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I cannot understand why the current I expect is so different from what I am measuring.

I found a very low resistor, 15 ohm and connect it to 5 volts. Why is the current I measure so low? Shouldn't it be 0.333 amperes?

Obviously, I am a noob, but I am tripped up by this. Are there good reasons why it is like this?

Why is the current so low compared to what I would expect from Ohm's law?

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  • 3
    \$\begingroup\$ Probably your 5 V power supply went into short circuit mode. \$\endgroup\$ – Jack Creasey Nov 15 '17 at 6:23
  • \$\begingroup\$ (1) The resistor colour code looks like bn-gy-bn which would be 180 Ω. Can you check? (2) Is the PSU rated for 500 mA. (3) Is it powering anything else at the same time? (4) Does the voltage remain at 5 V when you connect the resistor across the supply? \$\endgroup\$ – Transistor Nov 15 '17 at 6:24
  • \$\begingroup\$ The resistor colors are brown, green, black \$\endgroup\$ – Steven Koscho Nov 15 '17 at 6:27
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    \$\begingroup\$ You might also try putting two 15 ohm resistors in series and measuring the current as well as measuring the voltage across one of the resistors. The combination of these measurement should tell you what is going on. \$\endgroup\$ – Dean Franks Nov 15 '17 at 6:30
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    \$\begingroup\$ Ok, so I put the red and black power leads straight on each side of the resistor to test voltage across it. Whoa! It got hot and I unplugged it. It dropped to about 4.8V and the overall voltage has been measuring 5.1V. So, it must be my meter. Because the resistor wasn't heating up when I had the meter in series. \$\endgroup\$ – Steven Koscho Nov 15 '17 at 6:37
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The user manual of your multimeter tells us why:

In the current measuring mode (DC or AC): input impedance approximately 3 kΩ.

Which in all honesty is just a joke.

So with 5 V the most current you can get is 1.67 mA, but it doesn't even tell us the range of the input impedance, so the value you measured is "fine".

It says to be able to measure up to 4 mA (which is another joke), you need at least 12 V to get close to the measuring range by creating a "short" with your test leads.

Actually we can calculate the input impedance for your meter to be: 5.12 V / 1.367 mA - 15 Ω = 3730 Ω.

A good meter has something in the range of 10 Ω or less, depending on the selected range. The µCurrent, for example, has 0.02 Ω for mA measurement, 10 Ω for µA measurement and 10 kΩ for nA measurement.

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    \$\begingroup\$ @pipe that's ... why you don't buy a cheap meter. But at least it has a manual which doesn't lie. \$\endgroup\$ – Arsenal Nov 15 '17 at 7:53
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    \$\begingroup\$ @Arsenal - I've used cheap meters in my time. My current meter is a £5 chinese brand meter. None of them have had this kind of ... I was about to call it a limitation, but it's actually a fundamental design flaw that renders the device almost useless. \$\endgroup\$ – Jules Nov 15 '17 at 13:44
  • \$\begingroup\$ @Jules there are various other reasons why I'd steer clear of cheap meters, this one is just an addition to the list. Just have a look at the various teardown videos available. \$\endgroup\$ – Arsenal Nov 15 '17 at 13:51
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    \$\begingroup\$ Note that if you are characterizing a real constant-current source, like the current output from a photodiode, the input impedance of the meter doesn't need to be small. If you ever have access to a nice picoammeter, stick your cheap ohmmeter across its input; an input impedance of 10 kΩ isn't uncommon. Agreed that 3kΩ is unusually high for a cheap multimeter, and not really useful for this sort of low-voltage, low-impedance circuit. \$\endgroup\$ – rob Nov 15 '17 at 14:10
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    \$\begingroup\$ @PeterMortensen you are not really more precise, you just state a different quality of the µCurrent. What you've given correctly is the scaling factor of the µCurrent. But that involves an opamp, the resistance introduced into the circuit are the values I stated in my answer (you have to take the burden voltage figure (20 µV / mA for example to get that value). The maximum output is 1,25V. PS: thanks for proofreading my answer, kohm still doesn't seem right for me - but then I'm a German guy after all... \$\endgroup\$ – Arsenal Nov 15 '17 at 20:21

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