Microsoft



Pulse Width Modulation Power AmplifiersA linear power amplifier has an output voltage proportional to its input voltage, usually with a gain much greater than one. What if we have a linear power amplifier being run off a 12 volt power supply, and it needs to output 6 volts into a motor? The voltage drop across the amplifier is (12V – 6V) = 6V. What if the motor resistance is 2 ohms? The current into the motor will be 6V/2ohms = 3 amps. The power dissipation in the amplifier is P = V*I = 6V * 3A = 18w. The power dissipation in the motor is 6V*3A = 18w. So the dissipation in the amplifier is equal to the dissipation in the load. This is extremely inefficient and wasteful, and forces us to build the amplifier with a big heat sink, so it doesn’t overheat.Figure 1. Simplified schematic of a linear power amplifier driving a 2 ohm load.What if we build an amplifier with an output stage made up of two switches: one to ground, and one to the +12V power supply?Figure 2. Simplified schematic of switching power amplifier driving a 2 ohm load.If SW2 is on and SW1 is off, Vout = 0V. Current into Rload is 0. Currents through SW1 and SW2 are 0. Voltage across SW1 is 12V. Power dissipation in SW1 is 12V*0A = 0w. Power dissipation in SW2 is 0V*0A = 0w. Now what if we close SW1 and open SW2?Figure 3. Simplified schematic of switching power amplifier driving a 2 ohm load.If SW1 is on and SW2 is off, Vout = 12V. Current into Rload is 12V/2ohms = 6A. Current through SW1 is 6A. Current through SW2 is 0. Voltage across SW1 is 0V. Power dissipation in SW1 is 0V*6A = 0w. Voltage across SW2 is 12V. Power dissipation in SW2 is 12V*0A = 0w. So in both cases there is no dissipation in SW1 and SW2. The waveform at Vout:Figure 4. Waveform at Vout with 50% duty cycle. A?duty cycle?is the percentage of one period in which a signal is active.?When Vout = 0V, dissipation in Rload = 0w. When Vout = 12V, dissipation in Rload = V*I = 12V*6A = 72w. Since Vout = 12V for 50% of the time, the average dissipation in Rload is 72w/2 = 36w. So we are getting 36w into the load with no dissipation in the amplifier. The amplifier is 100% efficient. In the real world, you can get 90 – 95%, because the switches have a little resistance.Note that the average power is not equal to the average voltage times the average current. This is because power is a nonlinear function of voltage. So the power at the 12V peak of the pulse is very high.What if we want to send less power to the motor? We can reduce the positive pulse width, which gives a lower average output. To get more power, we can increase the positive pulse width, which gives a higher average output.Figure 5. Waveforms at Vout with 25% and 75% duty cycles. How does the motor respond to this waveform? We don’t want the motor to respond to each individual pulse. We want it to respond to the average over many pulses. A motor that takes 1 second to respond to an input will filter out pulses that are a lot faster than 1 second.Figure 6. Green: motor voltage step. Red: motor speed response (several tenths of a second). Figure 7. Green: motor voltage (PWM, 50Hz). Red: motor speed response. Pulses are mostly filtered out.Figure 8. Motor voltage (PWM, 500Hz): not shown. Red: motor speed response. Pulses are almost completely filtered out. So the motor is responding as if a linear amplifier were driving it.The PWM radiated noise problemPWM waveforms have very fast edges. So they create radiated noise which can interfere with sensitive systems nearby. If PWM signals cannot be tolerated on the wires going to the motor, the waveform can be filtered.Figure 9. PWM amplifier output filtered by lowpass filter L1C1 before the motor wires.Vmotor is the voltage on the motor wires. It will look like the DC average of the PWM waveform Vout, with a little ripple at the switching frequency. The inductor L1 has to be a high enough value so that ripple current through it is acceptably low. Otherwise, the PWM amplifier can be damaged, or it might blow a fuse or current-limit. A resistor can be used in place of the inductor, but then a lot of power will be lost as heat. ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download