# Half wave rectifier with RL load

Discussion in 'Homework Help' started by notoriusjt2, Sep 30, 2010.

1. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
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all of these equations are taken straight from the textbook

thanks to anyone who can help because this problem is driving me mad

Last edited: Sep 30, 2010
2. ### Georacer Moderator

Nov 25, 2009
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1,266
Why do you substitute $\omega=377\ since\ \omega=2 \cdot \pi f=2 \cdot 3.14159\cdot 400=2513.272$?

Also the $d(\omega t)=$ is the differentiable variable of the integral, that is, a notation of what variable should be integrated. It is the same as if you said $\int_0^\beta \sin(x) dx$.

Last edited by a moderator: Oct 2, 2010
3. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
thanks for pointing that out, i definitely screwed up the omega value.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Also on the 6th line of the attached solution you appear to have 100*√2 as Vm.

Shouldn't Vmax be 100 - as stated for V peak in the question? Why the √2 term?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You would also expect the function

$sin(\beta-\theta)+sin(\theta)*\epsilon^{-\frac{\beta}{\omega \tau}=0$

at β=0.

Since ...

$sin(0-\theta)+sin(\theta)*\epsilon^{0}=0$

You need to find the condition for the next zero value for the relationship which presumably occurs for some β>0.

6. ### Ghar Active Member

Mar 8, 2010
655
72
The inductor doesn't matter for average current.
Average means DC, and with DC all you need is the average (DC) voltage.

For a half-wave rectified voltage the DC value is $\frac{1}{\pi}V_{peak}$, or 31.8 V.
If you subtract the diode drop and divide by the 100 ohm resistance, you get 310 mA.

Impedance calculations don't make sense because they only work for steady state sinusoidal circuits. The diode is non-linear. What you could do is calculate the fourier transform of a half wave rectified voltage and then do impedance calculations but you'd need to do it for a lot of harmonics...

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It seems the issue in this case is to do with the circuit transient behavior during the circuit (diode) conduction / non-conduction intervals. The time constant is 150usec which is short compared with the base AC period of 2.5msec. Each current pulse is therefore a composite of a sinusoid and an exponentially decaying term. It seems credible that one might solve for the transient expression of the current pulse and then determine the mean of the pulse, taking into consideration the pulse occurs over a yet to be determined interval. If the inductor were sufficiently large to give continuous current then the solution will probably (?) be different.

Like you, I initially thought it was just a trivial exercise and then I took a closer look. I'm thinking of doing a simulation to see what the real story is. I'm not sure whether all this work is really worth it. A "ball park" approach will probably do.

I wouldn't be surprised if 310mA is the nearest solution!

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Did the simulation - no surprises there.

So questions arise.

1. Why bother with the analysis? - Just to get the student to do it ....?

2. The question is a multiple choice so one would expect a quick estimate based precisely on Ghar's method. If the question was only worth a few marks in an exam then it's a 'no brainer'.

3. Was the OP told to use this method in solving the problem?

4. Why are teachers so mean to their students?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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As suggested, increasing the inductance does have an influence.

With an inductance of

15mH: Iav=308mA
30mH: Iav=285mA
60mH: Iav=240mA
120mH: Iav=177mA

It appears the current does not become continuous however much L increases ....

Clearly this highlights the instructive nature of the analysis. I was a bit harsh on teachers!

10. ### Ghar Active Member

Mar 8, 2010
655
72
You're right, my assumption breaks down with larger time constants.

It looks like the inductor can never have continuous current and it makes sense - if the diode were to be on the whole time the inductor always sees Vsrc - Vf - I*R

Only Vsrc changes polarity so you always get more voltage reducing the current than increasing it and then the diode stops conducting.

If that's the case than the current can never actually build up between cycles. That means the inductor's only function is to limit the current peaks.

Looks like you have to be very careful applying DC models... you're right it was instructive but somehow I doubt that was the intent of the exercise

11. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
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where did you come up with the diode drop?

12. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
0
a follow up problem, they added a DC source into the series circuit..

would I just subtract the DC voltage from the 120Vrms, and then solve for current as in the previous problem?

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Basically the same approach but there will be an additional transient term arising from the DC source. The general current function will be something like this ...

$i(t)=\frac{V_{m}}{L\sqrt{\frac{R^2}{L^2}+\omega^2}}$e^{- \frac{R}{L}t}sin(\beta t)+sin(\omega t-\beta)$-\frac{V_{DC}}{R}(1-e^{- \frac{R}{L}t})$

$\beta=atan(\frac{\omega L}{R})$

This makes solving for the i(t)=0 condition somewhat harder but "do-able" as per your method.

14. ### Ghar Active Member

Mar 8, 2010
655
72
I just subtracted some small voltage which could justify the 310 vs 318 mA you'd get otherwise.