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how to calculate the dc output voltage after a diode rectifier?

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W_Heisenberg

W_Heisenberg

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diode.png

as shown in ther graph, the input ac voltage is 60V rms, and with 40V dc offset

according to Vdc=Vpeak/pi

how do we calculate the output voltage? maybe its rms value in this case.
 

Hello Dear W_Heisenberg
We have an AVG voltage formula : vave = 1/T integral vmax dt (from zero up to starting off time ) .
So the T will be 20ms for 50HZ input and t=10ms (on ).
And i think with these considerations you'll be able to find it simply . vmax is sinusoidal function. ( and you can consider the effect of that DC offset at this formula or after that .
Good luck
Goldsmith
 
Last edited:

I am not sure I understand your input sinusoid ... 60Vrms with 40V offset. Does the rms include the 40V offset? Is the input voltage the following?
\[
v_{in} (t) = 60 \sqrt{2} \sin (2 {\pi} f t) +40
\]
If the input was a pure sinusoid with no offset, the output would be a half-wave rectified waveform and you could use the formula that you have introduced in your post. Therefore, the average value of the output voltage would be
\[
< v_{out} > = \frac{60 \sqrt{2}}{\pi}
\]
This calculation ignores the forward voltage drop of the diode (approximately 0.7V) when forward-biased.

Best regards,
v_c
 
Last edited:

v_c said:
I am not sure I understand your input sinusoid ... 60Vrms with 40V offset. Does the rms include the 40V offset? Is the input voltage the following?
\[
v_{in} (t) = 60 \sqrt{2} \sin (2 {\pi} f t) +40
\]
If the input was a pure sinusoid with no offset, the output would be a half-wave rectified waveform and you could use the formula that you have introduced in your post. Therefore, the average value of the output voltage would be
\[
< v_{out} > = \frac{60 \sqrt{2}}{\pi}
\]
This calculation ignores the forward voltage drop of the diode (approximately 0.7V) when forward-biased.

Best regards,
v_c
Click to expand...

yes, but what if there is a dc offset 40V?
 

as shown in ther graph, the input ac voltage is 60V rms, and with 40V dc offset
Click to expand...
Despite of the unclear words, I assume that you mean a sine waveform with superimposed DC (offset), as shown in post #4 (with different DC offset).

Vin = 40 + 84.85 sin(ωt)

The circuit involves a nonlinear operation, thus the sine average isn't maintained in the output voltage. For an exact calculation, you need to calculate the phase angle for which the diode conducts (Vin > 0) and integrate Vin over the respective angle range.
 

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