# Question about Diode "Voltage Drop"?

Discussion in 'General Electronics Chat' started by WillYum, Oct 6, 2016.

1. ### WillYum Thread Starter New Member

Nov 21, 2011
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Although I have done it for close to 50 years, I now have a question about the term "voltage drop" when applied to a PN junction. Voltage drop across a resistor is simple - current x resistance = voltage drop. Ohm's Law pertains and everything is sweet BUT when we look at a PN junction, what we so easily term voltage drop (forward or reverse) does not follow Ohm's Law as current flow in a forward biased diode is determined by circuit impedance NOT the so called "resistance" of the PN junction. I have no problem understanding how the PN junction works, just a little problem explaining the term "voltage drop" when I am asked. I was wondering if anyone knows the proper terminology to use when explaining PN theory as I have done some searching and some sources use Forward (or Reverse) Voltage, other sources use Forward (or Reverse) Bias Voltage, while others use Voltage Drop. There also seems to be a lot of confusion over the non-linear "resistance" of the junction.

2. ### dl324 Distinguished Member

Mar 30, 2015
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Welcome to AAC!

When referring to the forward voltage of a diode, it can also be referred to as a voltage drop. A voltage drop is simply the voltage "dropped" across any element in a circuit.

The resistance of a diode varies with current and is not linear. A diode's dynamic resistance is 1/slope of the IV curve.

3. ### MrChips Moderator

Oct 2, 2009
14,255
4,174
Voltage drop is still voltage drop regardless of the properties of the device.
Voltage drop is simply the voltage difference between two points A and B. Hence the voltage drop is simply VA - VB where VA and VB is the potential at A and B respectively, (while paying attention to the sign of the voltage drop).

A diode is a non-linear device and hence does not obey Ohm's Law.
However, at a given operating voltage and current, the formula R = V/I can still be applied to determine the effective resistance of the device at that singular operating point. It just so happens that R is not constant over all values of V or I.

What is the confusion with the non-linear resistance of the junction?

Note that instantaneous resistance and dynamic resistance are not the same.
Instantaneous resistance is V/I.
Dynamic resistance is ΔV/ΔI and is equal to the 1/slope of the I-V diode characteristic curve.

4. ### nsaspook AAC Fanatic!

Aug 27, 2009
3,545
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A simple way to explain the PN junction forward voltage drop is to see it as a barrier (internal electric field “diode potential” ) created when the junction is formed as drift and diffusion currents become exactly equal and opposite in the depletion region. In forward bias (p-side positive) the applied potential opposes the internal electric field. This affects the drift(reduces) and diffusion (increases in majority carriers) currents causing the depletion region to shrink and flow current (seen as decreasing resistance with increasing voltage) across the junction. As the applied potential increases the depletion region shrinks until it substantially offsets the original unbiased diffusion potential and the junction turns ON (low resistance of the bulk semiconductors only) but the potential across the diode needed to offset the built-in voltage always must be present so we see that as a 'voltage drop' plus the voltage drop from bulk resistance.

Last edited: Oct 6, 2016
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5. ### crutschow Expert

Mar 14, 2008
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4,324
Not really.
As noted, the forward voltage drop (or bias voltage) of a diode is a non-linear function of current (it's logarithmic) or alternately, the current is an exponential function of this voltage.
This is discussed here.
This is also a bulk ohmic resistance of the diode material, which adds to this drop and shows up at higher current levels; the level where it becomes significant depends upon the diode current rating.