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

The Phantom · May 19, 2005

Tell us what happened.

Phil Hobbs · May 26, 2005

jpopelish@rica.net wrote:

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

John Popelish · May 26, 2005

Phil Hobbs wrote:

jpopelish@rica.net wrote:

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.

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

Jim Thompson · May 26, 2005

On Thu, 26 May 2005 18:03:01 -0400, John Popelish <jpopelish@rica.net>
wrote:

Phil Hobbs wrote:
jpopelish@rica.net wrote:

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.

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: <5ukc91pt19s3uo33kt595vnpicdes25223@4ax.com>

...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

John Popelish · May 27, 2005

Jim Thompson wrote:

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.

Jim Thompson · May 27, 2005

On Thu, 26 May 2005 21:16:50 -0400, John Popelish <jpopelish@rica.net>
wrote:

Jim Thompson wrote:

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.

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.

480)460-2350 | |
| E-mail Address at Website Fax

John Popelish · May 27, 2005

Jim Thompson wrote:

On Thu, 26 May 2005 21:16:50 -0400, John Popelish <jpopelish@rica.net
wrote:
(snip)
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.

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.

Jim Thompson · May 27, 2005

On Fri, 27 May 2005 09:23:09 -0400, John Popelish <jpopelish@rica.net>
wrote:

Jim Thompson wrote:
On Thu, 26 May 2005 21:16:50 -0400, John Popelish <jpopelish@rica.net
wrote:
(snip)
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.

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.

480)460-2350 | |
| E-mail Address at Website Fax

John Popelish · May 27, 2005

Jim Thompson wrote:

On Fri, 27 May 2005 09:23:09 -0400, John Popelish <jpopelish@rica.net
wrote:

Jim Thompson wrote:
John Popelish <jpopelish@rica.net>wrote:

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.
(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.

Jim Thompson · May 27, 2005

On Fri, 27 May 2005 17:03:31 -0400, John Popelish <jpopelish@rica.net>
wrote:

Jim Thompson wrote:
On Fri, 27 May 2005 09:23:09 -0400, John Popelish <jpopelish@rica.net
wrote:

Jim Thompson wrote:
John Popelish <jpopelish@rica.net>wrote:

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.
(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.

480)460-2350 | |
| E-mail Address at Website Fax

John Popelish · May 28, 2005

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.

Jim Thompson · May 28, 2005

On Fri, 27 May 2005 14:09:37 -0700, Jim Thompson
<thegreatone@example.com> wrote:

[snip]

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

...Jim Thompson

480)460-2350 | |
| E-mail Address at Website Fax

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

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