I am just getting into electronics and been reading about voltage dividers, I wanted to build one myself. I have built the below schematics on a breadboard and I am not getting the voltages I want.
below are my calculations.
R3:
bleed = (20mA + 30mA) * .1 = 5mA
R3 = 3.3/0.005 = 660Ω (really used a 680Ω)
R2:
R2 = (10 - 3.3) / 25mA = 268Ω (really used a 27Ω)
R1:
R1 = (12 - 10) / 55mA = 36Ω (really used a 33Ω)
After R1 I am getting a reading of 11.59V, and after R2 I am getting 8.26V.
I can't seem to figure out what I got wrong in my calculation.
Picture of breadboard:
Best Answer
I suggest you test your voltage divider without motor M1 and LED1 connected. You will be able to satisfy yourself that your multimeter reads correctly and that Ohm's Law is exactly true when you deal with resistors only. Now connect the motor and observe that it does not behave like a fixed resistor, as The Photon points out. A motor resembles a low value resistor when it starts but becomes like a higher value resistor as it speeds up. This is because the armature movement inside the motor internally generates a back e.m.f. that opposes the applied voltage. (You may correctly conclude from this that the voltages in the divider will not be stable or predictable when a motor is in the circuit, and that a DC motor can also work as a generator.)
And so to the LED: LEDs (and other forward-biassed diodes) do not obey Ohm's law because their current depends non-linearly on voltage. I expect 3.3V and 20mA is the operating point that the data sheet recommends. That means the LED can be driven simply by a 435 ohm resistor to a 12V supply. I don't advise connecting it into a circuit that may drive it with more current unless you are sure how much it can tolerate.