# Can I use unregulated power supply for LCD display backlight LED?

I am designing a PCB for a university project that will include an LCD display which is powered by 3.3VDC regulated supply. Unfortunately this display has backlight LEDs in series that have total forward voltage of 6.4 volts. My board includes 3.3VDC and 5VDC regulated power supplies which unfortunately cannot satisfy the required 6.4VDC. I am hesitating if I can use my 24VDC unregulated power supply for this purpose. The power supply would have a 2 volts voltage ripple(100Hz) with a max. peak of around 24 volts. I was thinking if it would be viable to put a resistor in series and supply the Display LEDs with this PSU.

1. Is this unregulated voltage going to present a problem for the display LEDs or the display in general?
2. Will the voltage ripple be visible through the LCD display brightness?
3. Do you have an advice for a better approach considering the above mentioned PSUs?
• Any idea how much current you require? May 10, 2021 at 16:36
• The LEDs are designated as 20mA@forward voltage, however the unregulated power supply will be used for additional stuff which could draw as much as 250mA. The transformer would be sized as 555mA. May 10, 2021 at 16:54
• Is the 24V unregulated supply (sounds like a linear transformer + rectifier + filter) powered from 50/60Hz mains? If so, it might be possible to perceive flicker in the backlight. Would be fix-able by placing a filter capacitor across the backlight; perhaps 100µF/25v. May 10, 2021 at 17:09
• Linear transformer + rectifier + filter is the right guess. The distance between PSU and display includes inter board connection with a wire, so close filter cap could definitely improve things. Thank you. May 11, 2021 at 20:51

If you're going to use a series resistor then it has to drop 24 - 6.4 = 17.6 V. If you want 20 mA then $$\ R = \frac V I = \frac {17.6}{20m} = 0.88\ \mathrm{k\Omega} \$$ or 820 Ω for the nearest E12 value. Recalculating for that gives us $$\ I = \frac V R = \frac {17.6}{820} = 21.5 \ \text{mA}\$$.
Power dissipated in the resistor will be given by $$\ P = I^2 R = 0.0215^2 \times 820 = 0.38 \ \text W \$$. Make up the resistance using a few series or parallel with twice the required wattage total so that it all runs cool.