Ohmmeter & Diodes

  • Thread starter JeeebeZ
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  • #1
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Homework Statement



Midterm 1-Q1.jpg


2. The attempt at a solution

So, I know that you can test a Diode with a ohmmeter.

Based off of the R you get Very large reversed biased and fairly low forward biased. But I don't know why it would show these values.

The Large one, revered biased. Kinda makes sense to me since it would have to have a large current to breakdown the diode. potentially...? I'm not quite sure really.

Since Rd is generally about 7ish Ohms, I assume both have to do with the internal current required to drive the diode.

any ideas would be helpful, THX
 

Answers and Replies

  • #2
NascentOxygen
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Hi JeeebeZ. can you estimate the current (roughly) that the circuit will send through the diode? (To do this, you need to estimate the voltage, E.)
 
  • #3
rude man
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For the forward direction your ohmmeter will measure V/i where i is the current it sends thru your diode and V the voltage across it. The relationship for your diode is (p-n junction assumed) i = i0exp(V/VT) where i0 = reverse saturation current and VT ~ 26 mV at room temperature. V will be around 0.7V for a Silicon and 0.3V for a Germanium diode.

From this formula you can see that the current for any negative V (reverse voltage) is ~ i0 which is maybe 2e-15A or 2e-12 mA. This is called the leakage current since for an ideal diode the reverse-voltage current = 0.

However, when you hook up the reverse connection so that V is negative, depending on your battery voltage used for the high - resistance settings, the diode MIGHT "break down" and conduct a lot more than i0/SUB]. This is unlikely; for most diodes like the popular 1N4148 the breakdown voltage is about 100V.

Referring to the question, what is di/dV which is 1/(dynamic resistance), vs. i/V which is 1/what the ohmmeter measures?
 
Last edited:

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