Halfwave Diode Rectifier w/ RL load (Current Conduction Angle)

Discussion in 'Homework Help' started by jegues, Nov 9, 2012.

1. jegues Thread Starter Well-Known Member

Sep 13, 2010
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Hello all,

Consider the circuit in the figure attached for a source voltage of,

$v_{s}(t) = V_{m}sin(\omega t)$

Where,

$V_{m} = 50\sqrt{2}V$

$R +j\omega L = 400 + j300 \Omega$

Where,

$\omega = 2\pi 60 rad/s \quad R = 400 \Omega \quad L = 0.8H$

This was a circuit we built in the lab and we had measured the voltage Vd and the current Id, and using the calculations from the theory I was able to obtain comparable results.

We also measured the current conduction time and current conduction angle using the cursors in the lab where the current is the purple trace. (As shown in the second figure, we observe and measure a current conduction time of 7.60ms)

For the current conduction time and current conduction angle I was not able to use the theory and recreate values close to our measured values.

My work is attached for calculating what I understand is the current conduction angle β. I am obtaining ~10ms instead of 7.6ms and the angle is quite far off from the measured angle.

Does anyone spot any mistakes or can explain why I am not obtaining similar results from the theory?

Thanks again!

EDIT: My gut feeling says there must be something wrong in the measurement. At 60Hz we expect a peroid of 16.6ms, so half of that is 8.3ms, the current conduction time should be larger than half the peroid so 7.6ms doesn't make any sense.

There must have been some error making the measurements or in the equipment.

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Last edited: Nov 9, 2012