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

20/.3 = 66.66V

You know that this answer is wrong because the units are messed up. You are dividing a voltage by a voltage, which gives you a dimensionless number, and then you are tacking on a V to the answer because you want the answer to be in volts. But all you've done is guaranteed that you have a meaningless and wrong answer.

You seem to be struggling trying to figure out what information is really being given and what information is being asked for. To help you figure it out, I would recommend drawing (or sketching) clearly labeled diagrams that clearly define the quantities you are working with and that you are looking for. If nothing else, it will greatly help those trying to help you to understand how you are viewing the problem so that, where needed, they can help clear up misinterpretations.

 

View attachment 273152

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.

Was any kind of figure, such as a schematic, provided for this problem?

A good place to start is to ask what the answers would be for some idealized versions of the problem.

What would the max load voltage be if the diode were ideal?
What would it be if an ideal Ge diode were used?
What would the max load voltage be if the diode has a 0V turn-on voltage but 0.25 Ω of resistance?

What would the efficiency be if the diode were ideal?
What would the efficiency be if the diode has a 0V turn-on voltage but 0.25 Ω of resistance?
What would the efficiency be if the diode were an ideal Ge diode?

You don't need to come up with exact solutions for these, just something close. Many of these should be answerable very quickly and should give you answers that will be close to the final answers, so if your final answers are significantly different, you can suspect that something has gone wrong.

Also, don't just go looking for equations to throw at the problem. Draw the circuit, identify what it is you are trying to find, and then analyze the circuit to find it.

 

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:

View attachment 273385
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:

View attachment 273386

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:

View attachment 273388

Hi,

That looks much better, but it does not look like an 'ideal' diode is being used. It looks like the forward voltage drops with increased current which an ideal diode would not do, or possibly you used a diode with higher forward voltage.

In other words we should be seeing something like this:
Vout=A*sin(w*t)-Vd

and solve for t by setting Vout=0 to find the starting time, and subtract that from the half wave time to get the ending time.

I tried A=10v and Vd=1v and got around 16ms for a frequency of 1Hz (i didnt want to solve with the actual problem specs).

 

That looks much better, but it does not look like an 'ideal' diode is being used. It looks like the forward voltage drops with increased current which an ideal diode would not do, or possibly you used a diode with higher forward voltage.

Hi Al,
The question states that the diode has an internal resistance of 0.25 Ohms, so the measured voltage drop will increase as the current increases.

The Ge diode forward voltage has been assumed to 0.2V

E

 

Hi Al,
The question states that the diode has an internal resistance of 0.25 Ohms, so the measured voltage drop will increase as the current increases.

The Ge diode forward voltage has been assumed to 0.2V

E

Hi Eric,

Oh sorry i missed that part
Yes, so the voltage would be more like vD=0.2+iD/4.

 

Hi Al,
The question states that the diode has an internal resistance of 0.25 Ohms, so the measured voltage drop will increase as the current increases.

The Ge diode forward voltage has been assumed to 0.2V

E

Hi again Eric,

Guess what, it looks like the diode current and resistance only changes the peak by 0.025 percent. That's 0.00025 decimal So the wave should look very close to the peak of the input sine wave.