Possibly broken multimeter - need help

Thread Starter

freeman3

Joined Jul 23, 2013
7
Wasn't sure where to put this so please move this if it is wrong. This is my first post, haha.


Anyway, I'm at a loss with what to do with my multimeter. I have a Greenlee DM-45. I've been trying to do simple example circuits but I've lost hope in building them because I can't get good readings from my multimeter.

Basically, I have a 9V battery (measures 9.12V at the moment) connected in series with a 470 ohm resistor. By ohm's law I should be drawing 19 mA of current at any given time.

If I place the multimeter in series with the circuit I get a reading of 0.016A using the A function and a reading of 0.16 mA using the mA function. Nothing is right. the A function shows 16 mA but the mA shows a 16/100s of a milliamp? That makes no sense.

Figuring I blew a fuse at one point or another I decided to switch over to resistance measurement and touch the probes together. 0.8 ohms. Seems like the fuse is okay.

My next test was to put an LED in series with the resistor. Without the multimeter in series the LED will light up just fine, if I place the multimeter in series with the resistor and the LED the LED will not light up. That's odd.

My final test was to try a 2.2 KOhm resistor alone in series with a 9V battery. I should be getting around 0.004A. Instead I get the same reading as above - 0.016A.


So all I've got from this is that my multimeter is acting like there's some internal resistance, but when I measure the internal resistance using the probe-touching technique I don't get a "fuse blown" reading. I wouldn't think a fuse was bad anyway because I'm getting something out of the current measuring function.



I'm at a loss. All other functions work (voltage, resistance, etc). This thing is junk as far as I'm concerned. I'm about ready to just shell out for a new multimeter because this thing is ruining any fun I was ever planning on having with my breadboard. Can anyone help me here? I don't know where to go from here.

Thank you!
 

MaxHeadRoom

Joined Jul 18, 2013
28,702
It may not be the cause of your problem, but I know with the Fluke meter I have, if the battery is getting low, before it shows low battery symbol, the readings go haywire, for example, measuring 120vac will display 240v.
Evidentally this is common with some Fluke meters.
And resistance and other reading will be off.
Max.
 

panic mode

Joined Oct 10, 2011
2,761
there is a chance that indeed something is wrong with your multimeter, however i don't think that is the case, you are just overlooking inherent properties of components you use.

for example you only have one multimeter and you don't use it to your advantage. as a result you didn't see how battery voltage changes under load and then you draw wrong conclusions. yes, you measure 9.12 V under no load which means battery with considerable wear/depletion. in other words it's internal resistance is already grown to a quite considerable value.

when you measure current, resistance in the circuit is sum of resistor, internal battery resistance and your multimeter (check burden for your multimeter in the datasheet, could be significant on particular mA range). this means that your theoretical voltage remains at 9.12 V but total resistance is some 560 ohms (resistor is biggest part of that resistance but the other two cannot be ignored, they add up to some 90 ohm).

ideally while measuring current, you would use another multimeter to measure voltage across different part of the circuit (battery for example). alternatively you should check battery voltage under load too.

Rinternalofbattery = (Vbatterywithnoload - Vbatteryunderload)/Iunderload
 

Thread Starter

freeman3

Joined Jul 23, 2013
7
there is a chance that indeed something is wrong with your multimeter, however i don't think that is the case, you are just overlooking inherent properties of components you use.

for example you only have one multimeter and you don't use it to your advantage. as a result you didn't see how battery voltage changes under load and then you draw wrong conclusions. yes, you measure 9.12 V under no load which means battery with considerable wear/depletion. in other words it's internal resistance is already grown to a quite considerable value.

when you measure current, resistance in the circuit is sum of resistor, internal battery resistance and your multimeter (check burden for your multimeter in the datasheet, could be significant on particular mA range). this means that your theoretical voltage remains at 9.12 V but total resistance is some 560 ohms (resistor is biggest part of that resistance but the other two cannot be ignored, they add up to some 90 ohm).

ideally while measuring current, you would use another multimeter to measure voltage across different part of the circuit (battery for example). alternatively you should check battery voltage under load too.

Rinternalofbattery = (Vbatterywithnoload - Vbatteryunderload)/Iunderload

That makes sense. If you're right that would account for the 3 mA difference. What's the best way to get the voltage of the battery under load without a second multimeter? Secondly, why does 9.12 indicate a battery with considerable wear and depletion?

