Electronics > Beginners
How can this be? (picture)
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billbyrd1945:
Riddle me this: My adjustable power supply, and, my Fluke 15B+ both indicate a draw of 0.005A while the math is 0.00634A. And, in the math I used measured values rather than color bands for the resistors. I realize we're talking third decimal point to the right, but isn't something wrong here? It doesn't bother me to admit that my $60 PS and my Chinese Fluke are not top of the line. And, if that's the cause of the discrepancy, I'm okay with it. I just need to know. Anyone?
mikerj:
The Fluke 15B+ has a resistance accuracy of 0.5% + 2 or 3 digits depending on the range.  Did you also measure the actual supply voltage (again 0.5% +3 digits) and use that in your math?
billbyrd1945:
I was released from high school on a plea bargain. Can you show me the math you just suggested. I know how to figure 5% of the displayed power supply values. You lost me on the "+ 2 to 3 digits". The Fluke is on auto-range and displays the same values as the PS. (Then I have a follow-up question regarding series vs parallel.)

0.005A + 0.00025(5%) = 0.00525A vs 0.00634A
Dave:
Are you taking into account that the Fluke itself has internal resistance when measuring current?
ArthurDent:
Any measurement you make on the resistors in your circuit using your multimeter has an effect on the circuit so knowing how the multimeter loads the circuit can determine how to use it to get the most accurate readings.

First, the basic accuracy of your Fluke meter is about 0.5% on DCV and 1.5% or 3 times worse on DCMA So if you can just take voltage measurements your readings will be more accurate.

If you’re measuring the voltage drop across a resistor the ‘burden’ a voltmeter puts on the resistor you’re measuring will form parallel resistors (say R5, 1K and the meter input resistance of 10M ohm) so the 1K resistor is now the equivalence of a 999 ohm resistor giving a slightly lower voltage drop across that resistor and slightly higher voltage drop across the other resistors in the circuit. Because this ratio is about 1000:1, the high impedance voltmeter will not have much of an influence on the reading accuracy.

Now if you wire your multimeter set to Ma in series with R5, 1K, the equivalent value could be 1020 ohms because the burden or series resistance on the Ma range could be 20 ohms. On uA ranges the burden could be 1000 ohms making R5 2000 ohms and this would influence the reading to a far greater extent. This added to the DC current accuracy being 3x worse than the DCV specs means measuring values this way will give far less accurate readings and the readings will not agree with the math.

So in this case with the circuit values you have it is best to measure voltage drops and use Ohm’s law to calculate the voltage and current values. If you had a much higher voltage and all the resistors were in M ohms and not ohms, then measuring the voltage drops with a 10M input impedance would cause fairly large errors and using the Ma meter in series wouldn’t have as large an effect. Some better meters have an input impedance of 10G ohms that make them useful for voltage measurement in very high impedance circuits.   

This video does an excellent job of explaining the current measuring problem in detail.


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