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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?
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  • \$\begingroup\$ Any idea how much current you require? \$\endgroup\$
    – Transistor
    May 10 '21 at 16:36
  • \$\begingroup\$ 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. \$\endgroup\$
    – Phill Donn
    May 10 '21 at 16:54
  • \$\begingroup\$ 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. \$\endgroup\$
    – rdtsc
    May 10 '21 at 17:09
  • \$\begingroup\$ 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. \$\endgroup\$
    – Phill Donn
    May 11 '21 at 20:51
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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}\$.

The voltage will remain fairly constant across the LEDs over a wide range of currents so we can just calculate the change in current caused by the resistor voltage varying. The 2 V ripple will give a 2/17.6 = 11% ripple in the current.

The fact that the LEDs are on bright in the first place, the fact that the ripple will be mains or double mains frequency (depending on the rectifier) and the persistence of human vision will make it most unlikely that you will observe any flicker.

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.

I chose 820 Ω. You could go for 1 kΩ and reduce the current a little.

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