# Thread: Led power supply

1. Member
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Originally Posted by gar
180225-0843 EST

karan:

Using 347 V, 0.6 A, and 0.9 PF the actual input power is about 347*0.6*0.9 = 187 W for 150 W out or an efficiency of 80%. Seems reasonable.

As a ballpark figure we can guesstimate at 150 W output from the power supply and 240 V input to the power supply that the input current to be about 0.6*347/240 = 1.44 A. Way over its rating.

You may not have a 150 W load. When you operate the LED in combination with the power supply with 240 V into the power supply what is your measured current into the power supply? Should use an RMS ammeter, but at a PF of 0.9 an average reading meter may be adequate. A Fluke 27 is average reading, and an 87 is RMS.

.
Same for both L1 and L2, 0.554A

2. Senior Member
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Originally Posted by GoldDigger
The one thing that you can be relatively sure of is that the input components (diodes, switching transistors, or whatever) will be carrying twice the current to get the same power on the output side going to the LEDs. Only the manufacturer knows whether they can handle that double current or not. Long term thermal damage to semiconductors is quite possible even though they may appear initially to work fine.
OP did feed the switching transformer with 240 volts which is lower than nameplate rating of 347 volt minimum.
By OHM's Law, we could infer that current will go up because of low input voltage. . . theoretically correct.

Most newer design switching power supplies have both constant voltage (CV) and constant current
output capability. Depending on the condition of the input voltage, the TTL (transistor Logic) could be deployed in the circuit to utilize either CV or CC in order to work in a closed loop control of the switching transistor.
The output is constantly monitored for any changes outside the design premise. If for instance there is an anomaly that will cause instability of the output, either CV or CC will selectively (automatically) switch to proper mode in order to correct the problem.

The low voltage (240 Volts) instead of 347 v that OP is trying (or had already tried) evidently worked.

The reason that it works is; the switching transistor enables the charging capacitor to charge and discharge due to the opening and closing of the switch. Charging when open, discharging when closed.
This charging/discharging routine is what raises the voltage to almost in parity with the needs of the electronic components.
Since most switching power supplies are designed for universal use, the wide range certainly make economic sense.

Some countries operate at 250 volts. The 347 v volt name plate rating is measured at the emitter and base junction of the power transistor. So, when fed from a 240 volt source the peak reading would be 339.0 volts.

The 347 v volt nameplate is the theoretical PP (250 volt X Sq rt of 2) which is roughly 353 v peak.

Bottom line is, it will work but at a reduced luminescent. You have to go through the power supply and NOT hook the LEDs directly to a 240v wall plug. This way you still get the advantage of a modulated power source.

One way to check if it operates in CV mode is; if at this low input voltage the LEDs flicker.. . .usually when in constant current mode flickering is unnoticeable because of the rapid response in controlling the output.

3. gar
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karn:

In my post #11 there is a mistake.

0.6*347/240 = 1.44 A
should be 0.87 and not 1.44 . The 0.6 got lost somehow in my original calculation, possibly I never entered it. 0.87 A is still a higher current than the power supply input rating.

I believe your power supply is a two wire input. Thus, both your L1 and L2 currents should be the same.

You are only reading 0.55 A at 240 V input. That is a VA input of 132. Actual power input will be less and might be about 100 W.

Your 0.55 A input is below the power supply input rating and you may be OK, but we know virtually nothing about the power supply, and a substantial input under voltage might cause some other kind of problem rather than just heat related.

I would really like to see output measurements as well as input, and from 240 to 480 V input.

Is there any sort of voltage, current, or power marking on the actual LED bulb. You can easily determine if the bulb itself is be driven with AC or DC with your meter.

.

4. what does the PDF spec sheet say ??

https://products.currentbyge.com/sit...M-347-480v.pdf

or call GE
GE Lighting Solutions NELA Park – 1975 Noble Road East Cleveland, OH 44112 1-888-694-3533 (888-MY-GE-LED)

5. gar
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FionaZuppa:

The reference you provided is very useful. However, GE provided no information on what the product is or how it works. One has to work backwards from little tidbits of information.

The output is DC seen by a very small + and - in a drawing.

The output is a constant currrent within some DC voltage limit range. Meaning that within the specified voltage range the currrent is approximately constant. The value of the constant current is programmable by some unknown means.

The input is an AC sine wave within the specified range, and input current is expected to be within the range indicated when fully loaded.

.

6. gar,
almost all LED driver PS's are DC output. some wonky landscape lighting stuff uses AC PS and the lamps have small controller boards that change AC to DC as well as do the dimming/color functions, and the control is addressable signals on the AC output lines of the PS.

but, i also did note their address and #, your observations of weak/poor PDF should prompt the reader to call GE.

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