Calculation of max load voltage, dc load current and efficiency half-wave rectifier

Thread Starter

studyhard24

Joined Jun 20, 2022
35
halfwave.JPG

Please don't give direct answer. I need some help to understand this questions, youtube/website link which has detail information about this topic is needed. I'm giving an improvement exam and this topics were not taught to us back then. Thank you for your valuable time in helping me out.
 
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Ian0

Joined Aug 7, 2020
5,945
Don’t forget that it is a half wave rectifier, the voltage for half the time is zero, and there is no AC power supplied for half the time.
 

Ian0

Joined Aug 7, 2020
5,945
hi 24,
Part (ii) looks OK, how about Part (i) of the question.?
E
Part (iI) doesn’t look OK to me, the only losses are the voltage drop of the diode, and 40% efficiency would require the diode to dissipate one and a half times as much power as the load.
I think that the TS has overlooked the fact that power is only being supplied to the load half the time.
[edit] The efficiency is part 3 not part 2. Part 2 (average DC voltage) is OK.
 

The Electrician

Joined Oct 9, 2007
2,903
hi 24,
This is what LTS shows for Part(ii), close enough DC average of 6.36V
E
What is the forward voltage drop of your "ideal" diode? The problem specifies a germanium diode. Does anybody here remember what the forward drop of a germanium diode is? I seem to remember something like .2 to .3 volts. It won't make much difference in the final answer, but the student should take it into account to demonstrate understanding of what's going on.
 

ericgibbs

Joined Jan 29, 2010
16,037
hi Elect,
Of course, you are correct in that the precise answer should include a Ge Vdiode forward drop of approx 0.2V.
E
My Ideal Diode LTS model: [Used as a reverse voltage blocking component in some LTS sims]

.model IdealD1 D(Ron=0.001 Roff=1Meg Vfwd=0.001Vrev=100 Ilimit=1 RevIlimit=1)
 

The Electrician

Joined Oct 9, 2007
2,903
The image of the clipped top of the waveform is interesting. I would think the clipped part would be horizontal; instead it's sloped, and one of them is sloped differently than the other three. I wonder why? This image shows the simulated answer to the question of maximum load voltage.

A purely mathematical caluclation of the average gave me 6.26495

Your simulation average result did increase with fewer cycles and increased resolution.
 

Ian0

Joined Aug 7, 2020
5,945
The image of the clipped top of the is interesting. I would think the clipped part would be horizontal; instead it's sloped, and one of them is sloped differently than the other three. I wonder why? This image shows the simulated answer to the question of maximum load voltage.

A purely mathematical caluclation of the average gave me 6.26495

Your simulation result did increase with fewer cycles and increased resolution.
That's just limits of resolution in SPICE. Reduce the step time and it will be sinusoidal.
Number of cycles will make no difference provided that they are complete cycles
 

The Electrician

Joined Oct 9, 2007
2,903
That's just limits of resolution in SPICE. Reduce the step time and it will be sinusoidal.
Number of cycles will make no difference provided that they are complete cycles
Eric, when you re-ran the simulation with fewer cycles, you said you increased the resolution. Did you mean the resolution of the dispaly, or did you also change the step time?

I'm not convinced that changing the number of cycles will make no difference in the calculated average in all the decimal places. That's why I asked Eric to change the number of cycles--to find out just how good LTspice is.
 

Ian0

Joined Aug 7, 2020
5,945
Eric, when you re-ran the simulation with fewer cycles, you said you increased the resolution. Did you mean the resolution of the dispaly, or did you also change the step time?

I'm not convinced that changing the number of cycles will make no difference in the calculated average in all the decimal places. That's why I asked Eric to change the number of cycles--to find out just how good LTspice is.
LT Spice is as good as the information you give it.
 
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