Something I thought I knew, but am left puzzling... Take a simple class B output pair driven directly by an opamp. Result is na= sty crossover distortion. Now, the nfb is taken from the circuit's output t= erminal, not from the opamp's output, so surely an ideal opamp should corre= ct for the 0.6v Vbe drops. Question is why doesn't it? The other question is what strategies exist to tackle it. - bias the output trs to class AB - helper resistor from opamp output to circuit output, so the opamp powers = the output directly from -0.6 to +0.6v - low Vbe transistors, ie germanium - a diode drop in the feedback path seems to reduce it to a degree Any techniques I've missed? NT
Class B distortion
Started by ●December 20, 2014
Reply by ●December 20, 20142014-12-20
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
Reply by ●December 20, 20142014-12-20
meow2222@care2.com wrote:> Something I thought I knew, but am left puzzling...> Take a simple class B output pair driven directly by an opamp. Result > is nasty crossover distortion. Now, the nfb is taken from the > circuit's output 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?The op amp may not have the high slew rate needed to switch rapidly between vbe drops. This is especially true at higher frequencies.> The other question is what strategies exist to tackle it. > - bias the output trs to class AB > - helper resistor from opamp output to circuit output, so the opamp > powers the output directly from -0.6 to +0.6v - low Vbe transistors, > ie germanium - a diode drop in the feedback path seems to reduce it to > a degree> Any techniques I've missed?> NTAdd 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.
Reply by ●December 20, 20142014-12-20
On Sat, 20 Dec 2014 00:55:39 -0800 (PST), meow2222@care2.com wrote:>Something I thought I knew, but am left puzzling... > >Take a simple class B output pair driven directly by an opamp. Result is nasty crossover distortion. Now, the nfb is taken from the circuit's output 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? > >The other question is what strategies exist to tackle it. >- bias the output trs to class AB >- helper resistor from opamp output to circuit output, so the opamp powers the output directly from -0.6 to +0.6v >- low Vbe transistors, ie germanium >- a diode drop in the feedback path seems to reduce it to a degree > >Any techniques I've missed? > > >NTThis is class B, I guess. https://dl.dropboxusercontent.com/u/53724080/Circuits/Amps/Opamp_Boosted.JPG -- John Larkin Highland Technology, Inc picosecond timing laser drivers and controllers jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
Reply by ●December 20, 20142014-12-20
>wrote in message >news:9795867e-6869-495b-9078-bf72384ef4da@googlegroups.com...>Something I thought I knew, but am left puzzling...>Take a simple class B output pair driven directly by an opamp. Result is >nasty crossover distortion. Now, the nfb is taken from the circuit's output >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?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 = 0. (open loop distortion) / loop gain, with loop gain -> zero is a big number. Kevin Aylward www.kevinaylward.co.uk www.anasoft.co.uk - SuperSpice
Reply by ●December 20, 20142014-12-20
On a sunny day (Sat, 20 Dec 2014 19:59:46 -0000) it happened "Kevin Aylward" <ExtractkevinRemove@kevinaylward.co.uk> wrote in <roadnTcI5ZqxSAjJnZ2dnUVZ7sidnZ2d@bt.com>:>>wrote in message >>news:9795867e-6869-495b-9078-bf72384ef4da@googlegroups.com... > >>Something I thought I knew, but am left puzzling... > >>Take a simple class B output pair driven directly by an opamp. Result is >>nasty crossover distortion. Now, the nfb is taken from the circuit's output >>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? > >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 = >0. (open loop distortion) / loop gain, with loop gain -> zero is a big >number.+ | |/ 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.
Reply by ●December 20, 20142014-12-20
On Saturday, December 20, 2014 12:11:02 PM UTC, Phil Hobbs wrote:> 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 HobbsDoes 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
Reply by ●December 20, 20142014-12-20
On Saturday, December 20, 2014 2:38:14 PM UTC, Tom Swift wrote:> meow2222@care2.com wrote: > > > Something I thought I knew, but am left puzzling... > > > Take a simple class B output pair driven directly by an opamp. Result > > is nasty crossover distortion. Now, the nfb is taken from the > > circuit's output 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? > > The op amp may not have the high slew rate needed to switch rapidly > between vbe drops. This is especially true at higher frequencies. > > > The other question is what strategies exist to tackle it. > > - bias the output trs to class AB > > - helper resistor from opamp output to circuit output, so the opamp > > powers the output directly from -0.6 to +0.6v - low Vbe transistors, > > ie germanium - a diode drop in the feedback path seems to reduce it to > > a degree > > > Any techniques I've missed? > > > NT > > 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 driveI'm not managing to work out what you mean there> 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.NT
Reply by ●December 20, 20142014-12-20
On Saturday, December 20, 2014 4:06:02 PM UTC, John Larkin wrote:> On Sat, 20 Dec 2014 00:55:39 -0800 (PST), meow2222@care2.com wrote:> >Something I thought I knew, but am left puzzling... > > > >Take a simple class B output pair driven directly by an opamp. Result is=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?> > > >The other question is what strategies exist to tackle it. > >- bias the output trs to class AB > >- helper resistor from opamp output to circuit output, so the opamp powe=rs the output directly from -0.6 to +0.6v> >- low Vbe transistors, ie germanium > >- a diode drop in the feedback path seems to reduce it to a degree > > > >Any techniques I've missed?> This is class B, I guess. >=20 > https://dl.dropboxusercontent.com/u/53724080/Circuits/Amps/Opamp_Boosted.=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
Reply by ●December 20, 20142014-12-20
On Saturday, December 20, 2014 8:57:22 PM UTC, Jan Panteltje wrote:> + > | > |/ 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.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
