Need help testing AC/DC power supplies

[wasman]

I am trying to create a test method and jig to test new power supplies we get to confirm that are indeed capable of the rated Current and Power.

For example we have a new power supply with output of 15-17V 5.00A and rated at 90Watts.

So far I connected the power supply to a few different resistors (3.2 Ohm, 3.6 Ohm and 2.6 Ohm) and connected a multimeter in series to read the current. I got 4.3A, 3.93A and 4.92A respectively. I am not sure how to best confirm if the power supply is capable of delivering 90Watts of power.

Any help would be appreciated.
Thanks

 

[Klaus]

You need to measure the current AND the voltage across the load. Multiply the two and you have the power consumed by this load, VxA=W.
Car headlights would make nice loads for your power supply, 90W resistors are quite big unless you water cool them.
Klaus

 

[Roff]

According to my calculations, you can only get about 70 watts out of this supply. Here's how to calculate it:

Think of your supply as a battery of Vs volts with an internal resistance of Rs ohms.
Take the two measurements you made with R1=2.6 ohms and R2=3.6 ohms. This yielded currents I1=4.92 amps and I2=3.93 amps respectively. Now,

I1=Vs/(Rs+R1), I2=Vs/(Rs+R2)

You have 2 equations and 2 unknowns, Vs and Rs.

Using simple algebra, Vs=19.53 volts and Rs=1.37 ohms.

You can verify that your 3rd measurement will give you approximately the same results. You should also get about 19.5 volts if you measure the output with no load.

With an internal resistance of 1.37 ohms, you will get maximum power transfer (load power) when the load is equal to the internal resistance, i.e., Rload=1.37 ohms.

Using Rload=1.37 ohms and calculating the output current, I=Vs/(Rs+Rload), or I=19.53/(1.37+1.37).

I=7.13amps.

Output power P=I^2*R, or P=(7.13^2)*1.37

P=69.6 watts

This won't be exact, but I don't think you'll get 90 watts from this supply. If your optimum load is not 1.37 ohms, you'll get even less than 70 watts.
Bummer.

 

[john1]

W = (IxI) x R

Eye squared Arr

W = (4.3 x 4.3) X 3.2
W = 18.49 X 3.2
W = 59.168

W = (3.93x3.93)X 3.6
W = 15.68 X 3.6
W = 56.45

for fifty watts, the current would be three and an eighth Amperes.

(V)16 x (I)3.125 = (W)50

Best of luck with it, John


ive just seen the last post, and i agree, you may not get 90 watts from
that supply, unless there are tappings you could try .... but it might be
a switching supply ... ?