Do we include the voltage drop across a diode when calculating the current through the diode, taking into account breakdown has occured?

So, am doing this question paper for my electronics, and I stumble across a question that asks me to calculate the current through a diode, with a breakdown voltage of 1.7, an Imax current of 20ma, when the diode is hooked up to a resistor of unknown value (calculated on a further question). There are 5 pairs of these LED's/resistors in parallel by the way, but only asks me to calculate the voltage of one diode. My intuition told my to subtract 1.7V from 9V, which would enable me to calculate currents and resistances. I was right. However this is only due to previous experience where I got it wrong.

I don't understand WHY we don't include the forward voltage drop, surely that's part of the overall resistance, as V=R*I and the diode cannot conduct until it reaches 1.7V due to the resistance faced by the PN junction... so, wouldn't excluding the fact that 1.7V is being dropped due to the PN band at the diode lead to false results?

You can't use Ohm's law to calculate the current through a diode.

However, the resistor is in series with the diode and, as you've learned, series connected circuit elements have identical currents.

Thus, you can use Ohm's Law to find the current through the resistor and diode but, to do that, you need the voltage across the resistor.

When you subtract the diode voltage from the source voltage, by KVL, what remains is the voltage across the resistor $V_R$.

Now, if you know the desired current, e.g., $20mA$, you can calculate the resistance required for that series current.

Thus, by Ohm's Law, the desired resistor value is:

$$R = \dfrac{V_R}{20mA} = \dfrac{V_S - V_D}{20mA} = \dfrac{9V - 1.7V}{20mA}$$

• Well, technically you can use Ohm's law, but only after you've applied too much current and converted your LED into a DED (dark-emitting diode). At that point, it looks like a very large value resistor. Dec 15, 2013 at 15:32