Comparison between bulk and ac resistances of a diode.

Godwin Kessy

Hellow guys!!
I find this concept really confusing, if the bulk resistance of a diode is an approximation of the I-V characteristic curve for a diode, Why then is that bulk resistance compared to ac resistances of the same diode at a particular operating in calculated values, seem not to be approximation of one another at all..
Example, there was a solved problem, whose answers of bulk and ac resistances where 1 ohm and 26.53 Ohms, can this really mean approximation. What I use to believe, resistance values don't depend on whether the source is A DC or AC..
Can someone prove me wrong pleaseee!!!

carlgrace

The bulk resistance is an actual resistance related to the resistivity of the semiconductor used to construct the diode. The ac resistance is not a physical resistance but a small-signal model parameter that is used to "linearize" the diode at the operating point in order to make hand calculation possible.

So one isn't an approximation of the other, they are quite different things.

NascentOxygen

It is confusing, but only until your understand it!

(BTW I think you probably mean DC or static resistance, rather than bulk resistance which is something different again.)

I'll pose two questions for you to answer. Here is the first one:

Imagine your lab work centres on a circuit involving a battery and some resistors, a light bulb, and a diode. You measure all voltages and currents, and find that the diode has a current of 0.08A and its forward voltage is 0.72V. Being DC these levels are fixed and steady.

If you replace the diode by a resistor of a particular value, the circuit will continue to operate identically and no one will detect that you have made the substitution, not without they themselves making some other changes and these are not allowed*. So, what value of resistor will give this equivalent DC (or static) resistance?

Next question will follow later.

Godwin Kessy

Okay, the value of the resistance, will be equal to 0.72V/0.08A=9 Ohms... And this is what is the Static resistance of the diode i get it.. Isn't it?

Godwin Kessy

Guys can you please view the image I have attached!!!
Isn't one graph the approximation of the other? and why? Okay, leave alone that, for pure resistors, their resistances doesn't depend on whether the applied voltage is ac or dc. Why is it for diodes??

Attached Thumbnails

 

NascentOxygen

For diodes, it isn't, either. But to explain the difference between the two, including showing it on a graph, most authors believe it is easier to illustrate by considering the outcome of a small AC signal sitting on a larger DC level.

NascentOxygen

Okay, that was for the static condition (best known as the DC resistance or large signal resistance).

Now, imagine that you find the battery in that circuit is getting old, so you replace it by the nearest model you can find, and this brings a slightly higher voltage to your circuit. You re-measure the voltages and currents, and find that the voltage across the diode is now a tiny bit higher, its 0.015V higher than before, and the current is 0.01A greater. If you were again to substitute a resistor for the diode and not tell anyone, what value of resistor should you use, i.e., what resistance would best describe the relationship whereby an increase in voltage by 0.015V causes current to rise by 0.01A? This is the dynamic resistance, better known as the AC resistance or small signal resistance, or incremental resistance. (Memorize all those alternative terms and you can't go wrong. )

The two resistances are different because the diode is a non-linear device. If you must model the non-linear diode by a linear element (a resistor), then what measurements you are interested in determines whether you should model its behaviour using the DC resistance or the AC resistance.

Godwin Kessy

Nascent.. I think am getting you, for the new reading am supposed to replace the diode with a different resistor, not 9 ohms as before, it's gone be (0.72+0.015)/(0.08+0.01)=8.167ohms

Okay.. And we have this concept.. total dynamic resistance, which is bulk resistance plus the dynamic resistance, how does this become valid... This has been done during the analysis of voltage across the diode, when a DC supply is super imposed by a small ac signal... See image below;

Attached Thumbnails

 

NascentOxygen

By calculating it the way you have, you are again calculating the static resistance, but at a new Q point. (Your diagram has the operating point labelled the Q point.)

I intended you to say resistance = .015/.01 = 1.5Ω
This is the appropriate value, because only if the resistance is 1.5Ω will an increase of .015V bring about a current increase of .01A. It won't happen if you use 9Ω as was appropriate for the earlier large signal case.
Are you quoting something you got in your lectures? Or is this what you think I've been saying? You'd better explain what you understand by "total dynamic resistance, which is bulk resistance plus the dynamic resistance."

Godwin Kessy

Ahaa.. I think am getting you even better.... Of course the concept is from one of the lectures, that the question-attachment i have posted was solved. And it was like this, Using the superimposition theorem, the dc and ac voltage across the diode was calculated as follows. With the dc equivalent circuit the bulk resistance and the forward current across the diode was calculated, using the forward current the ac/dynamic resistance of the diode was calculated(Dynamic resistance=thermal voltage/forward current).

The lecturer continued and he used the summation of bulk and dynamic resistance to get the ac voltage across the diode.. He called it total dynamic resistance. Got it? Actually it's from here where all my confusion raised from the beginning...

NascentOxygen

Diode action takes place at the junction of P and N silicon. Away from that active region the doped silicon acts as a conductor exhibiting some Ohmic resistance, the bulk resistance. This doped silicon in turn is contacted by copper to provide connectors to the outside world. The copper, too, has Ohmic resistance, but usually much less than the doped silicon.

Summing Up ....

If we cause a small increment in the current through the two-terminal diode component, the voltage across its terminals increases incrementally due to a number of factors: the resistance of the copper, the bulk resistance of the P silicon, the PN diode junction's exponential V-I characteristic curve, and the bulk resistance of the N silicon.