0
\$\begingroup\$

I have two TENMA 72-10415. One is set to measure ohm and the other to µA I am trying to measure the resistance between the 10A_MAX <-> COM and mAµA <-> COM. Measuring between 10A_MAX <-> COM I get as I expect 0.8 ohm and 0A. However when I measure between mAµA <-> COM I get 1 kohm and 118µA on the ammeter.

I want to measure the current output of my device with a load of 1 kohm. Does this mean the I can simply load my circuit with the ammeter or is something completely wrong with my multimeter?

\$\endgroup\$
4
  • \$\begingroup\$ Your multimeter test shows no fault found. \$\endgroup\$ – Andy aka Dec 8 '20 at 14:56
  • 1
    \$\begingroup\$ I want to measure the current output of my device with a load of 1 kohm. - why not simply measure the voltage on the load? \$\endgroup\$ – Eugene Sh. Dec 8 '20 at 15:00
  • \$\begingroup\$ Low current circuits tend to have a very high impedance so 1K is a short-circuit relative to them. It is difficult to measure a voltage drop across a low resistance at low current. \$\endgroup\$ – DKNguyen Dec 8 '20 at 15:10
  • \$\begingroup\$ Thanks a lot. I am silly of not thinking of just loading with 1kohm and measure voltage across to double check my result. I got the same current result from measuring with ammeter as voltage across 1kohm added resistor. Feels good I got the double confirmation :) \$\endgroup\$ – hansky Dec 8 '20 at 15:14
0
\$\begingroup\$

Does this mean the I can simply load my circuit with the ammeter

Yes, but only on that scale, and only if the applied voltage does not cause current to exceed the meter's full scale reading.

The meter may have over-current protection consisting of back-to-back diodes which conduct when the meter voltage goes over eg. 0.6 V. The resistance will then be much lower than 1k, causing high current flow that could damage the meter or the device under test.

On higher current scales the meter resistance is lower, which could also be dangerous as it is closer to a short circuit.

\$\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.