Are you just measuring the current across the open terminals of the power supply not connected to the computer? If so, that’s part of your problem. For one thing, the power supply probably has some sort of short-circuit protection (which is pretty much what you’re doing if you’re just connecting the meter across the output). If you want to measure the current draw of the laptop, you need to put the meter in series with the power supply, and to do that you need to either cut one of the wires, or rig up some kind of adaptor. If this IS what you’re doing, your reading isn’t necessarily wrong–the 3-amp figure is merely the maximum rated current draw; in actual use, the laptop’s consumption will vary considerably, depending upon what it’s doing.
Did you connect the multimeter (in its current setting) to the same two terminals that you used when measuring the voltage? If that’s the case you have short-circuited the power supply’s output and you are measuring the current that the power supply outputs to a short circuit - a meaningless measurement at best, and a way to damage the power supply or blow your multimeter’s fuse at worst.
The 3 A rating of your power supply is not a quantity to be measured simply by attaching a multimeter to the output.
ok this makes sense to me, maybe I shoud be asking the question this way, how best to determine if the power supply is capable of generating sufficent power to drive the device in question.
I would just go ahead and plug it into the laptop, given that the voltage reading is pretty well dead on. The worst that’s likely to happen is that power supply will be overloaded and fail, which means it was on the way out anyway. It’s within the realm of possibility that the PS could fail in such a way as to take the laptop out with it, but this is pretty unlikely, IME. You could rig up a simulated maximum load (3 amps @ 20 V is about 6.6 ohms–a 20 W, 120 V light bulb should do nicely), if you were really worried about it.
To determine current you would need to hook the multimeter up to the battery in series with the actual load as was mentioned by Q.E.D.. Unfortunately this would be quite difficult with the battery of a laptop. Doesn’t your computer tell you if your battery is insufficient, or are you trying to do this before hooking up a battery?
Actually, I remember now, you could take a known load and hook that and your multimeter up in series with the battery. Some multimeters include a battery tester function, and this is what that function does (I just read a book on using multimeters). So, you need a load of the right impedance. Hmm, now I need to figure out what would be appropriate.
No, it won’t. I got amps and watts confused in my head. A 6.6-ohm load at 120 V will draw about 18 amps! That’s a heck of a light bulb. A couple of 1000-W hair dryers in parallel will do it; as I said, though, it’s probably more trouble than it’s worth.
Sure. All that matters is the DC resistance, which is provided by the heating element. The low voltage input won’t run the motor, but that makes no difference here.
To test a regulated power supply, the best method is to measure the output voltage with a “light load” and measure the output voltage with a “heavy load.” A light load is defined as a load that doesn’t draw much current (around 10% of its maximum output current) and a heavy load draws between 90% and 100% of its rated maximum output current.
To do this, you simply need some *known * resistors and a voltmeter. You do not need to measure current… simply connect each resistor (one at a time) to the output of the power supply and measure the output voltage for each. The voltage should remain fairly consistent.
So let’s say you have a regulated power supply with a fixed nominal output voltage of 19 VDC, and it is capable of sourcing between 0 and 3 ADC. (If it’s a switching supply, it might have a problem sourcing current in the milliamp range, but I won’t get into that.) You would need two resistors. The light load resistor would need to be around 60 Ω and capable of dissipating at least 6 W. Here are some ways to do it:
Use a 60 Ω / 10 W resistor, e.g. www.mouser.com part number 71-RH10-60. Price is $2.37.
Stick two 30 Ω / 5 W resistors in series, e.g. www.jameco.com part no. 660498CK. Each resistor is $0.36. Total price is $0.72.
String six 10 Ω / 1 W resistors in series, e.g. Radio Shack part no. 271-151. A package of two 10 Ω resistors is $1.29. Total price (minus tax) is $3.87.
The heavy load resistor is a little more difficult; you need a 7 Ω resistor that can dissipate at least 52 W. Here are some ideas:
Use a light bulb. But you’ll have a rough time finding one with just the right resistance (7 Ω) when powered w/ 19 V. (Light bulbs are very nonlinear.)
Use the computer as a load. But this has a couple drawbacks: a) The current may be fluctuating and/or noisy, b) It may not draw the maximum current.
Stick a 4 Ω / 50 W resistor in series with a 3 Ω / 50 W resistor, e.g. www.mouser.com part numbers 71-HL50-06Z-4.0 and 71-HL50-06Z-3.0. Each resistor is $3.38. Total price is $6.76.
Stick eleven 0.62 Ω / 5 W resistors in series, e.g. www.jameco.com part no. 660084CK. Each resistor is $0.36. Total price is $3.96.
Use 40-feet of 32-AWG copper wire.
Another idea altogether is to use an adjustable slide resistor or power rheostat. But those are even more expensive.
If we assume each element is around 14.4 Ω it should work O.K. The only problem I can think of is that most heating elements have a pretty significant PTC, which means the resistance for each element is likely to be less than 14.4 Ω when operated at a lower temperature.
As long as the resistance of an element doesn’t dip below 12.67 Ω (which is a 12% decrease) then it should work O.K.
Jeebus Kwist on a Kwutch, Crafter_man how many times do I have to tell you to quit it with the driveby posts, already? How about putting some effort into it? Maybe a bit of actual research?