real noob question. measuring current

Ok I got 4 LED's in series with a 100ohm resister and a 12v battery. If I want to measure the current through the LED's;

1) Do I simply just put my tester over 1 LED, like my tester being in parallel with 1 LED?

Or

2)Do I need to break the circuit somewhere and connect it up with my tester in series?

Since you have the known resistor already, all you need to do is measure the voltage across it. Using a voltage divider to bring the voltage down to the 5 V level of the arduino, simply measure the voltage at the 12 V point and the other end of the 100 ohm resistor. The current is then (V12 - V100ohm) / 100.

If the 12 V is well regulated and you know it is 12 V, you don't need to measure it.

To measure current you need to break into the circuit and connect your meter in series.

To measure voltage you woukd connect in parallel.

Make sure you have your meter on the correct setting. In particular do not connect in parallel with your meter set to ammeter as it has a very low input impedance in that mode and coild easily draw enough current to fry your Arduino pins and possibly other components.

In your case, the best approach would be to measure the voltage across the resistor and calculate the current using [u]Ohm’s Law[/u] (Current = Voltage/Resistance).

Since the LEDs and resistor are in series, the same current flows through the resistor as the LEDs. Note that you can’t use Ohm’s Law directly with the LEDs because the LEDs are nonlinear… The resistance of the LED changes when the voltage changes.

We rarely measure current directly. Mainly because you have to break the circuit to make the current go through the meter. I work in electronics, and I measure voltage & resistance every day. I can’t remember the last time I’ve measured current. (Except there are voltage & current meters built-into my bench power supply, and it’s often helpful to see if a board under test is drawing excess current.)

There is also a fuse inside the meter, and if you have a short (or something else causing an unexpected amount of current) you can blow the fuse in your meter.

Electricians use a “clamp on” current meter to measure “house current” without breaking the circuit or actually touching the wires. (I think these may work magnetically, but they might have a hall-effect sensor.)

You can make a pretty good estimate of what the current will be before measuring: (12V - (4 x Vf))/100 = current For ~2.2V Red LEDs: (12 - (4 x 2.2))/100 = 32mA For ~3.3V Blue, Green LEDS: (4 x 3.3) is >12V so likely only 3 will turn on. Same for 3.7V White LEDs.

Measure the voltage across the resistor: Vr/R = current. Say 2.1V/100 ohm = 21mA

Ammeter does the same - with your meter in series with the circuit, the meter is adding a really low resistor in series and calculating the current by measuring the voltage across the resistor.

Awesome thanx for the good explanations guys I understand now

(I think these may work magnetically, but they might have a hall-effect sensor.)

The Hall effect is a magnetic effect.

100 Ohm resistor may not be exactly 100 Ohms. 5%-10% tollerance? Not much delta, but how accurate do you need to be?

Accuracy needed depends on the situation. If you put together the worst-case path: High Vcc (5V +/-5% can be 5.25), Vf of LED 2.2 to 2.7V, 10% tolerance Resistor so 100 % is really 90: (5.25-2.2)/90 = 33.9mA. If Absolute Max is 30mA continuous thru your LEDs, then your looking at an LED that's not gonna last long. More than likely, 5V will only be 4.3, Vf will be closer to 2.5, and 100 will be closer to 100: (4.3-2.5)/100 = 18mA, so things are cool.

Well im done with my display now but when i measure the voltage drop across a resister i get 0.8v. But the LED's are bright so i dont know why it shows such a low voltage. 0.8v/100ohms=9mA? is this maybe because the display is multiplexing and the tester is reading it incorrectly?

If the current isn't steady DC then your meter won't be able to read it, no. You would need to use an oscilloscope and measure the peak voltage from it.

majenko: If the current isn't steady DC then your meter won't be able to read it, no. You would need to use an oscilloscope and measure the peak voltage from it.

Thank you, that put my mind at ease now