# 240w Power Supply at 93.5% efficiency = 256w or 271w at wall?

#### EdHo

Joined Jun 10, 2016
10
I have a Meanwell HLG-240H-C2100B power supply (119v, 2.1a, CC, powered by 240v AC)
powering 3 Bridgelux Vero29 LEDin series. The output from the PS reads 115v, 2.1a = 241.5w.

Shouldn't the AC Watts "in" measure 258w? (241.5/.935=258)
I get a reading of 271w in - which suggests an efficiency of 89% not 93.5%.

Should I be checking the test equipment or is this difference to be expected?

#### mcgyvr

Joined Oct 15, 2009
5,394
93.5% is "typical" and based on a specific load.. Did you read the charts in the datasheet?
Specifically the "Efficiency vs load" chart on page 7

Not sure what you are using to measure output/input wattage either.. Could be accuracy issues there too,etc...

#### EdHo

Joined Jun 10, 2016
10
If I read it correctly, 100% load is 250w at 93.5% efficiency. Since this load is 241.5w, which is close to full load (96.6%).
According to the chart, efficiency shouldn't drop below 90% until load is below 35%. As far as "typical", I have checked 6 of these and they yield similar results +-2%.

I think you confirmed that I need to check the accuracy of the meters.
Thank you!

#### RichardO

Joined May 4, 2013
2,270
I think you confirmed that I need to check the accuracy of the meters.
Does this mean that you are using two meters -- a volt meter and a current meter?
If so, you are also measuring the power factor since the voltrage and current are not going to be perfectly in phase. Even a power factor corrected power supply is going to have a less than 100% power factor.

#### AnalogKid

Joined Aug 1, 2013
10,778
The tolerance for an efficiency spec rarely is stated, and typically is 5-10%. Also, peak efficiency might be achievable only with a resistive load.

ak

#### EdHo

Joined Jun 10, 2016
10
Does this mean that you are using two meters -- a volt meter and a current meter?
If so, you are also measuring the power factor since the voltrage and current are not going to be perfectly in phase. Even a power factor corrected power supply is going to have a less than 100% power factor.
A Fluke79 measures the dc output voltage and current. A separate in-line meter measures the AC input voltage and current. (Bayite AC 80-260V 20A).

#### AlbertHall

Joined Jun 4, 2014
12,293
Does this mean that you are using two meters -- a volt meter and a current meter?
If so, you are also measuring the power factor since the voltrage and current are not going to be perfectly in phase. Even a power factor corrected power supply is going to have a less than 100% power factor.
The data I found on the Bayite meter didn't say whether it was measuring true RMS or not. However it did say " This module is only suitable for pure AC 50-60HZ the mains." which rather suggests it isn't <surprised>

#### EdHo

Joined Jun 10, 2016
10
The tolerance for an efficiency spec rarely is stated, and typically is 5-10%. Also, peak efficiency might be achievable only with a resistive load.

ak
Wouldn't LEDs be a resistive load?

#### EdHo

Joined Jun 10, 2016
10
The data I found on the Bayite meter didn't say whether it was measuring true RMS or not. However it did say " This module is only suitable for pure AC 50-60HZ the mains." which rather suggests it isn't <surprised>
I am not the sharpest tool in the outlet. I "thought" that my municipal power didn't need true rms to be measured correctly...
Is that not so?

#### AlbertHall

Joined Jun 4, 2014
12,293
I am not the sharpest tool in the outlet. I "thought" that my municipal power didn't need true rms to be measured correctly...
Is that not so?
The voltage should be OK, but the current (input to SMPS) is unlikely to be pure sine.
[EDIT] What are the voltage and current readings from The Bayite?

Last edited:

#### Alec_t

Joined Sep 17, 2013
14,005
Wouldn't LEDs be a resistive load?
No. They have a very non-linear voltage-current relationship.

#### AlbertHall

Joined Jun 4, 2014
12,293
No. They have a very non-linear voltage-current relationship.
But they are being fed with a constant current (I think that's what the "CC" in the original post means).

#### EdHo

Joined Jun 10, 2016
10
The voltage should be OK, but the current (input to SMPS) is unlikely to be pure sine.
[EDIT] What are the voltage and current readings from The Bayite?

Bayite numbers for 3 Power Supplies
247v - 3.39a - 811.9w

3 Power supplies output
#1 114v 2.077a = 236.77w
#2 113.5v 2.097a = 238w
#3 115v 2.1a = 241.5w
716.27w total

#### EdHo

Joined Jun 10, 2016
10
Sorry,
Bayite numbers for the AC going into 3 power supplies.

#### EdHo

Joined Jun 10, 2016
10
So do I need a true RMS meter for the AC going into the Constant Current power supply, the DC out, neither or both?

#### crutschow

Joined Mar 14, 2008
33,339
So do I need a true RMS meter for the AC going into the Constant Current power supply, the DC out, neither or both?
For the AC power you need a real power meter that compensates for the effect of the power factor, such as a the Kill A Watt meter.
Measuring the voltage and current separately, even with an RMS volt/amp meter and calculating the VA product, will give you the apparent power, not real power.

For DC power any DC meter will work since they all measure real (RMS) power.

#### crutschow

Joined Mar 14, 2008
33,339
No. They have a very non-linear voltage-current relationship.
The may be non-linear but they are still resistive.
Being non-linear doesn't make it non-resistive, i.e. the power is still current times voltage drop.

#### EdHo

Joined Jun 10, 2016
10
For the AC power you need a real power meter that compensates for the effect of the power factor, such as a the Kill A Watt meter.
Thank you.
Can't find anything like the Kill a Watt for 230v.

#### crutschow

Joined Mar 14, 2008
33,339
Thank you.
Can't find anything like the Kill a Watt for 230v.
Bummer.
It's curious that they don't make one since 220V-240V is commonly used in much of the world.
Perhaps one of these would work.

#### Techno Tronix

Joined Jan 10, 2015
139
For the AC power you need a real power meter that compensates for the effect of the power factor, such as a the Kill A Watt meter.
I agree. Even Kill a Watt meter can really help to cut down on costs and find out what appliances are actually worth keeping plugged in.