next topic »

[on:]

In a single direction configuration such as below I can see how placing the diode in parallel will cause a short with the motor.


But in an H-bridge configuration below, how does the diode configuration allow current to flow from one end to the other end of the motor? The polarity of the back emf is shown on the picture below

[Reply #1 on:]

Here is one example where D1-D4 perform the same function as the single diode in a simple switched unidirectional motor driver.

http://www.robotroom.com/HBridge.html

Lefty

[Reply #2 on:]

I still don't understand especially about the part where he talks about the motor voltage being below gnd.

[Reply #3 on:]

I still don't understand especially about the part where he talks about the motor voltage being below gnd.

 A collapsing magnetic field in coil of a motor generates a voltage of reverse polarity from that which developed the field. Saying below ground is just another way of saying a negative voltage in a positive voltage circuit.

Lefty

[Reply #4 on:]

Using your diagram of the H-bridge, you have to consider the battery as an effective short circuit (the battery internal resistance being extremely low).  The voltage generated by the motor then has a circuit route via the first series diode , through the battery and back to the motor via the second series battery.  It works irrespective of which direction the motor may be turning.

In effect this current flowing through the battery in a reverse direction actually performs a recharge of the battery and so provides regenerative braking to the motor.  However as the motor slows, its voltage output falls and hence the regenerative braking effect diminishes.

jack

[Reply #5 on:]

I still cannot visualize this, can you someone draw a current path in diagram above showing this? I would appreciate it.

[Reply #6 on:]

--
The Rugged Motor Driver: two H-bridges, more power than an L298, fully protected

[Reply #7 on:]

Ah, I see it now. Thank you for your help.

[Reply #8 on:]

I may have a stupid question, but I'm wondering : could all voltage sources sink a current like this ?

[Reply #9 on:]

No,  Some can and some can't.

[Reply #10 on:]

Ok. I would like to make a H-bridge and a logic circuit to control a stepper motor. I would like to supply the H-Bridge with a linear regulator and I wonder if this chip could sink the coil switch off current as described on the last scheme ?

[Reply #11 on:]

I know I'm probably to late to the party, but I feel like chipping in since I found this site googling for h-bridges and freewheeling current paths.

I'm currently taking a course in Power Electronics and we have been studying H-bridges, or 4 Quadrant Converters as we call them.
The ones we calculate are in a somewhat different legue when it comes to power (in the vicinity of 0.5kV and ~150A) but the ideas are the same I would presume.

Anyway, we do not connect the switches in pairs but instead have 4 control pins, one for each transistor, FET, IGBT. That way you could go from TR1,TR4 conducting to (turning off TR4) TR1,D3 conducting and have a path for the current in the top of the bridge. Then turn on T4 and in the next cycle turn off T1 to have a path for the current in the bottom of the bridge, TR4,D2 conducting.

That way you share the load between the two freewheeling diodes D3 and D3 and you don't have to rely on your voltage source to be able to handle the freewheeling current.

[Reply #12 on:]

Ok. I would like to make a H-bridge and a logic circuit to control a stepper motor. I would like to supply the H-Bridge with a linear regulator...

That's a very odd thing to do, given that linear regulators are very inefficient, and motors are not very fussy about the exact voltage they get. Why do you want to use a linear regulator?

... and I wonder if this chip could sink the coil switch off current as described on the last scheme ?

Definitely not. However, if you put a large enough capacitor across the regulator output, then the capacitor will absorb the motor switch off current.