Does rectified DC have Frequency??

ericgibbs

Joined Jan 29, 2010
9,561
hi,
If you full wave rectified a 60Hz sine wave, what would expect the shape of the output waveform to be and at what frequency.?
E
 

wayneh

Joined Sep 9, 2010
16,254

In this diagiam why is it showing 120HZ at output and why R(out) is used??
R(out) is your load. The relationship between R and C determine the amount of ripple on the output voltage, which does indeed cycle at 120Hz (see the references). With a big C and a big R (meaning high resistance, low load), there will be very little ripple. The C will be acting much like a peak detector and the voltage will be fairly clean DC at the peak voltage coming off the rectifier. Reversing the situation to low C and low R, the output voltage will track the rectifier output. This will be almost 100% ripple, from 0V to the peak voltage. We still call the output voltage DC because the current does not change direction, even though it's changing widely in amplitude. It would also be accurate to call it DC with an AC component, or vice versa.
 

ScottWang

Joined Aug 23, 2012
6,873
R(out) is your load.
The R(out) looks like a false load and it is not the real load, the real load should be connects on the Vout, The R(out) was used for discharge when the real load is disconnect from the Vout, and the AC120V is turn off, so the c has a path to discharge and it will make the Vout has a more quickly discharge to 0V, and keep the Vout and ground safe to avoid any accidently short-circuit.
 

MisterBill2

Joined Jan 23, 2018
5,211

In this diagiam why is it showing 120HZ at output and why R(out) is used??
As others have mentioned, R out is a substitute for an external useful load. AND, also important, if there were no Rout, the capacitor would indeed charge up to the peak voltage of the transformer secondary and there would not be much ripple, or probably none if the capacitor had very low leakage. But under those conditions there is no useful power being delivered, so why bother? In a real application current flows into some load element and so the capacitor delivers that current until the voltage of the next cycle rises to the point of supplying current and re-charging the capacitor.
 
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