You're welcome. I forgot to mention that that was for the maximum peak to peak ripple current although I edited that into that post now too. That's usually the desired units for these kinds of problems.

Okey, maybe you can help me with my second question about the same circuit:
https://forum.allaboutcircuits.com/...ith-a-capacitor-in-a-bridge-rectifier.202414/

 

Hello,

You could approach this from the standpoint of harmonics which turns this problem into a simple AC circuit with no rectifiers.

The output of a regular full wave bridge has major 2nd harmonic content which is around 70 percent of the fundamental. That means the major AC component the inductor sees is going to be 100Hz with a 50Hz power line input. The next harmonic is the 4th, which would be much lower maybe around 12 percent. You may be able to ignore that or just factor that in with the 2nd harmonic. The 6th will be about half of that, and the higher order harmonics will be reduced each by about 1/2 of the previous, so you may get away with just considering the 2nd harmonic and maybe the 4th if you want to be more accurate.

What this means is that you can use the same techniques you use for pure AC circuits just with a higher frequency.
Of course you also have to consider the DC component for the wire size and saturation characteristics if you have to think about that too.

The simplification that most readers use is:
Z=sqrt((w*L)^2+R^2)

and here w=2*pi*f, and f would not be 50Hz it would be 100Hz for that single 2nd harmonic. You can see how fast this gets you to an answer, but you can also include the 4th harmonic if you like.

The real question here though is how you were taught to do this in the past. Were you taught to do this all in the time domain (such as with differential equations) or in the frequency domain, or some other technique? If you want to get an answer that matches some other result you'd have to use that preferred method.

There are other more precise ways to do this based on the actual bridge output waveform, and of course you should check your results with a simulator.

Yes inductors like this can be big but that's just a given. I've worked with inductors that weigh over 100 pounds and use 1/4 inch diameter enameled copper wire. Power electronics is not anything like your typical Japanese portable radio. You might even see diodes as large as a hockey puck, or stud mounted parts as large as salt shakers.

Doesn’t that calculate the ripple current at maximum output (as if the thyristors had been replaced by rectifiers)?

 

Doesn’t that calculate the ripple current at maximum output (as if the thyristors had been replaced by rectifiers)?

Yes that's why I mentioned the ideas about the average toward the end. If the average is not 0.6366 times the peak then we have to figure that into the calculation as reducing the peak. We can look into this deeper if you like though.
The dominate harmonic should still be the 2nd, but we have to realize at some point we'd have to do a full time domain analysis, which is probably not too difficult because there is only one storage element the inductor. It also looks like the conduction angle is much greater than 90 degrees although still short of 180 degrees.

Doing a few quick tests however shows that the averaging technique may not work out as well with the conduction angle change. This would complicate the calculation as we'd have to consider other harmonics. The 2nd harmonic would still be dominant, but I do not think it would be enough even with a conduction angle of 120 degrees which is still fairly large.
Thus, for the time being we will have to consider this technique mostly for use with a 180 degree conduction angle (no cutback).

 

Yes that's why I mentioned the ideas about the average toward the end. If the average is not 0.6366 times the peak then we have to figure that into the calculation as reducing the peak. We can look into this deeper if you like though.
The dominate harmonic should still be the 2nd, but we have to realize at some point we'd have to do a full time domain analysis, which is probably not too difficult because there is only one storage element the inductor. It also looks like the conduction angle is much greater than 90 degrees although still short of 180 degrees.

Doing a few quick tests however shows that the averaging technique may not work out as well with the conduction angle change. This would complicate the calculation as we'd have to consider other harmonics. The 2nd harmonic would still be dominant, but I do not think it would be enough even with a conduction angle of 120 degrees which is still fairly large.
Thus, for the time being we will have to consider this technique mostly for use with a 180 degree conduction angle (no cutback).

I made a mistake when I did the test. This actually works better than I thought. What happened was I had forgotten to consider a lower series resistance for the diodes used in the test setup. I was using the old 1 amp diodes for a 30 amp circuit

Ok, so that error corrected, it works better now using the averaging, but it's still not quite as good as I think we could get it. I'd like to find a solution that's even better.