Please note: although this question involves a Raspberry Pi (hereafter RPi), it is really a pure robotics question at heart!
I am trying to connect my RPi 1 Model A to a breadboard with a single, simple LED on it. The LED is rated at 1.7V and 20mA. The RPi's GPIO pins are providing 3.3V of power and are not really meant to exceed 16mA. So my circuit needs a resistor, and I believe its calculation is as follows:
R = V differential / I = (3.3V - 1.7V) / .02A = 1.6V / .02A = 80 ohms
If I'm understanding this calculation correctly, that means that providing my LED circuit with 80 ohms or higher will safely limit the current on the circuit.
Problem is, I only have 47-ohm resistors available to me. So I'm wondering if I can daisy chain two 47-ohm resistors (for a total of 94 ohms, which is greater than the 80-ohm requirement) and not fry the RPi and/or the LED?
Here's my wiring:
In the pic above:
- A red jumper connects a GPIO pin to a place on the breadboard; this feeds power to...
- The first 47-ohm resistor; this then feeds power to...
- The 2nd 47-ohm resistor; this then feeds power to...
- The LED's anode; the LED is powered and power flows out of the cathode to...
- A pair of jumpers (small yellow one and then a brown one) that lead back to the
GNDon the RPi
Here's a slightly better angle of the breadboard:
So I ask: will I fry my pi with this setup? Have I calculated resistance correctly (80 ohms)? Can I daisy chain two 47-ohm resistors together? Anything look "off" about my wiring setup in general? Thanks in advance!