Electronics > Beginners
Getting confused with the basics and using my power supply
Caliaxy:
Can you use your other DMM to measure the voltage across the resistor (while your Uni-T Ameter is still in series with the resistor)? What happens to the current value (as displayed on the power supply meter and on the Uni-T meter if you short the Uni-T meter (while in series with the resistor)? Try it with the Uni-T meter set to mA then on A.
HobGoblyn:
--- Quote from: Caliaxy on February 06, 2020, 04:27:29 pm ---Can you use your other DMM to measure the voltage across the resistor (while your Uni-T Ameter is still in series with the resistor)? What happens to the current value (as displayed on the power supply meter and on the Uni-T meter if you short the Uni-T meter (while in series with the resistor)? Try it with the Uni-T meter set to mA then on A.
--- End quote ---
Sorry for being thick, I presume when you say "what happens ....if you short the Uni-T meter ....", by short, you are saying measure the current (as if I understand things correctly, that is in effect shorting)?
When the Uni-T is set to A, it measures 0.145A, if I measure the voltage across the resistor (at the same time as measuring the amps) , it's 1.45v, the power supply says 144mA
If I change the Uni-T to mA, it measures 74.50 mA, the voltage across the resistor is 74.28 and the power supply says 73mA
ArthurDent:
Saying you're getting the 'wrong' results is misleading. The meters are accurately displaying what they are measuring but you haven't understood what they are actually measuring. You are using a 10 ohm resistor and when your DMM is on the higher current range the burden resistor is a very small resistance so the drop across the meter is very small-close to zero volts. Once you switch to the MA range you change the value of the burden resistor from almost nothing to 10 ohms. Now instead of having a 10 ohm resistor across the supply you have a 20 ohm load and the current will be 1/2 what it was on the A range. The values of current you got were about 150 and 75 which verifies that the MA range indeed has a 10 ohm burden resistor equal to your load resistor and thereby giving you 1/2 the current.
HobGoblyn:
--- Quote from: ArthurDent on February 06, 2020, 05:03:52 pm ---Saying you're getting the 'wrong' results is misleading. The meters are accurately displaying what they are measuring but you haven't understood what they are actually measuring. You are using a 10 ohm resistor and when your DMM is on the higher current range the burden resistor is a very small resistance so the drop across the meter is very small-close to zero volts. Once you switch to the MA range you change the value of the burden resistor from almost nothing to 10 ohms. Now instead of having a 10 ohm resistor across the supply you have a 20 ohm load and the current will be 1/2 what it was on the A range. The values of current you got were about 150 and 75 which verifies that the MA range indeed has a 10 ohm burden resistor equal to your load resistor and thereby giving you 1/2 the current.
--- End quote ---
Sorry, I misread your previous post with the vid, to be saying the problem was using two meters at once. hence my reply saying I'm only using one meter at a time.
After reading this post, I re-read the previous post and watched the vid, and it makes a lot of sense and explains a lot, but as is always the case, the more I learn, the more I get confused :)
I understand everything that the vid (and you) are saying. But that opens up a dilemma for me, when going through a beginners book. I was expecting to simply do a few experiments to prove to myself I understand whats happening. And it's been harder than I thought. Then again, I have learnt a lot by them giving unexpected (to me) results.
I sort of feel that while starting out (in that I want to learn electronics properly from the beginning), having a mA or even uA setting on my DMM is almost pointless.
But I'm learning a lot, thats the main thing.
Caliaxy:
As ArthurDent says. A-meters don't measure the current directly, they actually measure the voltage drop across a small internal resistor (shunt or burden resistor, r) they put in series with the circuit. They display the results as current (using Ohm's law, I=u/r, where you measure u and know r).
The current through the circuit changes when you insert the A-meter, from I=U/R to I=U/(R+r), where U is the voltage of the power supply and R is your external resistor (10 ohms). Ideally, the internal shunt resistor r has a very small value, much smaller than that of the resistor R under test (through which you want to measure the current), so (R+r) can be approximated as R. The bigger the shunt (r), the bigger the error. From your measurements, r is ~10 ohms on the mA scale and much smaller (negligible) on the A scale, hence the "instrument error". In the experiment I was suggesting, shorting r (forcing it to be 0 ohms) brings the current back to its original value and the voltage drop over r (as displayed by the A-meter) becomes, of course, 0V (displayed as 0 A).
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