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Luis,
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Here's a datasheet for the 7812A:
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On page 8, it says that Vdrop when supplying 1 Amp out is typically 2 Volts, could be somewhat more on your device.
So if you put the 7812A on a 13 Volt battery, the you will typically only get 11 Volts out when the transmitter is drawing 1 Amp.
It is only when the battery is above 14 Volts that the output is a well regulated 12 volts when drawing one Amp.
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The IRF4905 that Don showed in that schematic has an RDSon of 0.02 Ohms, so when fully on it looks like a 0.02 Ohm resistor.
The two NPN transistors are very sensitive, they slam that IRF4905 fully on if the output voltage is a little bit below 12 Volts.
So with a 12 Volt battery at 1 Amp, the output would be 12v - 0.02*1 = 11.98 Volts, not the 10 Volts you would get out of the 7812A.
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Again, that's a typical drop of 2 Volts for the 7812A, they don't spec a worst case, it could be only 9 Volts.
And your diode would reduce it by another 0.7 volts, to 8.3 Volts when transmitting and drawing 1 Amp from the 7812A.
So with the 7812A you might be wasting a quarter of the power coming out of a 12.0V battery, with Don's PFET you lose almost nothing.
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Also, you can adjust the value of R3 and R4 to give exactly the voltage you want, instead of using a diode.
If you really want the 11.3 Volts that you might get with a diode following a 7812A (when the battery is at 15 Volts),
then adjust the resistors such that there is 5 Volts across R4 and (11.3 - 5.0) = 6.7 Volts across R3.
The ratio you need is then R3/R4 = 6.7/5.? ? ?Having R3=6.7k and R4=5k would work.? Or R3=67k and R4=50k.
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For exactly 12.0 Volts out, that ratio should be (12-5)/5 = 7/5 = 1.4
Don's schematic shows a ratio of 30k/22k = 1.36, not exact because he is using standard resistor values.
The current through R4 will be? 5v/22k = 0.2273 ma, and the voltage across R3 will be 0.2273ma*30k = 6.819 Volts.
So the output voltage is 5 + 6.819 = 11.819 Volts, slightly below the target of 12.0 Volts.
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Jerry, KE7ER
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On Sat, Mar 15, 2025 at 06:10 AM, Luis Pacheco wrote:
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