Diode numerical (calculation of current and voltage across diode)

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Discussion Overview

The discussion revolves around calculating the current and voltage across a diode in a given circuit, specifically addressing the application of the superposition theorem and the characteristics of the diode's V-I graph. The scope includes theoretical and practical aspects of diode behavior in response to both DC and AC signals.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant questions the appropriateness of using the superposition theorem for calculating output voltage with both DC and AC sources.
  • Another participant suggests that the fixed slope of the diode's V-I characteristic should be considered, as it differs from a typical silicon diode.
  • Some participants express confusion regarding the application of the cut-off voltage and forward resistance of the diode when using superposition.
  • A later reply clarifies that the AC component should be treated as a small-signal, implying that it does not affect the cut-off voltage but rather the dynamic resistance at the DC bias point.
  • One participant provides a method for solving the problem by summing currents and writing an expression for diode current based on the input voltage.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and confusion regarding the application of the superposition theorem and the characteristics of the diode. There is no clear consensus on the correct approach to take in this context.

Contextual Notes

Some limitations include the potential misunderstanding of the diode's dynamic resistance and the conditions under which the AC component can be considered small compared to the DC component. There are also unresolved aspects regarding the specific characteristics of the diode in question.

lazyaditya
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The Diode in the circuit shown below has the non linear terminal characteristics as shown in the figure.Let the voltage be "coswt" V . Question 3.1.33 and my attempt is shown in the figure.
 

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The formula ηVT / Id is the slope of a Si diode's exponential V-I characteristic, and gives the Silicon diode's small signal resistance. You don't have a Si diode for this question, you have some clunky diode whose V-I graph has a fixed slope regardless of Id.

I would much prefer that you typed out the text of a question, rather than forcing me to jockey back and forth between multiple graphics (and wasting my 3G data quota unnecessarily in the process).
 
Last edited:
Sorry for inconvenience.
 
Can i calculate the voltage at output using superposition theorem . Considering dc voltage and ac voltage sources separately.
 
lazyaditya said:
Can i calculate the voltage at output using superposition theorem . Considering dc voltage and ac voltage sources separately.
Yes, that should work out.

Did you get the right answer for 3.1.33?
 
No i didn't , i get very confused with this concept. When using superposition theorem and considering DC signal first i took cut off voltage and forward resistance of diode into consideration to get output voltage 1,when i considered AC signal then also should i take cut off voltage of diode into consideration ?
 
But i did understood that i shouldn't use dynamic resistance of diode in this case .
 
lazyaditya said:
No i didn't , i get very confused with this concept. When using superposition theorem and considering DC signal first i took cut off voltage and forward resistance of diode into consideration to get output voltage 1,when i considered AC signal then also should i take cut off voltage of diode into consideration ?
AC is synonymous with "small-signal" and also "incremental", it imposes what amounts to just a tiny wobble in the DC. So this small wobble does not see the 0.6 V cut-off, it sees just how the current increases for a tiny increase in voltage: the dynamic resistance or incremental resistance, and this is the slope of the diode's graph at the DC bias point.

Of course, you should from the outset confirm that the AC component really is small compared with the DC. There may be occasions when an examiner deliberately sets a trick question to catch anyone not fully awake.
 
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Since you're given multiple choice, you can proceed as follows:
Let Vin = 2 + 1 = 3V (highest input voltage)
Sum currents to zero at the anode
write expression for diode current assuming v > 0.5V. This should be obvious from the diagram.
Solve for diode current, then diode voltage.
 
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Thank you for help. Problem solved :)
 

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