AC to DC Conversion: Determining Expected Q Factor

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First off, I am sorry this post doesn't match the template guideline. I didnt realize that it was mandatory until I went to post this. I tried resetting it to get the template back, but it just restores my post.

I am not understanding this topic whatsoever. We used an oscilloscope to measure the DC voltage and ripple voltage of a circuit (there was a diode, resistor, and capacitor on it).

With these values, I calculated the Q factor. The formula for this was:

Q = (DC voltage - ripple voltage) / DC voltage

I need to perform a consistency test on the experimental Q factor with the expected Q factor for an ideal AC to DC conversion.

...But I don't know what the DC voltage and ripple voltage would be for the ideal conversion.

Any help would be appreciated, whether its an explanation of the concepts or the values for the ideal conversion.

Thanks
 
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I am not sure what the ripple means. Our class discussion doesn't line up with the lab portion, so this stuff has not been covered. I have read over the manual several times but it isn't clear.

All I know is that the ripple voltage is the voltage from peak to peak of the ripple, and it is obtained when the oscilloscope "coupling" is set to AC
 
Can anyone please help me out?
 
When AC is rectified by a diode and capacitor and it feeds a load, the capacitor must supply the current to the load between the peaks of the AC waveform. As current is drawn from the capacitor it discharges according to the usual RC exponential curve. Thus the output voltage (across the load) "sags" between AC peaks. The voltage sags until the next positive going peak of the rectified AC "catches" it and restores the charge on the capacitor:

Fig1.gif


[EDIT: Oops. I accidentally posted an image for full-wave rectification rather than half-wave. But the principle stands.]
 
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Thank you gneill.

So in an ideal conversion, there would be no ripple.

Also thank you mfb, since your hint would have also been useful if I understood this.