I have the following circuit where both D1 and D2 are ideal diodes and Vin will work from 0V up to 150V:


simulate this circuit – Schematic created using CircuitLab

What I know:

0V - 25V -> only D2 will work (polarized positive)
25 - 100V -> both D1 and D2 will work
100V - 150V -> only D1 will work

What I don't understand:

In each of the three ranges I mentioned what the corresponding Vout will be? Ultimately, what I need is to plot the output voltage as a function of the input voltage.

  • 1
    \$\begingroup\$ Hint: If D2 is reverse biased your output is 100V. If D1 is reverse biased and D2 is ON you have to solve the 2 source 2 resistor circuit to find Vout. If both diodes are forward biased you can replace them with shorts, so what is Vout then? \$\endgroup\$ – John D Aug 18 '16 at 23:50
  • \$\begingroup\$ Why don't you just run the simulator and find out for yourself? \$\endgroup\$ – EM Fields Aug 18 '16 at 23:52
  • \$\begingroup\$ What you put down as "What I know" is incorrect. For example, the range for only D2 conducts is not 0 - 25V. You need to assume D1 does not conduct, then solve for the voltage at junction R1-D2 to deduce the range. The answer given by Wesley Lee has to be modified accordingly to be correct. \$\endgroup\$ – rioraxe Aug 19 '16 at 3:38

I'll consider ideal diodes except for their forward voltage (for the sake of the exercise):

"0V - 25V"

D1 does not conduct. Current on R1, R2 and D2 will be:

100V - 25V - 0.7V = (100k + 200k) * i, so:

i = 74.3/300k = 0.000247666A ~= 0.25mA

So the voltage at Vout will be:

100V - 200k * 0.247666mA = 100V - 49.5332V ~= 50.5V

As Rioraxe pointed out, if Vout is at 50.5V, this means the node between D1, D2 and R1 is at 49.8V and D1 will be reverse biased until Vin >50.5V. So this range should be 0V to 50.5V.

50.5V - 100V

Voltage at the node between D1, D2 and R1 will be Vin - 0.7V, so:

Vout = Vin - 0.7V + 0.7V = Vin

100V - 150V

D2 does not conduct, so Vout = 100V

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    \$\begingroup\$ Thanks for the detailed explanation. One last question, when a diode is ideal doesn't that mean that when it's on it has 0V and acts like a wire or am I wrong? \$\endgroup\$ – DimChtz Aug 19 '16 at 0:02
  • \$\begingroup\$ yes, but for the sake of the exercise I added a 0.7V forward voltage. In this scenario the first two ranges would actually be 0V to 25.7V, 25.7V to 100V I think. Its easier to remove the 0.7V from the calculations than to add them if I didnt put them there. \$\endgroup\$ – Wesley Lee Aug 19 '16 at 0:07
  • \$\begingroup\$ @WesleyLee Please see my comment to OP. You need to adjust the ranges of your answer for it to be correct. \$\endgroup\$ – rioraxe Aug 19 '16 at 3:40
  • \$\begingroup\$ @rioraxe -- thanks for pointing that out. Do you mean the 25.7V voltage I mentioned on the previous comment or some other aspect I'm still missing? \$\endgroup\$ – Wesley Lee Aug 19 '16 at 4:03
  • \$\begingroup\$ As you have calculated with D1 not conducting, Vout = 50.5V. So D1 would start to conduct at around the same, so the first range should be 0 - 50.5V using your assumptions. \$\endgroup\$ – rioraxe Aug 19 '16 at 4:13

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