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# diode voltage drop clarification

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#### electronicslearner77

##### Junior Member level 2
sorry but iam asking a very basic question. if you take an ideal diode we say some voltage drop. what exactly is voltage drop? why should it drop the voltage? For diodes we say some almost fixed voltage drop, but for resistor we say it will vary with current why? It will be of great help to me if you can answer this question.

Hi,

A diode is a semiconductor. The voltage drop is caused by the pn junction.

With resistors the voltage drop depends on current. The voltage drop follows Ohm's law.
U = R x I.

Klaus

yes I understand but why is it causing the drop?

Its to do with the amount of energy an electron needs before it can jump from the P material to the N material at the junction of these zones which are the heart of the diode. Until the electrons flow there is no current, once they have enough energy and the current starts to flow, the junction becomes a low resistance path. It could be that at .55V the current flow is 1µ A, at .6V the current could be 1mA and at .8V the current could be 1A.
Frank

Can I say something like the diode uses around 0.6v of the power supply to do something. As of now I don't know what is that something. Sorry for my layman's language.

The diode automatically changes its own resistance, to generate the voltage drop.

The value is very high ohms with light load.

At heavy load it drops to a very low value. As you increase the voltage across it, the Ampere level starts to be limited by a separate factor, namely its internal ohmic resistance.

The diode automatically changes its own resistance, to generate the voltage drop.

This sounds as if the current through the diode is "first" - and the voltage the result.
I rather think the opposite is true (always!): No current without voltage.
Hence, the term "voltage drop" is somewhat misleading. We need a voltage to "open" the diode (physically: To neutralize the diffusion voltage at the pn junction) and let a current flow, dont we?

LvW's explanation is the correct one although the others state similar things in other words.
The PN junction is just that, P doped material meets N doped material. The effect is to push a barrier of electrons across the junction. Before current can flow, the electrons have to be 'pushed back'. Different semiconductor materials exhibit different energy levels and hence require lesser or greater voltage to overcome the barrier threshold..

Brian.

Betwixt - thanks.
I thought some additional explanations/corrections were necessary because the questioner was asking
"why is it causing the drop ?"

This sounds as if the current through the diode is "first" - and the voltage the result.

Since you mention it, I suggest you picture two scenarios, in which we know the voltage:

(a) 10V applied through a 100 ohm resistor, then the diode.
and
(b) 10V applied through a 1k ohm resistor, then the diode.

Notice in both situations 10V is applied. Current flows, because this overcomes the diode's Vfwd.

What voltage is dropped by the diode? The difference is slight between the two cases above (in agreement with post #4).

And now what about the dynamic (internal) resistance? Are they the same in each case?

No, they are very different. How do we calculate the values? We calculate these based on the available current, not the applied voltage.

We call any semiconductor an "active" device not just because it amplifies but because they are voltage or current control devices that may be linear or non-linear depending on bias , polarity and range.

Diodes are also active and used primarily as switches in logic or clamps to suppress overvoltage or reverse blocking switches to prevent current flow when polarity switches. In some cases they are used to rectify AC to DC with some voltage drop due to the physical junction type and size.
When they conduct with 10% of their rated current, or so, we call it saturated and the voltage drop only rises above this small voltage due to a small series resistance. So essentially all diodes are current rectifiers and they are often referenced in schematics as CR1,2,3 or D1,2,3

Each chemistry has a different threshold voltage.

Germanium, Schottky, Silicon, Gallium Arsenide, GaAs, Silicon Dioxide, Gallium Nitride , GaN, and some of these are Light emitting diodes or LEDs. They can also be light detecting photo diodes when reverse biased and voltage controlled capacitors.

With time, you may get to know how these are the basic PN junctions of many components.

This sounds as if the current through the diode is "first" - and the voltage the result.
I rather think the opposite is true (always!): No current without voltage.
Hence, the term "voltage drop" is somewhat misleading. We need a voltage to "open" the diode (physically: To neutralize the diffusion voltage at the pn junction) and let a current flow, dont we?

But if you pump a current into a diode through a current source, it will develop a voltage. I think, it is the current that generates the voltage drop, not the other way round.

Since you mention it, I suggest you picture two scenarios, in which we know the voltage:

(a) 10V applied through a 100 ohm resistor, then the diode.
and
(b) 10V applied through a 1k ohm resistor, then the diode.

Notice in both situations 10V is applied. Current flows, because this overcomes the diode's Vfwd.

..... because "this" overcomes the diode`s Vfwd

May I ask: What is the quantity behind the word "this"? A voltage or a current?

- - - Updated - - -

But if you pump a current into a diode through a current source, it will develop a voltage. I think, it is the current that generates the voltage drop, not the other way round.

I think, we cannot "pump" a current into the diode because there is no ideal current source. The source we call "current source" is always a voltage source with a large internal source resistance (each current needs a driving voltage; no current without voltage).
Then, the effect can be described as follows: After connecting the diode to the source the voltage (at t=0, much larger than 0.7 V) opens the diode - and before the flowing current can destroy the diode the voltage loss at the source resistor causes the current to reach an equilibrium which fulfills the typical diode characteristics with a voltage in the region 0.6...0.7 Volts.

This sounds as if the current through the diode is "first" - and the voltage the result.
I rather think the opposite is true (always!): No current without voltage.
I think that the validity of this statement is context depedent. I agree with LvW if the statement is talking about diode modelling. In network theory, there's a duality between voltage and current, one or the other becomes the indepent variable depending on the used circuit description. Saying one of it is primary would turn out as a hen-egg-problem.

In general physics, I believe that charge is the primary quantity (based on elementary charge), both current and voltage are derived quantities.

I think, we cannot "pump" a current into the diode because there is no ideal current source.
Sorry, but that's a poor reasoning. An ideal curren source isn't harder to design than an ideal voltage source. Strictly speaking neither of it exists, but both can be fairly approached under circumstances.

Talking about real electronic circuits, there are many cases where the current through a diode is determined by an about to ideal current source, and only the recourse to semiconductor theory holds me back from considering the diode a current controlled device...

An ideal curren source isn't harder to design than an ideal voltage source. Strictly speaking neither of it exists, but both can be fairly approached under circumstances.
Yes - I agree. There are no ideal sources - neither a voltage nor a current source.
However, what I wanted to make clear was the following: A source we call "current source" (real or nearly ideal) always needs - as a core element - a device able to cause charge separation as a precondition for a corresponding current. And in electronics such a device is called "voltage source".

Weall know the graphical method for finding the operating point of a diode which is connected with a series resistor Rs to a voltage source:
The V-I characteristic of the series resistor Rs is superimposed to the classical pn-diode V-I characeteristics. In this drawing the Rs-line has a negative slope.
Both curves meet at a point which gives us the diode´s voltage and current. I think, this picture clearly shows what I mean with "equilibrium".

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