I want to test my power supply at the 12 V 10 A specification. I had tried with connect a 100-watt 0.7-ohm load resistor across it, but by that time the voltage went down to 3 to 4 volt. How does it happen? And how can I test this power supply?
Electronic – How to test the power supply across the load
current measurementpower supplyvoltage
Related Solutions
It depends greatly on the exact power supply model, but in general I would not use it unless you example the schematics of your PSU.
Example 1: http://www.smpspowersupply.com/ATX_power_supply_schematic.pdf
In this schematics, +12V and -12V actually come from the same winding on the transformer. The only difference is -12V has PR1507 diode rated 1.5A, while +12V has U20C20 high speed, dual diode rated 20 amperes. So if your PSU looks like this, you can get +24V as long as you make sure the diodes do not overheat (it is quite possible that -12V diode has no proper heatsink, and thus it cannot handle full 1.5A for extended periods of time)
Example 2: http://www.smpspowersupply.com/atx-power-supply.html
In this schematics, -12V comes from a separate transformer using 7912 regulator. If you try to draw too much current, either L4 will not be able to provide it, so the voltage will sag, or 7912 will overheat. D5 will then start conduting and your -12V line will measure around +0.7 volts. If your PSU looks like this, you should not try to get +24V from it.
Here's a way of learning about the LEDs and improving your electronics knowledge.
LEDs are diodes and they exhibit a non-linear relationship between the voltage applied to them and the current through them. See the graph below. What this means is that as the voltage increases the current increase exponentially and we are in danger of smoking the LEDs.
To get around this problem we limit the current through the LEDs either with a resistor or a constant current power supply. Either way the current is limited to a safe value.
The next problem to address is that if the LEDs are paralleled as in your diagram uneven current distribution will occur. Due to manufacturing variation some will pass more current at a given voltage (and they all share the same voltage in your circuit). The one that's passing most current will heat up more than the other, allowing even more current through it, etc., until it fails. The pattern will repeat and you may blow all the LEDs.
If you wish to pursue this the best thing would be to get your hands on a bench power supply - borrow or go somewhere where you can use one for an hour and proceed as follows:
- Solder wires to the lugs of the LED.
- Mount the LED on a heatsink. 10 W is huge power for such a small die and you need to carry the heat away. You'll need thermal paste.
- Set the power supply voltage to 12 V. Set the current to zero.
- Connect the LED observing polarity.
- Switch on the power supply output. The voltage should drop to zero since you have set a zero current.
- Turn up the current slowly. You should get plenty of light by 100 mA and the voltage should have risen. Make a note of V and I.
- Repeat in 100 mA steps up to 1000 mA checking your heatsink temperature.
- If you have time, repeat the procedure for one or two more of the LEDs.
- Draw a graph and post it here (as an update to your original question).
You should get help then.
Related Topic
- Electronic – Why is the regulation of the 5 volt power supply so poor and how to address it
- Electronic – Dual power supply gives higher voltage as soon as load is connected
- Electrical – Understanding DALI power supply
- Electronic – Making a bench power supply to test car & bike starter motors at no load
- Electronic – Why is the voltage across the nichrome wire so low
Best Answer
Start by using the right load. The supply is rated for 10 A maximum at 12 V. By Ohm's law, that means the smallest valid load resistance is (12 V)/(10 A) = 1.2 Ω.
By connecting a 700 mΩ resistor to the supply, you violated it's current spec. Again by Ohm's law, (12 V)/(700 mΩ) = 17 A. The supply dropping its output voltage when you attempt to draw more than rated current from it is a totally reasonable thing for it to do.
You really need to look up Ohms's Law and understand what it means.
Also consider the power the load resistor must dissipate. If you manage to load the supply to its maximum rating, then the power into the resistor will be (12 V)(10 A) = 120 W. Even if your resistor were the right resistance, it doesn't have enough power handling capability.
You could get a second 700 mΩ 100 W resistor and put them in series. That effectively makes a 1.4 Ω 200 W resistor. That is within what the supply can drive and the combined resistance can handle. By Ohm's law again (yes this comes up a lot and is really useful), (12 V)/(1.4 Ω) = 8.6 A. That's what the current will be thru the combined resistor. It doesn't test the supply to the limit, but it's a good start.