How to improve efficiency?
Hi,
I'm designing a 187W power supply using the TOP261YN. The schematic file is as attached. I also wanted to attach the PIExpert file but wasn't able to.
I'm getting very low efficiency. At low line ~58% and at high line ~55%.
By the way, the tranformer was ordered from China. Generally it seems okay, but one thing that concerns me is the leakage inductance. PIExpert specifies 5uH, but I've got ~15uH with primary inductance of 330uH. Would this affect the efficiency?
Any suggestion how can the efficiency be improved?
Thanks.
Comments
Hi Crusher,
Thanks for your reply.
1) I've measured the efficiency again without L1 & L2 and the efficiency is about the same.
2) I've also measured the temperature of some components, while the power supply was running for 5 minutes and these are the observations:
Ambient 23°C
Snubber resistor 75°C
TOP261 (with heatsink)65°C
Transformer 55°C
Output diode (heatsink)70°C
Input rectifier 40°C
NTC 65°C
Comparing these to data in Section 12 of DER-204, I don't see that they are too bad. Which made me suspect that the measurement method is wrong..? ~50% efficiency means ~100W is lost within the power supply, that's a lot of heat!
3) I then used the measurement method recommended in another forum with topic 'Measure efficiency without an AC power meter' dated 22/8/08, to measure the current and voltage after the bridge rectifier and before the bulk capacitors. The resulting efficiency is even less.. ~45%!
Please help.
Hello juinbeh,
I have to recommend you again to optimize the transformer for your application. Use our PIExpert for designing the transformer.
You’re right, you have to identify where the power is dissipated as heat. I would recommend you to remove R6. D3 is supposed to clamp the reset voltage to 200V over DC rail and R6 is actually preventing voltage rising.
Cheers,
PI_Crusher
Hi Crusher,
The transformer was designed with PIExpert!
As I mentioned previously, we don't think there's 100W dissipated at the unit. The unit is just warm, most of the heat is dissipated through the output diode & it's heatsink.
I still suspect that there's error in the input current measurement.
We're looking at reducing ~100W, would removing R6 improve it by that much?
Cheers,
juinbeh
Hi Crusher,
I did a measurement of the input current through a 0.5Ohm. I've attached the waveform as below.
I tried to do a quick calculation to work out the average current based on this waveform. By treating the curves as rectangulars,
Vresistor = 2 x 1.66 x 2.4ms/20ms = 0.4V
Iresistor = 0.4/0.5Ohm = 0.8A
P_in = 240 x 0.8 = 192W
P_out = 22 x 7.22 = 158.8W
Eff = 82.7%
Is that correct?
Cheers,
Juin
Hello Juin,
For efficiency evaluation I would recommend you to check our forum and support:
http://www.powerint.com/en/forum/ask-pi-engineer/power-supply-efficiency
http://www.powerint.com/en/design-support/appstv#6
Cheers,
PI_Crusher
Hi Crusher,
I did the measurement again with a domestic power meter. It's saying that at full load the full load power is 191W. According to my earlier calculation using the waveform it's 192W. The error makes sense as the curves was assumed to be a rectangular to simplify calculation.
That means the efficiency is ~82% which is great!
Thanks,
Juin
Hello juinbeh,
A good reference for your design is DER-204:
http://www.powerint.com/sites/default/files/PDFFiles/der204.pdf
The very low efficiency you reported is definitely not normal. One first change I would suggest is to test efficiency without L1 and L2, eventually replace this differential filter with a common mode transformer. It is possible your two inductors are saturated, just double check them. You have to run EMI tests to design the proper filter.
Also, it is very important to minimize transformer losses: for minimal copper losses use maximum wire diameter allowed, for minimum leakage inductance use minimum number of turns allowed, and for minimum capacitive losses use isolation layers (trade off against leakage, the more isolation you use, the higher leakage).
Cheers,
PI_Crusher