Electronics > Beginners
Zener regulator circuit, and power ratings.
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Simon123:
Its just ohms law, and transistor hFE and some kirchoffs voltage and current laws, but if you have questions ask :).
sureshot:
Thanks again.  :)
sureshot:
So I've been doing the maths, i didn't want to start another thread as this maths is to do with what I'm trying to achieve. If someone could look over it i would be very grateful.
My maths as follows: 13 volts zener doide of 5 watts..
5/13 = 0.385 mA. Current the zener can deliver. Series resistor for the zener doide, supply voltage 16.92 volts - 13 = 3.92 volts / 0.385mA = 10.18 ohms. Adding a series pass transistor with a minimum hfe of 25. So 0.385mA × 25 = 9.62 Amps. Total current avaliable.  I'm not sure that is right. I would appreciate someone to see if I've got that right. Thank you for reading.

These are the formula I've used in these images.
The images show a 0.5 watt zener diode and an example transistor, also the voltage was just an example i followed using my voltage and power rating parameters.
CJay:
Your figures look right, haven't checked but it's important to note, the zener dissipates power, it doens't source current, that's what the 5W rating is, it's also important to realise that's a maximum dissipation and it's likely based on ideal circumstances, ambient temperature, cooling etc. etc. so it's not a good idea to design to those figures unless you want to fry zeners, build in some headroom.

Note that you'll need to design for maximum and minimum current drawn, the zener will dissipate maximum power when current drawn from your supply is lowest, if the output of your PSU is open circuit then the Zener will need to be able to deal with all the current that can be supplied via that resistor.

While it's not a problem with the simple design you have, zener's aren't guaranteed to be 'in spec' until they're passing a specified minimum amount of current so regulation can suffer, they're also noisy, it's common to use a zener as an RF noise source (remember, I know what you're using these PSUs for :) ) so they might not be ideal for your needs.

All in, Zener diodes are useful things but if you're trying to design a PSU, there are better ways to do it and a circuit that simple is going to perform worse than your others that use a 7812 with pass transistor.
Simon123:
Yes, they appear to be correct, but you want to design it from Output (your goal) towards the input. For example you want like 3A and 12.4V

So for example, you decided you need 3A max output.
And you have a transistor with hFE(min)=15. (i suggest you to use darlington for higher hFE).
This means base current needs to be 3A/15=200mA.
Now you said your supply is about 17V (this value needs to be of a loaded supply).
Now the output will be 12.4V, which means you need 12.4V+0.6V=13V zener.
And voltage across resistor is 17V-13V=4V

Now since you know base current, which will flow threw the zener, when there is no load ... which is 200mA
You can calculate resistor for base. 4V/0.2A=20ohms.

Oh yea, and at the end you can check zener maximum power dissipation; 0.2A*13V=2.6W, but as i suggested use darlington pair, because then zener will dissipate less, temperature wont change as much and it will be more stable.

So yea your calculations are correct, but they are more of circuit analysis then designing.
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