Quote:
Originally Posted by Clutz
You're missing jarthel's point, which is that even if you only had a voltmeter (which has more limited functionality than a multimeter) you are capable of measuring current using ohm's law: V=I*R. Using a bit o' algebra magic, we can restate Ohm's law in terms of current as: I=V/R
Assuming the resistance is 1Ohm (as per jarthels suggestion, although you'd probably need a pretty freaking beefy 1Ohm resistor - something in the range of 900 Watts, which I'm imagining doesn't actually exist), then the above equation for current simplifies down to: I=V.
Hence, any voltage measure is equal to the current running through that device.
Thus, if you place a 1 ohm resistor between the +ve and ground the secondary of your transformer and measure the voltage across it, you're also measuring it's current output capacity.
(You could also use a resistor with a higher resistance, say, 1KOhm, and then get away with using a 1 Watt resist, in which case, I=V/1000)
Additionally, I'm willing to bet that if what you have is truly a multimeter, that it is actually capable of measuuring current, or it's probably not a multimeter, and just a voltmeter (multimeter meaning it is a meter capable of measuring multiple things, voltage being one of them).
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You’re confusing the thread even more here, and with a lot of nonsense at that
What this basically describes is a shunt resistor, where you insert a small known resistor inline with the power supply line, measure the voltage across it and as ohms law, you get I=V/R
But Clutz, you have the wrong idea on what is a suitable shunt resistor. You want something that is going to have minimal effect on the rest of the circuit. as you know, if you put a current through a resistor you are going to have a voltage drop across it, now with our shunt in the power supply line, that is going to cut into the voltage used to supply the circuit itself, and therefore change the current draw as it is receiving less voltage now. so you have to pick a shunt resistor that is not going to have much effect at all, one with a small V drop. Using the dynahi which when working right will use about 600mA, picking the 1 ohm resistor you said first, will measure about 600mV across it (1ohm * 600mA) which while high is not going to severely effect the operation of the amp, now for power loss in the resistor, you said it would be 900W, but I think you got your formulas a bit mixed up, P = I^2*R, = 0.36w which is no problem
Now you go on to use a 1K resistor, with 600mA through that we get 600V!! Which is 10 times more that what dynahi uses - we have gone completely wrong here
A 0.1 ohm resistor would be a very suitable choice, with a voltage drop of 60mV across it @ 600mA and we can see that lover R we have the lower V we get across the resistor and less effect on the circuit we want to measure.
But after all that electrical basics, the dynahi is a well known diy project and the power requirements are pretty well given as +/-30V @ 600mA and using the power supply designed for it, 30V AC * 2 is a good AC supply for it, gives our transformer requirements of 30 * 2 * 0.6 = 36VA, we always add a bit more onto that and one of the most common values for a power transformer is 80VA so that is most recommended