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I am able to understand the use of flyback diodes when a basic low side unidirectional motor switching is done with an N channel MOSFET for example. I also understand the working of H-Bridge configuration. However, I could not understand how those diodes are acting as flyback diodes in this configuration and also why we don't need them while using MOSFETs but do need while using a darlington pair for example.

enter image description here

enter image description here

Here, it is obvious that for Q1 & Q4 are ON(closed) situation, the current will flow in this path:

+12V - Q1 - M - Q4 - GND

And then, when we open(OFF) Q1 & Q4, the path for current flow has been corrupted. However, there is still magnetic energy stored in the inductor of the motor, but, I don't understand with which path it will be discharging and how this will protect the transistors.

Thanks in advance.

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2 Answers 2

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However, I could not understand how those diodes are acting as flyback diodes in this configuration and also why we don't need them while using MOSFETs but do need while using a darlington pair for example.

Most MOSFETs, by their configuration of bulk-connection, have an in-built diode that does the job of fly-back voltage suppression: -

enter image description here

So, you "get them for free" with nearly all MOSFETs.

As for BJTs (Darlington or otherwise) more care is needed and regular flyback diodes need to be employed.

I could not understand how those diodes are acting as flyback diodes

Consider the picture below. SW1 and SW4 are activated causing current to flow through the load (a motor in this example). If SW4 was momentarily turned off, in order for the back-emf from the motor's inductance to discharge its energy safely there has to be a return path for the current - that flows via D3 to the positive rail and back through SW1.

enter image description here

If both SW1 and SW4 opened at the same time then the current returns via D3, then through the power rails to GND then via D2: -

enter image description here

Thank you @transistor for the images. If you need any guitar or piano fills, just let me know!

there is still magnetic energy stored in the inductor of the motor, but, I don't understand with which path it will be discharging and how this will protect the transistors.

The transistors are protected from inductive stored energy because no-matter what transistor opens, the inductive fly-back voltage that generates current, will flow safely back to the other terminal of the motor load without generating a voltage that can cause harm.

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  • \$\begingroup\$ Very nice answer, thanks. So, the key point I was missing is the possible path inside the power source(battery) itself. Two questions now, first, while this current flows inside the battery, is it recharging the battery? If so, does it have to be a rechargable power source? For example cannot I use a DC power supply? Second, so, from your answer, can I conclude that not using any external flyback diodes with MOSFETs will not create a problem for most situations at least? \$\endgroup\$
    – muyustan
    May 9, 2020 at 11:36
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    \$\begingroup\$ Yes, it flows into the battery. It does recharge it but, because this is fairly uncontrolled, good designs opt to have a fairly large capacitor across the rails and the impedance of the capacitor is usually much lower to the rapid current flow and that current bypasses the battery in the main. As for not using diodes when using MOSFETs, you need to check that the body diode spec (always in the data sheet) is sufficient for the maximum load current. \$\endgroup\$
    – Andy aka
    May 9, 2020 at 11:41
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    \$\begingroup\$ @muyustan It does return energy to the bus. Bus filter capacitors may be designed to handle it and/or the supply may support regeneration. For a motor the net energy transfer is into the motor so storage needs be only short term while powering the motor. Under braking (eg electric vehicle down a hill) regernerated energy may be significant and must be dealt with by design in some manner - either return to the or a battery or dissipation. Look at the disks in a formula 1 car under braking - white hot - the same electrical energy handling situation applies. \$\endgroup\$
    – Russell McMahon
    May 9, 2020 at 11:41
  • \$\begingroup\$ @Andyaka ok, on the images you provided(images of user Transistor), the voltage labelings on the two sides of the motor are clear for the first picture. However, for the second one, is that correct? By looking the voltage labels, both D2 and D3 looks like they should not be conducting to me. \$\endgroup\$
    – muyustan
    May 9, 2020 at 12:07
  • \$\begingroup\$ @muyustan when both SW1 and SW4 open simultaneously, the current flow through the motor in the direction shown has to keep flowing in the same direction in order to release the inductive stored energy. That's how inductors work. So, the only viable path is via D3, the supply and D2 (as I said in my answer). Were you having problems with the current direction when the switches opened possibly thinking that it reversed and wanted to flow through D1 and D4? \$\endgroup\$
    – Andy aka
    May 9, 2020 at 12:13
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Complementing Andy's answer:

Under "flyback" the energy is returned to the bus.
Bus filter capacitors may be designed to handle it and/or the supply may support regeneration.

For a motor the net energy transfer is into the motor so storage needs be only short term while powering the motor.

Under braking (eg electric vehicle down a hill) regenerated energy may be significant and must be dealt with by design in some manner - either return to the or a battery or dissipation. Look at the disks in a formula 1 car under braking - white hot - the same electrical energy handling situation applies.

In electric vehicles the power able to be generated by electric braking may equal or exceed the peak power levels when the motor is driven. Typical LiIon batteries have limitations on charging rate which may make transfer of power to the battery challenging under heavy regeneration. Some electric vehicles us an LTO (Lithium Titanium Oxide) "front end battery that is able to handle very high charging rates in conjunction with the main LiIon battery to store the majority of the energy.

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  • \$\begingroup\$ thanks, about those bus filter capacitors, are they connected between GND and Vcc? \$\endgroup\$
    – muyustan
    May 9, 2020 at 12:08
  • \$\begingroup\$ @muyustan Yes. | Ground and V+ - whatever the source rails are that deliver the energy to the motor via the "switches" \$\endgroup\$
    – Russell McMahon
    May 9, 2020 at 12:32
  • \$\begingroup\$ ok, but, won't the capacitor be at the same potential with Vcc? How current will choose to go through capacitor instead of the battery? \$\endgroup\$
    – muyustan
    May 9, 2020 at 12:35
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    \$\begingroup\$ @muyustan What happens must be designed. Regenerated energy will be delivered to the "bus" - here +12V and ground rails. If no cap or battery present the voltage will rise until "it goes somewhere". May be turned into magic smoke. | If the supply will not accept energy - eg psu diodes block energy return, then the bus capacitors will be "pumped up". How much is up top the designer. If psu CAN accept energy and there also are caps the voltage will rise somewhat and what goes where is up to the designer. | eg if supply is a 4S LiIon battery at say 14V (part charge) and able to accept 10A ... \$\endgroup\$
    – Russell McMahon
    May 9, 2020 at 14:29
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    \$\begingroup\$ ... then if motor returns 1A the battery will probably accept it with little V rise. If the motor returns 100A then Vbus may rise somewhat as battery is pushed past its limits. || Again - its up to the designer what happens - there is no single simple "this always happens". \$\endgroup\$
    – Russell McMahon
    May 9, 2020 at 14:31

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