First of all, although I couldn't find any similar question(s), excuse me if one has been asked before.

I want to make an LED driver for 2 different LED strips. One needs about 45 VDC and the other needs about 54 VDC. The manufacturer recommends using 150 mA constant-current driver.

So I'll make a single driver operating from 18-32 VDC for driving both. Note that only one of them will be driven at a time, not both at the same time.

Anyway, I have an LM2577 on hand and I want to use it as a constant current driver despite it is a voltage regulator. Of course there are many ICs on the market for this purpose, but I want to use this IC if possible. So here's my concept design:


simulate this circuit – Schematic created using CircuitLab

Since the IC will keep the FB pin at 1.23VDC, I think the strip will be driven with a constant current of 1.23V / 8R2 = 0.15A.

Will this work? Is there anything wrong with this? Any disadvantages or potential faults? Needs any improvements?

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    \$\begingroup\$ You need cap before L1. It looks OK on principle, but that thing is also kind of low frequency for this sort of thing. \$\endgroup\$ – Trevor_G Dec 6 '17 at 17:01

This will work with some modifications. First, the inductor needs to be increased to 680uH according to the datasheet, or else system may be unstable. Output cap should be rated for 100V. If you get flicker, you may need to increase size of output cap.

You need to select compensation resistor value of about 1K based on your load, min input voltage and max output voltage. Feedback cap will need to be computed according to the datasheet.

Also, since Vf of LEDs varies from one lot to another you need to verify if 54V is typical, or max. If typical, than you need to determine max. Even with typical value, you are closing in on the maximum operating voltage of 60V for the IC, as LEDs will require minimum of 54V plus 1.23V across Rsense plus 0.5V across diode. Also keep in mind that Vf of LEDs increases with increased temp, so you may get very close to the operating limit of the IC. If you exceed it, this may result in damage to the internal switch, or flicker if the output voltage does not get high enough.

Good luck and enjoy.

  • \$\begingroup\$ Thanks for noticing about the values. They were random in the original question, because I've not read the datasheet deeply. I edited the schematic. I was aware, too, about the maximum output voltage, but the datasheet of the strip states that the load voltage is 54 VDC (typ.) and 56.4 VDC (max.) so it will not be a problem. The strip will be attached to an aluminum profile, so it will act as a heatsink thus the thermal conditions will be acceptable. Thanks again. \$\endgroup\$ – Rohat Kılıç Dec 7 '17 at 9:16
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    \$\begingroup\$ A problem with using step-up regulators as current sources is that if the load goes open circuit the regulator can produce excess voltage and destroy itself. The solution is to put a Zener (or string of Zeners) across the output, so that it conducts if the load fails. However, here that would dissipate 8.5W, so you may need heatsinks and/or a shunt power transistor, but it is still a good safety precaution. \$\endgroup\$ – henros Dec 7 '17 at 13:33

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