How do I start this problem? (diode circuit)

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    Circuit Diode circuit
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Discussion Overview

The discussion revolves around a homework problem involving a diode circuit, specifically focusing on finding the output voltage (Vo) and drawing the corresponding waveform. Participants explore the implications of diode characteristics, transformer properties, and the relationship between input and output voltages in the context of AC signals.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants express uncertainty about how to find Vo and the waveform, indicating a lack of clarity in the problem statement.
  • One participant notes that the diode voltage drop of 0.7V applies as soon as current flows through the diodes, suggesting that this is a critical factor in determining Vo.
  • Another participant highlights that the secondary voltage of 1V rms is specified without the diodes, prompting a discussion about how this affects the output waveform.
  • Some participants propose that the output voltage Vo could simply be 1Vrms, while others challenge this by stating that the diodes limit the voltage to 0.7V.
  • There is a suggestion that the problem may involve unclear assumptions regarding transformer impedance and its effect on the output voltage.
  • One participant suggests that the diodes are connected back-to-back, which influences how the voltage drop is calculated.
  • Another participant summarizes that there are at least two correct answers depending on whether the transformer is ideal or practical, indicating the complexity of the problem.
  • There is a discussion about whether the output waveform would resemble a rectified sine wave or another form, reflecting differing interpretations of the circuit behavior.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as multiple competing views remain regarding the assumptions about the transformer and the behavior of the diodes. The discussion reflects uncertainty and differing interpretations of the problem.

Contextual Notes

Participants note that the problem lacks clarity regarding the transformer impedance and whether it should be considered in the analysis. The assumptions about the ideal versus practical behavior of the transformer and diodes contribute to the complexity of the discussion.

servehover
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Homework Statement


dl6J4GV.png

Find Vo and draw the waveform. The two diodes are silicon diodes with a voltage drop of 0.7V.

Homework Equations

(given by my teacher)[/B]
Id = Io * (e^(qv/nkt)-1)
Vp = Vrms * sqrt(2)
Vrms = Vp/(sqrt(2))

The Attempt at a Solution


I have no idea how to find Vo or the waveform.
 
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servehover said:

Homework Statement


dl6J4GV.png

Find Vo and draw the waveform. The two diodes are silicon diodes with a voltage drop of 0.7V.

Homework Equations

(given by my teacher)[/B]
Id = Io * (e^(qv/nkt)-1)
Vp = Vrms * sqrt(2)
Vrms = Vp/(sqrt(2))

The Attempt at a Solution


I have no idea how to find Vo or the waveform.
Welcome to Physics Forums!

What is the input ac voltage? What is the turns ratio of the transformer?
 
cnh1995 said:
Welcome to Physics Forums!

What is the input ac voltage? What is the turns ratio of the transformer?
Neither are given
 
servehover said:

Homework Equations

(given by my teacher)[/B]
Id = Io * (e^(qv/nkt)-1)
Vp = Vrms * sqrt(2)
Vrms = Vp/(sqrt(2))
Not really relevant here. You are given the diode voltage drop as 0.7V which applies in this case as soon as some current goes thru the diodes.
I have no idea how to find Vo or the waveform.
It's a bit misleading. For a hint, realize that ther secondary voltage of 1V rms is WITHOUT THE DIODES. Why? You need think about this, then you shoulsd be able to draw the output waveform.
 
servehover said:
Neither are given
Oh..actually neither is required. I didn't read "1V rms" written on the secondary. So the input to the diode circuit is 1V rms.
Hint: There is no series resistance between source and the diodes.
 
rude man said:
Not really relevant here. You are given the diode voltage drop as 0.7V which applies in this case as soon as some current goes thru the diodes.
It's a bit misleading. For a hint, realize that ther secondary voltage of 1V rms is WITHOUT THE DIODES. Why? You need think about this, then you shoulsd be able to draw the output waveform.
cnh1995 said:
Oh..actually neither is required. I didn't read "1V rms" written on the secondary. So the input to the diode circuit is 1V rms.
Hint: There is no series resistance between source and the diodes.
Would the output voltage v0 simply be 1Vrms? Is it because the voltage across the 1k resistor is also 1Vrms?
 
servehover said:
Would the output voltage v0 simply be 1Vrms? Is it because the voltage across the 1k resistor is also 1Vrms?
Yes.
 
cnh1995 said:
Yes.
Any hints for something like this?
BoBbBxf.png
 
servehover said:
Would the output voltage v0 simply be 1Vrms? Is it because the voltage across the 1k resistor is also 1Vrms?
No.
The diodes are limited to 0.7V as given in the problem. So the voltage across the diodes can't possibly be 1V rms = 2.82V peak-to-peak when the peak-to-peak is limited to 2 x 0.7V = 1.4V.

