# Regenerative Braking circuit

I am reading this great website.

I don't understand the topic #4. I don't see how the current will flow back to the battery, and how the Q2 enabling changes the circuit. As far as I can see, the current will use the diodes to flow, so why turn q2 on? Why just don't use Q1 and D2?

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The last diagram in section 6.0 explains it a bit better. In fact, take the two diagrams where current is flowing through Q2 and Q3 in that section and delete Q1 and Q4. You now have the half-bridge of topic 4. – pjc50 Nov 14 '12 at 16:18

Summary: If you use a position encoder wheel you can manage the commutation of generation and regeneration according to the difference between supply and load for each phase of the motor. The H bridge allows you control polarity of drive voltage to or from the battery depending on which is higher and demand of acceleration or braking.

Two methods are show below.

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The motor (as a generator) can only deliver power to the battery if its terminal voltage exceeds the battery voltage.

The page you reference skips over a lot of detail, but in section 4, the author is actually using Q2, D2 and La (the inductance of the armature) as a boost (switching) converter to raise the voltage sufficiently to charge the battery.

Note that the situation in section 6 is different. Here, he seems to be ignoring the armature inductance and simply applying reverse voltage across the motor to slow it down more quickly. This does not result in any regeneration, since no energy is actually returned to the battery. It is possible to do regeneration with a full bridge, but it's more complex than what is shown there.

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I would think regeneration with a full bridge would be pretty simple if the switching rate is fast enough relative to motor inductance. If the motor is shorted half the time and tied across the supply half the time, it will want to turn at half its full-voltage no-load speed; if it's actually turning faster than that, it will power the supply. – supercat Jun 16 '14 at 23:32