Sure, here it is.
+5V>-+-----------+---+---------------+-----+---+ | | | |R1 | | | | COM | | 10K] | | | +------+ | | | | CR1 | | | |K | O | | +-[ | | | | [200] | | | +------+ | | | | | |Rd | | | NO | | | | [10K]-+---G NCH | | >-+--[2.2M]---+---|----|-/ LM393 S VIN>---[10K]-+--|-/ LM393 |+ | | U1B | | | | U1A [51µF]| | | [200] [1N4148]| C1| | | | | |A | | | | | GND>-+-------+---+---------------+---+-----+---------+
Also, I've set the input threshold to switch at 0.1V, so that eliminates one pot. You need the other one (Rd) to adjust your amplifier turn-off delay time because of the tolerance and leakage current of C1. Also, I've decreased the value of C1 to 51µF in order to decrease its potential leakage current.
--
John Fields
Professional Circuit Designer
.
"John Fields" wrote in message
news:0ca0t15aajektac2j7rja4gnuau4np7tr4@4ax.com...
> On Wed, 18 Jan 2006 21:28:52 GMT, "Mike DeAngelis"
> wrote:
>
> >John , back in December you gave me a circuit for a remote turn on for an
> >amplifier. I am just getting to test this circuit and I have some
> >oscillation on the relay when it drops out. I suspect that some positive
> >feedback (hysteresis) is needed. What is the best way to implement this?
>
> ---
> Without being refreshed as to what the circuit looked like, I\'d have
> to guess. Can you post the circuit?
>
> --
> John Fields
> Professional Circuit Designer