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Old 22nd September 2011, 04:05 AM   #181
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Thanks Sheldon, that straightforward response was what I was hoping for. Referencing it to a CCS helped immediately. So the voltage across R2 forms the Vref for Q1 to balance through R1. What I wonder about next is the zener above the base of Q2. Does it mean that the change in output voltage when adjusted is all at Q2's Vce ? - And if so does that mean that Q1 is able to do that by simply controlling the current into the emitter of Q2?

Sorry to ask all this - I'll build it anyway, but I am not yet a spice user and can't just sim it to see what happens.

Thanks again !
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Old 22nd September 2011, 04:54 AM   #182
Sheldon is offline Sheldon  United States
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Quote:
Originally Posted by Hearinspace View Post
Does it mean that the change in output voltage when adjusted is all at Q2's Vce
If I understand the question, no. The zener sets a constant voltage (which, of course, has to be lower than the minimum voltage drop required at the collector of Q1) at the base of Q2. This sets a constant voltage at the collector of Q1. This improves the performance of the current source, because it works with a constant Vce.

The output voltage of the gyrator is set by the tubes response to the the current from the gyrator. With a regular CCS, as the tube ages and drifts over time, the plate voltage can change quite a lot. However, in this set up, as the tubes response to current changes over time, the voltage divider adjusts the current to maintain a constant plate voltage. The time constant of C1 and R2//R3 must be long enough so that signal frequencies are not seen at the base of Q1.

Sheldon

Last edited by Sheldon; 22nd September 2011 at 04:57 AM.
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Old 22nd September 2011, 07:09 PM   #183
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Thanks Sheldon. I knew that necessarily the output of Q2 had to follow the bottom of the voltage divider but was unclear about what the transistors were doing to make it happen. Without confidence in the basics of transistor function, Mickey started conjuring and it all got out of hand. Now that you've explained it (and I see it) it's so simple.
Many thanks !
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