Hello, First time posting here. I'm an amateur at best with electronics.

I have this Velleman MK103 sound to light kit. I have replaced the red LED's with white LED's. I put 3 whites in a series to replace 4 red's that were in a series and adjusted the resistor (R9) accordingly. I have tried removing that resistor altogether among other things. I'm trying to get more current (brightness) flowing through my LED's. When I connect them straight to power they are probably twice as bright as when going through this circuit.

Can someone take a look at this schematic and tell me what I need to change or alter to make the LED's brighter? The package says 8-15v will work with this circuit. So I replaced my 9v with a 12v and that helped a lot.

Thanks for your help!

Tony

 

Decrease the resistance of R5, R12, R11 and R14 proportionately.

For example, if you reduce R5 to 51k (or two 100k resistors in parallel), then change R11 & R12 to 750k, and R14 to 24k.

 

What do you think would happen if I just shorted all 4 of those?

 

Shorting all those resistors will make the circuit non-functional.

It sounds like you may have already determined that the Vf for the white LEDs is greater than the Vf for the red ones - which is probably why you used three in series instead of four.

You really should try to determine how much current you want to put through the LEDs to get the brightness you want, and protect them from overcurrent and burning them out.

Let's assume you want more current than you had for the red ones. Lowering R9 will help provide more current assuming T4 is saturated. Lowering R8 will also make more current available. To account for the increased current through T4, you'll likely have to reduce the value of R5 by the same ratio. For example, if you cut R8 in half, then cut R5 in half. You'll then have to cut R12 by half to keep something close to the same operating point.

 

With a 9V regulated supply the circuit can light only two 3.5V white LEDs in series because three 3.3V LEDs need 10.5V plus a couple of volts in the current-limiting resistor.

A little 9V battery quickly drops to 7.2V then two 3.5V white LEDs with a series current-limiting resistor will be very dim.