(In "real life" either the transformer impedances are high enough to limit the current or the poor little things would burn out, in which case you really would have 2.82 V pk-pk. The 1K resistor does nothing except get hot).

(In
 
  • #10
servehover said:
Any hints for something like this?
BoBbBxf.png
Same basic idea. Vo is limited to 5.6V. You need to think graphically about Vo as it rises from zero to the max. voltage . That also applies to your previous problem BTW.
 
  • #11
rude man said:
No.
The diodes are limited to 0.7V as given in the problem. So the voltage across the diodes can't possibly be 1V rms = 2.82V peak-to-peak when the peak-to-peak is limited to 2 x 0.7V = 1.4V.

(In "real life" either the transformer impedances are high enough to limit the current or the poor little things would burn out, in which case you really would have 2.82 V pk-pk. The 1K resistor does nothing except get hot).

(In
So V0 would be 0.4949Vrms? (0.3535 * Vpp)
 
  • #12
rude man said:
So the voltage across the diodes can't possibly be 1V rms = 2.82V peak-to-peak when the peak-to-peak is limited to 2 x 0.7V = 1.4V.
But the source (transformer secondary) is connected directly across the diodes. There is no series resistance given in the problem (like secondary winding resistance). If the diode voltage is less than 1V rms, where did the remaining voltage go? Doesn't that violate KVL? A diode clips the voltage across it to 0.7V only when there is an impedance in series with it.
Edit: I see you already explained it in the last para of your post, but if the transformer impedance is not mentioned, should we assume some nonzero impedance or should we ignore the transformer impedance and write the diode voltage as 1V rms?
 
Last edited:
  • #13
As the answer is dependent on unclear assumptions there is surely an error in the diagram.
I suspect the 1k resistor should be in series with the transformer secondary.
 
  • #14
cnh1995 said:
But the source (transformer secondary) is connected directly across the diodes. There is no series resistance given in the problem (like secondary winding resistance). If the diode voltage is less than 1V rms, where did the remaining voltage go? Doesn't that violate KVL? A diode clips the voltage across it to 0.7V only when there is an impedance in series with it.
Edit: I see you already explained it in the last para of your post, but if the transformer impedance is not mentioned, should we assume some nonzero impedance or should we ignore the transformer impedance and write the diode voltage as 1V rms?
I agree the problem is confusedly stated. But since it's stated that the diode voltage is 0.7V the only reasonable assumption is that the 1V rms is the no-load voltage across the secondary of the transformer.
 
  • #15
servehover said:
So V0 would be 0.4949Vrms? (0.3535 * Vpp)
Assuming sufficient transformer impedance that the diodes don't burn out, what you need to do now is draw a picture of what Vo looks like, then compute the rms voltage accordingly.
 
  • #16
rude man said:
Assuming sufficient transformer impedance that the diodes don't burn out, what you need to do now is draw a picture of what Vo looks like, then compute the rms voltage accordingly.
I am still kind of lost as to how I get Vo.
The two diodes are in parallel, so would the voltage drop across both diodes be 0.7V?
Is Vo simply 0.7Vpp?
 
  • #17
servehover said:
I am still kind of lost as to how I get Vo.
The two diodes are in parallel, so would the voltage drop across both diodes be 0.7V?
Is Vo simply 0.7Vpp?
The diodes are connected "back-to-back" so only one is "on" at a trime.
Ask yourself what Vo would look like without the diodes, then figure out what function the diodes perform. Do they alter the sinusoidal shape of the unloaded Vo? How?
 
  • #18
rude man said:
The diodes are connected "back-to-back" so only one is "on" at a trime.
Ask yourself what Vo would look like without the diodes, then figure out what function the diodes perform. Do they alter the sinusoidal shape of the unloaded Vo? How?
If the diodes were removed, Vo would simply be 1Vrms. Voltage drop across a diode is 0.7V so the output voltage has a peak of 0.7V. Would the output waveform simply look like a sine wave with a peak of 0.7V?
 
