I plan on building the following circuit with TIP142 and TIP147.

enter image description here

Source: Circuits Gallery.

Can you use 12V for the Vcc terminals? Thanks.

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    \$\begingroup\$ Welcome to EE.SE. Please include enough information in the question so that we don't have to follow links to know what you're talking about. It also keeps the question useful to others when the links die. I've added it for you. \$\endgroup\$
    – Transistor
    Commented Aug 8, 2017 at 9:06
  • \$\begingroup\$ The transistors are fine with 12V what does your motor require? \$\endgroup\$ Commented Aug 8, 2017 at 9:08
  • \$\begingroup\$ You can use 12V for Vcc if your motor is a 12V motor. \$\endgroup\$
    – Steve G
    Commented Aug 8, 2017 at 9:08
  • \$\begingroup\$ Thanks for the replies. The motor uses 12V. I wanted to know if it is ok to use 12V to switch on the transistors. I am new to quite new to electronics so I apologise if the question is not correct. \$\endgroup\$
    – HP1000
    Commented Aug 8, 2017 at 9:45
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    \$\begingroup\$ Uh, DON'T BUILD THIS! ... as drawn, both bases are tied together via 200 ohms, turning ALL transistors on HARD, shorting out the power supply and destroying the transistors. \$\endgroup\$
    – user16324
    Commented Aug 8, 2017 at 12:07

2 Answers 2


To use this circuit with a 12V supply R5 & R6 must be rated for at least 0.25W power and the 100 ohm resistors must be rated for at least 2W power. Depending on the transistors used these 4 resistors (100 ohm) may need recalculation to avoid excessive base current and power dissipation.

However this circuit will not work. One problem is that it produces large cross-conduction currents while switching from one state to the other. When you press SW1, the voltage drops and Q3 starts to turn on a long time before Q1 is completely turned off. In this moment, which may be from microseconds to a couple of milliseconds depending on the transistors used, both transistors are turned on and the current flows from Vcc to GND. To partialy avoid this, you may add a small electrolytic capacitor in parallel with each 100 ohm resistor.

The other, much bigger problem is the resistor values. With 100ohm base resistors connected this way and a 1kOhm pullup resistor from SW1 to Vcc when the switch is open both transistors will be turned on for a steady state. The circuit will short its power supply and if it can deliver enough power (like a battery can) may even produce a fire. For the circuit to work properly with this configuration, the pullup resistors should be at most 5-10 ohm, which will lead to very poor efficiency and a lot of power dissipation.

DO NOT USE THIS CIRCUIT! They claim that it is tested, but it can't be true.

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    \$\begingroup\$ They claim it's tested ... they don't claim it passed! \$\endgroup\$
    – user16324
    Commented Aug 8, 2017 at 12:08

That circuit is a disaster. Run away!

Analyze what happens with Vcc = 12 V and both switches open. Let's say the B-E junctions of the transistors drop 700 mV. We only need to analyze what happens on one side. The other side will do the same by symmetry.

Since the emitter of Q1 is grounded, the base of Q1 will be at 700 mV. The left side of R2 (use component designators next time!) is driven by 1 kΩ to 12 V, and 100 Ω to 11.3 V. Find the voltage at the bottom of R6. That will then allow you to find the currents thru each of the resistors.

Basically, the problem is:

Applying basic voltage divider math, the answer is 6.48 V. The current thru R1 is therefore 52.2 mA, and thru R2 57.8 mA. Note that these are also the base currents of Q3 and Q1, respectively.

Now think about what over 50 mA of base drive for both the top and bottom transistors does. Both are basically on simultaneously. That means there will be large current directly thru Q3 and Q1, effectively shorting the supply.

Again, this circuit is a mess. No, don't apply 12 V to it.

  • \$\begingroup\$ Thanks, however the voltage source and the motor are fixed at 12 V. So can you please suggest the necessary changes to this circuit or a different idea altogether. \$\endgroup\$
    – HP1000
    Commented Aug 8, 2017 at 13:32
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    \$\begingroup\$ @HPE: This is a case where you need to multiply by zero and add the appropriate quantity. Surely there are plenty of real H-bridge circuits out there. By the time you "fix" this one, you will have designed a real one from scratch. \$\endgroup\$ Commented Aug 8, 2017 at 13:34

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