Raspberry Pi Lab

Current Draw In My Three LED Circuit

by on Jan.04, 2014, under Electronics

ohms-lawI an previous post I described a Three LED circuit and Python code to drive the LEDs.  A reader posted, via Facebook, the following question…

“You decided to drive them directly through a resistor instead of going through a transistor. What is the current draw on the GPIO pins?”

Well, I am certainly not an electrical expert, but I think I can attempt to talk about the math behind this circuit.

As you recall,  I drove the LED but connecting a GPIO pin to a 560 Ohm resister.  The resistor was connected to the positive leg of the LED.  Lastly, the negative leg of the LED was connected to ground.

I used a 10mm Red LED purchased from Ada Fruit.  The datasheet for this LED can be found at on the Ada Fruit website.  In short, the forward voltage of the LED is a typical 2.0 v with a maximum of 2.4 v.  It draws 20mA.

Ohm’s Law reads the current (I) = Voltage (v) / Reistance (R).   I=V/R

Well, we already know that the LED wants to run at 2.0 V and draw 20mA.  We just need to know what size resistor to place on the LED so as not to blow up the LED.  A simple algebraic re-arrangement of Ohm’s law tells us that R=V/I

The GPIO pins on the RPi all output 3.3v.  Since the GPIO is giving us 3.3 V when we only need 2.0 V, we will have to remove (resist) 1.3V.  Now we can calculate the minimum resistor size.

R = (3.3 V = 2.0 V) / 0.02A = 65 ohms.

That means the minimum size resistor for this circuit using this LED is 65 Ohms.  Using a larger resistor just means the LED won’t shine as brightly as it would with the minimum sized resistor.

So why did I decide to use the 560 ohm resistor?  Simple… That is what I had available in my toolbox.

I think I can now answer the reader’s actual question of how much current was actually drawn from the GPIO pin.

Because I used a 560 ohm resistor and I know the voltage provided by the GPIO and the voltage used by the LED, I can calculate the current draw via Ohm’s Law.

I = V/R

The LED needs 2v of the 3.3 v provided on the GPIO and we know the resister is 560 ohms.

I = (3.3 V – 2.0 V) / 560 ohms = 0.0023 A = 0.23mA

The spec sheet says we can draw up to 16mA from a GPIO (as long as we don’t do that much for more than one pin at a time).  So our 0.23mA is well within spec..

 

 

 

 

 

 

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