Resistor Dilema

[BrownOut]

Oops, your right

 

[Voltz]

Oops, your right

About what

 

[BrownOut]

You don't need the FET's. Just connect the LED cathode to the output and the anode thru a resistor to V+.

 

[Hero999]

Assuming the comparators are powered from 5V, the LEDs will hardly light because the voltage drop across the MOSFETs will be too high as they're configured as source followers.

 

[Voltz]

Assuming the comparators are powered from 5V, the LEDs will hardly light because the voltage drop across the MOSFETs will be too high as they're configured as source followers.

But the output from the 5V is only 1.2V as a reference for the voltage of a Ni-Cd/Ni-MH rechargeable battery, so the output of the comparator will be 1.2V won't it? And how do I light an LED from 1.2V

EDIT: - **broken link removed** is this a comparator and how would I connect it up

 

[BrownOut]

No. The output of the comparator will drive all the way to 5V ( or very near it anyway, if the output was connected correctly ) Get rid of those FET's. There are problems with the way you're using them, but rather than discuss it, just crap-can 'em and things will be much better.

 

[Hero999]

What IC are you using for the comparators?

If it's the LM339, it won't work at all because the outputs are open collector and need a pull-up.

If it's the LM324 then it will work but the LEDs will be very dim because the threshold voltage of the MOSFETs and the voltage loss in the output stage in the op-amp will add together. If the LEDs are blue, white or violet so have a forward voltage of >3V, they might not even light.

As mentioned above, get rid of the MOSFETs.

 

[Boncuk]

Your first calculation is correct, but I would probably use a value like 560Ω, 1/2 watt.

Why waste space? The 560Ω resistor dissipates 178.6mW (rated 250mW). Using a 680Ω resistor the LED current is 14.7mA (still bright enough) and the power dissipation is reduced to 147mW.