Multimeter readings do not match math

I'm using a multimeter to measure the ohms, amperes, and volts coming from the the Duemilanove's 5V pin. Connecting one probe of the multimeter to the 5V pin, and the other to the ground pin, the multimeter reports a resistance of 0.9kohms, 5V, and 0.55A. I convert 0.9kohms to 900ohms and plug the values into Ohm's Law:

V = I * R
5 = 0.55 * 900
5 = 495

Obviously this isn't right... so what am I doing wrong?

Thanks,
Jordy

You're not understanding basic physics.

If you really did measure 550mA from an AVR output pin, you've probably wasted that pin, if not the entire chip.

You're right, I don't understand basic physics. I started with electronics a couple days ago.

It does still measure .55A, 5V, .9kohms. What is it supposed to measure? What am I not understanding?

... and plug the values into Ohm's Law:

Ohm's law applies to linear devices that exhibit a characteristic called 'resistance'. Ohm's law does not apply to non-linear devices such as microprocessors.

I started with electronics a couple days ago.

It took most of us several years to get some of this straight.

Don

Ohm's law applies to linear devices that exhibit a characteristic called 'resistance'. Ohm's law does not apply to non-linear devices such as microprocessors.

So is Ohm's Law useless whenever working with a microprocessor? Can it be used for calculations of a circuit on a breadboard connected to the microprocessor?

Thanks.

So is Ohm's Law useless whenever working with a microprocessor? Can it be used for calculations of a circuit on a breadboard connected to the microprocessor?

Ok... I think I understand... so I could use it on a circuit where the wire doesn't branch/fork, but not where it does?

Here's the deal. AVR pins are not meant to source more than 40mA TOPS, and you are sourcing nearly 14x that much. I'd be very surprised if it still works.

Odds are, because you're not measuring I and V at the same time, V dropped significantly to account for the excess current being drawn.

No, I'm not an expert, but I think that's what's going on.

In the future, ALWAYS use a current limiting resistor (preferably 300 Ohm+).

One thing to be very careful about in the future. If you did indeed put your meter in current mode and then measured between ground and the +5vdc power pin you were in fact short circuiting the +5vdc to ground. Something saved your ass, depending on how you are powering the board, either the USB thermofuse or the on-board +5vdc regulator going into overload protection mode.

Until you gain more knowledge and experience, be very careful measuring current. Measuring ohms (only with power off) and voltage (only with power on :wink: ) are much less intrusive and dangerous, but current measurement can be very destructive if done incorrectly.

Lefty

Here's the deal. AVR pins are not meant to source more than 40mA TOPS, and you are sourcing nearly 14x that much. I'd be very surprised if it still works.

So... I should use external power source to power anything greater than 40mA? Alright, thanks. :slight_smile:

Until you gain more knowledge and experience, be very careful measuring current. Measuring ohms (only with power off) and voltage (only with power on Wink ) are much less intrusive and dangerous, but current measurement can be very destructive if done incorrectly.

Will do, thanks. :slight_smile:

So... I should use external power source to power anything greater than 40mA? Alright, thanks

Actually 40ma is a AVR maximum never exceed rating, 20-25ma is a better maximum limit to stick to. Anything above that one uses a switching device (npn transistor, mosfet, etc) to switch the higher current load. You don't always have to use external power, say if the load device is a +5vdc device you can use the Arduinos +5vdc power to a overall current draw of less then about 500ma or so, but you do need to stay under any digital output pin current limit with switching devices mentioned.

Lefty

Can it be used for calculations of a circuit on a breadboard connected to the microprocessor?

Yes, Ohm's law can be used for the linear devices, typically resistors, that are connected to the microprocessor. Let's say you want to drive a red LED from a microprocessor pin using a series resistor to limit the current. A typical red LED might require about 10 mA and at that current have a voltage drop of around 2 volts. This means that your series resistor will have 3 volts across it (the 5 volt output from the microprocessor pin less the 2 volts across the LED), and 10 mA through it (the same as the LED). In order for these conditions to exist you should use a series resistor of 300 ohms (3 volts divided by 10 mA).

Note that the LED is a non linear device so you can't use Ohm's law on it and say that it has a resistance of 200 ohms (2 volts divided by 10 mA) since its calculated "resistance" will be different with currents other than 10 mA.

Don

Thanks everyone for the help. :slight_smile: