# Current drawn from USB C port to Arduino board

Hey guys,

I am new to the electronics world and currently following this tutorial by Jeremy Blum.

There’s a few things that are confusing me.

First of all, I tested connecting an LED directly to the 5V and GND pins on the Arduino board in order to test the amperage reading with a multimeter and learn why resistors are necessary. I know you’re not meant to do this, but Jeremy said it would be fine for a short period of time. But as soon as I plugged the Arduino into my computer’s USB C port (via an adapter) the LED burnt out, and I didn’t even get a chance to hook up the multimeter.

I then attempted the same circuit but included a 1000 Ohm resister this time. This worked fine and gave me a medium-lit LED, but when attempting to measure the amperage on my multimeter I found the value started really high and then continuously lowered until around 300mA (yes, you heard me correctly). Connecting the multimeter to the circuit also seemed to drain all power from the Arduino as both onboard LEDs and my external LED turned off while the reading was taking place.

All of this has now lead me to two ideas which are probably completely wrong, but that’s where you guys come in! Here goes:

• My multimeter is causing more current to be drawn through the Arduino, and is hence not giving me an accurate reading of the amount of current that was flowing before connecting it up.
• My 20V/5A USB C port is somehow providing too much current which is why the LED burnt out straight away.

Lastly, I am confused about the calculation that Jeremy does in this video to determine how many Ohms of resistance is required to run the LED at 20mA. This is his calculation:

V = IR
R = V/I
R = 3V/20mA (3 out of 5 volts because 2 volts drop over the LED)
R = 150 ohms.

This confuses me because despite a 1000 Ohm resistor, my circuit appears to be drawing well in excess of 20mA, and why wouldn’t it? How does the circuit know to only use 20mA with 150 Ohms resistance when there is 5A to play with on the USB C port? Is it because the voltage is regulated to 5V through the Arduino board?

I’ve included a picture of my circuit for context.

I hope that all makes sense (despite me being a n00b) - thanks in advance for your help! I'd bet you put the meter in PARALLEL with the LED, ( a SHORT CIRCUIT), reading DC current requires the meter to be in SERIES with the load being tested.

Jeremy said it would be fine

Look for another tutorial. This kind of tips could kill your computer when attempting to measure the amperage on my multimeter I found the value started really high and then continuously lowered until around 300mA

How did you connect your multimeter? Please provide a hand-drawn schematic or a photo. Seems you are shorting the port with your multimeter. Read about how to use a multimeter to measure current.

How does the circuit know to only use 20mA with 150 Ohms resistance when there is 5A to play with on the USB C port?

How does your ceiling light know to only draw 100W from the power plant providing several Gigawatts? Read about Ohm’s law.

edgemoron:
I’d bet you put the meter in PARALLEL with the LED, ( a SHORT CIRCUIT), reading DC current requires the meter to be in SERIES with the load being tested.

Oh my god, That is EXACTLY what I was doing. My bad… Thanks so much for your help! I’m getting accurate readings now.

olf2012:
How does your ceiling light know to only draw 100W from the power plant providing several Gigawatts? Read about Ohm’s law.

Yes I think I’m getting a much better grasp on this now. If I understand this correctly, the Arduino regulates the voltage to 5V, so without factoring in how much resistance is on the Arduino board itself, the amperage drawn would be equal to whatever resistance is in the circuit. So with a simple circuit containing only a 1000 Ohm resistor the amperage would be somewhere around 5mA, as per the following calculation:

V = IR
I = V/R
I = 5V/1000 Ohms
I = 0.005A/5mA.

This does beg the question. How much resistance is on the Arduino board itself/How many amps would be drawn if I removed the resistor from this circuit? I tried to test this out, and measured around 280mA. So using the following calculation I get a resistance of 17.86 Ohms (rounded to 2 decimal places):

V = IR
R = V/I
R = 5V/0.28A
R = 17.86 Ohms.

Is this correct?

I should note that when I carried out this test the onboard Arduino lights faded off as the amp measurement lowered, so they were completely off by the time I hit 280mA. Is this because the multimeter became the path of least resistance? And therefore 17.86 Ohms is actually the resistance on the multimeter rather than the Arduino?

Sorry for the n00b questions… Appreciate you guys helping me.

Hi,
When you go from measuring voltage to current and back to voltage, you do change the leads plug connections on the DMM?

Tom... Hi Tom,

Yeah, with no resistor I had to switch to the unfused connection on my DMM because the fused connection only measures up to 200mA.

Measuring Voltage or Current is simply a case of turning the dial however.

Is that what you're asking? 