As for the burden voltage -

The datasheet says for the range I'm in it's around 2.7 mV/mA. How do I factor this in?

Thanks again for all your help.
 

GopherT

Joined Nov 23, 2012
8,009
Do it in three steps,
1) Measure the voltage of the battery when the resistor is connected
2) and when it is disconnected.
3) Then switch to amps and measure in series of the current through the resistor.
 

Thread Starter

freeman3

Joined Jul 23, 2013
7
I decided to forgo the battery in order to get to the bottom of this. First, I changed the battery in my multimeter to a battery that tested at around 9.18V. As a side note, I tested the previous multimeter battery and it read at 9.21V so that shouldn't of been the problem...

I got out my arduino and used it's 5V pin to output a constant 5V to my circuit. At least I can guarantee voltage. My multimeter agrees and settles on 4.99V.

Now with the 2.2 KOhm resistor I should be getting around 4.99/2200 = ~0.002A. Around 2 mA of current. Given the tolerances of my multimeter that should sit me around 1.91 mA at the low end and 2.09 mA at the high end. I click over to the mA measuring feature and get a 0.8 mA measurement. That makes no sense.

I did manage to inspect the fuses when I opened it up but unfortunately they're opaque so I can't see through them. They didn't look burned or anything though.


Will the 2.7 mV/mA burden voltage have anything to do with this?
 

absf

Joined Dec 29, 2010
1,968
I dont think there's anything wrong with your meter. If the duse is blown, you wont get any reading on the 10A range. There are 2 inputs for measuring current: - 10A and INPUT.

Are you using the 10A input for measuring mA and uA?
Are you using the AC or DC range for your current?

Allen
 

Thread Starter

freeman3

Joined Jul 23, 2013
7
I dont think there's anything wrong with your meter. If the duse is blown, you wont get any reading on the 10A range. There are 2 inputs for measuring current: - 10A and INPUT.

Are you using the 10A input for measuring mA and uA?
Are you using the AC or DC range for your current?

Allen
I'm using INPUT for measuring current and the DC range. The meter itself has a mA function and that's what I'm using.

If you look at the below link you can see it:

http://www.greenlee.com/products/DMM%2540c-600V-AC%2540FDC%2540c-10A%2540c-CAP%2540c-TEMP--%28DM%2540d45%29.html?product_id=20431


Could it be because im not using clip probes and the needle probes aren't getting the right reading?
 

MrChips

Joined Oct 2, 2009
30,824
Do the math.

5V/2.2kΩ = 2.27mA

If you are not getting 2.27mA reading on the meter then something is wrong.

I think your meter is faulty.
 
Last edited:

Thread Starter

freeman3

Joined Jul 23, 2013
7
Got a hold of a Fluke 179 for testing purposes and it reads 2.3 mA. Looks like it's my meter.


Is there any chance this could be caused by a fuse or do I just have a bad meter altogether?
 

MrChips

Joined Oct 2, 2009
30,824
Usually the fuse just blows and it is open circuit.

Take the fuse out and check it with an ohmmeter. It should read close to 0 ohms.

Check that the fuse holder is clean and making good contact with the fuse.
 

panic mode

Joined Oct 10, 2011
2,761
What's the best way to get the voltage of the battery under load without a second multimeter?
use your multimeter as voltmeter and monitor battery voltage with and without load. once you know battery voltage under load, measure current...

why does 9.12 indicate a battery with considerable wear and depletion?
check few different batteries and compare. if i remember correctly. last time i measured fresh 9V battery it was some 9.75V or so. when it got to 8.5V it was practically toast.

similarly, 1.5V battery shows some 1.68V fresh and once it hits 1.49 it is a good sign that not much is left... if you monitor voltage and add some load (resistor for example) you will notice that in this case even small current will result in notable drop of voltage.

even my car battery is labelled 12V but measures close to 14V. guess what the voltage is measured when battery is considered empty. no, it is not 7V,,,

finally, 9V battery is quite weak, i try to avoid pulling more than 10mA from it.
 

Gdrumm

Joined Aug 29, 2008
684
I had an occassion once where I was using my meter and getting unexpected readings. It turned out that I was holding the electronic component in my hand(s), and my body was creating the unexpected variation. Once I isolated the component (laid it on a wooden workbench, and didn't touch it), the variations disappeared, and I was able to obtain the readings I expected to see.
 
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