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Getting confused with the basics and using my power supply

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HobGoblyn:
I've done a few tests as I like to have proof :)   Same simple circuit, 1.5v,  each time a different resistor



1)  20.2 ohm   1.5/20.2 = 74mA  Actual amp meter reading = 49mA

Meter on mA has internal resistance of 10 ohm

20.2 + 10 = 30.2 ohm    1.5/30.2 = 49mA. Same as amp meter



2)  5.1 ohm (this was five 1 ohm resistors in series)

1.5/5.1= 294mA  Actual amp meter reading 114

5.1 + 10 = 15.1      1.5/15.1 = 99mA  close to amp meter  (only test that wasn't spot on)



3) 47.1 ohm        1.5/47.1 = 31mA   Actual amp meter reading = 26mA

47.1+ 10 = 57.1 ohm.    1.5/57.1 =  26mA  same as amp meter



4) 100.3 ohm    1.5/100.3 = 15mA   Actual amp meter reading 13.5ma

100.3 + 10 = 110.3 ohm    1.5/110.3 = 13.5mA  same as amp meter



5)  219.7 ohm     1.5/219.7= 7mA   Actual amp meter reading = 6.5mA

219.7 + 10 = 229.7 ohm.   1.5/229.7 = 6.5mA



6) 989.7 ohm     1.5/989.7 =  1.5mA  Actual amp meter reading 1.5

989.7 + 10 = 999.7 ohm    1.5/999.7 = 1.5ma


So it proves there is a 10 ohm resistance when meter is set to mA.  The only one that wasn't spot on was test no2, that was the only test that didn't use just 1 resistor, it had 5 in series.


If I set meter to uA  for test 6, instead of 1.5mA amp meter shows 877.u

This must mean the meter when set to uA has a 720 ohm resistance?

989.7 + 720 = 1709.7 ohm.  1.5/1709.7 = 0.000877 which I presume is 877uA ?

ArthurDent:
I think you've got it figured out.  :-+

Siwastaja:
Right, and do note that a 10 ohm burden resistor (also called "current measurement resistor", or "shunt resistor") for the mA range is just huge. Sure, they (multimeter designers) get a larger voltage drop over this resistor, making it easier to measure accurately, but such high value disrupts many actual circuits (as you have found out).

A higher-quality meter would use a smaller burden resistor (say, 1 ohm), and a better amplifying circuit (higher gain, lower offset, lower drift, lower noise...) inside to still measure the current accurately, even though it has to work with a smaller voltage drop over the measurement resistor.

Though, the principle stays the same: even if the burden resistor was just 1 ohm, you would still see a difference between your expectations and reality.

Voltage drops happen in almost all "high current" circuits, in wiring, contacts, current measurement resistors, etc. Hence, having another multimeter available just to read voltage at the point of interest, bypassing wiring losses etc., is a good idea.

AVGresponding:
I prefer to make my own shunts for some measurements, as I can make them 0.1 ohm or 0.01 ohm for example, and using many paralleled resistors also gives you greater current measuring capacity.

You do need a quite high precision mV reading meter to make use of these at lower currents, but they do have the advantage of having a low enough burden voltage not to interfere with most circuits operation.

Fungus:

--- Quote from: HobGoblyn on February 03, 2020, 09:19:19 pm ---...my experiments only seem to draw as many amps as they need.

--- End quote ---

Ohms law.  :-+

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