I understand the quoted statement differently.
I understand, by the quotation, that the zener does consume the same ammount of power, independently if it's loaded or not.
Here's an easy explanation:
The zener diode itself, in a shunt regulator, will get hot when the load is not applied and stay cool when the load is applied.
I'm not trying to be a stubborn.
But a zener that is submitted to a reverse voltage higher than the Zener Voltage, represents a very low impedance path to the current, lower than the load itself.
So it will get hot anyways. Unless the load resistence is low enough to cause a higher voltage drop enough to "cut" the zener diode.
Is this the circuit we are talking about?
-----R----+-----+
DZ LOAD
+--+--+
GND
With no load, all the current is drawn through the limit resistor and the zener. With a load the current after the limit resistor is shared by zener and load.Originally Posted by Hayato
The zener and load become parallel thus current is shared.
Wait, I take my last back. The current through the zener remains the same with load or no load. Current increases in the limit resistor with a load. I had to work it out on paper. Or maybe it is late and I am confused...
Last edited by Mikebits; 11th April 2009 at 07:11 AM.
When a load is connected across a zener diode, the current taken by the load is effectively taken from the zener diode.
The reason is this:
Suppose a 12v zener diode is connected to a resistor called the "delivery resistor." This resistor is connected between the high voltage being supplied to the ZENER SHUNT REGULATOR and the zener itself.
With no load, the zener voltage may be say 12.00v and the zener is drawing 100mA.
When a load of 50mA is connected to the circuit, the zener voltage may decrease to 11.95v but it will allow only 50mA to flow through it at this voltage. The other 50mA is allowed to flow through the LOAD.
What happens is this:
You apply a load. The load takes a current. This current is "robbed" from the zener diode. The voltage across the zener drops (slightly) until the "overflow current" is exactly what the zener will pass at the particular voltage. It’s the zener that determines the final voltage across the load because it has a much higher change in current-requirement for each millivolt rise or fall.
Ok, I got confused.
Nevermind, you guys got it sorted out before I posted.
Last edited by speakerguy79; 11th April 2009 at 08:18 AM.
Then you could always just add a pass transistor to keep the zener load constant =)
But that's neither here nor there.
"Because I be what I be. I would tell you what you want to know if I
could, mum, but I be a cat, and no cat anywhere ever gave anyone a
straight answer, har har."
At that point I think it would be easier (and way better) to just buy a 7812
Reply With Quote
