2
\$\begingroup\$

I'm not entirely sure if I have done the following equation correct or not, but based on the calculations, my multimeter result seems to be incorrect. I have a very simple circuit, containing a 9V battery, an LED (with 3.2 V forward voltage at 24 mA), and a 270 ohm resistor (as shown below).

schematic

simulate this circuit – Schematic created using CircuitLab

My question is, when I do the calculations to determine what the current should be in the circuit, I do (voltage / resistance = I), so I get (I = 5.8V / 270 = 0.0214 Amps). This is 21 mA, approximately. When I insert my meter into the circuit, (inserting the meter between + and the LED, breaking the circuit and properly measuring amperage) I obtain a reading of about 2.1 mA. Did I do the calculation wrong?

I am using a Craftsman 82344 autoranging multimeter. Also, when I insert the multimeter into the circuit, the LED no longer illuminates as it should (at least according to the articles I have studied). Any help would be appreciated, if any more details are needed let me know.

\$\endgroup\$
  • \$\begingroup\$ What is the burden voltage? Take that into account and all the numbers likely start to match \$\endgroup\$ – PlasmaHH Nov 30 '15 at 21:38
5
\$\begingroup\$

Since your circuit has a known resistance in it, you can avoid the meter burden voltage issue by calculating the current from a measurement of the voltage drop across the resistor measured with your meter in voltmeter mode, rather than by inserting the meter into the circuit in ammeter mode.

Of course, if your resistor doesn't have the value you think it does (manufacturing tolerance or human error in reading the markings) that determination will be invalid, but you can remove the battery and use your meter in ohm mode on the unpowered circuit to measure the resistor.

\$\endgroup\$
  • \$\begingroup\$ After measuring voltage across the resistor, it turned out to be correct. So the burden voltage of my meter was causing a drastic incorrect reading? \$\endgroup\$ – CodeNMore Nov 30 '15 at 21:50
  • \$\begingroup\$ Apparently. If you had a manual ranging meter, I would suggest trying a higher current scale, which would probably yield a lower burden voltage. It sounds like maybe the auto-ranging algorithm in your meter interacted poorly with the not-exactly-linear behavior of the circuit. Many practical tools don't turn out to match simple ideals, so understanding how to work around their nuances is a key part of the trade. The resistor trick is also really useful if you want to measure pulsed currents with a scope. \$\endgroup\$ – Chris Stratton Nov 30 '15 at 21:53
  • 1
    \$\begingroup\$ Even on autoranging meters the 10A range is usually completely separate (even to the point of having a seperate input connector). \$\endgroup\$ – Peter Green Dec 1 '15 at 2:12
  • \$\begingroup\$ Yes, I leveraged that fact myself last week, using the 10 A range to do a ballpark assessment of a battery charger where the burden voltage of the lower ranges was hopelessly confusing it. But I wouldn't expect a lot of accuracy when measuring a current expected to be less than 1% of the full scale range. In comparison measuring the voltage across the resistor could be quite sound, especially if the resistor is also measured. \$\endgroup\$ – Chris Stratton Dec 1 '15 at 2:15
2
\$\begingroup\$

The manual doesn't say what the internal shunt resistor is but it may be high enough to affect the circuit you're measuring. Try hooking up again and manually changing the range to the highest that will still give you two significant digits of current reading - probably using the 'select' switch on your meter. This will switch in progressively lower shunt resistors. Record the current reading for each setting.

Let us know how that works.

\$\endgroup\$
0
\$\begingroup\$

I suspect you had the meter connected incorrectly.

When measuring current, the meter must be inserted in the circuit - break the circuit at some point, and connect the meter leads to the two points that you disconnected, so all the current flowing in the circuit passes through the meter.

\$\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.