[sdiy] Current source for LM13700
shedsynth at gmail.com
Wed Apr 8 15:02:26 CEST 2020
I'm very definitely not an electronics engineer, but I found this very helpful when building my VCAs:
I built a mixture of Figure 7 and Figure 11, driven by an op-amp summing amp though a preset pot gingerly adjusted to limit the OTA input to 2mA measured on a meter.
You might like this idea: input CV on a centre-detent pot into a polarizing mixer (cf Matrix Mixer II at http://www.doepfer.de/DIY/a100_diy.htm) so it's easy to configure two VCAs as a panning control.
From: Synth-diy <synth-diy-bounces at synth-diy.org> On Behalf Of Richie Burnett
Sent: 07 April 2020 17:32
To: Tom Wiltshire <tom at electricdruid.net>; SDIY List <Synth-diy at synth-diy.org>
Subject: Re: [sdiy] Current source for LM13700
> Firstly, there’s this example:
> As I understand it, the current would be:
> Iout = Vin / Rin (so R88 / 33K in this example)
Yes, because no significant current flows into the inputs of an op-amp.
> With a 33K input resistor, you’d need 66V before you went over the 2mA
> that’d damage the 13700. So what does R87/6K8 do? Is it just
> belt-and-braces protection?
I guess it provides additional protection if the op-amp output went to the positive rail and risked reverse breakdown the B-C junction of the transistor.
> Similarly with D3. Testing the circuit in a simulator suggests that
> you don’t get any output current with negative input voltages (the
> transistor is switched off). So what’s the need for D3? Is it there to
> protect the transistor junction when it is reverse biased?
It's there to maintain feedback around the op-amp if the input was to go even the slightest bit negative. Otherwise the output of the op-amp would go to the positive rail and risk breakdown of the BC and BE junctions. It essentially just provides a current path to stop the op-amp going open-loop when the input goes negative.
> Secondly, there’s this example by Olivier Gillet:
> -mkII-v02.pdf What is the purpose of R7 / 2K2? Its value doesn’t
> affect the output current, so what does it do?
Again, for current limiting if something goes wrong. He just put it in the emitter path instead of collector. It doesn't matter since for high hfe collector current and emitter current are practically equal.
> C8/100n presumably provides LP smoothing (a -6dB rolloff) to the PWM
> input, but which resistor does it work in conjunction with? Is it R7
> or R8?
Phase-lead compensation, I would say. To make sure the op-amp is stable right up until wherever the loop gain eventually crosses unity, and it doesn't oscillate at some MHz frequency.
There is also a common-base (grounded base) single transistor method of providing Iabc that Roland seem to use a lot in their early analogue synths, that doesn't require an op-amp.
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