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I found this level shifting circuit online. When I simulated this on Multisim, it works fine, but how do I find the relationship between the output and the input. Here is the circuit.

Circuit Diagram

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    \$\begingroup\$ First step: Label one node as the input and another node as the output. \$\endgroup\$
    – The Photon
    Sep 18 '16 at 16:54
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    \$\begingroup\$ Be very wary of the left-hand side of that circuit. If you connect it to mains so that V1 is live then R1 will have 240 V across it and alternate 24 mA between D1 and D2. Your Multisim sees R5, 1M, between the AC and ground but in the real world neutral is normally grounded so that R5 does nothing. Try adding another ground to the bottom of R5 and see what happens. \$\endgroup\$
    – Transistor
    Sep 18 '16 at 17:06
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MultiSim uses input and output nodes to display signals.

However I recognize this circuit is simply a poor R divide by 2 that will dissipate too much power ( 100k ~1M is better) will clip the 700Vpp grid voltage via Schottky diodes with a unity gain Op Amp that (98% of) 10Vpp output.

A better way would Ac couple with 1M to 10k then use a comparator with 1% positive gain feedback for hysteresis using Vin- =0V and signal input via series 10K with 1M feedback on Vin+

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit You can string series 500V thru-hole parts for safety to achieve 3kV BDV rating. C shunt across 10K filters out transients that may cause false triggers.

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  • \$\begingroup\$ And don't forget to check the max voltage rating of R1 and R2. Most 240 V circuits have to use two resistors in series - 500k + 500k in your case. \$\endgroup\$
    – Transistor
    Sep 18 '16 at 19:15
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    \$\begingroup\$ Why are you commenting on your own post rather than editing it? \$\endgroup\$
    – Transistor
    Sep 18 '16 at 19:16
  • \$\begingroup\$ what comments? ? \$\endgroup\$ Sep 19 '16 at 0:48
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R2 and R3 make a Thevenin equivalent voltage and resistor. Apply this to your 10k input resistor and you should find the relationship between V1 and the output, since the op amp is configured as a voltage follower, and the output will be very close to the input.

That said, the circuit as shown will almost never produce anything other than a 50 Hz more-or-less square wave at 0 and 5 volts. The rise and fall times will not be terribly fast, since the linear portion of the output is the +/- 15 volt portion of the 120 volt line voltage swing (the input 240 VAC is cut in half by the 1M resistors, assuming no connection to ground at the AC).

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