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I am trying to design a electronic circuit which will only pass voltage between a threshold, like a dual clipper. The input is a varying DC input and the system (would) allow output only above 3.3V. Initially I tried to design a negative clipper with bias voltage but then the bias voltage powers up the output when the input voltage goes below the bias voltage, which would be detrimental as I would need to recharge the source.

circuit diag

Please suggest alternate approaches.

Edit 1: The source is a low impedance input ranging in 10 - 30 ohm. The load is a boost circuit which boosts the voltage to steady 5V. The voltage tolerance is not an issue because the load can withstand voltage from 2.5 V - 5.5 V. I would like the bias voltage to not be forwarded to the output, thus when the input falls below threshold, the output falls to zero.

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  • \$\begingroup\$ "black level clamp" \$\endgroup\$
    – user16324
    Oct 13, 2016 at 19:33
  • \$\begingroup\$ Define actual source impedance and load, voltage tolerance and can you use Op Amps? \$\endgroup\$
    – Tony Stewart EE75
    Oct 13, 2016 at 19:49
  • \$\begingroup\$ What do you intend to have at the output when your input falls below your lower limit point? Should it drop to 0V or should it stay at that lower limit? \$\endgroup\$
    – brhans
    Oct 13, 2016 at 21:35
  • \$\begingroup\$ The source is a low impedance input ranging in 10 - 30 ohm. The load is a boost circuit which boosts the voltage to steady 5V. The voltage tolerance is not an issue because the load can withstand voltage from 2.5 V - 5.5 V. I would like the bias voltage to not be forwarded to the output, thus when the input falls below threshold, the output falls to zero. \$\endgroup\$
    – Parth Singh
    Oct 14, 2016 at 21:19

1 Answer 1

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schematic

simulate this circuit – Schematic created using CircuitLab

  • Adjust R ratios to suit supply voltage , and 1 diode drop below threshold
  • Output equals input with very little offset.
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  • \$\begingroup\$ I hope you don't take this the wrong way, but this answer, along with some of your other recent answers, would be so much more informative if you explain more about the way the circuit works. You obviously have a vast amount of experience and it's good to share it, but maybe longer narrative is more accessible to less experienced people than just bullet points. If you see answers from some of the other big-hitters - like Olin, Spehro, Russell, Andy, etc. - you'll see what I'm trying to say. \$\endgroup\$
    – Roger Rowland
    Oct 14, 2016 at 5:13
  • \$\begingroup\$ Absolutely no offence intended but as a SW engineer of 40 years experience now getting into EE, I for one would relish a clearer look inside your head :-) \$\endgroup\$
    – Roger Rowland
    Oct 14, 2016 at 5:14

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