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.

enter image description here

below are my calculations.


bleed = (20mA + 30mA) * .1 = 5mA
R3 = 3.3/0.005 = 660Ω (really used a 680Ω)


R2 = (10 - 3.3) / 25mA = 268Ω (really used a 27Ω)


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: enter image description here

  • \$\begingroup\$ How did you determine that the motor current would be 30 mA? The motor current will probably depend strongly on how the motor is loaded (mechanically). \$\endgroup\$ – The Photon Apr 8 '15 at 21:33
  • \$\begingroup\$ Also, a picture of how you constructed the circuit might help us see a problem. \$\endgroup\$ – The Photon Apr 8 '15 at 21:34
  • \$\begingroup\$ Right now I am just testing the voltage using a multimeter. \$\endgroup\$ – Vitaly Babiy Apr 8 '15 at 21:39
  • \$\begingroup\$ In your test, there's no motor or LED present. So the current is ~12 mA through all three resistors. Then you expect 11.56 V at your upper test point and 8.2 V at the lower test point, which is pretty close to what you got. \$\endgroup\$ – The Photon Apr 8 '15 at 23:01

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.


Add all those resistors up it's 964 Ohms. 12V / 964 Ohms is 12.4mA which is the maximum current that will flow for that circuit. The voltage drop across R1 is 12.4mA * 36Ohms which is 0.448V. 12-0.448 is 11.55V, so it appears to be working exactly as you have it setup :) Ohms law always wins.


With the loads (LED and Motor) disconnected, the values that I calculate match the values that you have measured:

\$V_{out} = V_{in} [R_{part}/R_{total}]\$,

where \$R_{part}\$ is the amount of resistance to ground, and \$R_{total}\$ is the total resistance.

In the case of your circuit, \$R_{total} = 36+268+660 = 964\Omega\$.

Your motor node voltage is:

\$V_{in}[(R_2+R_3)/R_{total}] = 12[(268+660)/964] = 11.55V\$

Your LED node voltage is:

\$V_{in}[R_3/R_{total}] = 12[660/964] = 8.21V\$

Just as you measured.

However, there are further complications. Once you attach the motor and/or LED to the circuit, they each will have some resistance, which will throw off all the earlier calcs.

For a circuit like yours, it is common to use voltage regulators to avoid this confusion.


The only factor is the difference in the load resistance.Should the load resistance is in fact similar to the calculation of resistance.


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