Measuring Direct Current With Multimeter

allenpitts

Joined Feb 26, 2011
99
Hello ACC Forum,

This is pretty basic question.
Measuring the current flowing thru an LED. The specs say 20 mA. Need to
light 292 and interested in how many amps they will draw.
So a simple circuit was fashioned: a voltage source, a resister, and an LED.

The circuit was then broken and a connection to the multimeter inserted

But the multimeter says all zeros.

The multimeter is set to the lowest setting available
but no current is measured.

Could it be that 20mA is to small an amount of
current for the multimeter to measure?

Thanks.

Allen

Alec_t

Joined Sep 17, 2013
10,359
On the 200mA current range as shown, 20mA should register correctly. If the meter showed 0 then no current was flowing. Likely a wire disconnected.

Joined Apr 16, 2011
375
Does you multi-meter manual make any mention of an internal fuse. Maybe it is blown?

crutschow

Joined Mar 14, 2008
23,332
Since there already is a resistor in the circuit, you can just measure the voltage across the resistor, and calculate the current using Ohm's law (I=V/R).

AlbertHall

Joined Jun 4, 2014
8,355
Does you multi-meter manual make any mention of an internal fuse. Maybe it is blown?
Yes, check the connections first.
Then, still wired, short the meter leads together. If the LED then lights then check the multimeter fuse.

KeithWalker

Joined Jul 10, 2017
413
Hello ACC Forum,

This is pretty basic question.
Measuring the current flowing thru an LED. The specs say 20 mA. Need to
light 292 and interested in how many amps they will draw.
So a simple circuit was fashioned: a voltage source, a resister, and an LED.

The circuit was then broken and a connection to the multimeter inserted

But the multimeter says all zeros.

The multimeter is set to the lowest setting available
but no current is measured.

Could it be that 20mA is to small an amount of
current for the multimeter to measure?

Thanks.

Allen
The LED did not light with the meter in circuit so either something is not connected or the meter is open circuit (fuse?).

Last edited:

MrChips

Joined Oct 2, 2009
19,273
According to the photo posted, LED is not lit.

Fuse in the meter is blown.

This is a common newbie phenomenon.

Joined Jul 18, 2013
19,029
Confirm Fuse: Touch meter leads together and see if LED lights.
LED correct way around?
Max.

DickCappels

Joined Aug 21, 2008
5,879
To avoid blowing the fuse again, don't use the current scales again. Just read the voltage across a small shunt resistor. Even if the resistor goes up in flames it saves you the trouble of hunting down a replacement fuse.

KeithWalker

Joined Jul 10, 2017
413
To avoid blowing the fuse again, don't use the current scales again. Just read the voltage across a small shunt resistor. Even if the resistor goes up in flames it saves you the trouble of hunting down a replacement fuse.
Just measure the voltage across the 220 Ohm resistor. The current will be V/R.

bwilliams60

Joined Nov 18, 2012
1,178
Good for him to learn how to use an ammeter properly. You can test the internal fuses on your meter. Follow the manual but usuall if you put the red lead in the port for voltage, far right, place your meter on ohms or diode deoending on the meter, and stick the other end of the lead in your two amp ports, you should get some kind of reading. If no reading, blown fuse or internal circuit. This test depends on manufacturer but they all have a way of doing it. Don't give up on using ammeters.

KeepItSimpleStupid

Joined Mar 4, 2014
3,611
It's very common for the fuse to be blown as was said, but the multimeter (most anyway) inserts it;s own resistor depending on range, thus really messing up your head. usually the meter does not drop more than 0.6V at full scale. See your manual.

allenpitts

Joined Feb 26, 2011
99
Hello Forum,

There is a fuse in the multimeter. Will replace and retest.
Thanks.

Allen

WBahn

Joined Mar 31, 2012
24,692
I have that exact same Micronta meter and I blew the fuse in it the first night I had it (in 1989 or so) when I went to measure voltage on an AC outlet and still had it in current mode. Thought I had destroyed by brand new meter, but it still works fine, though I have never replaced the fuse (so the fused ammeter functions may or may not work, but the unfused 10 A range still works fine). I still use it to make low-value current measurements all the time, but I use a shunt resistor. You usually get a better measurement by measuring the voltage across a series resistor (which is often already part of the circuit) instead of inserting a milliameter into the circuit, which usually requires disrupting the circuit and often causes significant errors in the measurement (which you can compensate for if you understand their nature and source).

allenpitts

Joined Feb 26, 2011
99
Hello Bahn, KISS, KeithWalker, Mr Chips, et al and the ACC Forum,

Replaced fuse. Getting reading.

