I have two of these circuits on a board which are both behaving identically. ie. neither is holding a sampled voltage but rather just passing the input directly to the output (indicated as with the input switch closed).With the switch open and no charge on the cap I would expect to see ~-8V at the Anode of D2 in order for the FET to be completely off but it sits at 0V....is this right? The Voltages indicated are with the input switch open. The circuit is copied from an original schematic. Please help me get this working. Thanks:)


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    \$\begingroup\$ Does the trigger signal ever change to a voltage other than +5 V? What voltage does it change to? (and are you sure you want Q16's base tied to -8 V?) \$\endgroup\$ – The Photon Jan 22 '18 at 16:36
  • \$\begingroup\$ Sorry- The Trigger should be 8V peak. Its sits normally at just below 0V. Everything is as per original schem so unless there's a mistake on it I guess Q16's base is where it should be. \$\endgroup\$ – Sir Cute Jan 22 '18 at 16:41
  • \$\begingroup\$ If the "low" level for trigger is 0 V, Q16 will never shut off. Try connecting Q16 base to 0 V instead of -8 V. \$\endgroup\$ – The Photon Jan 22 '18 at 16:47
  • \$\begingroup\$ @ThePhoton Thanks so very much. That sorted it. How do I give you a thousand points? :) \$\endgroup\$ – Sir Cute Jan 22 '18 at 17:15

In comments you said that the trigger signal is 8 V when high and 0 V when low.

With this input, Q16 will never be shut off, so your FET switch will always be closed.

Try connecting Q16's base to ground instead of -8 V.

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  • \$\begingroup\$ Error was in the original diagram btw. I guess that's to fool people like me ;) \$\endgroup\$ – Sir Cute Jan 22 '18 at 20:31

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