line driver with automatic bias

It can't even output 0 volts, because the output PNP will zener. And its feedback is positive.

I think both PNPs are in wrong.

John

I compared the distortion of the output stages posted by Ban and by VLV. The idle current was set at 3mA for both. Rload = 600 Ohm. Both circuits were DC biased to set 0 V at the output. The opamps were taken out of picture so the negative feedback won't affect the results. Transistors were generic 3904s and 3906s. F = 1 KHz.

Vpp THD Ban's THD VLV's V % %

For both circuirs, the distortion strongly depends on the amplitude. Ban's is much worse at low amplitudes whereas VLV's is not as good at high swings.

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

How? It's connected to the collector. Relying on reverse operation? ;-)

...assuming your PNPs and NPNs are perfectly matched.

ge

I think the PNP output transistor is drawn wrong, the emitter should be connected to the output. And, either the PNP driver transistor is acting as a zener diode, or its emitter is drawn on the wrong end, too.

Is it really any different from this?

ftp://jjlarkin.lmi.net/SillyAmp.jpg

John

Both PNPs are drawn up-side-down, though it's still dumb if you flip them over.

Sure. Your PNPs are drawn the right way. ;-) ...but you have no feedback; crossover distortion is going to be *ugly*. ;-)

Even with feedback outside the followers, I think yours has worse crossover distortion as the opamp slews.

uploaded my simulation file, has some beefier driver in sziklay configuration. The PNPs are in fact upside down.

That's last one is not "mine"; I think it's equivalent to Ban's.

The little arrow thingie that says "FB" is where the feedback goes.

But even with no feedback, it has no crossover distortion, at least at low levels and low frequency. It's class A, beta biased. Horrible.

You may as well follow the opamp with a single NPN emitter follower, with a passive or constant-current pulldown. Works better, not beta-funny, less parts.

John

...

Well at least your PNPs are right way up, but circuit looks too busy, have you built this or just simulating? Looks to me as if you expect some matching that's not going to happen in real life?

Thanks for putting up a clearer circuit!

Grant.

Of course a simulated circuit looks a bit messy with a lot of additional placeholders. I have breadboarded the simple 4 transistor version and tried also transistors with different betas. Best is to take the PNP from the C-bin and the NPN B-suffix. The 4401/3 worked also. The circuit needs very little compensation, since there is no dead-time. And the transient behaviour is very good. I want to try breadboarding a headphone amp, when I have tested different output configurations in the simulation.

Ban

quiescent

I know. I used it 40 years back. It sucked then, too (but was good enough for the application). I don't think it's equivalent at all. I don't see where the OpAmp goes open-loop for Vbe X 2. Maybe I'm missing something, though.

You couldn't draw a couple of resistors?

If there is no crossover, you don't need two transistors. A follower will work.

More power.

It's one of those PhD circuits... Piled higher and Deeper... nice theory/simulation (?), but crap in practice :-)

(?) Which would catch the issues if the guy in the cockpit had any clue as to what to test for ;-) ...Jim Thompson

of 1.5mA

Thanks for the data. I guessed Ban's had better crossover performance because I didn't see a way to reliably set the idle current in yours. So, I assumed the idle current was effectively zero, causing a deadband, which I guess was wrong.

I also assumed Ban had a PNP or two upside down, which I assume you saw and corrected.

-- Cheers, James Arthur

I suppose if the op-amp is fast the cro

Nahh, think current mirror.

Tim

Ban only has a 15K resistor to beta-bias the outputs. That's going to limit the output swing capability. With a 600 ohm load, you may as well just use an opamp all by itself.

One thing I like to do is use multiple opamps in parallel. For example, two opamps each with a 100 ohm resistor to the load, net 50 ohm source. Or four opamps, same idea. If you want to go fast, into the MHz or 10s of MHz range, discretes get messy.

The base-to-base thing is sort of cute.

John

Why? VLV doesn't have a current mirror, just compound emitter followers. Idle current is set by Vbe drops and matching.

-- Cheers, James Arthur

With 1mA base current you can drive +/-10V into 50 Ohms, The O/P transistors have a beta of 300+ You can also reduce the bias setting resistors easily.

Yes, in other words, a current mirror ;-)

Compare to the conventional two-diode-biased emitter follower. Assuming the diodes are the same junction as the Vbe's (e.g., diode strapped 3904s, or even better, one NPN and one PNP, don't use big fat 1N914s which have much less drop), it's a current mirror.

Tim