Attn: John Popelish--what results from your tanh circuit?

Setting up the tanh amplifier with a maximum voltage gain of 5 was a failure, because the output saturated at 7 volts and I was trying for a

+- 10 volt peak capability. I had to go below about a peak gain of 3 before the output exceeded 10 voltes before saturating, and at that compression factor, the view wasn't worth the climb.

My latest version has a peak tanh voltage gain of 3.75 and parallels the tanh function with a linear gain of 1/4, for a peak voltage gain of

4 (to give me two extra effective A/D bits for small signals) so that once the tanh output saturates around +-9 volts, the gain falls to no lower than 1/4, effectively throwing away 2 bits of resolution for the largest signals. This looks quite good. This is a better approximation of a sort of symmetrical reciprocal gainfunction that would provide a constant relative resolution over the broadest voltage range.

I will post some graphs of calculated versus test data and the A/D resolution (both absolute and relative) on abse, tonight.

Came in late because I've been in a wheat field in northern France for a week. Diff amps can have the effects of extrinsic emitter resistance taken out by applying a little bit of positive feedback to the bases--but it's hard to adjust unless you have some definite null indication. This turns out to be very useful in laser noise cancellers, where it can get you another 20-30 dB SNR at high photocurrents.

There's a picture at

top of P. 912.

Cheers,

Phil Hobbs

Finally had time to add linear asymptotes, all nicely temperature compensated....

Newsgroups: alt.binaries.schematics.electronic Subject: Follow-up - TANH Compressor with Linear Asymptotes - TANH-Compressor.pdf Message-ID:

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

Not really applicable to my circuit, I think, but a good reference paper. Thanks.

I have redrawn and simulated your circuit and think I, at least superficially, understand what you are doing. I think your thermistor does all the temperature canceling. It looks like Q3 and Q4 contribute no temperature sensitive effect, because their emitter currents are controlled in a high gain feedback loop. If this is right, I assume I could replace Q3 and Q4 and their opamps with a linear addition of the input to a subtracter that is used to combine the currents from Q1 and Q2 and still have a compensated design.

If that is correct, I could parallel collector currents from several separately compensated pairs like Q1 and Q2, but run at different currents and with different divider gains at the front end, to generate various arbitrary but temperature compensated transfer functions.

In fact, I think I could do pretty well eliminate the U2 opamp and connect a compensated (but, perhaps lower impedance version of the) signal divider directly to Q1. But the position of use negative tempco thermistors would have to move to the input side of the divider.

Correct.

Provided each has its own TC'd divider.

Negative tempco thermistors aren't very linear, making good compensation nasty to attain. The QTI PTC thermistor I used is pretty linear, as is the TC of the diff-pair.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

One other thought: Eliminating OpAmps may get you into base current sensitivities, screwing your TC and linearity.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

Understood. I need fair correction over a fairly narrow range. Perhaps 15 to 30 C. I have also arrived at a better understanding of the actual transfer function I want to try for. It is essentially linear over the input voltage range of +-.1 volt, and has an incremental gain inverse to the amplitude (a 1% change of the input produces a 20 mV change in the output) over about .2 to 10 volts and

-.2 to -10 volts. Combining 2 or 3 tanh functions and a linear component, I think I can come very close to this response.

I have been using the two differential amplifiers in the LM13700 as my tanh generators, because I can temperature control the chip with the two darlington devices also on the chip and the two collector currents are already subtracted, but with your (also suggested by Ban) compensation scheme, I can simply subtract paralleled differential collector current pairs and add the linear component, though I do also have to come up with a few stable current sources. Especially handy if I need 3 pairs, instead of 2.

But they do have large tempcos, so I can add a series and parallel resistor to program one to a wide range of tempcos over some temperature range. The narrower the range, the better the fit. For instance, I could use one of these:

say the 10,000 ohm ERTD2FHL103S @ $1.06 from Digikey, parallel it with

9.1 k and series that pair with 24 k and achieve a resistor that has about 29 k at 25 C and has about the right negative temperature coefficient over a 0 to 50 C range, to act as the input side of a divider with a much lower ground resistance for my differential amplifier. If the grounded resistor was actually a string of low resistance values, I have a series of different divided ratio signals all with the about the same compensation, I think.

I think you can create a single linear temperature compensation, then scale it with additional dividers.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

(snip)

Got it. That is where the "perhaps lower impedance version of the signal divider" came from.

High beta transistors don't hurt, either.

Now for my next branch of thought:

What is the expression for the transfer function of a differential amplifier with emitter degeneration resistors? Does this take me into the realm of the Lambert W or is it still some form of the hyperbolic tangent?

I am suspecting that my best use of two differential stages in parallel with a small fixed gain to produce my desired transfer function will be most efficient with one undegenerated stage (high gain, low saturation current) and one with degeneration (lower gain and higher saturation current) but I am having trouble expressing this in Mathcad to have it show me what the optimum combination looks like.

I think it takes you into "Lambert W" land.

Did you not note that the extra differential pair I added is pure linear, because of the feedback from the emitters.

In Mathcad I'd sum TANH + linear with various weightings until I achieved "optimum", whatever that might be, since it's really just your choice.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

[snip]

I should have added... Design your single linear attenuator such that the extrapolated slope passes thru 0°K, to match diff pairs.

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

| 1962 | I love to cook with wine. Sometimes I even put it in the food.

So I have another educational experience to look forward to.

I have little problem setting up a figure of merit for the fit I want, so that Mathcad can search for the optimization I want, once I have an expression or even an implicit description of the functions involved.

I have already done quite a few different optimizations for two tanh functions (both in parallel and in series) and also in parallel with a linear gain. I also looked into adding positive feedback around one or both tanh functions. (That makes my curve fit considerably better, if I could put up with the tolerance exaggerations. I had a Duh! moment when I realized that if negative feedback improves linearity, positive feedback enhances nonlinearity.) So I have a pretty good idea what I can do with those cases.

Now I want to explore how adding emitter resistors to one or both differential pairs alters the optimizations. (My hunch is that one tanh function will be best with nonlinear enhancement [positive feedback] and one will be best with some linear enhancement and input range expansion [emitter degeneration]. But I am having difficulty producing an expression for this transfer function.

I am pretty sure I once knew how to do this. But I am having a senior moment about it, now.