# Is my wiring diagram correct?

I want to power 6 3watt led with 3 modes: bank 1, bank 2 and both banks on. I will use 1 linear pot to control the brightness, and a momentary switch for on/off. Also I'm not sure what transistor to use. I'm not electronically inclined. Thanks

The right hand end of the 1 ohm resistor that is in the collector circuit of the lower transistor should be connected to the right hand end of the other 1 ohm resistor and to the plus side of the battery instead of to Vin.

The wire from pin 7 should connect between the switch and the resistor, just like the wire from pin 8. You might be better off reversing the logic to use a pull-up resistor and have the switch pull the input low.

Don

The right hand end of the 1 ohm resistor that is in the collector circuit of the lower transistor should be connected to the right hand end of the other 1 ohm resistor and to the plus side of the battery instead of to Vin.Actually the drawing is a little hard to read but the positive on the DC is connected to the 1 ohm resistor, the other 1 ohm resistor and also to the Vin to power the board.

The wire from pin 7 should connect between the switch and the resistor, just like the wire from pin 8. Ahh Yes thank you for that, that was a mistake. You might be better off reversing the logic to use a pull-up resistor and have the switch pull the input low. So 5v to the switch, then to a resistor then to pin 8. When the switch is pressed the 5v goes to ground therefore pin 8 gets 0v or low? Correct? :-/

Gado, You can't control high power LEDs with resistors as current limiting devices. You need a constant current drive for each LED. As it stands if you try it this way you will fry something as the LED forward voltage will not hold tight enough to let you get away with resistors of so low a value.

Actually the drawing is a little hard to read but the positive on the DC is connected to the 1 ohm resistor, the other 1 ohm resistor and also to the Vin to power the board.

I understand what you did now. Your drawing is a little ambiguous.

So 5v to the switch, then to a resistor then to pin 8. When the switch is pressed the 5v goes to ground therefore pin 8 gets 0v or low? Correct?

No. Just swap the switch and the resistor. Starting at the positive terminal you go through the resistor, then the switch, to ground. Pin 8 gets connected to the junction of the resistor and switch. With the switch open pin 8 sees the positive supply voltage (there is almost no voltage drop across the resistor because the current is so small). When the switch is closed pin 8 now sees 0 volts (ground). The resistor is required to prevent short circuiting the power supply but it's value isn't critical. If it is too small then the power supply is overloaded when the switch is closed. If it is too large then you might not see a logic 1 at pin 8 when the switch is open.

Don

Mike:

I was going to get into that after the circuit was straightened out!

Don