Ohmmeter & Diodes Homework: Understanding Results

[JeeebeZ]

Homework Statement

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

 

[NascentOxygen]

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.)

 

[rude man]

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/V_T) where i₀ = reverse saturation current and V_T ~ 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 i<sub>0/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?</sub>