[SOLVED] Class AB amp, does this circuit even work?

[maark6000]

so below is a Class AB amp with a current mirror provided by diodes. I simply cannot get this circuit to work. I initially had the pnp in backwards, I forgot they flip it upside down, but now it's in right.

my input is a 1 kHz sine wave, 10v pp. my output is a capped sine wave, 1.4 v pp. it seems the circuit works until each diode reaches it's bias voltage, and then I don't know where the rest of the sine wave goes. could this be a grounding problem?

My power supply has a negative output, a ground, and a positive output. As such, I am hooking the negative to the -Vcc rail, the positive to the +Vcc rail, and not using the ground on the power supply. The AC signal has it's own ground rail, and the positive lead is at the input. Is that right?

Thanks.

 

[maark6000]

aha! figured it out. it was indeed the ground from my lab power supply not being hooked up to my ground rail.

I will say that it is my opinion that a great disservice is done to students of electronics by the shorthand that occurs in schematics, and this couldn't be more the case when it comes to ground. It is already confusing enough with all this DC voltages and AC signals, now we've got to just guess as to how these elements connect with regard to ground. anyway, enough of the soapbox.

so that this will be searchable via google, this schematic comes out of experiment 7 in the lab manual to "Electronic Devices," Thomas Floyd, 8th ed.

 

[BradtheRad]

Glad the obstacle was temporary. A similar grounding question often comes up with op amps too.

I was thinking your input signal (10V pk-to-pk) was too great amplitude. Even 7V pk-to-pk input causes slightly clipped output in my simulation.

Screen shot:

 

[maark6000]

hmm, interesting. The next step in this particular lab is to then crank up the input voltage to see where clipping does occur. with an 18 V theoretical swing possible between the rails, it actually starts clipping at around 14 V.

here is perhaps a dumb question, but one I'm going to ask anyway. If we are starting with a 10 v pp signal, and end up with (in my case) a 9.8 v pp signal at output... where is the "amplification?" Class A (common emitter) amps I totally get, you can start with one voltage and end up with a much higher swing with your AC signal. Even common-collecter amps make sense, because there is current amplification. But I could not detect any amplification going on with the Class B or Class AB amp. What am I missing?

 

[godfreyl]

... where is the "amplification?"...

It's current amplification, same as a simple common-collecter amp. The output current is much greater than the input current. It's difficult to measure the input current from the signal source though, thus difficult to calculate the gain.

 

[chuckey]

As Vout ~ Vin, gain = 5K/330 ~ X 15
Frank

 

[LvW]

But I could not detect any amplification going on with the Class B or Class AB amp. What am I missing?

Maark6000,
the task of this stage is NOT to amplify voltages but to deliver POWER to the load. Thus, the name should be power stage rather than power amplifier.

 

[maark6000]

got it... that would make sense as this is the chapter on power amplifiers. Frank, what does the ~ mean? And where is the 5K derived from?

one last question that may be better posted as its own topic... in computing the gain the lab book wanted us to try to calculate the impedance of the input signal. how does one do this? can it be measured easily?

 

[BradtheRad]

I misstated the peak-to-peak voltage in post #3. It was 14, rather than 7, where I saw clipping.

But I could not detect any amplification going on with the Class B or Class AB amp.

Voltage gain happens when the PNP is at the top and the NPN at the bottom. This is in addition to current gain. The emitters go directly to the supply rails, producing higher sensitivity.

I have spent hours experimenting with simulations, to find out how different one arrangement is from the other.

It would seem more efficient to combine voltage gain and current gain. However it gets unwieldy to do so much in one amplifier stage. Careful adjustment is needed, to minimize distortion.

It is easier to create voltage gain in class A, then current gain in class AB.

 

[keith1200rs]

got it... that would make sense as this is the chapter on power amplifiers. Frank, what does the ~ mean? And where is the 5K derived from?

one last question that may be better posted as its own topic... in computing the gain the lab book wanted us to try to calculate the impedance of the input signal. how does one do this? can it be measured easily?

The 5k is from the two 10k input resistors in parallel, I assume.

~ means "approximately"

Input impedance can be measured by voltage drop across a small, known, series resistor on the input.

To calculate it you must take account of the resistors on the input (the two 10k ones) but also the reflected load impedance through the transistors (the 330 ohms).

Keith

 

[maark6000]

oh, maybe I'm labeling this thing I'm looking for wrong. What you just described is R in (tot), correct? ... = (R1||R2||Bac(r'e+RL)) (that was from memory!)

I think what I'm trying to get at is that each signal has it's own inherent impedance. The AC signal coming from my function generator has some impedance, and the output of this circuit has some impedance... and what I'm wonder is how is that measured? Or are we indeed talking about the same thing here?

 

[Tunelabguy]

oh, maybe I'm labeling this thing I'm looking for wrong. What you just described is R in (tot), correct? ... = (R1||R2||Bac(r'e+RL)) (that was from memory!)

I think what I'm trying to get at is that each signal has it's own inherent impedance. The AC signal coming from my function generator has some impedance, and the output of this circuit has some impedance... and what I'm wonder is how is that measured? Or are we indeed talking about the same thing here?

No, the output impedance of the sig gen is not the same thing as the input impedance of the amp under discussion. But you can use the knowledge of that output impedance to calculate the input impedance of the amp even if you don't know anything about the amp and it is in a black box. Just measure the output voltage of the sig gen when it is disconnected (using a high impedance meter or scope). Then connect it to the amp and see how much the sig gen voltage drops. If the input impedance of the amp is equal to the output impedance of the sig gen then the voltage from the sig gen will drop in half when it is connected to the amp.

 

[LvW]

No, the output impedance of the sig gen is not the same thing as the input impedance of the amp under discussion. But you can use the knowledge of that output impedance to calculate the input impedance of the amp even if you don't know anything about the amp and it is in a black box. Just measure the output voltage of the sig gen when it is disconnected (using a high impedance meter or scope). Then connect it to the amp and see how much the sig gen voltage drops. If the input impedance of the amp is equal to the output impedance of the sig gen then the voltage from the sig gen will drop in half when it is connected to the amp.

I think it is rather problematic in this case (see the circuit diagram in post#1) to speak abut an input impedance because the input involves non-linear parts.
The same restriction applies to the question of "gain" (class A-B involves non-linearity).