## Summary

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.

## Power conversion

Basically, a PWM power stage is a power converter that transforms the input power of the supply *P _{in} = V_{cc} * I_{cc}* into an output power that is applied to the motor

*P*.

_{out}= U_{mot}* I_{mot}Since no energy is stored, we have (except for the losses): *P _{in} = P_{out}* or

*I*. The motor voltage depends on the required speed and the load and is controlled by the PWM duty cylce

_{cc}= (U_{mot}/V_{cc}) * I_{mot}*D=(t*.

_{on}/(t_{on}+ t_{off})The most important result is that **motor currents should be measured at the output of the controller's power stage**. Do not use the supply current for calculating the heating or torque of the motor.

## Deduction

Slightly simplified, a PWM power stage is essentially a step-down DC-DC converter or Buck converter.

The following correspondences apply:

- the input voltage
*V*-> the constant supply voltage_{i}*V*of the controller_{cc} - the output voltage
*V*-> the voltage applied to the motor_{o}*U*_{mot} - 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 correspondence (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. Beginning with the switch open (in the "off" position), the current in the circuit is 0.

### Switch *S* will be closed (MOSFETs conductive):

When the switch is closed (on-state), the current will begin to increase, and the inductor will produce an opposing voltage across its terminals in response to the changing current. This voltage drop counteracts the voltage of the source and therefore reduces the net voltage across the load. Over time, the rate of change of current decreases, and the voltage across the inductor also then decreases, increasing the voltage at the load. During this time, the inductor stores 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 (off-state), the voltage source will be removed from the circuit, and the current will decrease. The decreasing current will produce a voltage drop across the inductor (opposite to the drop at on-state), and now the inductor becomes a current source. The stored energy in the inductor's magnetic field supports the current flow through the load. This current, flowing while the input voltage source is disconnected, (..) totals to current greater than the average input current (being zero during off-state). The "increase" in average current makes up for the reduction in voltage, and ideally preserves the power provided to the load. During the off-state, the inductor is discharging its stored energy into the rest of the circuit.

If the switch is closed again before the inductor fully discharges (on-state), the voltage at the load will always be greater than zero." (from Wikipedia - Buck Converter , Interactive Diagram)

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