I'm using the following circuit: Applied to the motor is 12V, and either high (5V) or low (0V) to each transistor through a 1kΩ resistor. The transistors are TIP 142 and TIP 147. The diodes are 1N5817.

When I apply 0V to R2, R1, and 5V to R3, the motor runs fine; voltage across it is 8-10V. However, whenever I connect 5V OR 0V to R4, the voltage across the motor is 1-2V, and the transistor at R3, starts getting really hot and smoking.

Why does applying anything to R4 cause this?


  • 1
    \$\begingroup\$ What do you suppose happens when Q4 and Q3 are both on at the same time? \$\endgroup\$
    – user253751
    Aug 2, 2018 at 22:57

3 Answers 3


Q1 and Q3 turn ON when you connect +5V to their resistors. Q2 and Q4 turn OFF when you connect +12V to their resistors.

Connecting 5V to R2 or R4 will result in the transistor turning on.

So, if you connect 5V to R3 and less than 12V (or whatever the power supply voltage is) to R4, both transistors turn on creating a short circuit.

If you want to turn off Q2 or Q4, connect their resistors to +12V.

If you want to control the transistors with 5V (a microcontroller or whatever) then you can add two more ttransistors that control Q2 and Q4, like this: modified circuit diagram

Connecting 5V to R6 will turn Q4 on, connecting 5V to R5 will turn Q2 on.

  • \$\begingroup\$ I'd like to add that (if I understood correctly..) you also shouldn't forget to connect the ground also to the ground of your microcontroller. It's not part of the circuit drawing, so it's easy to forget. \$\endgroup\$
    – Niels
    Oct 2, 2015 at 14:03

Q4 is turned on by the current from its emitter to base. When you have the emitter at +10V and the base at +5 volts, current will flow, about (10 - 5 + 1.4)/1k or 3.6 mA. This is enough to turn on darlington Q4.

  • \$\begingroup\$ I don't understand, why does that mean applying 0 or 5V to R4 will cause the motor to have a lower voltage across it? \$\endgroup\$ Feb 4, 2012 at 4:53
  • 2
    \$\begingroup\$ @AmandeepGrewal What Markrages means is that for the low side (Q1 & Q3) you can use 0v and 5v to turn them on and off. But the high side (Q2 & Q4) needs +12v to turn OFF and something less than about 10v to turn ON. Since both 0V and 5V are less than 10v either voltage will result in the same thing: Q4 turning on. With Q4 and Q3 both on, the result is almost a dead short and lots of current flowing. That current is what causes the heat and smoke. \$\endgroup\$
    – user3624
    Feb 4, 2012 at 5:41
  • \$\begingroup\$ Thanks for that! I now understand the heat issue, and why Q4 turns on with both voltages. \$\endgroup\$ Feb 4, 2012 at 6:04

I see that your question is old, has already been properly answered and that you have understood why your transistor was burning up. But I felt like complementing the answer anyway.

While this doesn't answer your question, I wanted to point out that there is this BJT H-Bridge design that prevents such situations from occurring.

It gives you control over the motor to move it forward, in reverse, braking and coasting. It also lets you modulate the signal to control speed to a certain frequency. The page says it can handle currents near what you are expecting (5A), with proper heatsinks. But, in my opinion, the nicest feature about that bridge is that it doesn't allow dangerous input combinations that short battery contacts.

The design cleverly achieves that by wiring the control inputs in a way that the diagonal base transistors can never be turned on simultaneously.

See more details in this answer of mine.


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