Improving Slew-Rate Symmetry

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Samuel Groner

Well-known member
Joined
Aug 19, 2004
Messages
2,940
Location
Zürich, Switzerland
Hi

I played a bit with an opamp topology with balanced resistive load of the first stage: [removed]

I like this design but it has the flaw that the slew-rate is rather unbalanced: [removed]

Is there a simple fix for this without increasing complexity significantely (e.g. using an active load instead of the resistive)?

Thanks for your help!

Samuel
 
Don't throw away Q4's collector output but use it in a mirror (Q7 plus), so the output node has ~symmetry. It will also double the loop gain, which may require readjustment of compensation.

It will have the additional advantage of automatic current balancing of that second stage.
 
Thanks for your answer. Originally I used the current mirror and missed that the slew-rate is a tad more symmetrical than with the topology shown in the schematic. However with the current mirror it is still 21 V/us vs. 16 V/us--less balanced than with an unbalanced second stage...

Samuel
 
Well, probe and see where the currents are asymmetrical. Note that if you use the mirror, it is shorting out your one Q's compensation network, about as badly as with the collector tied to Vdd. If you make the input Z of the mirror high you lose signal swing, and you'll never get it up to the Z of the output node. For true symmetry and those R-C networks, you also must have equal voltage gains at each collector, since Miller strongly affects. Right now the left-hand one is not doing much.

I'd throw some R's in Q4-Q5 emitters for flexibility too.

See what the actual collector currents are for the two overload conditions at Q5. Maybe it is mainly a balancing problem based on the actual Q6 vs. Q7 magnitudes---after all, it doesn't take much to produce the mismatch reported.

EDIT: Actually the main problem is the response of the current source Q6, to the substantially greater voltage swing at the emitter of Q5 during the fall time. The nominal 5.924mA drops to 3.74mA for the duration of the slew event. This is about enough to account for the asymmetry I think.

You need a better current source, and maybe a bit of reduced swing on the bases of Q4 Q5---enough to fully switch. Maybe stiffen the impedance at the Q6 base too.

Actually there is a large asymmetry on the differential voltage at R3 R4. This is not helping. I think it's mostly to do with the large Miller effect at the base of Q5.

Further edit:

If you make the swing at Q4 significant by loading it with a large R (20k-ish) and feed its collector current from another 4403 (ground the 20k) you now symmetrize things much better, diffrential swings at R3-R4 are nearly equal, and the slew rate magnitudes are much closer. Both are a bit slower. So although not probably optimal at least it can be fixed.

I would probably find an alternate place to do the compensation that was inherently symmetrical, which probably entails a more inherently symmetrical topology as well.
 
You need a better current source.
Right, that's it! Unfortunately, it looks like it needs to be pretty good--the common two-transistor version is not good enough at least...

Both are a bit slower.
Er, they went from 21 V/us and 16 V/us right down to 9 V/us--or did I do something wrong?

Actually I think I've found the easiest solution, i.e. add a third resistor between R3/R4 and the negative supply. With a 1 nF bypass cap for the new resistor and R3/R4/Rnew = 1k6 this makes for a very clean impulse response. And noise went down as well--cool!

Do you see any disadvantages of this solution?

Samuel
 

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