I'm using some Meanwell power supplies and I'm a bit confused about the max output they can handle.

For an example I will mention one of the supplies to understand the logic.

Meanwell HLG-240H-48A

Well, that is an adjustable power supply, offering to adjust voltage and current through potentiometer.

Voltage adjustment range is 44,8 V to 51,2V

Current adjustment range is 2,5A to 5A

Now, it also rates a "constant current region" from 24V - 48V

Rated power is 240W

So here is my doubt. Voltage can be adjusted all the way up to 51,2V, but it rates 24 to 48V for constant current region.

For what I've read, that constant current region would be meaning the voltage the power can handle at full current? 5A being this case? in which max V would be 48V? is this correct?

If it's correct, does this mean I could drive more voltage as adjustment ranges are rated up to 51,2V when running less than full current and going in CV mode?

Going to the point of what I would like to do, given this specs, would it be possible to drive 4,6A at 50,5V? given that this would be in the 44,8V - 51,2V and 2,5A - 5A ranges it rates for adjustment? constant current region means the voltage ranges at full current (5A) when the driver goes in CC mode? (it is both CC and CV)

  • \$\begingroup\$ Are you driving LEDs? Nowhere I see you mention that. This part is good for driving LEDs too. \$\endgroup\$
    – User323693
    Commented Feb 9, 2017 at 6:58
  • \$\begingroup\$ Yes, driving LEDs \$\endgroup\$ Commented Feb 9, 2017 at 7:00
  • \$\begingroup\$ So, 4.6 A is the LED current I presume. The 48V is the standard voltage and you can tweak it a little (using a pot.). Hence 50.5 V is a valid output voltage. If total power dissipation is with in 240W, it should be fine. \$\endgroup\$
    – User323693
    Commented Feb 9, 2017 at 7:12
  • \$\begingroup\$ Thanks for your comments. Leds I would be driving can handle a lot more current, idea is to connect 4 in parallel (these are COBs, so each one can drive high currents) since this is an adjustable supply I though of adjusting current to 4,6A and 50,5V so to drive 1,15A and 50,5V to each chip. It would then dissipate a total of 232W aprox. What do you think? do numbers look healthy? Thanks \$\endgroup\$ Commented Feb 9, 2017 at 7:17

1 Answer 1


This device has 2 operating modes + protection modes against shortcuts, overvoltage and overheating

The operating modes are

  • constant voltage output for loads lower than the current limit (adjustable)
  • constant current output for loads that reach the current limit

The constant voltage output is for led strips that need some specified voltage and take the current that depends how many meters you have inserted that led strip to the output. The strip has the spec Amps/meter. Your device is intended for 48V led strips. The output voltage has about 6.4V wide range to adjust the output voltage. It affects to the intensity of the light. A slight overvoltage can be a good idea to compensate the voltage drop in very long thin wires to maintain the intensity, but you should know, what you do.

If you add more leds, the finally the overcurrent limit is reached. The device goes to the constant current mode. In this case the device drops the output voltage to prevent the overcurrent.

That makes in theory possible to use led strips that have lower voltage spec than 48V. For example 3 absolutely equal pieces of 12V strips can be piled to one 36V strip. 36V is well in the constant current area (=from 50% to 100% the adjusted output voltage, see the drawn curve in the datasheet) The length of the 12V strips must be so long that they take just the adjusted current limit at 12V. This kind of design is dancing on the tightrope because there's no easy expandability and all modifications need extreme care.

Parallel connection: Probably unstable - nobody knows how the current distributes between the units. The datasheet has no spec on it. If one of the supplies starts faster than the others, it gets to hiccup protection state due the overload. => the others also see overload. Constant blinking can occur. Needs intensive testing.

If you need more current, divide your load to different supplies. You still can have a common mains supplying cable and a common intensity control voltage. Bonus: In case of a fault the system can still be partially functional.

ADDITION about how to connect several led units in parallel to one power supply

As said each needs a balancing resistor. Without the resistors it's possible that one led unit takes much more current than the others. To analyze the situation formally let's have 2 led units in parallel:

enter image description here

Let at first the balancing resistors to be left out; ie. R1=R2=zero Ohm. In this case V1=V2=Vo ; the same voltage for both led units. Unfortunately they're never identical. One can take more current than the other at the same voltage. In the datasheet Charasteristics Table 2-1 is printed that 1,62A current can demand 8 volts more voltage for one led unit than for another. In Curve list 4-1 curve Vf/If shows that this large variation can cause one led unit to sink all the current and the other is without any current. This is extreme. More reasonable is to assume 4V probable difference between the fordward voltages. Even this makes a huge difference between the currents.

