# 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.

• "What am I missing?" A schematic maybe. Nov 19, 2020 at 18:00
• You are confusing back EMF of a motor with the dI/dt behaviour of an inductor. Switch in the resistor and current I will change; giving a very short V pulse from the inductor superimpoed on back EMF E. So, having changed R, I will change, and energy conservation will hold.
– user16324
Nov 19, 2020 at 18:01
• The inductor also has an E. Nov 19, 2020 at 18:05
• I am not confusing the back EMF with the di/dt, however both of you might be on to something: is the inductor providing the power for the additional resistor? Does this mean the energy dissipated in the winding cannot be reduced for a given mechanical energy to be dissipated? Nov 19, 2020 at 18:10
• @MisterMystère The inductor produces the amount of E which is needed to make the current follow the decay curve. Nov 19, 2020 at 18:26

## 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.

Lets say the motor provides 64W steady state and we use a 4Ω resistor to dissipate the energy:

The voltage across the resistor will be P=I^2R so sqrt(64W/4Ω)=4A and the voltage would be sqrt(PR)=V with sqrt(64W*4Ω)= 16V

16V times 4A across the resistor is 64W dissipated across the resistor

Now we put another 4Ω resistor in parallel for a total of 2Ω:

The voltage across the resistor will be P=I^2R so sqrt(64W/2Ω)=5.6565A and the voltage would be sqrt(PR)=V with sqrt(64W*2Ω)= 11.3137V

11.3137V times 5.6565A across the resistor is 64W dissipated across the resistor

Now we put another 4Ω resistor in series for a total of 8Ω:

The voltage across the resistor will be P=I^2R so sqrt(64W/8Ω)=2.82843A and the voltage would be sqrt(PR)=V with sqrt(64W*8Ω)= 22.6274V

22.6274V times 2.82843A across the resistor is 64W dissipated across the resistor

So the voltage and current change with the load but the power dissipated does not, this is assuming an ideal motor that is being driven with 64W of mechanical input. A real world motor would be slightly different because of losses and resistance of windings ect.

This is a little different from most circuits because the motor is like a constant power source and we are used to dealing with constant voltage or constant current sources.

Consider a brushed DC motor that self-commutates.

It accelerates with more current and voltage and visa versa. Yet when cutting power, it becomes a voltage source proportional to RPM with losses from winding resistance.

Yet compared to an inductor the time constant of Tau=L/R is very short compared to the time to accelerate or brake the much higher energy of rotational inertia. Therefore L is insignificant here and only V/(R+R2)=I controls the braking deceleration. Motors have an open loop V/RPM usually using kV/RPM so the stored inertial energy with 1/2 mv^2 for some mass and equivalent linear velocity (simplified analogy with a linear translator from RPM to v).

Thus the rotational energy is peaked with current initially then deceleration reduces with RPM or v and Voltage. And for a a fixed R+R2 current reduces the current and thus is slower to stop. Right? because you know opening the circuit only coasts to a stop by friction.

The surge current for starting is typically 10x the rated current because of V/R where R=DCR winding resistance. The same is true in the opposite case with full RPM no load is shorted via R then the same surge current with the opposite polarity.

In a regenerative motor / generator situation, you cannot restore all the energy needed full speed, only the rotational energy stored after it reaches any speed so the energy lost is gained by the work done. You can only save the energy you have when you decide to brake , which may be anything from 0 to full load at one instant but that is irrelevant and only the stored mechanical energy can be reclaimed as heat thru the winding heat dissipation or spread out to the external resistor R2, where R2 is now the ESR of your battery and the voltage difference is what counts. Since that difference is much smaller,the reclaimed energy must be converted down to a lower voltage then up-converted to the battery high voltage in order to improve energy recovery.

*Complicated eh? Not so much with some experience. It gets easier if your battery is already much higher in voltage then you down-convert with PWM, to run and rectify to brake, now using the coil inductance to integrate the digital clocking of power switches.( with x us of dead band to avoid shoothrough)

You don't need to think about the inductor for this, necessarily. The spinning motor always produces a back EMF. It is a DC source whose voltage is proportional to the speed of the motor. The windings have resistance. That is series resistance in the model. So whatever you connect you can analyze it that way most likely. If you short the windings, there is nowhere for the power to be dissipated except the windings. The power will be Vemf^2 / ESR.

The current is proportional to the torque, by the way.

If you apply a voltage to the motor smaller than the back EMF, current will flow from the motor into the smaller voltage. In other words, the motor will be in regeneration mode. The amount of current is based on the size of the series resistance and the voltage difference. It is not necessary that the applied voltage be opposite in direction to the back emf. It can be same polarity but smaller magnitude.

If you apply a resistor across the windings while the motor is spinning (and disconnect the driving voltage), power will be dissipated by the resistor. Current will flow backwards (regeneration) from the motor into the resistor. The amount of power dissipated in the resistor vs the windings is a function of the relative values of the resistor and the winding.

A larger resistor will produce less torque, but will also spare the windings. The faster you stop the motor, the more energy you dissipate in the windings.