Load test
Dear Sir,
I designed 9v 5amps smps. Now I am in a stage to do the load test for my smps. I did load test with resistors, but that does not worked, as I couldnt measure the maximum amps the power supply can produce. Actually I used, 9 numbers (18ohm resistor and 5 watts each) so totally i connected 45watts load to the power supply. I connected all the 9*(18ohm resistor,5watts) in parallel. As per calculation, the load resistor value should be 1.8ohms and 45watts(9(v)/5(I)= 1.8ohm(r)). To get this 1.8ohms and 45watts, I used 9 numbers(18ohms resistor,5watts) and connected all that in parallel so that it becomes 1.8ohms and 45watts. When i implemented this, the voltage got dropped when i gradually increase the load but it does not dropped right at the output side but it dropped at the load side, as the voltage got dropped it could able to measure the amps only till 2.23amps, after that amps not getting incresed as the voltage got dropped at the load side.
Please tell the right and perfect method to measure the full amps rating of my smps. What kind of load I should use to measure my 5amps output current.
Thanks,
V Kumar
Comments
Thanks for your reply sir,
I will update my measurements as soon as possible. In the mean time, I have attached my schematic and transformer design document. Please go through it.
Make sure when you meausre your VOUT while loading the power supply you measure it right at the output of the power supply
Dear Sir,
As per your advice, I have studied the kelvin connection and got to know full about it. Thanks for that.
Then, I did the measurements as you asked. By this measurements, I feel that my smps is working fine, anyway i need your final conclusion of my smps, Please verify my measurements and tell me the efficiency of the power supply.
Note: I have given +/-10 percent as my output tolerance and I measured the Vout right at the output of the power supply.
Here is the measurements,
Vout= 9.12v Iout=0amps
Vout= 9.12v Iout=0.51amps
Vout= 9.12v Iout=1.01amps
Vout= 9.11v Iout=1.47amps
Vout= (9.10v-9.11v) Iout=1.95amps
Vout= 9.10v Iout=2.40amps
Vout= 9.10v Iout=2.83amps
Vout= (9.09v-9.10v) Iout=3.28amps
Vout= (9.09v-9.10v) Iout=3.69amps
Vout= (9.08v-9.09v) Iout=4.12amps
Vout= (9.08v-9.09v) Iout=4.53amps
Vout= 9.08v Iout=4.93amps
Awaiting for your reply,
Thanks,
V Kumar
I used my input in between 200v-230v, For this input I have uploaded my measurements.
Pls go through it sir,
Thanks,
V Kumar
Your power supply is perferctly regulating, Right?
Yes sir.. Its regulating fine..

It is hard for me to understand the combinations of the resistor loads you are using. So I prefer you tell me only the total resistor value (RL) of your load and/or the power supply output current (IOUT). This way I think I can have a better understanding of the situation.
When you are measureing output voltage (VOUT), you should connect the voltmeter right at the output connector of the power supply. You do not measure VOUT at the load side because you have an extra voltage drop due to the resistance of the wires that connect the power supply to the load. This method is called Kelving connection. You may search in the web for more about it
You can take let say few measurements of VOUT at different IOUT, Let say 0%, 10%, 25% 50%...100%) of IOUT Then I will get a better feeling of the power supply load regulation. You can repeat the same measurements at VINMIN and VINMAX
Also please post your schematic and transformer design documentation