I am building the NOPPP Pic Programmer, and would like to make sure I don't explode my PIC or my printer port, so I think it's better to ask sooner rather than later (after I buy a new motherboard).

I am using a wall adapter rated 12v/300ma. Acutal output is 17.3v, so based on that I used a 2200ohm and 1000ohm 1/4watt resistors in parallel to give me 25ma. The parallel resistors give me a 1/2 watt rating (power is .43 watts). It reads correctly on my meter, but I am wondering why the supply rated for 12v is putting out 17.3v. I assume under load the supply will drop down to 12v. So which voltage should I use for calculating the resistor values?

I am using a wall adapter rated 12v/300ma. Acutal output is 17.3v, so based on that I used a 2200ohm and 1000ohm 1/4watt resistors in parallel to give me 25ma. The parallel resistors give me a 1/2 watt rating (power is .43 watts). It reads correctly on my meter, but I am wondering why the supply rated for 12v is putting out 17.3v. I assume under load the supply will drop down to 12v. So which voltage should I use for calculating the resistor values?

I suspect your supply will fall to 12v at 300mA. This means the internal resistance of the tansformer would then be (17.3v - 12v) / 300mA = 17.667 Ohms. Kinda high, but then again the wall packs are kinda cheap.

When modeling your circuit, you'd then have a 17.3V source, with a series resistance of 17.667 than in total is your 'wall wort' power supply. This is just a theoretical answer based on the ratings you've indicated. The best way is to do something like you've done but measure the voltage on the resistors and the current in the circuit (or implied from a precise restance measurement before hand).

I hope this helps.

Regards,

:NaN:

When modeling your circuit, you'd then have a 17.3V source, with a series resistance of 17.667 than in total is your 'wall wort' power supply. This is just a theoretical answer based on the ratings you've indicated. The best way is to do something like you've done but measure the voltage on the resistors and the current in the circuit (or implied from a precise restance measurement before hand).

I hope this helps.

Regards,

:NaN:

Okay, this makes much more sense to me now. Thanx Nan! :)