# How do I start this problem? (diode circuit)

1. Oct 11, 2016

### servehover

1. The problem statement, all variables and given/known data

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

2. Relevant equations (given by my teacher)
Id = Io * (e^(qv/nkt)-1)
Vp = Vrms * sqrt(2)
Vrms = Vp/(sqrt(2))

3. The attempt at a solution
I have no idea how to find Vo or the waveform.

2. Oct 11, 2016

### cnh1995

Welcome to Physics Forums!

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

3. Oct 11, 2016

### servehover

Neither are given

4. Oct 11, 2016

### rude man

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.

5. Oct 11, 2016

### cnh1995

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.

6. Oct 11, 2016

### servehover

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

7. Oct 11, 2016

### cnh1995

Yes.

8. Oct 11, 2016

### servehover

Any hints for something like this?

9. Oct 11, 2016

### rude man

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. Oct 11, 2016

### rude man

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. Oct 11, 2016

### servehover

So V0 would be 0.4949Vrms? (0.3535 * Vpp)

12. Oct 11, 2016

### cnh1995

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: Oct 11, 2016
13. Oct 11, 2016

### 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.

14. Oct 11, 2016

### rude man

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. Oct 11, 2016

### rude man

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. Oct 11, 2016

### servehover

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. Oct 11, 2016

### rude man

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. Oct 11, 2016

### servehover

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. Oct 11, 2016

### Staff: Mentor

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 modelled 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. Oct 11, 2016

### Staff: Mentor

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