using new technology on old machines

Brent Hilpert hilpert at
Wed Jun 17 03:12:46 CDT 2015

On 2015-Jun-16, at 3:35 PM, Dave G4UGM wrote:
>> -----Original Message-----
>> From: cctalk [mailto:cctalk-bounces at] On Behalf Of tony
> duell
>>> .. its also a nasty hybrid design with DC biased NPN and PNP
>>> transistors. I find it ugly and can see it being a pig to debug,
>>> though it simulates fine in LTspice...
>> I didn't find it that hard to basically understand in my head. After all,
> there
>> are only 4 transistors, and 2 of those are just an output buffer. Quite
> why
>> having both NPN and PNP transistors makes it harder to understand I do not
>> know.
> I am really used to RF circuits so am puzzled there is no inductor. It kind
> of looks like a Darlington Pair but it isn't.
> What I don't understand is why the emitter of Q1 is spliced in what I assume
> is a voltage divider in the collector of q2.
> I was expecting a multivibrator circuit...

I think you could characterise it as:
	- Q1 & Q2 form a two-stage direct-coupled non-inverting amplifier from Q1-B to Q2-C.
	- C5, the main timing cap, is thus effectively in AC positive feedback, as one would expect for an oscillator.
	- R9 (the R between Q2-C and Q1-E you were questioning) provides DC negative feedback,
	  presumably to help stabilise and fix the circuit characteristics, as some measure of stability is going to be
	  needed for a baud-rate generator. Note it also has it's own voltage regulator in D1 to D4.

At first glance I thought R9 might be there to provide some hysteresis in the switching thresholds for the RC charge/discharge but it looks like it acts in the opposite direction to that.

The base circuit of Q3 (the first stage of buffering) will draw current from the high-impedance side (R8,R9) of the oscillator output, pulling up the C5,R9 junction when Q2 is off, so it will probably affect the oscillator and be necessary to get the 'proper' functioning of the oscillator portion of the circuit.

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