4
\$\begingroup\$

I'm doing some extra-curricular learning about electronics before I know much. In the spirit of Platt's book, "Make: Electronics", I am "breaking and burning" things to help learn the fun way. I'm a 10 year old in a 74 year old body, what can I say.

My latest effort was hooking a multimeter (UNI-T UT89XD) with red lead plugged into 20 ampere jack, set the meter dial to 60 mA and touched the leads to the +/- of 9 volt battery terminals. The mA reading went down from around 35 mA to around 5 mA in roughly 10 seconds, and the battery got very hot. Its voltage was originally around 9.3 V and after my test it is now around 8.7 - I basically killed it.

I realize I shorted the battery doing this test. But besides this being no way to test a battery, is there anything to learn from this, namely why, as the battery heated up and the voltage went down from 9.3 to 8.7 volts, did the ampere reading go from 35mA to 5mA in 10 seconds? Did that tell me anything useful besides it was a dumb thing to do? Be positive now, and not just "I'm positive that was a dumb thing to do."

\$\endgroup\$
4
  • 1
    \$\begingroup\$ For similar meters, once you plug RED into the 20A jack, it doesn't much matter where you put the meter dial. Its unclear to me where the decimal point shows up, but you're likely reading amps. Your measurement of 35 mA seems suspect, especially considering that the battery got "very hot". \$\endgroup\$
    – glen_geek
    May 14 '20 at 2:47
  • \$\begingroup\$ 3.5A? Sounds high. 350mA sounds low -- but then, I haven't burnt up too many 9V batteries. \$\endgroup\$
    – TimWescott
    May 14 '20 at 3:03
  • \$\begingroup\$ Try it again with a fresh battery, and the meter set on its 20A range. Then you'll find out if it's 350mA or 3.5A or something else. \$\endgroup\$
    – TimWescott
    May 14 '20 at 3:05
  • 2
    \$\begingroup\$ Well done!!! This is the real way to learn. Seriously, carry on experimenting and do not be like the kids of today who (from the sound of it) won’t do anything unless a book tells them you can. I remember connecting two 9v batteries plus-to-minus, minus-to-plus. And taking them to bits as well. The zinc-carbon ones taste funny. (Just one thing though: your 35mA sounds terribly low). \$\endgroup\$ May 14 '20 at 8:53
3
\$\begingroup\$

The number one thing you should have learned is how easy it is to create a short circuit with a multimeter. Don't make the same mistake with house current -- it may turn your multimeter into a grenade! Literally!

Dave of EEVblog does an excellent multimeter input protection teaching on youtube (EEVblog #373 - Multimeter Input Protection Tutorial) where he explains the different levels of protection, including high rupture capacity (or HRC) fuses that should be in the current testing paths, along with the good physical shielding a quality multimeter incorporates. In order to test current, you are creating a short circuit...

If you watch the video, and you haven't heard him say "blow your hand off", you haven't watched the video long enough. There are certain things about electricity that we don't want to learn the hard way, so just make sure to stay with low voltage circuits while you are "burning and learning." And invest in a high quality multimeter (it's not just about precision and accuracy -- it must have the HRC fuses -- open the multimeter up!). Fluke is a revered name in the business, and recommended.

Here is a text-searchable list of Dave's EEVblog episodes -- watching the multimeter-related episodes now would increase the profitability of your recent learning experience. I still remember seeing multimeters blowing up, and I keep that in mind at times for my own self-protection.


EDIT:

See also: EEVblog #94 - Near Death Multimeter Experience

FYI, with a fresh "Heavy Duty" 9-volt battery on two of my multimeters, it puts out 1 Amp. And with a fresh "Alkaline" 9-volt battery, it puts out 5 Amps. I've experienced what you've done on other meters where the switch determines where the decimal point is and how it reads, but the numbers are otherwise correct. You likely went from 3.5 Amps down to 0.5 Amps. There's no harm in taking the same battery and, with the mode in the correct position, duplicating what you did, short-circuiting it for a few seconds and watching the current drop. I do this all the time as a quick test of the life in a battery. Just don't let the battery get too hot -- batteries can and do explode.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ I doubt modern 9V alkaline or carbon zinc batteries will explode, however I've heard a very credible report of ~90A short-circuit current and subsequent explosion from a 9V NiCd battery perhaps 20 years ago. \$\endgroup\$ May 14 '20 at 23:12
  • \$\begingroup\$ @SpehroPefhany -- Yeah. It's more the pattern I want to have in my brain, though. \$\endgroup\$ May 18 '20 at 15:00
1
\$\begingroup\$

why, as the battery heated up and the voltage went down from 9.3 to 8.7 volts, did the amp reading go from 35mA to 5mA in 10 seconds.

