Class B distortion

On 20/12/2014 19:55, snipped-for-privacy@care2.com wrote: ... an ideal opamp should correct for the 0.6v Vbe drops. Question is why doesn't it?

An ideal op-amp does. All you need to do is buy one of those, from the nearest unicorn.

Non-ideal op-amps have a finite maximum slew-rate.

Yup. And to get anywhere near it, they have to run a huge input error voltage--as much as a volt for FET types or bipolars with lots of input stage degeneration.

That's not a huge problem when the amp itself is reasonably linear, but it sur is in this case.

Cheers

Phil Hobbs

The op amp may not have the high slew rate needed to switch rapidly between vbe drops. This is especially true at higher frequencies.

Add a complimentary NPN/PNP emitter follower stage between the op amp and output to bias the output stage so it is barely on, and to add more drive capability. It may be desirable to add emitter degeneration resistors to each transistor to stabilize the bias current.

The driver stage can be especially useful at higher frequencies where the op amp may not be able to supply the high peak current needed.

This is class B, I guess.

No real op amp is ideal. i.e. has infinite gain. That's why a real one can not reduce x-over distortion in a pure class B stage. A class B stage has gain, essentially, zero. i.e. the loop gain is still, essentially, 0 X Av =

1. (open loop distortion) / loop gain, with loop gain -> zero is a big number.

Kevin Aylward

On a sunny day (Sat, 20 Dec 2014 19:59:46 -0000) it happened "Kevin Aylward" wrote in :

• | |/ c -------| NPN | |\ e in ---- +|\ | R | | >------===----| -- - |/ | |-------- out | | |/ e | | --------| PNP | | |\ c | | | | | - | --------------------------

For some value of R the cross over is eliminated as the opamp drives the load for small signals.

Does that mean if I ran the opamp at gain=1 it would at least reduce the problem to some degree? The bad news is it will be any old opamp, eg LM358, LM324 etc.

NT

I'm not managing to work out what you mean there

NT

nasty crossover distortion. Now, the nfb is taken from the circuit's outpu t terminal, not from the opamp's output, so surely an ideal opamp should co rrect for the 0.6v Vbe drops. Question is why doesn't it?

rs the output directly from -0.6 to +0.6v

JPG

I like the upsides, minimal part count and ok on linearity. The need to tai lor the Rs to the opamp is a big problem for people that have no testgear, and wouldnt know how to use it anyway. Tendency to kill transistors when mi sadjusted is also a problem. I think the failure rate would be too high.

NT

Yes, I tried that in spice. It seems the R has to be very low for an opamp though. 0.65v on 8ohm load = 75mA. 75mA 0.6v = 8ohms for the R... plus 8R load makes a 16R load. Can a random opamp drive that?

NT

Probably not. The slew rate limitation is a nonlinear phenomenon, changing the op-amp (or, presumably the whole-circuit) gain is a mostly- linear fix.

Biasing your output to class AB, and living with the extra dissipation, may be the road you'll have to take.

Yep. There's no OpAmp with a fast enough slew-rate (that you can afford ;-) ...Jim Thompson

Then let them use data sheets.

Could you use a large-ish resistor in the op amp power supply lead, with a small current sense emitter resistor on the power transistor and another transistor connected to pull base current? Bypass the emitter resistor with a capacitor. Then I think it would self-regulate and you wouldn't need to tailor the opamp sense resistor so much.

is nasty crossover distortion. Now, the nfb is taken from the circuit's ou tput terminal, not from the opamp's output, so surely an ideal opamp should correct for the 0.6v Vbe drops. Question is why doesn't it?

owers the output directly from -0.6 to +0.6v

ed.JPG

tailor the Rs to the opamp is a big problem for people that have no testgea r, and wouldnt know how to use it anyway.

Its an option. It would fit the plan better if possible to have one small d ocument that contains everything they need - I only want the moon on a stic k. Putting a table of R values in there might well work. Its certainly a si mple circuit, I like that.

NT

That sounds like a current limit. Then if the opamp lead R is too high, the op trs will sit drawing lots of current, just being limited to i_max. It doesnt sound good.

If one left out the 2nd tr that steals base current from the main one, I think I'll sim it, see if it does what I think, sounds interesting.

NT

I may do. But I would like to find a better option if it exists.

Total budget zero on this one. Its not like I'm tight or anything :)

NT

DC-wise, it's simple. Size the base resistors for, say, 0.3 volts drop at the opamp's max supply quiescent current. A 1 mA max Iq implies 300 ohm resistors. The opamp drives light loads without help but, when the load current gets up to 2 mA or so, the transistors turn on and help.

The AC situation is more complex.

My idea was that it's not a lot of current - just enough to keep the power transistors conducting a little so their bases are sitting at ~ 0.6 volts without having to match the opamp resistor to each op amp...you'd just need to select it for the minimum expected quiescent op amp current. If the current was higher the second transistor would pull away the excess. The capacitors bypass the emitter resistor on the power transistor for AC; the emitter resistors are just to set the DC operating point.

Maybe this won't work and I'm missing something?