|Feb12-12, 03:34 PM||#1|
In a simple zener shunt circuit consisting of a zener diode in parallel to the input voltage and load how does the regulator compensate for a drop in input voltage?
In other words the zener will conduct or 'shunt' when the source voltage exceeds the breakdown voltage of the zener diode. How does it compensate for an drop in source voltage below nominal?
|Feb12-12, 04:05 PM||#2|
It doesn't. It's up to the designer to ensure the source voltage is always greater than the zener breakdown, if you want it to be regulated by that zener. Also ensure some current limiting so zener power rating is never reached.
|Feb12-12, 04:19 PM||#3|
In shunt regulator using zener, say you have a raw voltage of 10V and you put the -ve to ground and and resistor R from +10V to the anode of the zener or 5.1V. The current draw would be I(total)=(10-5.1)/R. If you use this to generate the 5.1V to drive, as long as the load don't draw close to I(total), then you get regulation. Once you draw more than I(total), the zener will turn off and you totally loss regulation.
With this same idea, if you raw voltage dip, then I(total) will drop, using the explanation above, you can draw less current before you loss regulation.
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