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I've briefly looked at including an H-Bridge circuit in one of my designs and I see that most (if not all) have N-Type transistors (either Darlington or MosFets). This however needs two separate control signals to turn one half of the circuit on while the other is off.

Would the following circuit work? The theory (in my head at least) is that when the "Motor-Direction" signal is high N-Type Transistors T2 and T3 would be enabled while P-Type transistors T4 and T1 are disabled. Alternatively When the same signal is Low the opposite would happen. T5 would simply cut the power to the entire circuit and allow the motor to slow down and stop via the "Motor-Enable" line. I get that you would not have the Motor-Stop or fast reverse capabilities, but for a simple circuit that doesn't need that, would this not work just fine? enter image description here

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No, your circuit will not work at all, so don't even try it.

Let's just for a second leave Motor-Direction unconnected (open circuit) and inspect what will happen if T5 (Motor-enable) is turned on, the voltage at the top of the H-bridge would want to be at about 11V (T5 saturated).

However, if you examine T3 and T4 they represent a current path of 4 Base-Emitter junctions with a forward voltage of about 2.4V. Therefore T3, T4 would draw current based solely on the limitation of T5 and your power supply and hold the top of the H-Bridge at about 2.4V. If your power supply can deliver enough current then T5 would be held out of saturation.

T5 will get very hot and eventually be thermally destroyed .....and if T5 short circuits then you would likely destroy either T3 and T4 next.

The Base-Emitter junctions are simply acting as diodes and it makes almost no difference to the problem if you provide a current drive or not.

If you want a simple circuit like to work, you'd have to use FETs which are voltage driven.

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  • \$\begingroup\$ Actually his circuit will work, but is badly designed just like you said in your answer, which can cause the transistors destruction if Motor-Direction becomes an open circuit. \$\endgroup\$ – Flávio Alegretti May 8 '18 at 19:40
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    \$\begingroup\$ Motor-Direction has to be current limited ....but essentially it DOES NOT matter whether it is driven or not ....the Base-Emitter junctions limit the voltage over them. \$\endgroup\$ – Jack Creasey May 8 '18 at 21:03
  • \$\begingroup\$ How would T3 and T4 be the current path? T4 is PNP and T3 in NPN with their bases tied together. PNP turns on when the Base is at a lower voltage that it's collector and the NPN turns on when it's base is at a higher voltage than its emitter, so this means they can not both be on at the same time, No? The +12V at the top was arbitrary, would it make a difference if this was say, 5V? Motor-Direction will also never be floating (let's say 5V logic levels). \$\endgroup\$ – Gineer May 9 '18 at 7:17
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    \$\begingroup\$ @Gineer Look at the emitter arrows inside T3 and T4: now replace them with diodes (they are PN junctions) and you will see your problem. This happens whenever MOTOR-DIRECTION floats. Also, if people could have designed H-bridges with only one control input, they would have! \$\endgroup\$ – JvO May 9 '18 at 9:25
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    \$\begingroup\$ No, I don't think a resistor would help; there's just a fundamental flaw with the design... Besides, you're not saving any components: a regular H-bridge uses just 4 transistors, yours needs 5. But both designs have a weak point: if you enable both inputs on a regular H-bridge you effectively have a short; in your design, if Motor-direction is left open you also have a problem. \$\endgroup\$ – JvO May 9 '18 at 13:30

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