0
\$\begingroup\$

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?

\$\endgroup\$

3 Answers 3

2
\$\begingroup\$

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

\$\endgroup\$
7
  • \$\begingroup\$ i think your formula has a mistake in the denominator, it should not be sqrt(2)/pi. i am talking about Vavg. \$\endgroup\$ Dec 30, 2018 at 18:12
  • \$\begingroup\$ @DontAskTheEye I corrected the equation \$\endgroup\$
    – G36
    Dec 30, 2018 at 18:17
  • \$\begingroup\$ (10V - 0.6V)/2.22 = 4.23V Any more proof needed? \$\endgroup\$
    – G36
    Dec 30, 2018 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\$ Dec 30, 2018 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, 2018 at 19:26
2
\$\begingroup\$

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
\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$
3
  • \$\begingroup\$ Are you sure about it? \$\endgroup\$
    – G36
    Dec 30, 2018 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\$ Dec 30, 2018 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, 2018 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.