  • #19
Summarizing the discussion: there are at least two correct answers to this problem, according to the assumption being made that:
(1) the transformer is ideal and the diodes are logarithmic Silicon (i.e., characterised by the exponential current equation your teacher provided), or
(2) the transformer is a practical transformer having some resistance and the diodes can be modeled as having an approximately constant voltage drop while forward biased.
or (3) some combination of (1) and (2).

In the absence of clarifying information from your teacher, it may be worth your while preparing separate answers for both (1) and (2).

⏩ Is your teacher known for setting trick questions designed to make you really think?
 
  • #20
servehover said:
Would the output waveform simply look like a sine wave with a peak of 0.7V?
A peak of ±0.7V, certainly, but not a sinewave.
 
  • #21
NascentOxygen said:
⏩ Is your teacher known for setting trick questions designed to make you really think?
Yes

NascentOxygen said:
A peak of ±0.7V, certainly, but not a sinewave.
Would it be a rectified sine wave?
PqwoR.jpg
 
  • #22
servehover said:
If the diodes were removed, Vo would simply be 1Vrms. Voltage drop across a diode is 0.7V so the output voltage has a peak of 0.7V. Would the output waveform simply look like a sine wave with a peak of 0.7V?
Yoou're getting closer, but Vo would not look like a sine wave although it would indeed have peaks of 0.7V.
 
  • #23
servehover said:
YesWould it be a rectified sine wave?
No. Why rectified? How can there be one polarity but not the other when the diode connection is perfectly symmetrical as far as polarity is concerned?
PqwoR.jpg
 
  • #24
rude man said:
Yoou're getting closer, but Vo would not look like a sine wave although it would indeed have peaks of 0.7V.
No. Why rectified? How can there be one polarity but not the other when the diode connection is perfectly symmetrical as far as polarity is concerned?
Would it go from 0 to 0.7 or -0.7 to 0.7? I really don't understand the problem(s) and neither do any of my classmates.
 
  • #25
servehover said:
Would it go from 0 to 0.7 or -0.7 to 0.7? I really don't understand the problem(s) and neither do any of my classmates.
Assuming sufficient transformer impedance in series with the diodes, the diodes "almost maintain" 0.7V across them when forward biased. See NascentOxygen's post #19 (2nd point). Look up "diode clipper" circuit.
 
  • #26
cnh1995 said:
Assuming sufficient transformer impedance in series with the diodes, the diodes "almost maintain" 0.7V across them when forward biased. See NascentOxygen's post #19 (2nd point). Look up "diode clipper" circuit.
So it would basically be a sine wave but clipped at 0.7V?
Edit: this circuit has the resistor in series though. Mine is in parallel
diode62.gif
 
  • #27
servehover said:
So it would basically be a sine wave but clipped at 0.7V?
diode62.gif
Exactly!
But this is true only if sufficient impedance is assumed in series with the diodes (like the R1 resistor in your image). Without that, the diode voltage will simply be equal to the source voltage i.e. 1V rms.
 
  • #28
cnh1995 said:
Exactly!
But this is true only if sufficient impedance is assumed in series with the diodes (like the R1 resistor in your image). Without that, the diode voltage will simply be equal to the source voltage i.e. 1V rms.
Does the resistor in parallel changed anything vs the resistor in series?
My teacher is the kind of guy to give trick questions, so I'm really leaning towards the 1V rms answer...
Can you give me any hints regarding 2nd problem?
 
  • #29
servehover said:
so I'm really leaning towards the 1V rms answer...
Me too, especially after knowing about your teacher. Maybe he wants to see if you get confused with the parallel 1k resistor and 0.7V diode drop.
servehover said:
Does the resistor in parallel changed anything vs the resistor in series?
Parallel resistor doesn't make any difference here. The circuit would work the same way even if it weren't there. It's the series resistance that is required for the clipping action.
 
  • #30
cnh1995 said:
It's the series resistance that is required for the clipping action.
Can you explain this? If there is no series resistor in my problem (it is parallel), does that mean there is no clipping in my problem?

Do you have any hints for these two?
upload_2016-10-11_22-16-22.png
 

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