using new technology on old machines

[Jonas Otter]

On 2015-06-17 13:28, Dave G4UGM wrote:
> I found it easier to think of it in DC terms. So the Cap charges through R5
> + R3 and R9 + R8.
>
> As the Cap charges the voltage on the base of Q1 rises until it turns on,
> which then turns on Q2.
While the cap charges, it steals the base current which would otherwise 
have gone to Q1, thus keeping Q1 turned off. When the cap nears the end 
of the charge, more current goes to the base of Q1 which turns on, 
turning on Q2, which raises the voltage over R8 and R9. Since the 
voltage on a capacitor cannot change instantaneously, the voltage on the 
base of Q1 rises while the cap discharges through the base of Q1, 
keeping it hard on. As the cap discharges and charges in the reverse 
direction, the base current of Q1 decreases and ultimately Q1 turns off, 
turning off Q2 and lowering the voltage over R8 and R9, and the cycle 
starts over.

For the circuit to work, I think (I may be wrong) the base current 
supplied to Q1 by R5 and the pot has to be not quite sufficient to turn 
it on. Also the cap is reverse charged for one half cycle.

I believe this is a classical astable multivibrator circuit, but not the 
more common one with two cross-coupled transistors with capacitors from 
the collector of one to the base of the other. The DEC circuit I think 
can be seen a lot in old Siemens application books from the 1960s, such 
as may be found here (note German books): 
http://rainers-elektronikpage.de/SIEMENS-Fach---u_-Datenbucher/siemens-fach---u_-datenbucher.html

/Jonas
>
> At this point the cap is then charged (or discharged) in the reverse
> direction via Q2, D5 and R4 until Q1 turns off.....
>
>
>> 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.
> I included that in my LTSpice model....
> ... but it doesn't actually have that much effect...
>
> Dave
> G4UGM
>
>