Power electronics principle

Thread Starter

Peijie Xu

Joined Aug 18, 2015
4
Hi guys,
i am confusing about the mean value of |i| in this case.
Fs = 3.8kHz, Duty cycle(D) = 0.73, delta i = 4.9A
If the above waveform is VL, how to get the mean value of |i|?
I found Ton maybe equal 0.73*1/3800.
According to VL= L * di / dt, i got L?Untitled.png
 

MrAl

Joined Jun 17, 2014
11,496
Hi,

Are you tying to get the mean (or average) current through an inductor?

To get the average you usually integrate, but it may be simpler in some cases.
Did you ever do this before? For example, can you find the average of VL itself?
 

Thread Starter

Peijie Xu

Joined Aug 18, 2015
4
Hi,

Are you tying to get the mean (or average) current through an inductor?

To get the average you usually integrate, but it may be simpler in some cases.
Did you ever do this before? For example, can you find the average of VL itself?
No...I never did before. I think calculating VL need L*di/dt. Is it right?
 

MrAl

Joined Jun 17, 2014
11,496
Hi,

Then maybe you should concentrate on calculating the average of VL first since you have been given that information already. Once you master that, you can move on to calculating the average of the current which is what i think you are after.

Do you know how to integrate?
 

WBahn

Joined Mar 31, 2012
30,081
Since this is a repetitive waveform, you have to take into account that you don't know the initial state (unless you can determine it based on symmetry). But since you pretty much have to assume that the system is operating in AC steady state, you know that the state at the end of a complete cycle is the same as at the beginning of that cycle.

You haven't given much of a clue regarding what the circuit is that this applies to. Is the sketched waveform the voltage applied to an inductor?

Sketch what the current in the inductor looks like. For starters, go ahead and assume that the current starts at zero.

Once you have that sketch, take the absolute value of it.

Then find the average value of that.

Show your work as far as you can.
 
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