"....more voltage than input" ???

Output voltage across which nodes? Across the input nodes which are the only one that exist?

Input is at anode and output is observed across cathode

"Input is at anode and output is observed across cathode."

How can a voltage exist "across cathode" - that means: across one single node?

Definitely not. A diode requieres a kind of toll payment (threshold voltage) to let current passing through it. In consequence the input voltage will always be lower than the output. But amplifiers based on diodes circuit do exist and can be easily implemented. Low performances but it works

@ M. Blazquez:

"In consequence the input voltage will always be lower than the output"

Please can you explain this statement?

Between which nodes do you define the input and the output voltage, respectively?

Sorry for the mistake. The input voltage will always be higher than the output.

Let me correct ..input is applied between anode and ground ref......output is between cathode and ground node

@manuel i getting output higher than input

I am curious if I will get an answer to my question

"Between which nodes do you define the input and the output voltage, respectively?"

(As far as I remember, a voltage exist BETWEEN two nodes).

You can get amplification from an Esaki or tunnel diode which exhibits a negative resistance on its characteristic curve. Good article in Wikipedia on this component.

Lutz you r very right voltage exists between two nodes at different potential...so i taking output between cathode and ground node ,input at anode & gorund node

Julian thats is well documented thnx sir

Here's how, in 2008, my students made a perfect "diode follower" but with the help of an op-amp and a power supply

https://en.wikibooks.org/wiki/Circuit_Idea/Group_66a#Lab_4a:_How_to_make_perfect_circuits_by_series_NFB

The attached picture (a snapshot of the whiteboard during the lab) represents the Shaikh arrangement.

@Shaikh Imtiyaz: "Lutz you r very right voltage exists between two nodes at different potential...so i taking output between cathode and ground node ,input at anode & gorund node".

Thank you for clarification.

Now: I understand that your claim is that the cathode voltage vs. ground (for a normal pn diode) is or can be larger than between the anode and ground, correct?

And I assume that this voltage also can be measured and/or used to drive a current?

Is this also correct? Or do you refer to the internal diffusion voltage that - as far as I know - cannot be used as a voltage source?

It does not represent my arrangement sir

1) my input is d.c not a.c thats enough to change the behavior

2) i did not used any thing other than diode and d.c input

@lutz yes sir you got it right......."my claim that cathode voltage is more than at input node".. Sir at this stage i will not be able to tell much about current behavior...but yes definitely i thought it as diffusion voltage but problem in considering that is if you carefully go through my paper you will realize that at some point cathode vltg tends to get lower than anode input by difference of 0.02v..example for 9v input i get output 8.88v..so what to assume sir that the diffusion voltage have gone polarity reversal?

Dear Shaikh, I must admit that I found it difficult to understand the apparently simple experiment like some of the colleagues in above answers. But before going any further, whenever we talk amplification we immediately think of ac waveform and not dc voltages and currents unless of course we invert input dc (to ac) amplify and after that rectify the amplified ac voltage/current. A second note is that we always have to apply KVL (whether dc or ac). Hence, applying (dc or ac) as input to any circuit means that the measured voltage at any part of the circuit must be less than the input itself. Amplification is the process of replication of an input signal (voltage/ current) to get a higher output (voltage/ current) with the help of a power supply. In your case, what we have is a dc source and dc output and a diode in the model I can't see how come supply voltage is 0V while the output voltage is 2V (in the table)!. Thanks. @AlDmour.

@ismat sir that what even i do not know which is why i kept it here in front of dignities like you..but i must say if you get higher output than input for any signal it means amplification..but still i have tried justifying the behavior mathematically inn the paper

@ismat u right about k.vl sir in all those ckt where u apply kvl you also get practically verified output for kvl...now problem here is that u can not apply kvl to my model..that i will soon find out the reason for higher output vltg..insha allah..thnx one and all for effort

Cyril sir has given a very illustrative answer. I agree with his view.

I doubt entire discussion here is not concentrating of the problem depicted but it is circling around pole

Looking at the way the "ideal" diode follower is configured, if the amplifier is "ideal" i.e. infinite gain and no input offset voltage, then the diode voltage should be identical to the source. Given a non-ideal amplifier, the input offset voltage can cause the diode voltage to be smaller or greater from the source by the input offset voltage.

@james leitch thanx for response will get try to see the offset condition of diode

@james i am attatching data sheet of diode in4007 i will be glad if you could point me the input offset condition in it

Shaikh, the "input offset voltage" is a property of the op-amp in the circuit of a precise diode limiter but not of the diode...

@cyril sir the idea of this offset was proposed by kames leitch see above comments..thats what surprised me

But yet I think he consider the "input offset voltage" as an op-amp characteristic...

Shaikh, thanks for the data sheet of the 1N4007 diode.

Following on from Cyrils' comments.

When the diode is placed in the amplifier feedback loop, the total circuit must be analyzed. Given the diode is placed in the feedback loop, and under the control of the operational amplifier with negative feedback, the amplifier ensures the diode is

at the same voltage as the input +/- the input offset of the amplifier.

@james i request you to kindly see the paper attatched with this question that i have asked

James, I would like to illustrate your explanation by a funny Flash movie (choose the diode from the library on the left side of the screen)...

The local voltages (potentials) are represented by voltage bars and the current - by a loop. You can control the animation by buttons (you can change the input voltage by the arrows and choose where to take the output from by the buttons on the right side).

Cyril, Nice app for showing the diode follower!

Shaikh, I read your paper.

There are a few inconsistencies in the model you propose.

First: The resistance of the voltmeter is never infinite, it has a high finite resistance.

The resistance is not in series with the voltmeter, but in parallel with the voltmeter (otherwise the voltmeter would not work accurately).

Second: Your statement "No electrons move through the pn junction at equilibrium" is

not accurate.

The term in the diode equation, Is (the saturation current) , and as the name implies causes carriers to flow across the diode junction all the time, even in equilibrium. This is present when forward or reverse biassed. It is small , and is more visible when measured in reverse bias.

You might want to correct your diode model , and rework your results.

Even if i agree with the errors pointed by you ..the resistance of voltmeter is in parallel between cathode high potential and ground low-potential second that saturation current even if considered is no way responsible for the higher output voltage than input voltage... So i guess considering saturation current is not going to effect the result obtained above..secondly the voltmeter is also in parallel