# Reversed Bias Diodes in Series

#### WakelessFoil

Joined Apr 16, 2020
12
Just learned in class that if you put reverse bias diodes in series (D1, D2) the voltage will be split between the two. This makes no sense as the voltage across D1 should equal total considering that on one side you have source (+) and on the other you should have have 0V. So shouldn't D2 have no voltage? Can someone explain this better than my professor? Thanks

Joined Jul 18, 2013
28,528
AC or DC?

#### dl324

Joined Mar 30, 2015
16,731
As this is schoolwork, you'll learn more if we guide you to the solution.

What relevant characteristic(s) might a reverse biased diode have that would cause the voltage division described by your professor?

#### crutschow

Joined Mar 14, 2008
34,063
Can someone explain this better than my professor?
The professor is assuming that the diodes are ideally identical with equal reverse leakage currents.
That would generate an equal voltage across the diodes.

Since this does not occur using real diodes, a high value resistor can be connected across each diode to help balance the voltage drops.

#### WBahn

Joined Mar 31, 2012
29,883
Just learned in class that if you put reverse bias diodes in series (D1, D2) the voltage will be split between the two. This makes no sense as the voltage across D1 should equal total considering that on one side you have source (+) and on the other you should have have 0V. So shouldn't D2 have no voltage? Can someone explain this better than my professor? Thanks
A sketch would be helpful. I'm guessing you meant something like

We can declare Vc to be 0 V (we get to pick any one node in the circuit and declare it to have whatever voltage we want, which is almost always 0 V).

That means that Va = Vo, which we arrive at by tracking the voltage gains/drops through the voltage source.

But what is Vb?

You claim it is 0 V. Why?

Remember, whatever reasoning you apply to one of the diodes also applies to the other.

All you really know is that

Vac = Va-Vc = Vo

that

Vac = Vab + Vbc

and that

Vab and Vbc are both at least zero because they are reverse biased.

If all that the professor said is that the voltage is split between the diodes, then that is a good description, even in the real world. In an idea world, because the two diodes are identical, the voltage drops across them should be the same, and hence the voltage would split evenly. But in practice, no two devices are truly identical and so there will be some minor difference that will result in one of the diodes dropping the bulk of the voltage. We don't know which one and we don't know how imbalanced the split will be.

If the reverse breakdown voltage of the diodes is less than Vo (which might be why multiple diodes are being put in series), we can't guarantee that the split won't be such that one of the diodes exceeds its breakdown voltage. The same is true for capacitors put in series. So we often put large value resistors across them to force the balance to be close to what we want.