simulate this circuit – Schematic created using CircuitLab

So I have a DC motor that should turn with light on LDR and should stop when I cover LDR or turn off the lights.

Basically the transistor should work in saturation and cutoff modes.

Transistor will operate in cutoff mode when no current goes through the base, that means I have to put a resistor with high resistance in series with my photoresistor. This high resistance plus the high resistance of LDR due to darkness will make transistor to work in cutoff mode.

Transistor will operate in saturation mode when the base voltage is at its highest that means low resistance at the base.

This is my dilemma, so if I put a high resistance in series with my LDR that will help my resistor to operate in cut off mode but will prevent my transistor to achieve its max current in saturation mode because of the voltage drop across the large resistor at the base. So what should I do?

  • \$\begingroup\$ how many milliamps are you running through your motor (excuse the pun). hFE for a 3904 is pretty poor once you get above Ic of >>10mA. You could change to a Darlington configuration to multiply the base current (hence hFE) \$\endgroup\$
    – user201365
    Oct 27, 2018 at 22:05
  • \$\begingroup\$ If you're not fixated on NPN it would be easier I think to do this as a PNP, that way depending on the LDR light ON value you can drive the base pretty hard. with no numbers there's not much more to say. Remember the BJT is a current mode device, everybody tends to fixate on terminal voltages, but that's kind of backwards (unlike a FET). \$\endgroup\$
    – user201365
    Oct 27, 2018 at 22:11
  • 1
    \$\begingroup\$ An LDR's resistance spans several orders of magnitude change, light to dark. And unless the manufacturer is scrupulous, they also vary one to another. You may want a way to manually adjust where the transition happens. You may also want some hysteresis. It would also help to know something more about the motor you plan to use and it may also help if you could specify the LDR. (Not strictly necessary, but more detail is better than less.) And is this about buying a product, or about building a designed one, or about learning to design one? \$\endgroup\$
    – jonk
    Oct 27, 2018 at 22:33
  • \$\begingroup\$ Hope you're planning on putting something that would prevent flyback. \$\endgroup\$
    – user103380
    Oct 27, 2018 at 23:06
  • \$\begingroup\$ @jonk well I am trying to make a light following robot and most of the designs online are the same as the one I have shown above. The problem is that my motor don't stop only slow down at 10 k and when I increase the resistance to get a my motor to stop I loose some current in saturation mode. \$\endgroup\$
    – Jack
    Oct 28, 2018 at 4:47

1 Answer 1


A MOSfet control element might be more appropriate than a bipolar transistor:
Since the specification of the light-dependent resistor is missing, R1 might have to be chosen differently to set the light threshold where the MOSfet turns on, and the motor starts turning.
In the simulation, a small DC motor is assumed, that requires a few hundred milliamps. Its resistance is estimated at 30 ohms. The 9V battery may be too feeble to handle this load.


simulate this circuit – Schematic created using CircuitLab Recall that light-dependent resistance has an inverse relationship to light level. Dark resistance is high, often larger than 1 MEGohm. Bright light may reduce resistance to hundreds of ohms.

In this circuit, motor current starts flowing when LDR1 is about 10K ohms. Gain is fairly high - this MOSfet almost acts as a switch - its linear region is quite small, where LDR1 ranges between 10k-to-12k. This helps to keep M1 cool. But linear region is probably large enough that a flyback diode (across the motor) isn't necessary. motor current (amps) vs LDR resistance


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