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Determing output impedance of a Class-AB pushpull

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Adam Horden

Wed Mar 29 2006, 09:06AM

Registered Member #176 Joined: Tue Feb 14 2006, 09:35PM
Location:
Posts: 44

Hi All,

This output stage works in class-AB.

When T1 is conducting D1 adds 0.7V to the signal. 9V + 0.7 = 9.7V. D1 cancels out the voltage drop and restores the input signal. D2 works in the same way but adds -0.7V to the signal. -9V + -0.7V = 9.7V.

There is a slight overlap where both transistors are conducting.

The problem is this output stage feed an op-amp stage as well as the load. The output is syncing a 75R load to ground. At the junction of the emitters also feeds a precission rectifer circuit.

The precision rectifer is a inverting amplifer so the input impedance is set by the input resistor. Standard stuff. Now to select this resistor I need to know the output impedance of the push-pull output stage.

Are there any equations to calculate the output impedance of the transistor Class-AB pushpull?

Is the output impedance just the load impedance?

Adam

Steve Conner

Wed Mar 29 2006, 10:17AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Hi Adam,

What answer do you want

It's going to be of the order of R1/Beta, and half of that during the overlap when both transistors are conducting (if R1=R2)

You should also reckon the output impedance of the previous stage plus the dynamic resistance of the diodes, which can be quite high at low bias currents, so it would be

(R1 || (Rout of previous stage + Rd of diode))/beta when only one transistor is conducting, or

((R1 || R2) || (Rout+Rd/2))/beta when both are conducting

AFAIK

WaveRider

Wed Mar 29 2006, 10:22AM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Hi Adam,
Since your amplifier uses a complementary common collector final stage, the low-frequency output impedance is going to be very low (depending on the transistors' parameters, of course...on the order of an ohm or less). This is desirable for a driver/output stage. Simulation (using the appropriate transistor models) to find the equivalent source model for the output is probably the best way.. Back-of-the envelope approximation can be done by considering the large-signal load-line (where peak output current/voltage is encountered) with a shorted output. Ohm's law can be used compute an approximate output resistance based on the short-circuit output current and the known parameters of the driving signal source.

These things depend on transistor parameters, bias and driving signal impedance, so a bit more info is needed to give a definitive answer. Also, unless you are working at RF/microwave circuit frequencies, it is unlikely that impedance matching is needed for reasonable operation of the circuit.

Adam Horden

Wed Mar 29 2006, 11:06AM

Registered Member #176 Joined: Tue Feb 14 2006, 09:35PM
Location:
Posts: 44

Steve Conner wrote ...

Hi Adam,

What answer do you want

Thanks Steve,

Pratically I just used R1/Beta 4.7K / 150 and the beta was allready calculated using a load line graph. This gave me a input resistance of 5.6k on the precission rectifer to ensure the precission rectifer did not load down the output. Then I just used a scope to check for loading.

Steve I used 4 formulas from diffrent books (art of electronics and prinicples of transistor circuits) all gave a diffrent output impedance

I got it down to 470R I just did not think that was right. I know output stages are usally low impedance but not this low.

I will try calculating it tacking Rd into account.

I think R1 / beta is probally the correct equation.

Waverider,

Its running at around 800Hz.

Thanks for the help,

Adam

WaveRider

Wed Mar 29 2006, 11:49AM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

I think R1 / beta is probally the correct equation.

Hi again,

Your bias resistors are likely to have much higher resistance than the output resistance of the previous stage. This means that R_out is approximately R_source/beta.. For example: if the previous stage has output resistance of 100 ohms (including the dynamic resistance of the diodes) and your push-pull transistors have beta approx 100, the output resistance will be about 1 ohm...

In short, make sure your push-pull stage gets adequate low-impedance drive for low output impedance...

Adam Horden

Wed Mar 29 2006, 01:01PM

Registered Member #176 Joined: Tue Feb 14 2006, 09:35PM
Location:
Posts: 44

The full circuit is here http://www.adamhorden.co.uk/temp/alc_1403_18feb06/alc_181205_ver2_1400_18feb06.pdf

Page 3 is the signal source for the puspull.

Page 4 is the push pull.

An op-amp has a low output impedance but I am not to shure how you would determine the output impledance of the op-amp.

Thanks,
Adam

Steve Conner

Wed Mar 29 2006, 02:51PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

The output impedance of an op amp is usually less than an ohm (for small signals only and assuming the op-amp has feedback and the frequency is not too high) so it's common to assume that it's zero. If you want to quantify it, you use the datasheet value for open-loop output impedance and plug this into the standard equations for feedback theory.

470 ohms seems like a sensible value for the output impedance of the circuit, if anything it's a little high, considering you can get under an ohm with just an op-amp.

If I were building this, I would use negative feedback around the output stage to bring the resistance right down, and then a 10 ohm resistor in series with the output to bring it back up to a stable known value. 10 ohms is a fairly standard spec for the output impedance of line drivers in pro audio gear.

Waverider: Under quite a few conditions, the current through the diodes in that circuit is very low, so it's not valid to ignore their dynamic resistance IMO. It's really not a very good circuit. Putting capacitors across the diodes would help a lot.

Adam Horden

Wed Mar 29 2006, 05:04PM

Registered Member #176 Joined: Tue Feb 14 2006, 09:35PM
Location:
Posts: 44

Steve Conner wrote ...

The output impedance of an op amp is usually less than an ohm (for small signals only and assuming the op-amp has feedback and the frequency is not too high) so it's common to assume that it's zero. If you want to quantify it, you use the datasheet value for open-loop output impedance and plug this into the standard equations for feedback theory.

470 ohms seems like a sensible value for the output impedance of the circuit, if anything it's a little high, considering you can get under an ohm with just an op-amp.

If I were building this, I would use negative feedback around the output stage to bring the resistance right down, and then a 10 ohm resistor in series with the output to bring it back up to a stable known value. 10 ohms is a fairly standard spec for the output impedance of line drivers in pro audio gear.

Waverider: Under quite a few conditions, the current through the diodes in that circuit is very low, so it's not valid to ignore their dynamic resistance IMO. It's really not a very good circuit. Putting capacitors across the diodes would help a lot.

Thanks for all the help guys. I am working on a better version of this output stage but for the moment the one on the schem will be used as its to late in the project to change. I will design a ver2 and improve on the version I have allready. The unit is built and works but could be improved.

This was the first circuit I designed from the ground up. I guess with all circuit designs after you design them you find better ways to do the indevidual circuit functions.

I allways remeber this quote but I dont know where it orginated:

"For every problem there are multiple solutions"

In my case I would add feedback to the output stage from the opamp and design a better output stage. I really liked the linear votmeter circuit and the precission rectifer aspect that was one of the hardest challanges.

I plugged some numbers for the ouput impedance of the op-amp and I got 0.116 this is very close to 0.

Adam

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