# Voltage across reverse biased diode

## Main Question or Discussion Point

Hi all,
I think I know the answer to this question but I'm having trouble explaining why it is so.

If I have a circuit with a fixed resistor connected in parallel with a reverse biased diode, I believe the voltage drop across each will be the same. Is this correct? If so can someone explain the physics behind how this works? I get that there is a drop of voltage across the parallel section and that this means the voltage drop across both the resistor and the diode must be the same, but conceptually I have been thinking of a voltage drop as the energy that charge carriers "deposit" within a resistor as they pass through it, yet there should be no current flow through the reverse biased diode so how does energy drop across it?

• Alpharup

Related Electrical Engineering News on Phys.org
Back to circuit characteristics: VOLTAGE is same across paralleled devices; CURRENTS are (typically) different in each paralleled path, however.

berkeman
Mentor
I get that there is a drop of voltage across the parallel section and that this means the voltage drop across both the resistor and the diode must be the same, but conceptually I have been thinking of a voltage drop as the energy that charge carriers "deposit" within a resistor as they pass through it, yet there should be no current flow through the reverse biased diode so how does energy drop across it?
The voltage is the same across the parallel resistor and reverse-biased diode. The currents are different, though, so the power dissipated in each is different.

The power dissipated by the resistor is P=V*I=V^2/R. The power dissipated by the reverse-biased diode is much less, since the only current flowing is the reverse saturation current Isat. So for the diode, P=V*Isat.

• Alpharup and cnh1995
ok, this helps me understand power dissipated but in terms of the voltage drop, is the only way to explain it that the voltage on one side of the parallel component is x and on the other side is y so the drop across each resistor is x-y ? I can't conceptually understand how the voltage drops across an ideal reverse biased diode if no current can flow through it.

Thanks

Tom.G
Consider a back-biased diode to be an open switch. (That is if you ignore its non-ideal nature, such as a little bit of leakage current and breakdown voltage limitations.) You definitely get a voltage drop across an open switch!

Rigorously, the voltage drop is not the same. (It's zero across the diode.) If you look in pre-1960 electronics textbooks, a distinction was carefully drawn between an applied voltage (E), and a voltage drop (V).

berkeman
Mentor
Rigorously, the voltage drop is not the same. (It's zero across the diode.)
I don't understand that statement. Could you elaborate?

In the old days, E stood for EMF, which might be (for example) the voltage of a battery -- a source of electrical energy.
On the other side of the coin, when current flowed, energy was used up, which caused voltage to drop (symbol V).

So there is no voltage drop across the diode. Technically, there is only an applied voltage across the diode.

As decades rolled by, however, it seemed that keeping track of this fine point made things unnecessarily complicated, so E and V are synonymous and used interchangeably now, except for academic purposes,
which is the case at hand.

berkeman
Mentor
So there is no voltage drop across the diode.
Of course there is. You should probably qualify your statement by saying "ideal diode". There is most ceratinly a continuous V-I curve tor the diode, including forward and reverse bias regions.

I guess you are trying to make a DC distinction for the terminology for open circuit elements like switches and capacitors.

CWatters
Homework Helper
Gold Member
Perhaps think of a reverse biased diode as a high value resistor. An ideal reverse biased diode is like an infinite value resistor or open circuit. In which case the voltage drop is equal to the applied voltage.

...yet there should be no current flow through the reverse biased diode
Ideal diode behavior was one of the OP's simplifying assumptions.

berkeman
Mentor
Ideal diode behavior was one of the OP's simplifying assumptions.
I took the OP's statement to be an error of understanding, not that he understood the difference between real and ideal diode behavior and was asking about the ideal diode case, but I could be wrong about that.
yet there should be no current flow through the reverse biased diode so how does energy drop across it?

Hi all,
I think I know the answer to this question but I'm having trouble explaining why it is so.

If I have a circuit with a fixed resistor connected in parallel with a reverse biased diode, I believe the voltage drop across each will be the same. Is this correct? If so can someone explain the physics behind how this works? I get that there is a drop of voltage across the parallel section and that this means the voltage drop across both the resistor and the diode must be the same, but conceptually I have been thinking of a voltage drop as the energy that charge carriers "deposit" within a resistor as they pass through it, yet there should be no current flow through the reverse biased diode so how does energy drop across it?