Let's make another assumption: R1=R2 and more than zero. Let's abbreviate the common resistance R1=R2=Rb (b = balancing). By Ohm's law we can write exactly:

I1=(Vo-V1)/Rb I2=(Vo-V2)/Rb. By dividing we get I1/I2 =(Vo-V1)/(Vo-V2). This seems at first as a proof that the value of Rb has no effect. This is too fast thinking, because V1 and V2 still depend on I1 and I2. Let's retreat a little and have a simple linearized model for forward voltage. Let's write it:

V1=RS1I1+Vf1 and V2=RS2I2+Vf2 where RS1,2 and Vf1,2 are individual unit parameters that vary.Vf1 and Vf2 depend on semiconductor materials. They must be quite same in the same. Rs1 and Rs2 depend on material thicknesses. That can vary between different led units even in the same batch.

Let's assume all forward voltage differences being caused by different Rs parameters. By solving our equations with this assumption we get the current ratio I1/I2 = (Rb+Rs2/(Rb+Rs1)

Our reasonable variation =4V between V1 and V2 at 1,62A means that Rs can vary 2,47 Ohm. From curve for Vf vs If we get a linearized basic value for Rs = (60V-50V)/(4A-0,5A) = 10V/3,5A = 2,86 Ohm. If that's true for led unit 1, then led unit 2 can have Rs2 = 2,86Ohm + 2,47Ohm = 5,33 Ohm

By calculating I1/I2 again we get I1/I2=(Rb+5,33Ohm)/(Rb+2,86Ohm). If Rb is =0 then I1 is nearly twice the I2. If Rb gets bigger then the ratio gets more close to 1.

Let's try Rb=3,9Ohm. The result is I1/I2 = 1,37. This is 37% bigger current and luminance, too for led unit 1. The power dissipation is remarkable. The resistor must stand it and the cooling must be thought. For led unit 1 if the nominal current was 1,1A, the dissiparion would be 8,8W. More standard resistances:

If Rb = 4,7Ohm, then I1/I2 = 1,33 Power dissipation in unit 1 =10,0W

If Rb = 5,6Ohm, then I1/I2 = 1,29 Power dissipation in unit 1 =11,3W

If Rb = 6,8Ohm, then I1/I2 = 1,26 Power dissipation in unit 1 =13,0W

No use to go on because your voltage reserve ends.

Try these. The resistors are cheap. Don't underrate the Watts. You must accept that the current difference still can be 25...33%. That's barely visible, but not disastrous.

If the balancing resistors eat too much power, active circuits are needed. A well known method is to have a current mirror. In your case a triple mirror is needed:

enter image description here

The leftmost is the model that 3 others are copied to follow. If you can choose, put your most voltage for 1,1A demanding LU to the left.

In your case proper RE in maybe 0,5 Ohm...1 Ohm You will still dissipate total 6...8 watts in one 4 unit group where each unit takes 1,1A. Proper transistor is a high current gain 2 A type - not a darlington!!!!! - This circuit needs proper testing - try to find some local help. It really needs closely matched transistors and good thermal contact that keeps them in the same temperature. The transistors have the collector in the case that needs a special isolation.

It's possible that someone sells thislike circuits. The demand on them is big and grows as the LED LIGHT business grows. Here's a link to an early more advanced circuit for total 700 mA. It has a limiter added. It's easy to adapt into my drawing, too. Every branch can have an own limiter. It takes only 1 resistor and 1 transistor extra per a branch. propoer calculations and testing are a must.


Your ideas for protective circuits earn some investigation. You have an extra reason to find a partner.

Individual active current limiters for each load is one kind of brute force solution. No closely matched transistors are needed. The price is higher component count. Mosfet is a good component because it does not need much power for control in DC applications. Gate-source voltage controls the current. An example:

enter image description here

Some component suggestions for 1.1A current and 50...57V supply voltage:

  • R1 = 4,7kOhm 0,5W
  • R2 = 1kOhm 0,5W
  • Rs = 0.6Ohm ; two 1.2Ohm 1W resistors in parallel
  • Q1 = 100V Vds and over 2A Id type: Old IRFU 110 is good because it's bulky enough to be handled easily and can dissipate a couple of watts to the air by minimal heatsink.
  • Q2 = 2N3904

RS need a possiblity to change the resistor. The individual properties of Q2 affect a little if 1.1A limit must be accurate. Q2 starts to drop the Vgs of Q1 if the current reaches 1.1A. Here is assumed that Q2 has steep BE conduction treshokd at 0.65 volts. That has easily 10% variation.

Keep Vin as low as possible to keep the mosfets cool.It would be ideal to need no heatsink. I recommend 1 sq inch 1 mm aluminium. NOTE: the Drain is internally connected to the case of the mosfet.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Dave Tweed
    Commented Feb 12, 2017 at 13:06
  • \$\begingroup\$ @Martin Miquel One solution is added to the answer \$\endgroup\$
    – user136077
    Commented Feb 15, 2017 at 20:26

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