Voltage across a diode

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Nikša

Joined Mar 26, 2018
86
I just wrote a reply to this old thread but it was deleted. Anyway.
As you can see from the last post of the OP he was not helped.

The question was how come resistor in parallel with a diode does not see supply voltage but diode voltage instead.

We all know resistance, that is, width of the depletion layer of a diode changes with applied voltage so within operating current range voltage drop across it is always 0.7V. Now this apparently simple phenomena is pretty strange.

Those who tried to help him did not really know the answer themselves. They replied along lines diode provides low resistance path thus very little current flows through the resistor and thus small voltage across it.. But this is nonsensical.

Suppose we have a very low resistance resistor in place of the diode, of course, most current will flow through it and very little through a big resistor in parallel but if we measure the voltage across each it will be full supply voltage.

I think i was clear, hopefully we can explain this for real this time so someone does not have to open a new thread in few years.

Mod: Link to Old Thread. [Do NOT delete Moderation Text]

https://forum.allaboutcircuits.com/threads/resistor-and-diode-in-parallel.15646/post-99584
instrumentationtools.com_pn-junction-with-depletion-layer.png
 
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BobTPH

Joined Jun 5, 2013
3,306
We all know resistance, that is, width of the depletion layer of a diode changes with applied voltage so within operating current range voltage drop across it is always 0.7V.
No, we do not all know that, because it is not true. The voltage across the diode varies with the current. It is not linear, like a resistor, it varies as a logarithmic function of the current.

The reason the resistor sees the same voltage as the diode is, quite simply, because they are in parallel!

Bob
 

Thread Starter

Nikša

Joined Mar 26, 2018
86
It varies with voltage and current, don't make nonsensical claims.

I know well diodes are nonlinear, their resistance is parabolic, not linear.

Incandescent light bulb is also non linear, but resistor in parallel with it will see supply voltage.

As for your "explanation", i'll remain polite and just say it's ridiculous.
 
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BobTPH

Joined Jun 5, 2013
3,306
Yes, of course the current varies with voltage. I did not say anything to contradict that.

You, meanwhile, made the following two statements:

1. The voltage across the diode is always 0.7V.

2. The current varies with the voltage.

From which we may conclude that you apparently believe the current through a diode is constant.

As for my “ridiculous” explanation, please inform me how components in parallel can have a different voltage across them.

Bob
 

Thread Starter

Nikša

Joined Mar 26, 2018
86
Yes current varies with supply voltage, nothing contradictory about that.

As for your "explanation", it is nonsensical cause it does not address the question. Question remains, why does resistor in parallel "see" voltage across the diode and not the supply voltage.

Let's put it this way, if we add another resistor in series before the diode and resistor in parallel, we can say most of the supply voltage will drop across it. So that first resistor drops all the "missing voltage". That is the point.
 
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MrChips

Joined Oct 2, 2009
23,509
This discussion is going around in circles since it was already explained in the original thread.
Are you offering your own explanation or do you truly do not know the answer?
If you think you know the answer then we can close this thread and move on.
 

Thread Starter

Nikša

Joined Mar 26, 2018
86
It was not well explained in the first thread, that is why that person remained confused. If they explained it like i just did with a series resistor before the diode-resistor in parallel it would have been perfectly clear to him.
 

BobTPH

Joined Jun 5, 2013
3,306
Yes current varies with supply voltage, nothing contradictory about that.

As for your "explanation", it is nonsensical cause it does not address the question. Question remains, why does resistor in parallel "see" voltage across the diode and not the supply voltage.
May I ask you, then, why would it see the supply voltage, when it not connected across the supply?

If the two components were connected directly across the supply, and the supply was capable of enough current, then both the resistor and the diode would indeed see the supply voltage. In the real world, the diode would likely be destroyed by any supply much greater than 1V.

For the original question, no further explanation is needed. If one then asks what voltage it would see, we need to know the behavior of the diode. Then we can solve it. But that is not why the resistor sees the same voltage, that is caused by the fact that they are in parallel.

Bob
 

bertus

Joined Apr 5, 2008
21,362
Hello,

Here is the schematic from the old thread:
resistor-diode parallel circuit.jpeg
The voltage accross R2 is equeal to the voltage on the diode.

Bertus
 

Thread Starter

Nikša

Joined Mar 26, 2018
86
Not sure i understand first part of your reply " not connected across the supply ", they are connected directly to the supply as shown, R1 limiting the current.

You again repeat "because they are parallel"...we all know elements in parallel have same voltage across them, that is not the point. The point is R1 drops most of the voltage from the supply.
resistor-diode parallel circuit.jpg
 
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Thread Starter

Nikša

Joined Mar 26, 2018
86
Also from old thread, voltage drop across R1 with diode forward and reverse biased.
EDIT: i have now noticed he put different battery voltage and parallel resistor values, i don't see why, but his point was to show how diode when forward biased controls the voltage in the parallel circuit .
afaik_inquiry_18Nov08.png
 
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Thread Starter

Nikša

Joined Mar 26, 2018
86
I am polite. It is not enough to say "cause they are in parallel" when real answer is "diode provides low resistance path which makes most of supply voltage drop across R1 and only 0.7 across the resistor in parallel"
 

MrChips

Joined Oct 2, 2009
23,509
Here is my answer.

1) Remove the diode from the circuit. Calculate the voltage across R2 as per voltage divider formula.

Knowing this voltage, you can obtain an estimate of the expected voltage when the diode is inserted (whether forward or reverse biased). Since current is expected to flow through the diode one would expect the voltage across the diode and R2 to be lower than that calculated in (1).

The voltage across the resistor is equal to the voltage across the diode. Period. End of argument.

2) There are three ways to determine the voltage across the diode.

#1 - You can apply the diode equation and solve the equation for diode current and voltage.
#2 - You can apply a graphical solution (which relies on the diode equation or some estimate of the diode I-V characteristic curve).
#3 - You can make an assumption that the diode voltage has reached the diode forward turn on voltage and assume an approximate forward voltage of about 0.7V in which case the voltage across R2 is 0.7V.
 

Thread Starter

Nikša

Joined Mar 26, 2018
86
What prompted me to this was analysis of slayer exciter circuit. It is said purpose of the diode you see is to prevent base-emitter junction from excessive reverse bias during negative half-cycle. And we are talking kilovolts here, at very low current tho.

slayer.jpg
 

sarahMCML

Joined May 11, 2019
39
Also from old thread, voltage drop across R1 with diode forward and reverse biased.
EDIT: i have now noticed he put different battery voltage and parallel resistor values, i don't see why, but his point was to show how diode when forward biased controls the voltage in the parallel circuit .
View attachment 226471
No, supply voltage is 5 volts and R2 (R4) is 1K in both cases.
 
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