The specs on the LED being used:
http://www.taydaelectronics.com/leds/round-leds/3mm-leds/green.html
$.03 Emitting color: Green Diameter: 3mm Lens color: Water Clear Wavelength: 515 - 520 nm Forward voltage(V): 3.2 - 3.4 V Current(mA): 20 View angle: 20 - 25 Luminous intensity(MCD): 10000 - 12000 mcd Ligitek L8G2041-PF The LED tester built has a nine volt battery that measures 7.21 volts. The tester has a 220 ohm resistor. With 1 LED measures 16 mA 2 LEDs measures 5.2 mA (in series) 3 LEDs measures 0 mA (LED glows very dimly) Following Keith Walkers excellent idea measured the voltage across the resistor (that is, w the meter in DC voltage mode connected multimeter leads at either end of the resistor) and got 3.40 volts. V/R is 3.4/220 = .0154 Not sure what that means. 15 mA? If so, then the 5mA difference between the specs and the measurement is chalked up to design and actual or the multimeter's or the resistor's tolerances or all three. Right? Allen Pitts PS Did test on second LED tester: Batt V 8.59 1 LED measures 23.2 mA 2 LEDs measures 10.6 mA (in series) 3 LEDs measures 2.4 mA 4. LEDs measures 0 (LED glows dimly) At first this seems counterintuitive because the more LEDs placed in service the more current would be drawn and measured. Or perhaps the voltage source has a finite amount of current and the more used by the diodes the less available to be measured. WBahn Joined Mar 31, 2012 24,692 Hello Bahn, KISS, KeithWalker, Mr Chips, et al and the ACC Forum, Replaced fuse. Getting reading. The specs on the LED being used: http://www.taydaelectronics.com/leds/round-leds/3mm-leds/green.html$.03
Emitting color: Green
Diameter: 3mm
Lens color: Water Clear
Wavelength: 515 - 520 nm
Forward voltage(V): 3.2 - 3.4 V
Current(mA): 20
View angle: 20 - 25
Luminous intensity(MCD): 10000 - 12000 mcd
Ligitek L8G2041-PF

The LED tester built has a nine volt battery that measures 7.21 volts.
The tester has a 220 ohm resistor. With
1 LED measures 16 mA
2 LEDs measures 5.2 mA (in series)
3 LEDs measures 0 mA (LED glows very dimly)

Following Keith Walkers excellent idea measured the voltage across
the resistor (that is, w the meter in DC voltage mode connected
multimeter leads at either end of the resistor) and got 3.40 volts.
V/R is 3.4/220 = .0154
Not sure what that means. 15 mA? If so, then the 5mA difference between the specs
and the measurement is chalked up to design and actual or the multimeter's
or the resistor's tolerances or all three. Right?

Allen Pitts

PS
Did test on second LED tester:
Batt V 8.59
1 LED measures 23.2 mA
2 LEDs measures 10.6 mA (in series)
3 LEDs measures 2.4 mA
4. LEDs measures 0 (LED glows dimly)
At first this seems counterintuitive because
the more LEDs placed in service the
more current would be drawn and measured.
Or perhaps the voltage source has a
finite amount of current and the more
used by the diodes the less available
to be measured.

If your 9 V battery is measuring 7.21 V with no load, then it is pretty much dead and likely has a pretty high internal resistance at this point.

Assuming, for the moment, that the battery is able to sustain this terminal voltage under load, then you would expect a current draw of somewhere between

(7.21 V - 3.4 V) / 220 Ω = 17.3 mA
and
(7.21 V - 3.2 V) / 220 Ω = 18.2 mA

Assuming the forward voltage of the LED at the drawn current of about 15 mA is 3.3 V (just taking the average), then the total effective resistance would be

15.4 mA = (7.21 V - 3.3 V) / (220 Ω + R)

which would yield R = 34 Ω.

This is probably not unreasonable for a 9 V alkaline nearing its end-of-life.

As you put more LEDs in series, they drop more of the total voltage, leaving less voltage to be dropped across the current-limiting resistor, which means less current will flow. By the time you have three LEDs in series, you have no voltage overhead left and barely any current will flow (a bit surprised you are seeing even a dim glow).

One thing that you need to look into is the voltage-current characteristic of an LED. The spec may say that the current is 20 mA, but what they are talking about is "when operated such that the current is 20 mA". The more voltage you put across it, the more current. The less voltage, the less current. But the relationship is so sensitive to voltage that a tiny voltage change results in a significant current change -- so much so that at currents anywhere near the intended operating current the voltage will only change by a small fraction of a volt. That's why we usually rely on in external part of the circuit, such as a resistor, to limit the current.

dl324

Joined Mar 30, 2015
8,890
Following Keith Walkers excellent idea measured the voltage across
the resistor
That suggestion was mentioned a half dozen posts earlier by @crutschow.

If you insist on using a meter to measure current, do yourself a favor and learn how doing that can affect the circuit and give you a misleading reading.

garyr

Joined Sep 23, 2019
1
I have had this problem before....reverse the leads on the meter. Mine doesn't read negative current.