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I plan on making an ambilight project for my TV, using a ~4m 5V APA102 60leds/m LED strip, controlled by a Raspberry Pi. In order to power the LED strip, I bought a PSU rated 10A 5V, which in reality actually measures to ~5.54V. I know this shouldn't be a problem for the LED strip, which is supposed to endure a max of 6V.

The problem: I want to power the Raspberry Pi in parallel with the same PSU that will power the LEDs. However (although I couldn't find any official documentation) 5.5V may be too much for the RPi to handle and I'm afraid not to burn it.

I was thinking of putting a diode just before the RPi, which would drop the voltage to ~4.8V, but then this could be too little. I actually tried putting an SR1100 diode after the power supply to measure the resulting voltage, but it still is 5.5V. Could it be because there is no load?

A simple drawing of what I'm trying to do

  1. Would it really be a problem if I directly supply 5.54V to the Rpi?
  2. Would a diode solve problem and why I can't see a voltage drop when I measure it without a load?
  3. Is there a better solution to reliably power the RPi from the same PSU?
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  • \$\begingroup\$ The PSU's current rating is the maximum current that can be drawn by ensuring the specified output voltage. You shouldn't worry unless u draw more than 10A. If the PSU is rated for 5.0 V the higher voltage is probably caused because it has no load while measuring. Note that 5 V x 10 A is a lot of power, you should not try to drive as much as LEDs to exhaust the 10 A (not unless you distribute the currents over multiple wires). Simply put: 50 W on "fine" wires could cause fire. \$\endgroup\$ – try-catch-finally Jan 3 '16 at 21:34
  • \$\begingroup\$ No load is the issue. The diode drop depends on the current through the diode. Same is likely for the PSU. Does the PSU have a trim pot on it to adjust the voltage? \$\endgroup\$ – Passerby Jan 4 '16 at 1:19
  • \$\begingroup\$ @try-catch-finally: The LED strip's documented power consumption is 18W/m, so that makes 3.6A/m. Apparently 10A won't be enough for 4m, but I expect that the real consumption to generally be much lower, as it won't be constantly on full brightness. I intend to power the strip from the both ends to distribute the current flow, but what would happen if it tries to draw more than 10A? I imagine the light would just be fainter? \$\endgroup\$ – pmishev Jan 4 '16 at 1:59
  • \$\begingroup\$ @Passerby: No, it's a closed box, just like a laptop PSU (ebay.co.uk/itm/…) \$\endgroup\$ – pmishev Jan 4 '16 at 1:59
  • \$\begingroup\$ If you attempt to draw more than 10 Amps, it either safely resets, blows a fuse, or provides more than 10 amps and slowly or quickly burns itself out in an unsafe manner. \$\endgroup\$ – Passerby Jan 4 '16 at 2:08
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raspberry pi is designed to run off USB power which is 4.75 volts to 5.25V a Schottky diode (eg:1N5817) will give a 0.3 to 0.4 volt drop which should do what you ask.

However I'd be concerned about the compliance of the power supply, how stable is that power supply voltage under load, and under supply fluctuations. it might make more sense to employ a cheap buck-boost (SEPIC) DC-DC converter to supply the raspberry pi.

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  • \$\begingroup\$ What about just connecting a 1000uF or so capacitor in parallel after (or before) the diode? Wouldn't that stabilize the voltage in case of fluctuations? \$\endgroup\$ – pmishev Jan 4 '16 at 16:07
  • \$\begingroup\$ that depends on what problem needs to be solved, without detailed specifications it's impossible to predict with any accuracy what will happen, it looks like it;s possible to command the strip to overload the powersupply, if that happens all bets are off. all I can say us that you are unlikely to damage the raspberry pi. \$\endgroup\$ – Jasen Jan 4 '16 at 20:22

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