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As part of a graded assignment on diodes, I was given this question. Text I ended up designing this circuit :-

schematic

simulate this circuit – Schematic created using CircuitLab

However, I only got partial credit for the circuit. The official solution was :- Text

With each of the zener diodes having a breakdown voltage of \$2.5\,\text{V}\$. I want to ask before I challenge the grading. Will my circuit also work?

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    \$\begingroup\$ "There also is a resistor in series with the sources which isnt shown" - Why isn't it shown? You drew a circuit (+1) but then you didn't supply the whole circuit, which contains very relevant information to your question? (-1) \$\endgroup\$ – Arsenal Apr 9 at 12:35
  • \$\begingroup\$ Edited the circuit to add the resistor.I am sorry it wasn't there before, I forgot to add it while drawing and didn't know how to edit. \$\endgroup\$ – Mall Apr 9 at 14:49
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Zener diode works as a normal diode in forward biasing and works as voltage regulator in reverse biasing.

Since the diodes used in your problem are ideal, you can consider them as short in forward bias and as voltage drop of \$ 5 \, \text{V}\$ in reverse bias.

So I think you should remove the voltage drop you made, replace it with a short circuit and the circuit will work but you should notice that it will produce square wave which has value of \$5\,\text{V}\$ in the first half cycle and value of \$-5\, \text{V}\$ in the second half cycle.

Adding a generic diode for the output of your circuit will rectify the output signal and make the peak value to be zero in the second half cycle.

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