I'm trying to find out the transfer curve (Vin Vs Vout) of the circuit given below.

enter image description here

I approached this way: enter image description here

Is it the right curve?

  • 3
    \$\begingroup\$ Which pair of nets are Vin and Vout across? \$\endgroup\$
    – Andy aka
    Commented Jul 18, 2017 at 18:48
  • \$\begingroup\$ By looking at your approach, I can't tell how you came to that conclusion. Can you explain how your got your curve? \$\endgroup\$
    – user103380
    Commented Jul 18, 2017 at 21:15
  • \$\begingroup\$ im not sure what happened to your 10vpp sine... but even if youre considering ideal diodes you will have voltage drop across each 1k resistor. and in this case theres regions for vin where only D1, D2 and both D1/D2 are forward biased. \$\endgroup\$
    – apache
    Commented Jul 18, 2017 at 21:30

1 Answer 1




simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Redrawn circuit. (b) Rearranged but equivalent circuit.

  • If you redraw the circuit (Figure 1a) it should become clearer what the D1/D2 junction voltage is without the sine.
  • If you are using real diodes rather than ideal then rearrange as shown in Figure 1b.

Now you should be able to replace the batteries, D1, D2, R1 and R2 with a single voltage source and resistor.


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