Somebody explaining this design?

way to answer ????? attack ????

Jure Z.

ISTR seeing it in about 1972, in a high performance jfet input instrumentation amplifier that we made, (not my design). The jfet sources were fed by about 1mA constant current npn's, 500uA up the jfet and 500uA to the feedback pnp's. I think the pnp's had their emitters up on the +15v rail, with resistors from the jfet drains to their bases. I don't recall any series resistors between the drains and the inputs of the opamp.

Still no response after 24 hours.

Where are all the "experts" when a real question is asked ?:-)

...Jim Thompson

"Jim Thompson"

the posturing, PITA, fascist cretin cannot see the one that has been posted already.

........ Phil

Isn't it just the Pullup-R divided by the Tail-R, times the final Feedback-R divided by the Pullup-R, probably with a x2 in there somewhere?

So the Pullup-R cancels, leaving the overall Gain as the ratio of final Feedback-R/Tail-R.

Pullup-R's DON'T cancel.

...Jim Thompson

useless".

I worked in out with the sweep of the one functioning eyeball, and promptly announced that it was crap ;-)

...Jim Thompson

Well, let's do some sums. Admittedly on a simplified and idealised circuit (to make life easy for me). R2a and R2b are the Pull-up R's in question.

-------+-------------+---Vs | | \\ \\ R3b /R2a R2b/ +-/\\/\\---+-->Vout \\ \\ | | | | | _ | | +---------+--|- \\ | | | | >-+ Vx---> +-------------|---------+--|+_/ | | | \\|/I1 I2\\|/ \\ | | /R3a |/ \\| \\ Vin ---|npn npn|----. | |\\e e/| | | 0v--- | R1 | ---+--+---0v +----/\\/\\-----+ | | \\|/ I I \\|/ Constant current

I1 = (I + Vin/R1) and I2 = (I - Vin/R1).

First find out where I1 sets the opamp +ve input (Vx).

I(R3a) = (Vs-Vx)/R2a - I1 = (Vs-Vx)/R2a - (I + Vin/R1).

Vx = I(R3a)*R3a = [ (Vs-Vx)/R2a. - (I + Vin/R1) ]*R3a

The opamp negative feedback also sets the -ve input at Vx.

So I(R3b) = (Vs-Vx)/R2b - I2 = (Vs-Vx)/R2b - (I - Vin/R1).

So V(R3b) = I(R3b)*R3b = [ (Vs-Vx/R2b) - (I - Vin/R1) ]*R3b.

Voltage out, Vout = Vx - V(R3b).

Vout = [ (Vs-Vx)/R2a - (I + Vin/R1) ]*R3a - [ (Vs-Vx)/R2b - (I - Vin/R1) ]*R3b

Now make R2a=R2b=R2 and R3a=R3b=R3, and cancel things.

Vout = -2.Vin*(R3/R1).

It's simpler to look at the collector circuits as Norton current sources with shunt impedances the R2's. Then each is driven by a differential current of +/-Vin/R1 which is an equivalent voltage drive of

Probably. Except that I find it easier to get to it via that big sum for Vout, which can then be expanded-out to show the effects of the various mismatches, etc..... which affect the Gain, DC Offset or PSRR, or whatever.

Note that whilst the R2's do cancel out, their value has a marked effect on the front end Gain, and therefore on the equivalent input noise.

That's a different circuit than the original.

...Jim Thompson

I was trying to avoid even looking at those silly resistors in series with the opamp inputs. But lets drop them in. R4a and R4b below.

-------+-------------+---Vs | | \\ \\ R3b /R2a R2b/ +-/\\/\\---+-->Vout \\ \\ | | | | R4b | _ | | +--/\\/\\---+--|- \\ | | | Vx | >-+ +-------------|--/\\/\\---+--|+_/ | | R4a | \\|/I1 I2\\|/ \\ | | /R3a |/ \\| \\ Vin ---|npn npn|----. | |\\e e/| | | 0v--- | R1 | ---+--+---0v +----/\\/\\-----+ | | \\|/ I I \\|/ Constant current

Doing similar sums as before, I get...

Vout/Vin = -2(R3/R1)*( R2/(R2+R4) ).

Now put some spin on it..... Call R4 = K*R2.

Vout/Vin = -2(R3/R1)*( 1/(1+K) ). :-)

nope, r2a,b now soak up some of the differential current.

Colin =^.^=

oops missed the fact that r2 is included via K, didnt realise youd changed your mind about them cancelling.

I didn't change my mind as such because in the first instance I didn't realise that anyone was considering those series resistors seriously.

I must admit though that the slippery quickstep to the 1/(1+K) was in the hope that JT would read it with his bad eye and not realise what was going on. :)

Right, it's a slick single-stage amplifier, compared to Phil's two-stage amplifier.

Tony, thanks loads, for taking the time to put up an ASCII drawing, where everyone can be on the same page in the discussion. This business of posting web page links, or placing pdfs on a.b.s.e., etc., simply does not lend itself well to a conversation.

