PWM power stages act similar to power transformers. Thus it is wise to measure the motor current at the controller output and not at the power supply.
Basically a PWM power stage is a power converter that transforms the input power of the supply Pin = Vcc * Icc into an output power that is applied to the motor Pout = Umot * Imot.
Since no energy is stored, we have (except for the losses): Pin = Pout or Icc = (Umot/Vcc) * Imot . The motor voltage depends on the required speed and the load and is controlled by the PWM duty cylce D=(ton /(ton + toff).
The most important result is that motor currents should be measured at the output of the power stage. Do not use the supply current for calculating the heating or torque of the motor.
Slightly simplified, a PWM power stage is essentially a step-down DC-DC converter or Buck converter.
The following correspondances apply:
- the input voltage Vi -> the constant supply voltage Vcc of the controller
- the output voltage Vo -> the voltage applied to the motor Umot
- the switch S -> the MOSFETs of the power stage
- the diode D -> the diodes of the power stage
- the inductance L -> the total inductance (in power stage, motor choke and motor)
- the capacitor C -> no direct correspondance (but you can think of EMI filter and CLL)
- the load (not shown) -> the motor (with resistance and induced voltage)
"The conceptual model of the buck converter is best understood in terms of an inductor's "reluctance" to allow a change in current.
Switch S will be closed (MOSFETs conductive):
Beginning with the switch open (in the "off" position), the current in the circuit is 0. When the switch is first closed, the current will begin to increase, but the inductor doesn't want it to change from 0, so it will attempt to fight the increase by dropping a voltage. This voltage drop counteracts the voltage of the source and therefore reduces the net voltage across the load. Over time, the inductor will allow the current to increase slowly by decreasing the voltage it drops and therefore increasing the net voltage seen by the load. During this time, the inductor is storing energy in the form of a magnetic field.
Switch S will be opened (MOSFETs not conductive):
If the switch is opened before the inductor has fully charged (i.e., before it has allowed all of the current to pass through by reducing its own voltage drop to 0), then there will always be a voltage drop across it, so the net voltage seen by the load will always be less than the input voltage source.
When the switch is opened again, the voltage source will be removed from the circuit, so the current will try to drop. Again, the inductor will try to fight against it changing, which it does by reversing the direction of its voltage and acting like a voltage source. Put another way, there is a certain current flowing through the load due to the input voltage source: in order to maintain this current when the input source is removed, the inductor will have to take the place of the voltage source and provide the same net voltage to the load. Over time, the inductor will allow the current to decrease gradually, which it does by decreasing the voltage across itself. During this time, the inductor is discharging its stored energy into the rest of the circuit.
If the switch is closed again before the inductor fully discharges, the load will always see a non-zero voltage. The capacitor placed in parallel with the load helps to smooth out voltage waveform as the inductor charges and discharges in each cycle." (from Wikipedia - Buck Converter , Interactive Diagram)