LT Spice is as good as the information you give it.

That's not an answer to my question. With the information Eric gave it, will it give a tiny different result with fewer cycles? I would like to see the result with two runs, differing only in the number of cycles.

 

Is max load voltage and maximum secondary voltage the same thing? I am unable to understand how to determine max load voltage

 

None of the lecturers with whom I've contacted in my varsity responded, My improvement is tomorrow, And I really want some guidelines on this chapter. Any website link/youtube suggestion will be highly appreciated, Thanks

 

Is max load voltage and maximum secondary voltage the same thing? I am unable to understand how to determine max load voltage

hi s24,
The question asked is:
(i) Find Max load voltage
(ii) DC load voltage

Where has maximum secondary voltage been asked for.?

You have calculated the DC Load Voltage
So what do you think 'Find Max load voltage' means.???

E

 

So what do you think 'Find Max load voltage' means.???

is it gonna be Vrms/.3(for Ge)

20/.3 = 66.66V

 

For the efficiency

 

hi s24,
How can this be possible.????

is it gonna be Vrms/.3(for Ge)
20/.3 = 66.66V


(i) Find Max load voltage
Ask yourself, is Maximum voltage the same 'term' as the Peak voltage.?

E

 

 

hi s24,
As this homework, I cannot give you the actual answer, just hints.
Look at this plot and tell me what you think the Vmax load is.??
E

 

Thanks a bunch, Sir, I've understood now,
The given voltage is gonna be the maximum voltage.

Why didn't I think of that?
And how on earth it's a question given in exam!

 

hi,
So what is the
The given voltage is gonna be the maximum voltage.???
E

 

hi,
So what is the
The given voltage is gonna be the maximum voltage.???
E

20 V

 

hi,
NO
What about the internal 0.25R and 0.2Vfwd drop.????
E

 

V1000 = (1000/(1000+.25))*20 = 19.99 V

Vm = 19.99-.2 = 19.79 V

 

V1000 = (1000/(1000+.25))*20 = 19.99 V

Vm = 19.99-.2 = 19.79 V

studyhard24, regarding AC voltages and their waveforms, you obviously know what a peak-to-peak voltage is, but do you know what RMS voltage is, and how to calculate it (have you studied calculus)?

It's Sunday, and I'll keep an eye on this thread, and I will be able to respond quickly.

 

I have studied calculus, but I'm sure this question is not in my league, I'm not understanding this topic at all. It seems to be pretty interesting, but unfortunately, I don't even have my basic clear

 

hi s24,
That post #34 answer looks OK, now calculate the efficiency
E

 

hi s24,
That post #34 answer looks OK, now calculate the efficiency
E

We helpers need to take note of something that I just now realized when looking at the initial post. The applied voltage is 20 volts peak-to-peak, not 20 volts peak. The applied voltage is 10 volts peak. So, Eric, your simulations are using 20 volts peak, when you should be using 10 volts peak.

 

@ Hi @The Electrician,
Many thanks for the heads-up on the misread of the question.

@studyhard24
Please adjust your calculations to give the correct answers for 20Vppk, ie: 10Vpeak !!!!

E

 

studyhard24, in post #3 where you were calculating the efficiency, you have the basic concept correct--efficiency equals DC power output/AC input power. You used a symbol that isn't defined; I'm referring to the symbol Im. My guess is that it stands for max current, and you calculated the power in terms of this Im. However, the most basic power calculations in AC power circuits involve integrating the instantaneous power. The next level up involves using the RMS value of voltage in the formula E^2/R or the formula I^2*R. You can do what you did, using the max current if it's allowable to do so for both the DC power and the AC power. When is it allowable? When waveshape for the AC power is the same as the waveshape for the DC power. Therein lies your problem.

Here is an image showing the waveform of the output of the transformer in blue, and the waveform in red after subtracting the .2 volt drop of the Ge diode. They appear to be nearly the same:


But what the diode is doing to the waveshape is hidden in this image. If I reduce the applied voltage from 10 volts peak to 1 volt peak, and show the 1 volt peak waveform in blue, and the waveform in red after the Ge diode's .2 volt drop has it's effect, it can be seen more clearly what's going on:



The red waveform is no longer the positive half of a sine wave; it's the cut off top part of a sine wave, and its RMS value is no longer just equal to the peak voltage times .707; you will have to calculate the RMS value of the red waveform using calculus.

To make it more clear why the red waveform is not sinusoidal any more, here's the waveform (blue) directly out of the transformer, and here it is (red) after the diode (actually 2 diodes to get the effect in both directions) subtracts its forward voltage, on a scale where what's going on is obvious. The red waveform is no longer a sine wave, and the positive half lobe is not part of a sine wave: