0
\$\begingroup\$

I know that with DC you can measure current by putting your meter in series with the positive supply.

I have a project running off a low voltage AC supply. Can I do the same?

\$\endgroup\$

2 Answers 2

2
\$\begingroup\$

Many (most inexpensive?) multimeters can only measure DC current - in that case, no, you can't.

However, if your meter can measure AC current, you would connect it in series with the circuit, as you would with DC.

\$\endgroup\$
2
\$\begingroup\$

AC current ranges are often not available on cheaper meters.

A "work around" that works with almost any DMM and is sometimes superior is to use a small series resistor and to measure the voltage drop across it with an AC voltage range on the DMM.

Most cheaper DMMs are "three and a half digit" meaning they display values from 0000 to 1999. Some meters have a 200 mV AC lowest AC voltage, range and others 2 VAC. A few very cheap ones may only have 20 VAC minimum.

What value series R to use depends on current and tolerable voltage drop. For eg a load drawing up to 100 mA and with a say 12VAC supply then 0.1 VAC drop in a sense resistor would usually be tolerable and higher may be.
0.1 VAC is 0.1/0.2 = 50% of a 200 mV range and gives 1000 counts (pout of 2000 possible) so in this case woukld allow a 100 mA/1000 = 0.1 mA resolution with the right resistor.
To do this select R so V_r = 0.1V at max crrent. So R = V/I = 0.1/100 mA = 1 Ohm.

If only a 2VAC minimum range was available then a 1 Ohm resistor will give 1 mA resolution in the above example.

A 20 VAC minimum range gives 10 mA resolution - use a larger resistor for more voltage drop or (better) get a slightly better meter.

Usually sense resistor value of 0.1 Ohm and 1 Ohm are useful. These allow easy conversion from current to voltage. If doint this regularly use a good precision resistor and perhaps select one from a batch that is most accurate. Note that power dissipation is usually not an issue but if you do overheat the sense resistor its value will often be affected.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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