How can I drop the voltage of a 5V DC power supply from 5.5V to 5.1V?

Question

I have some LED matrix panels (64x64) that appear to be super sensitive to voltage. Higher voltages introduce noise. I have a handful of 5 V, 10 A "brick" power supplies like this one.

The problem is that they output around 5.3 to 5.6 V which is causing problems with the LED matrix.

From various tests using a bench supply, the best voltage is 5.08 to 5.1 V. At that voltage, the LED matrix panel works perfectly. At 5.3 V the panel has ghosting and glitching effects; at 5.6 V the panel barely displays anything (it's a garbled mess).

I already have a handful of brick power supplies, and finding one to buy at an exact voltage (eg: 5.1 V) seems impossible. It really is pot luck what the supply will output.

I'm looking for a way to reduce the voltage by a small amount so that the output is, for example, 5.1 V. From research, I'm guessing a diode and resistor of some sort will probably be the best solution, or some kind of Zener shunt regulator, but my knowledge of these is limited and I'm struggling to find what I need, mainly because the various diodes are confusing me. Also, it seems that most Zener diodes have a drop of 0.6 to 0.7 V, which would be too high.

Source

Things of importance:

• Input voltage 5.3 to 5.6V
• Current draw could be as high as 10 A although that'd be very rare.

In most cases, the single 64x64 panels I have (and I'm not chaining) from their "spec" state:

• Maximum power consumption 18 W (3.6 A)
• Average power consumption 6 W (1.2 A)

UPDATE 22/07/2022: Thanks to all that answered and also the great comments. I got some 10A Schottky Diodes, ranging from 450 mV drop, to 600 mV drop. I also added a capacitor and now I'm getting a nice stable 5.15v and my panels behave perfectly when used with these random power supplies.

Answer 1

There are a few options here.

1. This is probably the cheapest and simplest method. Put a Schottky diode in series with the output of one of your power supply bricks. You say the output was between 5.3 and 5.5V. A Schottky diode will drop the voltage by about 0.3~0.4V depending on the specific diode and current.

2. Find a low drop out linear regulator and attach it to the output of one of your 5.5V power bricks. Many distributors including Mouser and Digikey offer them. You need to drop like 0.3~0.4V @ 1.2A, so you need the regulator to handle 0.36W ~ 0.48W of dissipation, which is quite doable.

3. This is probably the most efficient method (but also most expensive). Find a small DC-DC converter brick that has an accurate (or trimmable) output voltage. You can find modules for a few dolars from distributors like Digikey.

Answer 2

A big high-power Schottky diode as suggested by @user4574 is a simple approach that should be considered. In a similar approach, a saturated power-PNP transistor might work too:

^{simulate this circuit – Schematic created using CircuitLab}

TIP42 is a bit puny for these high currents. You can possibly adjust the small saturated voltage drop across the CE junction by varying R1. To reach saturation, ratio of IC to IB should be in the 20-50 ballpark (considerably smaller than transistor HFE).
The disadvantage of this circuit is the series resistance added to wiring resistance...the voltage seen by the load will vary with current - the OP's requirement here is rather strict.

Answer 3

some kind of Zener shunt regulator

Zener's idea is not really useful with heavy loads.
I should use preferably the idea of using automatic step-up-down boards.
Ok, not so simple a circuit as Zener.

Comment.

I already tested lower voltages. 5.08 - 5.1 seems to be the sweet spot! Any lower and the panels don't work at all.

So, only something as this could be used, to be tested at a limited current ...

Wide input voltage 5V ~ 32V;
Wide Output Voltage 1.25V ~ 35V ( with automatic buck, any voltage inputs, can be quasi arbitrarily regulated voltage output -by trimmer-) ;
Built- 4A efficient MOSFET switches enable efficiency up to 94%; High switching frequency of 400KHz
Add-on:
Output Ripple：50mV ...
Load Regulation：± 0.5%
Voltage Regulation：± 0.5%

Price: $2.45. Local added filtering is needed.

Answer 4

In comments, after suggesting this:

Why don't you try supplying the panels using battery power, say 9 volts of AA's, followed by an LM317T voltage regulator having a potentiometer voltage adjustment. That will remove the switching noise out of the equation, and you'll be able to see if you really have only the 20mV window, or if it's really noise that's your problem. –

You said:

@MicroservicesOnDDD I have an LM317T amongst my stuff so will test with a battery pack as you suggest and report back! Thanks for the suggestion –

and

After some testing, 5v - 5.35v works perfectly. At 5.40v I get very light ghosting/fazing (Not sure how else to describe it but there's the odd LED flicker that shouldn't be there). At 5.50 or higher, it becomes a garbled mess. I have about 10 5v 10A supplies bought from both the UK (via Amazon) and China (AliExpress) and the voltage output varies quite a bit. As an example, one is 5.38v, another is 5.65v (wtf!?) If I use a same brand 12v 6A supply I have, and connect a MP1584en inline (Bucking 12v to 5.04v (measured)) they work perfect also, So realistically, we're talking about a safe 300mV (5v - 5.3v) window, but anything over causes issues. Maybe I should buy 12v 6a (0r 10a, whatever) supplies and use buck converters with them! ha

Yes, isn't a 300mV to 350mV window much better than what you were telling us before? It's still not the best, but you can create your own intermediate electronics (or use the buck supply you've found that works) to achieve the tight regulation that you will need (So 5v to 5.35v means you're shooting for 5.175v +/- 175mV absolute max, +/- 100mV to have some margin). But I would still worry about ambient temperature moving the supplied voltage and causing a problem, or moving the electronics of the panel and causing a problem.

Having the 12V supply and then using the buck is a flexible pattern that does well and also leaves room for adding other things. I like the pattern because you can have less voltage droop and possibly save a lot of money on the copper to distribute the 5V, especially if you have a long distance to distribute the 5V. Some use a 24V, 36V, or 48V bus, and is called a Point-of-Load usage pattern.

You could also use the 5V power supplies with a buck-boost converter in place of the buck converter if you want to use what you've got.

Also, think about this. Think about if there is a heat wave. If a brown-out happens where you are, and you can only get 9V out of your 12V supply, that's still enough for the buck to be able to deliver 5.1 volts to your finicky panel. The 12V to buck is a good idea. And a 5V to Sepic or Buck-Boost should work similarly.

Answer 5

I think noise suppression is your friend here. As demonstrated in the comments, using batteries (a noiseless supply) significantly improved the voltage range that the panels could handle. They, given a high refresh rate, are also probably putting a lot of noise into the circuit.

I suggest adding some capacitors inline. It's common practice to use both a large electrolytic capacitor, since they have such high capacity and can be found in low-ESR variants, as well as add a ceramic capacitor as they are better at handling higher frequencies. Both capacitors should be rated for a higher voltage than they will see, and should be as physically close to the noise source as possible (maybe do this twice, once at the output of the power supply, and once again at the input to the LED panel).

As other answers suggest, handling the last stage of regulation yourself (perhaps combined with this noise suppression) will hopefully provide you the control to stay within the tight restrictions.

Answer 6

I would say that this answer by "The Photon" could solve your problem. In my opinion the best option for you would be a buck converter due to the high current demand. Something to be aware of is the power of the converter and to have enough capacitance at the output for the peak demand of the LEDs.

The zener diode approach that you mentioned is commonly used in low power applications to get a stable voltage. The main problem for your application is the required power. The resistor before the diode would have to dissipate P = I^2*R Watts of power and the zener diode P = V_zener *I Watts if I'm right.