[sdiy] Help understanding Serge 73 Oscillator

Guy McCusker guy.mccusker at gmail.com
Wed Feb 21 13:29:17 CET 2018


Thanks for the excellent replies! I am pretty sure I understand it
now. I couldn't get past the fact that what I think of as a current
input was being driven by a voltage. The TI application note does not
contain any suggestions that this sort of thing could be done. Serge
is very sneaky.

Dmitri is right that part of the circuit is included in the Sonica
patent: https://patentimages.storage.googleapis.com/eb/3d/bc/2b178942fd0730/US4306480.pdf
Figure 4 shows the oscillator core, but the non-inverting input is
driven by a current (voltage through resistor) there so the sneaky
part on the 73 Oscillator design is not being used.

As a follow-up question: the schematic hosted on electro-notes makes a
point that some of the transistors are matched. Is there a certain
amount of temperature compensation at play here, because the voltage
coming from the first stage is in part determined by the response of a
transistor which matches the one being used to generate the
exponential current that goes through the integrator? If so it is
natural to wonder how well that works.

Thanks again,
Guy.

On Wed, Feb 21, 2018 at 3:28 AM, Dmitri SFC <dankedout415 at gmail.com> wrote:
> Funny I think Serge possibly did patent that circuit as part of the Sonica.
> http://www.keytarcentral.com/tag/sonica/
>
> On Tue, Feb 20, 2018 at 2:30 PM, Ben Bradley <ben.pi.bradley at gmail.com>
> wrote:
>>
>> I haven't seen this or any circuit using an LM3900 as an exponential
>> converter before, but I recall using the LM3900 in circuits based on
>> the datasheet circa 1980. But understanding the chip's operation (and
>> the exponential nature of Ic vs. Vbe) I figured it out.
>>
>> The LM3900 inputs "respond" to current, rather than voltage as in
>> regular op-amps. The output, though, is still a voltage. The input
>> voltages pretty much stay at a diode drop above the negative or
>> "ground" power input to the LM3900.
>>
>> If the + input has more current going into it than the - input does,
>> the output goes toward the positive rail, otherwise it goes toward the
>> negative rail. If it has a resistor in the feedback path as R17 in
>> U1A, the output voltage will change to make the current into the -
>> input the same as the current into the + input.
>>
>> The + input goes to the first two transistors that make up a current
>> mirror, to change the direction of the current. This goes directly to
>> the - input, so the inverted current from the + input is subracted
>> from the - input. The third transistor and the gain stage handle the
>> amplification and voltage output.
>>
>> U1a is wired as a voltage output at the connection of R17 and R19. It
>> appears R19 doesn't "do" much here except limit the maximum current
>> from U1a so it doesn't destroy the input of U1b.
>>
>> So the input of U1b is a voltage, in the range of about 0.4V to 0.7V
>> or so. It's the SECOND transistor that does the exponential conversion
>> (the first one just pulls current from the input voltage). It has a
>> VOLTAGE input across the base and emitter, and this translates
>> directly through the transistor's exponential function into a CURRENT
>> output through the collector.  This current of course is matched by
>> the - input from C3, which generates a ramp at the output and is reset
>> by Q3 at a rate based on collector current.
>>
>> This is such a neat circuit, I have to ask - is it patented?
>>
>> On Tue, Feb 20, 2018 at 4:27 PM, Guy McCusker <guy.mccusker at gmail.com>
>> wrote:
>> > Hello list!
>> >
>> > I'm having a hard time figuring out how the exponential response is
>> > generated in the Serge 73 oscillator design. The schematic is here:
>> >
>> > http://electro-music.com/wiki/pmwiki.php?n=Schematics.OldSergeVCOUsingLM3900
>> >
>> > Like a lot of Serge designs there is heavy use of LM3900s which I do
>> > not understand properly. It seems to me that the first stage (U1a) is
>> > a more or less regular summing stage, the second (U1b) is an
>> > integrator, and the third (U1c) is a comparator to reset the ramp via
>> > the transistor across the integrating cap. So far so straightforward.
>> >
>> > But how (if at all) is an exponential CV response generated? The notes
>> > on the schematic hint that somehow the transistors within the LM3900s
>> > are being pressed into service as an expo converter but I can't wrap
>> > my tiny brain around it.
>> >
>> > Can anyone help please?
>> >
>> > Cheers,
>> > Guy.
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