Electronic – Energy conservation when adding braking resistor to H bridge


This may be a stupid question but I'm missing something fundamental about energy conservation of electrical machines – I am sure I have forgotten something from basic machine theory but I haven't found out what thus far:

Let L, R, E be the equivalent model of a DC brushed motor winding (inductance, resistor, back-EMF).

Say we are braking the motor either by short-circuiting the windings, or applying the reverse voltage, let I be the current value. The mechanical power is -EI right? Energy conservation dictates this power is dissipated somewhere, and here it's RI^2.

Now, I close a switch and insert in series with the winding an additional braking resistor (for example when freewheeling the current goes back to the supply and into a braking resistor when the DC link regulator switch is closed). The braking power is still the same, -EI, since the current has not yet changed thanks to the inductor. However, now both the windings and the braking resistor dissipate more power than before the switch closed: (R+R2)I^2.

What am I missing? Is the inductor providing the power for the additional resistor – meaning the energy dissipated in the winding cannot be reduced for a given mechanical energy to be dissipated?

This is ultimately what I am trying to achieve, my winding cannot handle the mechanical energy I have to dissipate from the rotor when braking.

Best Answer

This is for inductor stored energy

So, if you discharge a charged inductor onto a 1 Ω resistor, the instantaneous current will be the same as you connecting the inductor to a 1000 Ω resistor but, the whole show will be over much more quickly. In other words, with a 1 Ω resistor, it might take 1 second for the current to reach a level of one-tenth whereas for the 1000 Ω resistor it will take much, much less time and...

...the total energy taken by both resistors over a long time period will be the same. They will both get warmer by the same amount (assuming perfect heat insulation).

The energy that either resistor receives is \$\dfrac{1}{2}\cdot LI^2\$ where \$I\$ is the instantaneous current at the point when the resistor is connected to the inductor and, that current is defined only by the inductor (as we know).

For energy due to mechanical momentum

This is different and there is no correlation as implied here: -

The braking power is still the same, -EI, since the current has not yet changed thanks to the inductor.

So, don't mix up the two.