AC to DC Conversion: Determining Expected Q Factor

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The discussion centers on understanding the Q factor in AC to DC conversion, particularly how to calculate it using measured DC and ripple voltages. The user is confused about the ideal values for DC and ripple voltage in a perfect conversion scenario, as their class discussions have not aligned with practical lab work. It is clarified that in an ideal conversion, there would be no ripple voltage, meaning the output voltage remains constant without sagging. The ripple voltage is described as the fluctuation in voltage due to the capacitor discharging between AC peaks. Overall, the user seeks clarity on these concepts to perform a consistency test on their experimental results.
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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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Do you expect ripples in an ideal conversion?
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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