How to drive a LED strip from AC power supply?

Question

I have a LED strip (5050) and a sine-wave AC power supply of a sufficient wattage and the voltage regulated in a sufficient scale. What would be a good design of driving the LED strip from this supply? I want to keep it simple.

My idea was to:

• setup the voltage of the power supply to the level required by the output luminosity.
• include a rectifier to keep "no-power" time as short as possible.
• connect a capacitor parallel to the LED strip to prevent flicker and not to shorten LED strip life (because it's 50 Hz AC power). Is there a way how to calculate the capacity of the capacitor based on the LED strip power consumption? I know it has to withstand the max. voltage it connects to.

Is there anything else to make it better or more sustainable?

Edit:
This is a minor importance, but the final goal is connecting several LED strips in series and the power supply would be 220V AC electrical grid. The circuit will be with circuit breaker and an Earth leakage circuit breaker.

Answer 1

There are more than a few ways to drive LEDs directly from AC mains, among them:

In 1 and 2, half of the LEDs will be driven during the mains' positive half-cycle, and the other half during the negative half-cycle.

With 50 Hz mains, each string will half-wave rectify the mains and flash at 50Hz (i.e. be on for around half the time), which is visually detectable and annoying, so that approach will be abandoned.

In 3 and 4, the mains is full-wave rectified, so the string will flash at 100Hz, which is visually undetectable and, therefore. acceptable.

For 24V 5050 strips to be connected across 240V mains, no current limiters will be needed and all that'll be required is to connect 10 of then in series across the 240V mains like this:

Likewise, for 12V strips to be connected across the main, 20 will need to be connected in series then connected across the 240V mains.

Answer 2

This is the same problem as any "how do I get DC from AC?" question. You rectify it:

^{simulate this circuit – Schematic created using CircuitLab}

You can perform the rectification with four discrete diodes, as shown here, or you can purchase pre-made bridge rectifiers that are exactly this, in an integrated package.

C1 is optional, and serves to reduce the output ripple. Without it, the output bounces between 0V and some positive peak value, close to the peak-to-peak voltage of the AC input but reduced slightly by the forward voltage of the diodes in the rectifier.

from hyperphysics

If you omit C1, then your LEDs will flicker at 100Hz (twice your AC input frequency), but you probably won't be able to see this.

If you want to include C1, to calculate its value, decide how much voltage decrease is permissible between cycles. Say that we decide the output voltage can ripple between 12V and 10V: C1 must then be able to supply the current necessary to run the LEDs between each half-cycle without decreasing the voltage by more than 12V-10V=2V. We can make some simplifying assumptions: that the LEDs are a constant current sink, and that the capacitor will fully recharge at each half cycle, and must supply all the current between cycles. Start with the fundamental behavior of a capacitor:

$$ i = C\frac{dv}{dt} $$

Replace the LED current (let's say 20mA) for \$i\$. Our peaks happen 100 times per second (twice the input frequency) so the capacitor must go 0.01s between recharges (\$t_{dis}\$), and we decided the allowable voltage drop in this period was 2V. That gives us a value for \$\frac{dv}{dt}\$:

$$ \begin{align} 20mA &= C \frac{2V}{0.01s} \\ \frac{20mA \cdot 0.01s}{2V} &= C \end{align}$$

We can check that a farad is equal to an ampere-second per volt:

$$ F = \frac{A\cdot s}{V} $$

and these are the units we have above, so we can conclude:

$$ C = 0.1mF $$

Answer 3

setup the voltage of the power supply to the level required by the output luminosity.

From what I can see, the required voltage of commercial strips of 5050 SMD LEDs is generally 12 V. Example. You do not alter luminosity by varying the voltage. You need a 12 V DC supply designed for driving LED strips.

include a rectifier to keep "no-power" time as short as possible.

DC power supplies generally contain a full-wave rectifier (four diode rectifies in a bridge arrangement).

connect a capacitor parallel to the LED strip to prevent flicker

You should provide the LED strip with DC. A DC supply generally contains a smoothing capacitor.

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If your question is about designing your own PSU, you'd need to give some more details of the LED strip (or a link to a data sheet for it). You might want a switched-mode or linear PSU with a 240 V AC 50 Hz input and a 12 V DC output of (say) 150 W.

I'd just go out and buy one, but if you are experienced and aware of the safety issues, it shouldn't be difficult to design and build one for the purpose.