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edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0my set of question is like this, (question had specified it to be halfwave rectification) my expectation for the waveform from the output V1 is something like this: dw:1340782227032:dw

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0but i got this waveform which looks like clipping to me o.o

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.1During the positive half cycle of the input voltage the cathode of the diode is positive with respect to the anode. Therefore the diode cannot conduct and the positive half cycle will appear at the output. Durin the negative half cycle of the input voltage the diode conducts, clamping the output at about 0.7 volt. The diode needs a forward voltage of about 0.7 volt to enable it to conduct.

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0ouh so it means i will still get 0.7v if i measure across the diode, and flat for the negative region for measuring across the resistor?

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.1Do you intend to use an oscilloscope to measure these voltages?

edr1c
 2 years ago
Best ResponseYou've already chosen the best response.0yea i used an oscilloscope to measure the output waveform.

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.1You already have the voltage waveform across the diode. This waveform is 0.7 volts only during the negative half cycle of the input waveform. During the positive half cycle of the input voltage there is no conduction through the resistor and therefore no voltage across the resistor. The negative half cycles of the input voltage will be seen across the resistor.
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