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Simon Fitch
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Your circuit is able to switch load current on and off very well, since the MOSFET is behaving like an open or closed switch. If you were to plot current \$I_L\$ through the load, be it a motor, a LED or a resistor, that plot woueld bewould show a nice clean rectangular wavezero-level, alternating between 0A and some maximum.

Voltage across a motor is a different story, because it is rotating and generating an EMF \$V_L\$ roughly proportional to its angular velocity, even though current \$I_L\$ is zero. This is manifest in your plots as \$V_D\$ failing to rise to the full +12V when the MOSFET is off, and \$I_L\$ is zero.

Your circuit is able to switch load current on and off very well, since the MOSFET is behaving like an open or closed switch. If you were to plot current \$I_L\$ through the load, be it a motor, a LED or a resistor, that plot woueld be a nice clean rectangular wave, alternating between 0A and some maximum.

Voltage across a motor is a different story, because it is rotating and generating an EMF \$V_L\$ roughly proportional to its angular velocity, even though current \$I_L\$ is zero. This is manifest in your plots as \$V_D\$ failing to rise to +12V when the MOSFET is off, and \$I_L\$ is zero.

Your circuit is able to switch load current on and off very well, since the MOSFET is behaving like an open or closed switch. If you were to plot current \$I_L\$ through the load, be it a motor, a LED or a resistor, that plot would show a nice clean zero-level, alternating between 0A and some maximum.

Voltage across a motor is a different story, because it is rotating and generating an EMF \$V_L\$ roughly proportional to its angular velocity, even though current \$I_L\$ is zero. This is manifest in your plots as \$V_D\$ failing to rise to the full +12V when the MOSFET is off, and \$I_L\$ is zero.

Where is KE lost
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Simon Fitch
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In the push-pull arrangement of your function generator, when its output is \$V_D=+12V\$ (SW2 closed, SW1 open, above right) the motor is effectively short-circuited, having +12V applied somewhat brutally at both ends, for a potential difference \$V_L=0V\$. In this state the motor's kinetic energy is being drained rapidly (energy lost as heat in its own internal resistance), and the motor experiences hard braking. Therefore, with push-pull PWM, the motor alternates between phases of active acceleration and active deceleration.

In the push-pull arrangement of your function generator, when its output is \$V_D=+12V\$ (SW2 closed, SW1 open, above right) the motor is effectively short-circuited, having +12V applied somewhat brutally at both ends, for a potential difference \$V_L=0V\$. In this state the motor's kinetic energy is being drained rapidly, and the motor experiences hard braking. Therefore, with push-pull PWM, the motor alternates between phases of active acceleration and active deceleration.

In the push-pull arrangement of your function generator, when its output is \$V_D=+12V\$ (SW2 closed, SW1 open, above right) the motor is effectively short-circuited, having +12V applied somewhat brutally at both ends, for a potential difference \$V_L=0V\$. In this state the motor's kinetic energy is being drained rapidly (energy lost as heat in its own internal resistance), and the motor experiences hard braking. Therefore, with push-pull PWM, the motor alternates between phases of active acceleration and active deceleration.

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Simon Fitch
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There is another quirk of behaviour for a push-pull drive that distinguishes it from a single-MOSFET arrangement. In you own circuitthe single switch/MOSFET scenario, Whenwhen the MOSFET is off (SW1 open), the motor is effectively disconnected at one end, and is permitted to free-wheel under its own momentum.

In your own circuit, the motor is actively accelerated, but the only cause for deceleration is friction.

This has interesting consequences for motor speed control, in a closed loop. A mechanically unloaded motor will accelerate very rapidly, even at low PWM duty-cycle, and without any means to actively decelerate it (such as you have when using push-pull), it is disturbingly easy for motor speed to overshoot, and take an age to slow down to the desired set-point speed. This is an argument in favour of a push-pull driver, mostly referred to as a "half-bridge".

There is another quirk of behaviour for a push-pull drive that distinguishes it from a single-MOSFET arrangement. In you own circuit, When the MOSFET is off (SW1 open), the motor is effectively disconnected at one end, and is permitted to free-wheel under its own momentum.

In your own circuit, the motor is actively accelerated, but the only cause for deceleration is friction.

There is another quirk of behaviour for a push-pull drive that distinguishes it from a single-MOSFET arrangement. In the single switch/MOSFET scenario, when the MOSFET is off (SW1 open), the motor is effectively disconnected at one end, and is permitted to free-wheel under its own momentum.

In your own circuit, the motor is actively accelerated, but the only cause for deceleration is friction.

This has interesting consequences for motor speed control, in a closed loop. A mechanically unloaded motor will accelerate very rapidly, even at low PWM duty-cycle, and without any means to actively decelerate it (such as you have when using push-pull), it is disturbingly easy for motor speed to overshoot, and take an age to slow down to the desired set-point speed. This is an argument in favour of a push-pull driver, mostly referred to as a "half-bridge".

Active braking
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Simon Fitch
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Simon Fitch
  • 47.1k
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  • 135
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