How do I start this problem? (diode circuit)

[servehover]

Homework Statement

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.

 

[cnh1995]

Homework Statement

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?

 

[servehover]

Welcome to Physics Forums!

What is the input ac voltage? What is the turns ratio of the transformer?

Neither are given

 

[rude man]

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.

 

[cnh1995]

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.

 

[servehover]

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.

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?

 

[cnh1995]

Would the output voltage v0 simply be 1Vrms? Is it because the voltage across the 1k resistor is also 1Vrms?

Yes.

 

[servehover]

Yes.

Any hints for something like this?

 

[rude man]

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

 

[rude man]

Any hints for something like this?

Same basic idea. V_o is limited to 5.6V. You need to think graphically about V_o as it rises from zero to the max. voltage . That also applies to your previous problem BTW.

 

[servehover]

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)

 

[cnh1995]

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?

 

[Carrock]

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.

 

[rude man]

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.

 

[rude man]

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.

 

[servehover]

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?

 

[rude man]

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?

 

[servehover]

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?

 

[NascentOxygen]

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?

 

[NascentOxygen]

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.

 

[servehover]

Is your teacher known for setting trick questions designed to make you really think?

Yes

A peak of ±0.7V, certainly, but not a sinewave.

Would it be a rectified sine wave?

 

[rude man]

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 V_o would not look like a sine wave although it would indeed have peaks of 0.7V.

 

[rude man]

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?

 

[servehover]

Yoou're getting closer, but V_o 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.

 

[cnh1995]

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.

 

[servehover]

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

 

[cnh1995]

So it would basically be a sine wave but clipped at 0.7V?

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.

 

[servehover]

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?

 

[cnh1995]

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.

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.

 

[servehover]

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?

 

[rude man]

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.

The thing you need to see is that the diodes are "not there" if V_o is between - 0.7V and + 0.7V. So for that range of V_o you get just the sinusoidal output voltage 1.414sin(wt). But as soon as Vo > +0.7V or < -0.7V, Vo has to stay ("is clamped") at those voltage levels. Now try to reconstruct what V_o must look like over an entire period.

 

[cnh1995]

If there is no series resistor in my problem (it is parallel), does that mean there is no clipping in my problem?

Yes. There is no clipping.

Can you explain this?

Say your input is 1V rms. If the diode is maintaining 0.7V across it, where is the remaining voltage? It has to appear across the series resistance. That means, without the series resistance, all the voltage would appear across the diodes and there will be no clipping. If there is a series resistance, the diode almost maintains 0.7V across it and the remaining voltage appears across the series resistance.

 

[servehover]

The thing you need to see is that the diodes are "not there" if V_o is between - 0.7V and + 0.7V. So for that range of V_o you get just the sinusoidal output voltage 1.414sin(wt). But as soon as Vo > +0.7V or < -0.7V, Vo has to stay ("is clamped") at those voltage levels. Now try to reconstruct what V_o must look like over an entire period.

So it'd basically be a sine wave that is cut off beyond 0.7 and -0.7.
If this question is a trick question, the answer would simply be Vo = 1Vrms right?

 

[servehover]

Yes. There is no clipping.

Say your input is 1V rms. If the diode is maintaining 0.7V across it, where is the remaining voltage? It has to appear across the series resistance. That means, without the series resistance, all the voltage would appear across the diodes and there will be no clipping. If there is a series resistance, the diode almost maintains 0.7V across it and the remaining voltage appears across the series resistance.

So just to be confirm:
There is no series resistor in my circuit (it looks parallel). This means that the output voltage is simply 1Vrms...?

 

[cnh1995]

So just to be confirm:
There is no series resistor in my circuit (it looks parallel). This means that the output voltage is simply 1Vrms...?

I would say yes..