hello all, This is my first time posting in here, I tried using the search bar to see if there was an answer to a similar question but I did not find one.

I just finished one problem where I had an ideal diode connected to a square wave with a peak to peak at 6 volts. Then the diode splits to an out put Vo and also down to a 100 ohm resister which is grounded.

The question asked for both the peak and avg output voltage and also the peak and average current running through the diode, then finally asks what the maximum reverse voltage across the diode can be.

This was a simple task where I answered all the questions in short order, however the next question asked me to repeat the process if the square wave now has an average of 2 V but keeps the same peak to peak. I assume the max values stay the same but I am not sure at all how to change their averages. I have looked on google and not found a problem dealing with this kind of thing...any help would be appreciated, thank you.

 

You get the average voltage by the ratio of off and on times.

 

Im sorry, but what do you mean? I am not given any on or off times. All I was given was what I wrote...

 

A square wave, by definition, spends some part of the period at, say, 0 volts, and the rest at the peak voltage. Since the waveform is only at those extremes, an average voltage has to come from the ratio of the peak and zero voltage durations.

 

how do I know how long it stays at 0 and its peak? Im not given any frequency or period. only that the average of it is 2V and its peak is 3V...so I have to use those to find the average for the output voltage and and the average for the current across the diode. I dont mean to sound harsh, its just been frustrating because this is not covered by the book or the teacher.

 

If the duty cycle is 50%, then the ratio of 0 volts to 6 volts is equal, and the average voltage will be 3 volts. For a 2 volt average, the duty cycle will be less than 50%. What percentage of 6 volts is 2 volts? This does not depend on frequency.

 

I seems more likely to me from the OP that the entire square wave is given a DC offset, rather than the duty cycle changing. A 6VPP sq wave with an average a zero will range from -3 to 3 volts, and a 6VPP sq wave with an average of 2V will range from -1 to 5 volts.

 

That may be. We do need some clarification. Possibly the OP can post up a schematic of the circuit. That will resolve doubts.

 

Since the OP sounded a little befuddled by talk of a duty cycle, I thought that might not be involved.

It also sounds a lot like problems we give to the students in an electronics measuring techniques lab that I TA.