Half Wave Rectifier Problem

i have a half wave rectifier.......... and i am giving 12V AC to it through...... if i use oscilloscope here..... i have to find Vpp = peak to peak * volt/div....... then by multiplying in with 0.707 we will get 12V in oscillsope and in dmm.....


my question is when i connect the oscilloscope after diodes ...... how will i get the value....... am i required to convert it to rms.... or rectified dc voltage is already rms.....

Hi,

there are many discussions about voltage calculations like "RMS", "average" and so on.
--> do a forum search on your own.

****For full wave rectifier:

The rectifier just rectifies the direction of the voltage. To show this you need to connect a resistive load at the rectifier output, otherwise stryy capacitance will modify the signal.
RMS means Rout Means Square.

"Square" means the input signal becomes squared.
Therefore the input voltage direction (sign) is not important.
The square of a negative number is the same as the square of the positive number.

--> RMS voltage before a rectifier is the same as RMS voltage after a rectifier (for an ideal rectifier)
For a real rectifier the output signal will be less, because of the voltage drop in the diodes.

Klaus

- - - Updated - - -

*****Now for half wave rectifier:
12V sine AC input means:
* 12V RMS
* positive peak of the sine is 12V x sqrt(2) = about 17V
* negative peak of the sine is -12V x sqrt(2) = about -17V
* Vpp = 12V x 2 x sqrt(2)

now after the rectifier:
* everything that was negative before becomes 0
* positive peak of the sine is 12V x sqrt(2) = about 17V
* negative peak = 0
* Vpp = 12V x sqrt(2)

For RMS calculations:
The RMS value is equivalent as the DC value through a resistor for the same power dissipation.
or say:
P_DC = U_DC x U_DC / R
P_RMS = U_RMS x U_RMS / R
And P_DC should be the same as P_RMS.
X volts DC cause the same power dissipation in a resistor as X volts RMS.

Just imagine there is a resistor connected.
* for the positive halfwave the power dissipated in the resistor is the same with and without rectifier.
* but during the negative halfwave the power is 0
--> the average power is half than without a rectifier.

now as P becomes 0.5 as before U_RMS needs to become sqrt(0.5) than before. Sqrt(0.5) = 0.707
If there was 12V RMS without rectifier --> then there is 12V x 0.707 = 8.48 V RMS after the half wave rectifier.

Klaus