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- Newsgroups: sci.electronics
- Path: sparky!uunet!pipex!ibmpcug!robobar!stl!bnr.co.uk!bmdhh243!michthom
- From: michthom@bnr.ca (Michael Thomson)
- Subject: Re: How do I create a 2A load to test a power supply?
- Message-ID: <1993Jan24.130915.10811@bnr.uk>
- Sender: news@bnr.uk (News Administrator)
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- Organization: BNR Europe Ltd, Maidenhead, UK
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- References: <1993Jan23.235548.13731@selway.umt.edu>
- Distribution: sci.electronics
- Date: Sun, 24 Jan 1993 13:09:15 GMT
- Lines: 58
-
- Russell J. Pagenkopf (cs000rjp@selway.umt.edu) wrote:
- ..
- : The power supply uses 120VAC 0.40A grounded and outputs (supposedly) 20V,
- : 2.0A, 40VA Class 2 with four pins ( I have no idea what pins do what yet).
- ..
- : what does the 40va mean? 40VAC? I guess what this comes down to is, how
- : should I test this thing to make sure it is giving me what it should?
-
- Hokay - Here's my $0.02 worth...
-
- 1) A quick lesson on power delivery...
-
- If the load you are driving into is *purely* resistive (no capacitive /
- inductive elements) then the voltage and current are in phase, and the power
- consumed is simply the product of the two. e.g. 20V across a 10 ohm load gives
- a 2 Amp current, and you are dissipating 40 Watts of real power.
- If on the other hand the load has a COMPLEX impedance (there is some capacitive
- or inductive component) then the voltage and current will be out of phase. This
- means that the power dissipated in the load will be *less* than the answer you
- get if you multiply volts and amps together.
- WHY??? I hear you cry! The reason is that capacitors and inductors don't
- dissipate (much) power. They store it as electric charge and magnetic field,
- respectively. Thus the power is not all "used up".
-
- Summary: 1 Watt = 1 Volt * 1 Amp, in the special case where Voltage and
- Current are in phase (at DC, or AC into a purely resistive
- load).
-
- 1 VA = 1 Volt * 1 Amp, but here we are talking about the
- MAGNITUDE of the COMPLEX product of the voltage and current
- phasors. I realise this may be greek to you - a decent
- electronics textbook should explain about complex power (VA),
- real power (W) and the relationships better than I can.
-
- 2) Testing your supply...
-
- First measure the open circuit voltage at the outputs. You may see a bit more
- than 20 Volts. Don't worry unless it's a lot more (about 30V). This checks
- there is juice coming out. If no voltage - it's toasted, but make sure you
- check all of the outputs before you junk it, as there may be two wires
- connected to the same supply rail.
- I wouldn't recommend trying to measure the current with your meter alone, as
- this basically shorts the power supply, and the current may burn out the
- secondary winding on the transformer (or your meter!). Instead, get a 20 Ohm
- resistor rated for 20 Watts (This will get 1 Amp out of your supply) or a 10
- Ohm resistor rated at 40 Watts to get 2 Amps out of the supply. Remember, you
- can put two 20 Ohm, 20 Watts in parallel to do this. Hook this load across
- the supply and measure the voltage. This is the case where VA = Watts, and you
- should be able to check you get what you need out of this sucker.
-
- Hope this helps!
- --
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