lm7805 capacitors!!

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a single 2200uF should work fine, but a single 4700uF would be OK also, any more would be over kill and just take up more space. All the input filter needs to do is to keep the lowest peak of the input ripple voltage above the minimum input DC voltage specification for the regulators.

Your biggest problem will be the size of heat sink you will need, certainly for the 7805 that will be supplying 1 amp. That one has to dissipate 10 watts of heat which is a whole lot of heat. You might want to put a series power resistor in front of that 7805 such that the input voltage is closer to the 9-10volt range when supplying a one amp load.

Lefty
 
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I guess the idea is to use the Power Resistor is used to dissipate some of the power, helping the 7805 by not having to dump as much to regulate to 5v.

I have a few boxes of 10watt power resistors in various values, can you suggest a value that may be suitable.

I doubt that I will ever draw the full current from them, but as its a 'tool' im building, i dont really know what i'll be plugging into it - but it might as well be able to handle itss full rated capacity.

It is going in a rather large project box (A4 roughly, by about 100mm tall) so there is plenty space for heatsink and for a fan if neccisary.
 

Yes the idea is to spread the heat dissipation between the two components, the regulator and the voltage dropping resistor.

5 or 6 ohms should work. Just use ohms law, E= how much voltage you wish to drop and I = the max current to be drawn.

There are newer regulator chips that use switching methods for voltage regulation that dissipate a lot less heat at the same power levels. They cost somewhat more but sometimes that is worth the extra efficiency, especially if one is using batteries as the power source. For a bench supply old 78XX regulators work fine if you watch out for there heat sinking needs.
Lefty
 

The input voltage should stay about 2v above the output voltage. If the DC is unfiltered it goes all the way down to zero so the regulator would shut off every 8.3 mS.
Imagine troubleshooting that with a DC voltmeter.

12vdc equals a peak value of PI/2 x 12 = 19v. Your lowest desired value is 7V. Approximately, if your current is 0.5A you need a minimum of I ΔT/ΔV = 0.5 (.008)/(19-7) = 330 µF.
An exact solution is kind of messy, and caps have a wide tolerance anyway.

I wouldn't go with very large cap values; you could exceed the I squared T rating of the diodes in your adaptor, thereby giving you more troubleshooting tasks.

Do you have a scope?
 
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