I'm supposed to design a circuit for driving a DC motor with an opto-isolator and a microcontroller pin.

The motor is supplied with 12V, and its power is 5W. The opto-isolator has to isolate the motor and the microcontroller. The working current of the opto-isolator diode is 6mA, which produces a current of 2.2mA in the collector of the opto-isolator transistor.

So, the motor needs a current of $$ I_{M}=\frac{5W}{12V}.$$ The collector doesn't supply nearly enough current for the motor. So, I was thinking about adding another transistor to the existing one, making a Darlington pair. But, I have no idea how I should calculate it because I don't really have a base current to work with. I also don't know much about Darlington pairs or opto-isolators and I don't know how I can saturate the Darlington pair in this circuit.

Keep in mind that I'm not using any specific components. I'm only supposed to come up with an idea of how to accomplish the given task and calculate the values of the components I'm using. Here's a picture of what I've had in mind:


simulate this circuit – Schematic created using CircuitLab

Lets say the voltage drop across the opto-diode is 2V. R1 can be calculated as $$ R_1=\frac{5V-3V}{6mA}=330 \Omega .$$ I could use some help with the rest of the circuit from you guys (if the idea of this circuit is even correct).

  • \$\begingroup\$ Can you link a datasheet for the optocoupler? The CTR (Current Transfer Ratio) of the opto is needed to solve this. \$\endgroup\$ Commented Nov 7, 2016 at 23:31
  • \$\begingroup\$ This is just a concept I'm supposed to come up with, with no extra info besides the one that is provided. If there's no way around it, you can just take any average CTR. \$\endgroup\$
    – HarryHey
    Commented Nov 8, 2016 at 8:37

2 Answers 2


A darlington configuration - putting the optocoupler output Q1 in series with the load - will increase the voltage drop across Q2. Also a power BJT transistor might have a gain of 30-50 tops let's say. So with that circuit your maximum load current is probably no more than 60-100mA (30-50 * 2.2mA), which is pretty low unless you have a very small motor.

You have a load current of 416mA (5W @12V) so you need a current gain of 189. You are not going to achieve this driving a BJT power transistor directly (as shown in your diagram). You would require another transistor stage - which if using a darlington style configuration will result in a large voltage drop and loss.

Consider using a MOSFET to drive the load? You would connect the collector of Q1 to 12V via a small resistor and connect the emitter to the MOSFET gate and to ground via a larger resistor used to deactivate the MOSFET. If you select a MOSFET with sufficiently low Rdson it will have lower voltage drop than a BJT transistor. There is also no limit on current gain with MOSFETs, however...

2mA drive is not a lot, if your load is small and you are using a MOSFET with small gate capacitance then this could be ok. But you might find that the gate capacitance of the MOSFET is too high for the 2mA drive, resulting in very slow switching speed and high switching losses. Directly driving a suitable MOSFET with only 2mA gate drive is almost definitely no good for PWM control - you might get away with it for on/off control.

I would suggest either using your Q1/Q2 darlington configuration to drive a MOSFET. Or to look for a pre-packaged solution, such as an optoisolator with built in gate-drive, or a motor driver IC that can boost the signal for you.

Other things to look out for: Make sure the voltage rating of the parts is suitable. Can the optoisolator output handle 12V across it? The MOSFET gate might have a limit of 10V for example (although 20V is more typical), so you need to make sure you use a potential divider to keep the gate drive below that.

  • \$\begingroup\$ "A darlington configuration - putting the optocoupler output Q1 in series with the load - will increase the voltage drop across Q2. " Yes, maybe but that's negligible. \$\endgroup\$
    – arnisz
    Commented Jun 20, 2021 at 1:05
  • \$\begingroup\$ > but that's negligible -- not really. It's ~10 % loss. I would not call that negligible. \$\endgroup\$ Commented Oct 21, 2021 at 19:12
  • \$\begingroup\$ @arnisz for the OP's circuit you will have roughly 2 diode drops across the output transistor which in a good scenario might be 1.2V, but is likely higher than this as a saturated power transistor is going to need more than 0.6V on the gate. 1.2V on a 12V supply is 10% loss. As explained above, one would need an additional transistor stage to drive the required load, so in practice using a similar configuration with an extra stage would result in significant losses \$\endgroup\$ Commented Mar 14, 2022 at 16:50

Just few changes

  • Move the opto-isolator collector of the +12 V line (to reduce the voltage drop and improve efficiency)
  • Add a resistor in series with the opto-isolator emitter (to limit the current into the base of the transistor)
  • Add a resistor to ground (to speed up turn off and avoid spurious turn-on of the transistor)


simulate this circuit – Schematic created using CircuitLab


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.