If Iam not wrong, you are thinking like, 'Where the energy come from for the voltage drop'? It is simple. It is due to inherent nature of surrounding temperature(or other electromagnetic waves). The temperature causes the release of electrons and holes which constitute a reverse current. For example,
You may still may not be convinced. If so, It would be better if you can learn from a semiconductor physics book oriented towards electronics engineering.
You question seems to bend on energy aspect. The same concept of current flowing through a diode(connected in parallel with resistor) when exposed to sunlight is used in solar cells.

One point I wanted to make, there's a slight difference between voltage and voltage drop. If your power source was a standard 9V battery, measuring the battery connected across the reverse-biased diode would give you your standard 9V potential difference. However, since there is no current flow (assuming it's an ideal diode), there is no corresponding voltage drop. If we ignored the chemical reality of the battery and just defined it as a static 9V power source, and we allowed the diode to be ideal, then kept them in a strictly vacuum-controlled environment yada yada yada (you get what I'm saying; eliminate all outside factors) you could come back in 1000 years time and the battery would still be 9V. Now whether you want to define this as no voltage drop or a voltage drop of zero is strictly semantics. The voltage is energy potential across the load, the voltage drop is the energy potential lost as a result of current moving through the load. No current, no voltage drop. The 9V potential will always be there as long as the ideal diode is reverse-biased and is the only thing connected.

Just my .02

• Tom.G
While across the diode the voltage will be equal to Vcc(there is no voltage drop across it since no current flows, there is only applied voltage), the voltage drop across the resistor will be almost Vcc. Almost because the wire itself offers some resistance and so you got a voltage divider and the voltage across the resistor will be Vcc - some tiny fraktion of the voltage

Baluncore
2019 Award
Elements in parallel are subjected to the same electric potential.
Elements in series are subjected to the same electric current.

The voltage is energy potential across the load, the voltage drop is the energy potential lost as a result of current moving through the load. No current, no voltage drop.
Consider two identical ideal resistors in series, then subject them to a hypothetical voltage, v.
Will the voltage across each resistor then be v/2?
Will that be (1) an applied voltage, (2) a voltage drop, or (3) both ?

Consider two identical ideal capacitors in series, then subject them to a hypothetical AC voltage, v.
Will the voltage across each capacitor then be v/2?
Will that be (1) an applied voltage, (2) a voltage drop, or (3) both ?

Consider two identical ideal diodes in series, then reverse bias them by applying a hypothetical voltage, v.
Will the voltage across each diode then be v/2?
Will that be (1) an applied voltage, (2) a voltage drop, or (3) both ?

• jim hardy
jim hardy
Gold Member
2019 Award
Dearly Missed
What's in a name ?
Voltage is a potential difference, and it takes two potentials to have a difference between them.
Components have two ends so we can measure a voltage between those ends.
We should call it voltage across that component. Not voltage drop.
Here's why.

DROP as a verb infers to fall vertically,
https://www.bing.com/search?q=drop+define&pc=MOZI&form=MOZTSB
verb:
3. make or become lower, weaker, or less:
"he dropped his voice as she came into the room" ·
decrease · lessen · reduce · diminish · depreciate · fall ·
but it seems to me we're using it as a noun
noun:
2. an instance of falling or dropping:
"they left within five minutes of the drop of the curtain"
What drops across a device is the potential, or the voltage referred to our common measurement point.

So - i submit, the distinction suggested by XA293 in post 14 is an important one
as it highlights widespread abuse of a term .
Across any device in an active circuit that is absolutely time invariant ,
we have an increase or a decrease of potential but a constant voltage.

We should speak of "potential drop" not "voltage drop" (or rise).
The term "Voltage Drop(or Rise)" as meaning "Potential Difference" is too widely used by educated people to be called "Wrong" , but it is a grammatical redundancy because both words infer a difference.

NIce point, XYZ

• Tom.G