mrgizmo wrote:

>I am using a 2n3904 to drive a 12V relay off a 16F628 pic at 5V input, I'm 
>not sure how to figure out the resistor for the base? 
>
>  
>
The gain of most general purpose NPN switches (PN2222A, PN3904) is about 
100;
that means it will sink 100mA  if the base has 1 mA of current driving 
it. That's because
an NPN transistor is a CURRENT AMPLIFIER.

For a relay, I'd calculate the load at 2x the current needed to reliably 
close it. Let's
say that 100mA at 12V is needed to reliably actuate the relay. I would 
then need (for
a load of 200mA) a drive of 2mA minimum from the PIC through the base.

To calculate the resistor value, you will use ohm's law, I=E/R. The 
transistor switches
at 0.6V above GND, so the resistor will have imposed on it ((VCC of PIC) 
- 0.6).

To obtain the proper drive of 2mA, 0.002 = ((VCC of PIC)-0.6)/R.
Multiply both sides of the equation by R makes

0,002R = Vcc-0,6

for a VCC of 5V,  R is  2200 ohms... and slightly less is fine..

for a VCC of 3.6V, R is 1500 ohms... and slightly less is fine.

--Bob


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