I can't seem to decide which is the correct answer for this question.

Diode circuit

An AC voltage of 5 V (peak) is applied to the circuit shown. What is the peak inverse voltage (PIV) of the diode? Silion diode (0.7v)

a) 5v
b) -4.3v
c) 0v
d) -0.7v

I know it could be either A or B.

  • \$\begingroup\$ How do you know it's A or B? Expound on your thinking and see if that helps you come up with the answer - like think about just what PIV means. Also think about what the other values in the list of answers could represent besides PIV. All those values can be found in that circuit at different times and in different places. \$\endgroup\$ – Majenko Dec 6 '14 at 13:36
  • \$\begingroup\$ What do you understand by "inverse voltage" in this context? \$\endgroup\$ – Brian Drummond Dec 6 '14 at 14:02

Since the answer has already been spoon fed to you (although edited out now, so I'll take out my references too) I guess I can now give you the reasoning behind the answer, and also expand on my comment to the question.

Of the four possible answers, all four are present in the circuit in different places at different times.

The circuit can be seen, in this situation, as being in two possible states:


simulate this circuit – Schematic created using CircuitLab

In the left hand circuit the diode is reverse biased, and no current will flow through it. In the right hand circuit it is forward biased and current will flow through it. Let's take the right-hand, forward biased, circuit first:

Current flows through the diode, so there will be a voltage drop of -0.7V across it. It's -0.7V since ground is the +5V point of the battery, so all the voltages are below that, i.e., negative WRT ground. With -0.7V across the battery that leaves -4.3V left to be across the resistor R. So that is your answers D and B accounted for respectively.

So let's look at the left hand circuit looking at the same values.

The diode is reverse biased. No current will flow through it. Therefore no current will flow through the resistor either. Since V=R×I, and I is 0, then V must also be 0, regardless of the resistance. So the voltage across the resistor is 0V (answer C), which leaves the battery's full 5V left across the diode. That's answer A.

But which is the right answer for your question? Well, as the others have noted, the question is badly worded. What it should be asking is not what is the Peak Inverse Voltage of the diode, but the Peak Inverse Voltage that gets applied to the diode.

The Peak Inverse Voltage is the maximum it can cope with when reverse biased. So that means that it must be the left hand, reverse biased, circuit we are interested in. The voltage applied across the diode in that circuit, as we have already discovered, is 5V.

So which is the right answer to your question do you think?

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Tell your teacher that the question makes no sense because there's no way of determining the diode's PIV from the data given.

For instance, assume that in one case a diode with a PIV of 1000 volts was shown, and in another case a diode with a PIV of 100 volts was shown.

Ask your teacher how they'd determine the different PIV's from the circuit and data given.

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  • \$\begingroup\$ I agree. PIV is a diode max rating irrespective of its application circuit. \$\endgroup\$ – Andy aka Dec 6 '14 at 14:15
  • \$\begingroup\$ en.wikipedia.org/wiki/Peak_inverse_voltage has both device limit and voltage in a particular rectifier application as usages. \$\endgroup\$ – Pete Kirkham Dec 6 '14 at 14:50
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    \$\begingroup\$ @PeteKirkham It's all a matter of context. As a stand-alone value it's the absolute maximum. In a sentence such as "The Peak Inverse Voltage applied to ..." it's a value imposed by a circuit, as above. It's the context that's lacking from the original question. \$\endgroup\$ – Majenko Dec 6 '14 at 15:11
  • \$\begingroup\$ @Majenko: I disagree that the context was lacking, I just think it was applied incorrectly; perhaps in an attempt to make the question tricky as some instructors are wont to do. \$\endgroup\$ – EM Fields Dec 6 '14 at 19:21
  • \$\begingroup\$ @EMFields then he needs sacking - with a big sack. And some rocks. And a deep river. \$\endgroup\$ – Majenko Dec 6 '14 at 19:22

"EM Fields" is totally right on the fact that it is probably not the peak inverse voltage what is asked for.

However when your teacher means the voltage over the diode in this circuit when it is reverse biased? Ask yourself what kind of voltage drop exsists over the resistor when the diode is not conducting.

Simply said a diode conducts when it is forward biased. When it is reverse biased (and then the voltage over it, is called the inverse voltage), it does not conduct current.

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    \$\begingroup\$ Well done for just giving him the answer. You should help him to come up with the right answer by himself instead of just spoon feeding him. That way he will learn the theory behind it, not just regurgitate what he has been told. \$\endgroup\$ – Majenko Dec 6 '14 at 14:39
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    \$\begingroup\$ You were right. I edited my answer a bit. However, I most of all think that the teacher should make better questions because this is just a bad question. \$\endgroup\$ – Douwe66 Dec 6 '14 at 14:50
  • \$\begingroup\$ It's both an exercise in circuit analysis, and an exercise in context inference. Maybe it's a cross-over English Language / Electronic Engineering course...? \$\endgroup\$ – Majenko Dec 6 '14 at 15:21

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