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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.
 
P

Phil Hobbs

Jan 1, 1970
0
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
http://users.bestweb.net/~hobbs/canceller/noisecan.pdf, top of P. 912.

Cheers,

Phil Hobbs
 
J

John Popelish

Jan 1, 1970
0
Phil said:
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
http://users.bestweb.net/~hobbs/canceller/noisecan.pdf, top of P. 912.

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

Jim Thompson

Jan 1, 1970
0
Not really applicable to my circuit, I think, but a good reference
paper. Thanks.

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: <[email protected]>

...Jim Thompson
 
J

John Popelish

Jan 1, 1970
0
Jim said:
Finally had time to add linear asymptotes, all nicely temperature
compensated....

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.
 
J

Jim Thompson

Jan 1, 1970
0
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.
Correct.


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.

Provided each has its own TC'd divider.
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.

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
 
J

John Popelish

Jan 1, 1970
0
Jim said:
(snip)


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.

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.
 
J

Jim Thompson

Jan 1, 1970
0
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.

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

...Jim Thompson
 
Jim said:
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.

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:
http://www.panasonic.com/industrial/components/pdf/ARG0000CE2.pdf
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.
 
J

John Popelish

Jan 1, 1970
0
Jim said:
(snip)

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

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

High beta transistors don't hurt, either.
 
J

Jim Thompson

Jan 1, 1970
0
Got it. That is where the "perhaps lower impedance version of the
signal divider" came from.

High beta transistors don't hurt, either.

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

...Jim Thompson
 
J

John Popelish

Jan 1, 1970
0
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.
 
J

Jim Thompson

Jan 1, 1970
0
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 think it takes you into "Lambert W" land.
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.

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
 
J

Jim Thompson

Jan 1, 1970
0
On Fri, 27 May 2005 14:09:37 -0700, Jim Thompson

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

...Jim Thompson

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

...Jim Thompson
 
J

John Popelish

Jan 1, 1970
0
Jim said:
I think it takes you into "Lambert W" land.
So I have another educational experience to look forward to.
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.

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.
 
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