Operation of a Zenner diode

[tytower]

Looking at this circuit I got to thinking about the operation of the 12 V zenner.

Say I have 24 Volts input to this circuit . Say 200 milliamps flows  for whatever reason .
What current passes through the zzenner then . Is it 200 mA at 12 Volts or would the current be higher like a short ?
200 mA is passing around the circuit so is this all that flows through the zenner or does more or maybe less flow through the zenner.?

[Psi]

12V input and 12V zener = 0V left across the 100R resistor. Current through zener is zero. The zener cant conduct because you don't have enough voltage.

24V input and 12V zener = 12V left across the 100R resistor,  so current is 12V/100R = 120mA max
However the real world value will be less because the light sensor, or whatever that thing is, will draw some of the current away from the zener.

[Simon]

if your trying to make this work for a number of voltages you would be best to use a voltage regulator.

[Mechatrommer]

buffer it!

[tytower]

Thanks for the above but what doues below mean please?

buffer it!

[Mechatrommer]

Thanks for the above but what doues below mean please?

buffer it!

put the zener output on an opamp (buffer) input setup as voltage follower, or even with some gain to get another voltage reference level, that way you have lower output impedance, you may want to higher the R value (before zener) to save electricity but that will change the Vout alilttle bit i guess. and then for higher power Vref supply, you further can add pass transistor downstream the voltage follower. this will make more robust voltage reference/supply. fwiw.

[tytower]

OK thanks Psi  . Thats what I was trying to understand. The current flow in the thing as its cropped up in some light sensors and PIR's that I use .

[Mechatrommer]

yes to clarify Psi's post. once breakdown zener will act like a short, but maintaining its breakdown voltage ie 12V, any current will go through the zener, unless it got divided like your circuit. if your device is sucking too much power, the zener probably cease conduction and regulation fails, so its more of protection than regulation...
http://www.allaboutcircuits.com/vol_3/chpt_3/11.html

[MMDuino]

As i think you trying to use that Motion Sensor (which always used to be connected with AVRs or PICs or any Microcontroller), and for this you have to use 5V regulated by (7805) cause its better than Zener Diode otherwise you have to change the resistor from 100 ohm to 470 ohm and use 5v zener diode and no need to think about the current because that sensor will need only 12 mAmp=0.012Amp, and the output for this sensor will be 3.3v...
see next circuit with 12v Battery without regulation,
my suggestion is to use 7805 to regulate the source then put all your circuit after lovely 7805.

Looking at this circuit I got to thinking about the operation of the 12 V zenner.

Say I have 24 Volts input to this circuit . Say 200 milliamps flows  for whatever reason .
What current passes through the zzenner then . Is it 200 mA at 12 Volts or would the current be higher like a short ?
200 mA is passing around the circuit so is this all that flows through the zenner or does more or maybe less flow through the zenner.?

[tytower]

Sorry I did not see this picture and it has disappeared . Can you reload it perhaps and perhaps find a way to keep the images there . It only lasted 120 days .

Also just as a note the 7805,780x series all allow full supply voltage to the output when they fail.There are much better options available cheaply if you have a look around.

I imagine the 780x 's just have a zener in them that passes excess voltage to ground anyway.

[Nerull]

120 days? It's been almost 5 years since the last post in this thread.