# Understanding an exercise

#### drk

Joined Mar 8, 2008
41
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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#### steveb

Joined Jul 3, 2008
2,436
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?
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.

#### steveb

Joined Jul 3, 2008
2,436
The three sources represent an AC 3-phase source: http://upload.wikimedia.org/wikipedia/commons/c/cc/3_phase_AC_waveform.svg

The diodes will let the source with the higher voltage go through at each time. So the average voltage across the resistor will be:
$$\int_{-60}^{60} V_{max} \cos (\phi) d\phi -V_d$$
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.

#### drk

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

Thanks!

#### Georacer

Joined Nov 25, 2009
5,182
Yeah, I forgot to average the integral. Thanks Steve.

#### steveb

Joined Jul 3, 2008
2,436
Yeah, I forgot to average the integral. Thanks Steve.
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.