Register to reply

Zener Shunt

by fonz
Tags: shunt, zener
Share this thread:
Feb12-12, 03:34 PM
P: 81
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?

Phys.Org News Partner Engineering news on
Greater safety and security at Europe's train stations
Fingerprints for freight items
On the way to a safe and secure smart home
Feb12-12, 04:05 PM
HW Helper
P: 5,364
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
P: 3,898
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.

Register to reply

Related Discussions
Heat sink + a shunt copper tape. Why shunt copper not equal to heat sink temp? Mechanical Engineering 8
Do I need a shunt resistor? Electrical Engineering 7
Question about shunt motor Advanced Physics Homework 1
Dc shunt motors Electrical Engineering 4
Shunt motor Introductory Physics Homework 0