# Understanding an exercise

Discussion in 'Homework Help' started by drk, Nov 1, 2011.

1. ### drk Thread Starter Active Member

Mar 8, 2008
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0
Hi, I have an exercise which tells me to consider the load's inductance high enough, so that the load can be considered a current source (constant current).

I've simulated the circuit, and with a high inductance it does eventually ends up with a constant current.

My problem is, how do I calculate that current?
How to analyse the circuit?

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2. ### steveb Senior Member

Jul 3, 2008
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The approach is to consider average values in steady state conditions. The three sources with diodes will act as one source with the highest voltage getting through. This voltage will not be constant, but you can consider the average to be about 83% of the signal amplitude. Then subtract off the diode voltage. This average voltage will be across the resistor because the average voltage on the coil will be zero in steady state.

Nov 25, 2009
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4. ### steveb Senior Member

Jul 3, 2008
2,433
469
Hmmm, not quite, but that's the right idea. Averaging requires dividing by the time. I obtained the 83% figure by the following integral.

$v_{avr}={{3\;v_{max}}\over{\pi}}\int_0^{\pi/3} \cos\theta d\theta={{3\sqrt{3}}\over{2\pi}}v_{max}\approx0.83\;v_{max}$

Then the diode voltage can be subtracted from this value.

5. ### drk Thread Starter Active Member

Mar 8, 2008
41
0
I'm getting the value of current +- what I get on the simulation.

Thanks!

6. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
Yeah, I forgot to average the integral. Thanks Steve.

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
No problem. I tend to forget that too.

Also, that pi/3 ratio being nearly equal to one has tricked my mind a few times. The difference is just small small enough that it does not allow our engineering judgement of "reasonableness" to detect the error. So, I came close to making the same mistake.