I'm rebuilding optical printer (the machine which does Super8 to 16mm film blow-up) and designed new electrical circuit for it so I can control it with Arduino (or maybe later with Raspberry Pi). There are High Power 10W LED controlled by Arduino PWM signal (to expose the film), 3 Relays (each works with 5V and switches 110V circuit inside the Opt.Printer) controlled by Arduino's Digital pins and 5V Servo Motor (in diagram it wrongly said Stepper motor, sorry) controlled by Arduino as well. I decided to use one power supply for the entire circuit except for Arduino (it's powered via USB and communicate with computer via serial protocol). I'm going to use a wall wart which has 15V 0.84A because I couldn't find LED driver to power my LED 10V 10W. So for that case I pulled down the current and voltage from power supply with R 6.8 Ohm so it won't fry LED. Then with 7805 the original voltage is converted to 5V which is used for relays and Servo Motor. My doubts are if I designed it right or not because of sharing the same power supply within entire circuit. The working cycle of circuit consist of 2 steps: 1) All relays switched on and no power to LED neither to Servo 2) relays are switched off and LED with servo motor are powered for ~1 sec.

The parts of circuit are: 3 x relay 5v 1 x Darlington ULN2003A 1 x MOSFET n-chanel IRF1010E 1 x 7805 1 x Resistor 6.8R 1 x Power LED 10W (10V, max. 1050A) 1 x capacitor 46uF

Power Supply Wall Wart 15v 0.84A the ground of Arduino and Power Supply are connected

Thanks a lot. and sorry for my electronic amateurism

circuit diagram


1 Answer 1


Just a few elements:

  • if you have 1 A going through the LED the resistor will be dissipating 6.8 W. That is a LOT of wasted power, the resistor will need to be huge and will get very hot. Resistors are not appropriate for power LEDs. You can find many easy-to-use LED drivers that would work, eg. the LDD-1000H which also supports PWM dimming and costs less than 10 EUR. That one needs a power supply at least 3V above the LED forward voltage.

  • Your power supply is too weak. You have 10 W for the LED plus maybe a couple watts for the servo. Add some generous safety margin and you're looking at a 15-20 W power supply if you use the LED driver. If you use the resistor (not recommended) that's 17 W for the LED+resistor so a 25-30 W power supply would be needed.

  • [EDIT: ok, I misread you diagram, you do have a driver for the relays. That's fine then] You must add a protection diode across each of your relays to absorb the voltage spikes that are generated when the relay operates. Also the arduino pins are not capable of driving the relays directly (too much current, that would fry the arduino), you have to add a transistor and a resistor. See here for instance.

  • Make sure that your voltage regulator is big enough for the relays and servo.

  • Make sure that the 10W LED is well mounted (with aluminium PCB and thermal paste or tape) to a big heatsink. It will burn out in seconds otherwise.

Why not two separate power supplies?

  • \$\begingroup\$ Also note that a 10V, 10W LED pretty much needs 1 A, and a 0.84 A supply is going to have difficulties. \$\endgroup\$ Commented Jul 27, 2015 at 22:41
  • \$\begingroup\$ Thanks a lot for answer.I tried this set-up with standard 9v battery (minus current limiting resistor for LED) and everything worked.The strange thing is that when I measured the current which goes to LED it was 350mA and LED shined pretty bright. I don't understand the way how LEDs work with the brightness and current. Yes, LED is fixed on the brass plate which is already inside machine plus there is a cooling system. The one power supply is because of using only one plug, so it will be easy to other people to use. The relays are 0.36W and servo I have to measure the current in working cycle. \$\endgroup\$
    – labo doble
    Commented Jul 28, 2015 at 16:44
  • \$\begingroup\$ @WhatRoughBeast so, finally I ended up with two separate power supply. one regulated current for Power LED and second ordinary 9v wall wart. thanks for your advice. \$\endgroup\$
    – labo doble
    Commented Jul 30, 2015 at 12:02

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