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As you know, H-bridge circuits are used in order to control DC motors. Most primitive motor controller ever

Switches S1-S4 are closed to turn the motor in one direction, S2-S3 are closed for another. But at the time of changing motor direction, when switches (transistors) are opened, Back EMF kicks in and makes the switches blow up. That's the reason of using diodes parallel to switches (See figure below). enter image description here

On the other hand, these diodes consume significant amount of power. I am currently trying to solve this problem, and make a "diodeless" H-bridge circuit work.

I am using an mCU to control the motor with PWM signals. Is it possible to make a flawless timing for switches to not blow up?

For example, on the schematic above, assume that the situation is as this: S1 and S4 are closed and motor turns in that direction. When I want to change the direction, I open S1-S4; and close S2-S3. By this, I have burnt the switches several times. Now, my idea here is to open S1, and close S3 for a while. By this, I think I'll be able to let the remaining current be absorbed (on the circuit S3-Motor-S4). Then, I open S4 and close S2, which reverses the direction of the motor. As I said above, I am not sure about that if it is possible with an mCU. Also, I would need to analyze the characteristics of the motor and switches in order to find the perfect timing for the algortihm.

What do you think about this idea? Is there any method that you know or can suggest to overcome this power consumption issue? I have searched online but couldn't find anything.

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    \$\begingroup\$ You are trying to reinvent the wheel, but without deep knowledge. Just forget about it. \$\endgroup\$ Commented Jan 13, 2016 at 10:56
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    \$\begingroup\$ How do you conclude that the diodes consume a significant amount of power? That appears to be your motivation but I believe you to be fundamentally misled. \$\endgroup\$
    – Andy aka
    Commented Jan 13, 2016 at 11:10
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    \$\begingroup\$ In fact, when the diodes conduct, they're actually putting energy back onto the power rail, reducing net consumption by a tiny bit. Also, be careful of your terminology: "back EMF" and "inductive kick" are two very different phenomena. Back EMF is produced by the motion of the armature. By definition, the voltage generated by it cannot exceed the supply voltage. Inductive kick is produced by rapidly reducing the current through the motor's residual inductance, and can reach very high levels with fast switching. It is the latter that the diodes are protecting the switches from. \$\endgroup\$
    – Dave Tweed
    Commented Jan 13, 2016 at 12:45

5 Answers 5

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Well, what you're asking for has already been done for many years in a completely different field. We call "Diodeless H-bridges" class AB push pull amplifiers. You'll likely want some pretty fast transistors that have a low saturation voltage. Bonus, the less switching you do the more efficient this method is!

Schematic below:

schematic

simulate this circuit – Schematic created using CircuitLab

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What are you talking about with diodes in parallel with the switches? That is never done.

Anyway, what you are doing is called "Plug Reversing", and yes, you need heavy duty contactors to handle that. The typical and inexpensive solution is to put NC contacts with a low ohm resistor around the armature itself. When all of your 'switches' are open, the armature is shorted through the resistor, which acts to dynamically brake the motor to a stop. Then use a time delay between enabling the forward/reverse switching.

For the record, inexpensive (probably less then the cost of your switches) DC Drives can handle this easily. You would need a regenerative speed control.

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  • \$\begingroup\$ Check the edit (second figure). \$\endgroup\$
    – ddyn
    Commented Jan 13, 2016 at 11:26
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When I want to change the direction, I open S1-S4; and close S2-S3. By this, I have burnt the switches several times. Now, my idea here is to open S1, and close S3 for a while. By this, I think I'll be able to let the remaining current be absorbed (on the circuit S3-Motor-S4). Then, I open S4 and close S2, which reverses the direction of the motor.

  1. Probably the transistors burn because they can't turn off/on so quickly. When you switch from S1-S4 to S2-S3, in a certain period of time you have S1,S2,S3,S4 that are half conductiong each, the current passes trough pair S1/S2, S3/S4 directly from Vcc to GND and burns them, this is normal if you don't have a short dead time.
  2. So first close S1/S4 then wait certain ammount of time then open S2/S3. While in this dead time the diodes are conductiong, without diodes the back EMF will rise until the transitors or motor winding breaks down and game over.
  3. With your proposal to have S3/S4 turned on is nothing else than creating a short circuit between Vcc and GND, so byebye transitors.
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The diodes are required to protect the switching devices. The power "waste" is low and unavoidable.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1 and 2. Half-H-bridge (for clarity) with catch diodes.

In the Figure 1 we can see the switches are closed and current flows through the motor.

In Figure 2 we see that the switches have opened but due to the inductance of the motor current is still flowing. The current will flow no matter what else happens.

With the catch diodes in circuit current can be provided from the negative rail and fed out back onto the positive rail to be reabsorbed by the power supply. There is no waste other than the unavoidable VI loss caused by the diode forward voltage drop.

schematic

simulate this circuit

Figure 3. Circulating current through Q3 and Q4.

Your idea to circulate current through Q3 and Q4 won't work. One of them will be reverse biased - Q3 in this example.

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  • \$\begingroup\$ Regarding your last point: There is such a thing as "reverse-active mode" in a BJT, in which the roles of the emitter and collector are reversed. The transistor will conduct, but it won't have as much gain as it does in the normal "forward-active mode". That's why most medium-power drives use MOSFETs -- they conduct equally well in both directions when on. \$\endgroup\$
    – Dave Tweed
    Commented Jan 13, 2016 at 14:13
  • \$\begingroup\$ @DaveTweed: Thanks for the info. I've made AC switches out of back-to-back MOSFETs and they worked well. With a MOSFET bridge would it be normal practice to create low-side and high-side loops as the OP suggested? If so, I presume the diodes would still be required to cover the period between, say, S1 opening and S3/Q3 closing. \$\endgroup\$
    – Transistor
    Commented Jan 13, 2016 at 15:41
  • \$\begingroup\$ Yes, you would create the loops if you want to use the back EMF of the motor as a brake. And yes, I would keep the diodes, because in my experience, electronic commutation is never "perfect", and better safe than sorry. \$\endgroup\$
    – Dave Tweed
    Commented Jan 14, 2016 at 0:29
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The diodes are required.

Even wimpy old 1N4148s can handle more current (200mA) than those transistors (100mA).

Maybe you need a more capable H-bridge- that one is suitable only for the tiniest of motors- remember that motor current under startup conditions will be much higher than run current, and if you 'plug' the motor (reverse while it is spinning) double that again.

If your diodes are getting hot then your motor is too big for that design of H-bridge.

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