I imagine one of Jim's criticisms of the circuit, I'm guessing since he refused to spell it out, is that the R4 resistors take away from the potential gain of the overall amplifier, or, for a set gain, take away loop gain that could be used to reduce distortion, etc. Your first (top) drawing without the R4 resistors is better, but it still suffers from the fact that the R2 resistors will have to be smaller than the R3 resistors, shunting current. Even though R2 doesn't appear in the gain equation, they are degrading the performance. One solution is to replace R2 with current sources, Ix. These would have to be servo'd to match I1, with another opamp looking at the average value of Vx compared to some bias voltage, etc.

servo'd constant current | | R3b \|/ Ix \|/ ,-/\/\---+-->Vout | | | _ | | +---------+--|- \ | | | | >-' Vx---> +-------------|---------+--|+_/ | | | \|/I1 I2\|/ \ | | /R3a |/ \| \ Vin ---| npn npn |----. | |\e e/| | | 0v--- | R1 | ---+--+---0v +----/\/\-----+ | | \|/ I I \|/ Constant current

This would be a bit of a mess, for an arguable improvement, but possibly not something an IC designer, who generally has lots of silicon available, would shy away from.

Now, to change the subject to an item I'm not at all happy with, the huge ugly electrolytic in the input-pair emitter path, see C1 below.

-------+-------------+---Vs | | G = 2 R3 / R1 \ \ /R2a R2b/ R3b \ \ ,-/\/\---+-->Vout | | | _ | | +---------+--|- \ | | | | >-' Vx---> +-------------|---------+--|+_/ | | | | | \ | Q1 Q2 | /R3a |/ \| \ --+ --| npn npn |--. | Vin | |\e e/| | | --+-|-- | R1 C1 | --' -+---0v | | +--/\/\--||---+ \ \ | | Q1 Q2 composite / / \|/ I I \|/ Sziklai pairs | | Constant currents --+-+-0V

C1 is deemed necessary to prevent the offset voltage of the input NPN transistors, which could be 25mV or even 50mV, from saturating the output at G = 1000. R1 is only 22 ohms at the highest gain, so C1 has to be large, 1000uF for a 7Hz rolloff, in Phil's design,

But C1 is trouble for several reasons. One problem is the distortion created from the high dielectric absorption for electrolytics. I've read that this is an issue for frequencies up to 10x away from the -3dB frequency, which implies degradation up to 70Hz at the highest gain. I'd like to eliminate C1, and rely on the set of C2 capacitors at the opamp input to deal with the offset voltage. Phil used 47uF electrolytics, which gives a 1.5Hz rolloff compared to his R2 = 2.2k resistors, which nicely moves the 10x distortion region down to 15Hz.

And as a bonus, the 100ms time constant is fixed, independent of gain. This compares to C1 R1, which varied from 22ms to 10 seconds over the gain range.

-------+-------------+---Vs | | G = 2 R3 / R1 \ \ /R2a R2b/ R3b \ \ ,-/\/\---+-->Vout | | C2b | _ | | Vy--> +----||---+--|- \ | | | C2a | >-' Vx---> +-------------|----||---+--|+_/ | | + - | | | \ | Q1 Q2 | /R3a |/ \| \ --+ --| npn npn |--. | Vin | |\e e/| | | --+-|-- | R1 | --' -+---0v | | +----/\/\-----+ \ \ | | Q1 Q2 composite / / \|/ I I \|/ Sziklai pairs, | | Constant currents matched offsets --+-+-0V

I'd solve the dc-offset problem simply by selecting Q1 and Q2 for a maximum offset voltage of say 5mV. The input transistors are cheap, easily allowing for a small pile to choose from, and they should be checked for noise anyway, before use.* If we assume a 10mV worst-case input offset, this results in a modest +/- 1-volt deviation from nominal at points Vx and Vy. With C2 in place, that's acceptable.

What's more, thanks to the use of Sziklai pairs, the operating current of each input transistor is unaffected, because the current change is taken up by the second transistor in each Sziklai pair.

So that's three parts taken from Phil's microphone amplifier, arguably for an improved performance.

• Reading in the forum suggested by Martin Griffith
  formatting link
  some in the studio pro-audio crowd like the 2n4403 pnp transistor for low-noise input stages (most of us instrumentation engineers prefer other parts). Apparently Motorola made very quiet ones in the good old days, and Fairchild's parts are pretty good now. But the 2n4403 isn't specified or suggested for low noise, and surely each one should be individually vetted for such use.

In article , Winfield Hill wrote: [snip]

I still don't like the C2's where they are.

My first thought for solving the DC offset problem for an AC amplifier was to sense the output offset, integrate, and use it to unbalance the CC sources in the input tails. As sketched below.

-------+-------------+---Vs | | G = 2 R3 / R1 \\ \\ /R2a R2b/ R3b \\ \\ .-/\\/\\---+-->Vout | | | _ | | +---------+--|- \\ | | | | >-+ Vx---> +-------------|---------+--|+_/ | | | | | | | \\ \\ | Q1 Q2 | /R3a / |/ \\| \\ \\ --+---| npn npn |--. | | Vin | |\\e e/| +----+---0v | --+-|-- | R1 | | | | +----/\\/\\-----+ .----||----+ \\ \\ | | | _ | / / \\|/ I I \\|/ | / -|---' | | Constant currents

That's much better, Tony. I was sure you'd have a bright idea. BTW, why do you keep drifting back to a single-ended input? Do you have more to tell us?