exponential converter temperature stabilization

Doesn't look like it does anything useful. The op amp at top left is keeping the current through the top diode at zero, so if the PNP is passing any current, there'll be a full V_BE across the OTA's inputs, so it's output is always railed.

Cheers

Phil Hobbs

...its output,....

Grr, brain dead autocorrect on my phone!

Cheers

Phil Hobbs

There is a newsgroup...

alt.possessive.its.has.no.apostrophe

The leftmost 1M resistor is to pseudo-ground, and is an input signal terminal.

The current in the two OTA input diodes will mismatch according to the signal presented at the rightmost 1M resistor. It and the OTA output form a current summing junction, and the rightmost op amp keeps that current sum zero.

The central 1M resistor is a bias current source, probably not an input.

It appears that the multiplication is done by the matched input bias diodes of the OTA. Probably the temperature-sensitive gain of the OTA compensates it (but I'd want to scribble over a few sheets of paper to be sure).

On the forum where that came from there are a lot of wacky schemes to try to electronically compensate the Vt dependence of a differential pair. exponential converter.

If it works this seems like one of the cleanest of the bunch.

But it doesn't. The top left op amp needs to run its noninverting input at some nonzero bias voltage, because otherwise there's no current through the top diode, as I said earlier. It can be fixed up, but it's broken as posted.

Cheers

Phil Hobbs

On Friday, August 19, 2016 at 8:23:58 AM UTC-7, Phil Hobbs wrote: ...

...

I think the 1M resistor at far left goes to a negative supply - not ground. The schematic is a bit confusing and ambiguous.

kevin

input

it's

...

Well, I think it's just wrong, meself. returning the 1M at left to +5 would make it do something more sensible.

Cheers

Phil Hobbs

On Friday, August 19, 2016 at 9:54:08 AM UTC-7, Phil Hobbs wrote: ...

ld make it do something more sensible. ...

Surely not - returning it to a negative supply would forward bias the upper diode and cause the opamp output to go positive to maintain the inverting input at ground. In so doing it would make the OTA balanced when the curren t through the transistor is the same as the reference current through the 1 M.

The input current through the right 1M would unbalance the loop. The right opamp would then servo the base of the transistor to rebalance - the curre nt through the transistor (which is also the output) would then be exponent ially related to the current injected via the right 1M because of the diode s at the input of the OTA.

The output (To VCO) would have to go to a VCO that is referenced to a negat ive supply to provide a suitable voltage across the transistor.

kevin

I'm going from memory here, but I thought the 1M went to the OTA input and the op amp was driving the I_diode pin negative. The input 'diodes' are actually diode-connected NPN transistors nominally identical with the diff pair, so you have to dump current into the inputs and suck it back out of the I_diode pin.

So if the 1M is connected to the input, the other end has to be positive for the circuit to work.

Cheers

Phil Hobbs

I've probably got I_diode upside down--I vaguely recall that there's another current mirror on I_diode so you don't need a negative supply, but don't have time to look up the LM13700 datasheet and see. One way or another, the op amp is trying to put a constant bias of zero on one of the input diodes, which needs to be fixed.

Cheers

Phil Hobbs (scrambling madly to get a BOM done by COB today)

On Friday, August 19, 2016 at 12:03:11 PM UTC-7, Phil Hobbs wrote: ...

...

No, it's the other way round. The diode bias pin needs to be positive wrt the diff inputs.

They could be NPN transistors diode connected but they are in series with diff inputs with their collectors joined and brought out to the bias pin.

kevin

On Friday, August 19, 2016 at 12:14:09 PM UTC-7, Phil Hobbs wrote: ...

...

I disagree, I think it just needs a negative input to the 1M. The opamp output will then go positive to bias the input diodes and maintain the zero voltage at the diff input.

kevin

Except that the bias current has to flow into both + and - inputs. I'm on m y phone, so I'm still working from memory, but ISTM there are actually two mistakes there--the bias current polarity on I_diode and the lack of positi ve current into one input.

cheers

Phil Hobbs

e wrt the diff inputs.

because there's a current mirror in there. they aren't PNPs.

pin.

They aren't in series, they're in shunt. Have a look at the schematic in th e datasheet.

cheers

Phil Hobbs

Phil, It's an LM13700, so the diode polarity is bass-ackwards from normal Norton Amplifier thought, so the 1Meg _does_ need to be connected to some negative reference.

But, all-in-all, the scheme is _not_ well thought out. ...Jim Thompson

ive wrt the diff inputs.

s pin.

the datasheet.

I'll wait until you have a computer available.

I agree I was being inaccurate in saying they were in series, I meant that bias current could flow through the diodes and then into the bases of the i nput transistors - from the AC point of view they are in parallel as you sa y.

Here is the schematic for the LM13700 from

cheers

kevin

• ^ VCC

• /_\

• _____|____________________________________________

• | | | | | |

• ->

• |___| |

On Friday, August 19, 2016 at 2:56:09 PM UTC-7, Jim Thompson wrote: ...

...

Ahh, that's why Phil and me are talking at cross-purposes - we made different assumptions.

thanks

kevin

Yep. I was likewise snagged until I looked up an LM13700 datasheet. ...Jim Thompson