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half wave rectifier

Hint: in my calculations below, to make math easier, i assume that the
diode = ideal(there is no 0.7 voltage drop across the diode)

Vrms = 0.707Vp => Vp = 14.1v, therefore Vp(load) = 14.1v assuming diode = ideal and Vdc = Vp/π = 14.1/π = 4.49v, which is pretty close to my screenshot(it says Vdc = 4.22 V ).

I want to compute Vdc. There is a formula for half wave sine: form factor = Vrms/Vdc = 1.57 . My naive mind says: Vdc = Vrms/1.57 = 10/1.57 = 6.369 . The simulator says Vdc = 4.22. :-(

I would like to use the form factor formula and get a correct result. What is my mistake?

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Because you are wrong

\$ \Large \frac{V_{RMS}}{V_{AVR}} = \frac{\frac{V_{peak}}{\sqrt{2}}}{\frac{V_{peak} }{\pi}}= \frac{V_{peak}}{\sqrt{2}} \cdot\frac{\pi}{V_{peak}} = \frac{\pi}{\sqrt{2}} \approx 2.2214\$

As for the average value in the half wave rectifier:

$$V_{DC} = \frac{1}{2\pi}\int_{0}^{\pi}V_p sin(x) dx = \frac{V_p}{\pi}$$

https://www.symbolab.com/solver/step-by-step/%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Cint_%7B0%7D%5E%7B%5Cpi%7D%20Vp%20sin%5Cleft(x%5Cright)%20dx

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  • \$\begingroup\$ i think your formula has a mistake in the denominator, it should not be sqrt(2)/pi. i am talking about Vavg. \$\endgroup\$ – DontAskTheEye Dec 30 '18 at 18:12
  • \$\begingroup\$ @DontAskTheEye I corrected the equation \$\endgroup\$ – G36 Dec 30 '18 at 18:17
  • \$\begingroup\$ (10V - 0.6V)/2.22 = 4.23V Any more proof needed? \$\endgroup\$ – G36 Dec 30 '18 at 18:33
  • \$\begingroup\$ i am afraid i just noticed that the numerator is also wrong. I am talking about Vrms . It is not equal to Vpeak/sqrt(2). It should be : Vrms = Vpeak/2 for half wave sine. \$\endgroup\$ – DontAskTheEye Dec 30 '18 at 19:12
  • \$\begingroup\$ @DontAskTheEye This equation I showed in my answer is true if you use the Vrms for the input voltage (10V) not the RMS voltage across a resistor after a rectifier. But if you want the find the ratio between Vrms/Vdc at the output of a rectifier then the answer is \$\frac{\pi}{2} = 1.57 \$ But in this case, you cannot substitute the 10V into the formula. \$\endgroup\$ – G36 Dec 30 '18 at 19:26
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Conversion for rectified Sine

Wave  Vp  Vrms  Vavg        Vp    Vrms    Vavg
Full  1   1/√2  2/π         1    0.707    0.637
Half  1   1/2   1/π         1    0.500    0.318   

Full √2   1     2√2/π       1.414 1.000   0.9003  Vrms/Vavg= π / 2√2 
Half √2   1/√2  √2/π        1.414 0.707   0.450   Vrms/Vavg= π / 2
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The \$V_{DC}\$ of your simulator assumes a full cycle average voltage. Half a cycle is completely missing, when the diode is reverse-biased.

\$V_{rms}=10v\$

\$V_{peak}=14.14\$

\$ V_{average}={2V_{peak} \over \pi} \$ for one half cycle.

\$ V_{average}={V_{peak} \over \pi} \$ for one full cycle or multiple cycles.

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  • \$\begingroup\$ Are you sure about it? \$\endgroup\$ – G36 Dec 30 '18 at 18:04
  • \$\begingroup\$ i doubt that your idea is correct glen_geek. But for the sake of the argument let's see: Full wave sine form factor = 1.11=Vrms/Vavg. Thus, Vdc = 10/1.11 = 9.009 volts. If i can magically divide that by two 9.009/2= 4.5 volts which is what i want to see :p \$\endgroup\$ – DontAskTheEye Dec 30 '18 at 18:18
  • \$\begingroup\$ @DontAskTheEye We are in agreement. Your circuit is half-wave. Your formula applies to full wave, where both half-cycles contribute to the average value. So yes, divide by two (its not magic). \$\endgroup\$ – glen_geek Dec 30 '18 at 18:23

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