to Q1's base, and to subtract it from the input voltage and apply the difference to Q2's base. We can implement it by two diodes (D1 and D2) connected in series. They are forward-biased through the "pull-up" resistor R1 and "pull-down" R2.

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As you can see, the output voltage strictly follows the input voltage.

Overloading

But if the load is too heavy, another problem appears - at some value of the input voltage the output stops changing. Let's drastically reduce the load resistance (from 1 k to just 10 ohm) to observe this effect. It is convinient to use the DC live simulation - hover the mouse over the circuit components to find the reason. Aha... we see that the Q1's base current and accordingly, the voltage drop across R1, have significantly increased.

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Straightforward solution

We can simply decrease the R1 and R2 resistances, for example, to 100 ohm...

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... and the problem is solved. But now the current through diodes is very high (100 mA).

Elegant solution

We finally realize what the problem is - R1 and R2 are constant ("static") resistors. So, when Vin increases, the voltage drop across and the current through R1 decrease... and at some point the diode D1 turns off. The same happens at negative voltage with D2. So, the remedy is to make R1 and R2 "dynamic" - they have to decrease their resistance when the voltage across them decreases and v.v. In other words, they have to be "current-stabilizing resistors" (transistors).

Let's mimic their behavior using CircuitLab by manually changing their resistance as the input voltage changes. Here you will need even more agility than before because you will have to change three quantities at once - Vin, R1 and R2. For example, below I set Vin = 6 V; then decreased R1 by 500 ohm and increased R2 by 500 ohm. Thus, the total resistance R1 + R2 = 2 k remained the same and the current did not change accordingly.

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Formal presentation

These "magic dynamic elements" are formally presented as "current sources".

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