Chances are that the battery voltage didn't go down to 8.7V permanently -- now that it's had a chance to sit, try it again and see what voltage it is. If it's not up to 9.3V and if you really want to experiment, heat it in the oven at about 150F* for half an hour and measure again

Did that tell me anything useful besides it was a dumb thing to do?

If you can figure out the meter calibration when the wires are plugged into the 20A jack and the thing is set to 60mA, then you'll know the short-circuit current of the battery.

It told you, in general, how the short-circuit current of the battery diminishes over time (and that batteries get hot when you discharge them quickly).

It told you that whatever current you did pull, it wasn't too much to fry the meter, so probably less than 30 or 40A.

It told you that it doesn't take long to discharge a modern alkaline cell.

Given that the voltage across the meter and leads was very small (it wasn't 9.3V, for sure) it told you that when you pull lots of current from a battery you don't get any useful power out of it. If you have more than one meter and you want to burn up another battery, set one up to 20A and another up to measure voltage. Measure the voltage on the battery as you do your test and watch the current at the same time you're watching the battery voltage. I expect you'll see much less than a volt at the battery terminals.

* not degrees C -- if you're in one of those countries do the conversion.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Not a microwave oven :) \$\endgroup\$
    – HandyHowie
    May 14 '20 at 7:23
  • 1
    \$\begingroup\$ That would be another set of interesting lessons. \$\endgroup\$
    – TimWescott
    May 14 '20 at 15:06
1
\$\begingroup\$

Did that tell me anything useful

Another point that wasn't highlighted is you saw the effect of the battery's internal resistance. As noted in other comments and answers, electronics heat up with high current, and the internal resistance of a battery isn't large (thus relatively high current). It should be noted that this internal resistance also changes over time for batteries. As the temperature changes, so does the internal resistance.

http://www.learningaboutelectronics.com/Articles/Battery-internal-resistance

\$\endgroup\$
1
\$\begingroup\$

Connecting the 9V battery between 20A & COM inputs of the multimeter would have resulted in a high short circuit current through the 20A shunt.

The selector switch being in the 60mA position had no bearing on the readings '35' and '5' taken during the test since the short circuit current did not pass through the 60mA shunt.

Hence it would not be possible to declare with any certainty what the short circuit current actually was.

Overheating of the battery could have resulted in an increase in the battery's internal resistance and consequent decrease in the short circuit current from'35' to'5'.

\$\endgroup\$
2
  • 3
    \$\begingroup\$ The current wasn't 5 mA, just the reading, because the "60mA range" assumes the correct shunt is connected.. To know the actual current, the meter should have been set to the 20A range to match the current shunt used. \$\endgroup\$ May 14 '20 at 13:08
  • 1
    \$\begingroup\$ Hi Brian, Thank you very much for pointing that out. I have altered my answer accordingly. \$\endgroup\$
    – vu2nan
    May 14 '20 at 18:43
1
\$\begingroup\$

a multimeter (UNI-T UT89XD) with red lead plugged into 20 ampere jack, set the meter dial to 60 mA and touched the leads to the +/- of 9 volt battery terminals. The mA reading went down from around 35 mA to around 5 mA in roughly 10 seconds,

This multimeter has a "mA" jack and a "20A" jack, so if you use 20A range I guess you're supposed to plug the red wire in the "20A" jack, and if you use the mA range you're supposed to plug it in the "mA" jack. Check the manual.

Since you plugged it in the "20A" jack but selected the "mA" range, readings are probably bogus, so don't trust the numbers.

and the battery got very hot. Its voltage was originally around 9.3 V and after my test it is now around 8.7 - I basically killed it.

If it got hot then lots of power was dissipated, which means current was much higher than a tame 35mA.

9V batteries have pretty high internal resistance so short circuit current isn't that high, maybe a few amps maximum, that won't damage your multimeter 20A shunt or fuse.

I realize I shorted the battery doing this test. But besides this being no way to test a battery

Ackshually, if you want to test a CR2032 or other button cell which have quite high internal resistance, measuring short circuit current with a multimeter in amp-meter mode is the way to go. Just make it quick. If the cell has 3V on it, it may be good or bad, but short circuit current will tell the real story. A better way would be to measure voltage drop with a known current, but who has time to grab a resistor?

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.