Voltage Regulator?

Question

I need a way to get 5 mA of DC power at 9 volts out of 25 A of DC power at 12 volts. Can I just run leads off of it and use 3.1V Zener diodes or will the current blow them out? If so, how can I do this?

Answer 1

If all you need is 5mA, a zener will work for this. A LM7905 voltage regulator will work much better though, and it will be more efficient.
I wouldn't worry about regulating the current. Regulate your voltage to 9V and your load will draw whatever current it needs, dependant on its resistance. Your suppy is 12V @ 25A, so you have plenty of overhead to work with. Yeap, get a LM7905 regulator, connect 12V to input, and your load to output, and connect the ground of your 12V supply, and your load, to the ground of your regulator.

Answer 2

A Zener will work. You use a resistor to set the current. This link shows a good picture of the circuit (scroll down to Simple Zener Regulator).

R1 is the resistor, R2 is your 5mA load. It depends on the zener diode you choose, but R1 will probably be in the 100-500 Ohm range (exact value doesn't matter, but the lower it is, the more the diode will heat up).

The Zener will dissipate V^2 x I Watts of power. V^2 in this case is 3.1 x 3.1 = 9.6. For a small package, you don't want to dissipate more than about 100mW in the Zener, so that sets your maximum current at around 10mA, which means you'd want a 300 Ohm resistor to limit the current - (12V-9V)/0.01A = 300 Ohms.

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One more thing about the circuit in the diagram. You're actually better off swapping R1 and the diode. Then, you only need a single 3.1V Zener (12 - 3.1 = 8.9V). R1 sets the current, and as long as R1<
The guy who posted after me doesn't know what he's talking about. The lightbulb analogy doesn't make any sense, and the 9V Zener and 100mA of current is 0.9W of power dissipated in the diode, which is excessive.

Answer 3

9.1 volts is a common value for a Zener diode at the voltage that you want from the 12 volt source. The Zener drops the applied voltage to it's rated value, and more or less holds it there. You need a current limiting resistor to keep the Zener from turning into a cloud of smoke. Personally, I wouldn't try to hold the Zener output to 5 mA because that may be to low a current rate for the Zener to operate at. Set the current to at least 100mA, and size the resistor in terms of the value of resistance for that current level. Your circuit will only draw what it needs, so the higher current level will not hurt anything. To allay any fears, or concerns to the contrary, a standard household outlet has a potential of 115 Vac across the terminals, with a current capacity of 15 amps. That works out to about 1800 watts of electrical power. Does the 100 watt lamp get the full 1800 watts through it? No, of course not, it only gets what it needs. The same holds true for any other power source unless it has been specifically designed to force a certain current amount through anything attached to it's terminals.

Answer 4

using a series blocking zener, as you suggest will drop voltage but provide no regulation at all. you are totally at the mercy of the 12 volt source being always in regulation and if that handles 25 amp current swings, that is doubtful.

any attempt at zener regulation pays a heavy toll on current draw and works only if you take into account the voltage swings of the source as well as the current swings of the load. since you did not specify either, the above "designs" are totally bogus.

the cheapest solution, both money wise and power wise, is to use a 3 terminal IC voltage regulator. they will normally work at the 3 volts of headroom although 5 volts would be desirable. if there is a 15 volt source available that would be a perfect match.