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Getting a 5Volt output from 12V with minimum losses

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tytower

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I have to take 12Volts from acar battery and power a device with no more than 5 volts fairly steadily. I have always thought 7805 regulators to be avery energy inefficient way of doing it due to the wasted heat generated .

I have assumed that those regulators behave as a 5V zenner diode where the excess voltage is just shunted to ground almost like a light which gets brighter the more energy goes to ground. Perhaps thats my answer to use that shunted voltage to do something .

So does anbody have any suggestions as to an efficient wayto get 5V ?
 
In a zener regulator it isn't excess voltage that is shunted to ground, it is excess current.

A zener regulator will take a constant current, usually a bit more than the maximum current that the load will take.

I think of it like a fixed valve filling a cistern with an overflow.

A series regulator like the 7805 blocks the excess voltage. The current it takes is about equal to the load current, and varies as the load current varies.

I think of it like a ball valve filling a cistern.

A buck converter can output more current than the input current. The no-load current is often a bit more than the no load current for a series regulator.

The choice of regulator type depends on how much current the load takes.
 
A simple buck converter will work well for you.
Here's a link another user here provided to me ages ago that should be a drop in replacement, depending on your current needs.
Complete circuit and component listing for a 14V to 5V regulated buck converter. A zener is used as a voltage refrence so it can actually be set to any desired output voltage in range.
2-transistor Black Regulator
 
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If you want a quick easy solution, look at a Recom regulator.

**broken link removed**

I use loads of these, 26V In 5V Out @ 500mA max.


Jim
 
Thanks -things have changed . That RECOM device looks good . Now to see if I can get them in OZ
 
Phwew $15 each , I ordered 1
 
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