# Half wave rectifier

Discussion in 'General Electronics Chat' started by dalam, Aug 10, 2014.

1. ### dalam Thread Starter Member

Aug 9, 2014
58
6
Can someone solve this problem for me.

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2. ### ericgibbs Moderator

Jan 29, 2010
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hi,
Have you calculated the Vpeak and the diode conduction time for one cycle.?

E

3. ### phantomzz New Member

Sep 26, 2013
18
0
Figure out the conduction period of the diode during a cycle. Calculate current in diode during the non-conducting period.

Calculate the instantaneous current during the conduction period and integrate it over the time period.

Finally take the avg. of the current during conducting period and non-conducting to give you the average over an entire cycle.

4. ### MikeML AAC Fanatic!

Oct 2, 2009
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What peak voltage will the cap charge to?
(Hint: peak of sine - one diode drop)
How much will cap voltage sag until next sine peak?
(Hint: period of sine, use peak resistor current)
What is the ~average voltage across cap?
(hint: peak-sag/2)
What is the average current through the load resistor?
(Hint: Dr Ohm)
(Hint: Kirchoff)

Last edited: Aug 10, 2014
5. ### MikeML AAC Fanatic!

Oct 2, 2009
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Did you follow the hints? My posted method finds the both the peak and average voltage across the capacitor, therefore the inital resistor current is just (peak cap voltage)/R.

One simplification is to assume that the discharge curve of the capacitor is linear (as it would be if the cap was being discharged by a constant-current; not a resistor.)

6. ### MikeML AAC Fanatic!

Oct 2, 2009
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Lets see how the method does:

Peak cap voltage is 10-0.7 = 9.3V
Initial Discharge (resistor) current = 9.3/100 = 93mA
Sag (ΔV) = i*t/C= 93m*20m/4m = 0.465V
Av. Cap Voltage = 9.3-0.465/2 = 9.067V
Ave Resistor current = 9.067/100 = 90mA
Therefore average diode current = 90mA.

Now look at a simulation:
LTSpice says average diode current is 89.9mA. I'd say the method works...

Note the peak diode current; Note the duration that current actually flows through the diode...

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7. ### phantomzz New Member

Sep 26, 2013
18
0
I have confused myself here....

During the positive cycle diode conducts....

Impedance of resistor = 100 Ω

Impedance of capacitor = 1/(ω*C)= 1/(2*∏*50*4*0.001)= 1/1.25 Ω

Since they are in parallel, using reciprocity and addition

1/Z = 1/100 + j1.25 = 0.01 + j1.25

1/Z= mod(0.01+j1.25) ≈ 1.25

Z= 1/1.25= 0.8

Therefore current during conduction period is V sinωt/Z = 12.5sinωt.

If I take the intergral of the current flowing ∫12.5 sinωt dt between the period 0 to ∏ it is 12.5*2= 25.

During non-conduction cycle, I(diode)=0 from ∏ to 2∏

Therefore average current= 25/(2∏-0)= 12.5/∏≈ 4A

What am I doing wrong?

8. ### Dodgydave AAC Fanatic!

Jun 22, 2012
8,282
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I get capacitors reactance of 795 ohms, in parallel with 100r resistor is 88 ohms total.

9. ### blah2222 Distinguished Member

May 3, 2010
582
38
I think you may be assuming that "steady-state" is referring to using an AC analysis approach to this problem. The diode brings nonlinearity into the mix causing harmonics to appear, making this approach less useful.

The most straightforward way to approach this is to follow MikeML's approach and track what the voltage and current is doing from the perspective of the capacitor being charged and discharged. "Steady-state" in this context just means when the transient on startup finishes and the circuit behaves in a predictable manner.

10. ### ericgibbs Moderator

Jan 29, 2010
8,549
1,723
hi phantom,

IF its a print of an 'old' circuit diagram, 4mF code was sometimes used to indicate 4uF, the 'm' was assumed 'micro'.

It would be a good idea to clarify the capacitor coding.

Also it states an 'ideal' diode, which has a zero forward voltage drop.

E

Oct 2, 2009
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12. ### phantomzz New Member

Sep 26, 2013
18
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@ericgibbs

I am not the OP, so I can't really clarify on the question but I do believe mF over here stands for milliFarad.

@DodgyDave

How do you get the capacitive reactance as 795 Ω?? Did you take the Capacitance as 4 μF??

@blah2222

The question speaks of an ideal diode so I don't think factors such as harmonics and non-linearity of diode are required to be taken into account. Even if I did assume practical factors it doesn't explain the huge difference between MikeML's answer and mine.....

@MikeML

In your calculation of capacitor discharge you have taken t=20ms. The capacitor doesn't discharge during the entire cycle right?
Also if avg. resistor current is 90mA how does that make avg.diode current as 90mA ?

13. ### blah2222 Distinguished Member

May 3, 2010
582
38
An ideal diode still only conducts in one direction doesn't it? That smells highly nonlinear to me...

You are dealing with more than the fundamental frequency once you get past the diode.

14. ### MikeML AAC Fanatic!

Oct 2, 2009
5,444
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I use the approximation that the charge time (time the diode conducts) is so short that the discharge time is the entire duration until the next charge pulse comes along. At 50Hz AC, the period between successive pulses is 1/50 = 20ms. If you want, you could say that the charge time is ~2ms and the discharge time is ~18ms. This is borne out by the LTSpice sim... My way, you calculate that you need a slightly larger filter capacitor, so the error is not significant.

Note that if this was a full-wave rectifier instead of a half-wave one, we would be talking about 10ms, not 20ms. This explains why FW is easier to filter...

15. ### dalam Thread Starter Member

Aug 9, 2014
58
6
Your answer is correct for an diode which has an forward drop of 0.7 volts but here diode is ideal so I think peak cap. voltage should be 10 volts.
Rest all looks good.

16. ### MikeML AAC Fanatic!

Oct 2, 2009
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Which begs the question: What is an ideal diode? The one I linked to, or one that has a forward voltage drop of zero?

17. ### blah2222 Distinguished Member

May 3, 2010
582
38
I think for this question, 'ideal' is referring to the on-diode as a short. Most textbooks describe 'ideal' diodes in this way. The Shockley model seems a bit complex for this type of question.

With 0.7V or a short, the process is still the same.