This is a newbie question but I was trying to calculate the resistance required to ensure 3.3 V:
Max Specs: 40.0 mA, and 5V
V = IR ===> R = V/I
Resistance Required = (5V - 3.3V) / 0.040 A = 42.5 ohms
So why is this guy using 1.2 k Ohms?
I'm really confused...
You need two resistors not one to reduce voltage. They need to form a potential divider circuit. It is the ratio of the two resistors that control the voltage drop not the absoloute value.
He's not using a resistor, he using a resistor divider.
That 1.2k resistor is working together with the 2.2k resistor to provide a voltage in that middle that's about 3.3V.
(2k2 / (2k2 + 1k2)) * 5 = 3.23V
kat24385:
Max Specs: 40.0 mA, and 5V
No, the current is zero into the module, this is a logic signal and no current flows in the steady
state for CMOS logic (well less than a nano-amp for sure). If you wanted to overload an Arduino
pin you could connect it to 3.3V via a 42 ohm resistor I suppose! The resistor divider only lightly
loads the pin with a couple of mA.
The reason the 1k and 2k2 values are used rather than, say, 1M and 2M2, is so that the signals
can switch at logic speeds (tens of nanoseconds) - all circuitry and digital inputs have stray
capacitance (say 10pF), and when you charge a capacitor through a resistor it does so with
a time-constant of RC. So 10pF and 2k2 implies 22ns, 10pF and 2M2 implies 22us...