The one volt would be the diode voltage. The principle is the same accept that now the field is proportional to motor current instead of constant and you will have an even higher inductance.

As the motor speed goes down it's voltage goes down. As the voltage goes down the current and hence the field flux goes up.

The calculation looks correct. And under those conditions at 40KHz you will see a maximum ripple current in the motor of I=TV/L or 0.45A

 

Yes it was about the instant thing and the 1V would be the diode drop.

That type of motor is a bit unpredictable since the field flux varies with the current but the principles are similar. Problem is I am not sure how the field voltage behaves. The rotor voltage would be proportional to both the speed and the field current, it looks like there is a square law in there somewhere.

 

Thanks !

The type of DC motor is almost exact to a heavy truck starter motor, running as say, 36V, 50 A, 40KHz.

I do not understand how to use the 0.45 A ripple current to find the power wasted by the bemf. ¿ Is it 0.45 ripple Amperes x 36V = 16 wasted watts, being that 0.88 % of waste. Not much that deserves recovery with a lossless snubber

Instead of using a lossless snubber, will a simple capacitor in parallel with the motor greatly smooth out/average/integrate the pulsed voltage from the switching supply and thus the off instant will greatly diminish or disappear, and there would be nearly no lossy bemf 'generation' between cycles ?

Miguel

__________________
Abolish the deciBel !

 

BEMF does not waste energy, nor does the induction that keeps the current flowing. To improve efficiency reduce the drop across the FET and diode.

If you use a

IPB100N06S3L-03

for both the FET and the diode you can limit the dissipation to about 7W each (synchronous rectification). With a permanent magnet motor you could to regenerative braking, but I do not know how that motor responds when it freewheels.