# IB DP Physics: HL复习笔记11.2.9 Capacitors & Diode Bridges

### Capacitors & Diode Bridges

• In rectification, to produce a steady direct current  or voltage from an alternating current  or voltage, a smoothing capacitor is necessary
• Smoothing is defined as:

The reduction in the variation of the output voltage or current

• This works in the following ways:
• A single capacitor with capacitance C is connected in parallel with a load resistor of resistance R
• The capacitor charges up from the input voltage and maintains the voltage at a high level
• It discharges gradually through the resistor when the rectified voltage drops but the voltage then rises again and the capacitor charges up again A smoothing capacitor connected in parallel with the load resistor. The capacitor charges as the output voltage increases and discharges as it decreases

• The resulting graph of a smoothed output voltage Vout and output current against time is a ‘ripple’ shape A smooth, rectified current graph creates a ‘rippling’ shape against time

• The amount of smoothing is controlled by the capacitance C of the capacitor and the resistance R of the load resistor
• The smaller the rippling effect, the smoother the rectified current and voltage output
• The slower the capacitor discharges, the more the smoothing that occurs ie. smaller ripples
• This can be achieved by using:
• A capacitor with greater capacitance C
• A resistor with larger resistance R
• Recall that the product RC is the time constant τ of a resistor
• This means that the time constant of the capacitor must be greater than the time interval between the adjacent peaks of the output signal

#### Worked Example

The graph below shows the output voltage from a half-wave rectifier. The load resistor has a resistance of 2.6 kΩ. A student wishes to smooth the output voltage by placing a capacitor in parallel across the load resistor Suggest if a capacitor of 60 pF or 800 µF would be suitable for this task

Step 1:

Calculate the time constant with the 60 pF capacitor Step 2:

Compare time constant of 60 pF capacitor with interval between adjacent peaks of the output signal

• The time interval between adjacent peaks is 80 ms
• The time constant of 156 ns is too small and the 60 pF capacitor will discharge far too quickly
• There would be no smoothing of the output voltages
• Therefore, the 60 pF capacitor is not suitable

Step 3:

Calculate the time constant with the 800 µF capacitor Step 4:

Compare time constant of 800 pF capacitor with interval between adjacent peaks of the output signal

• The time constant of 2.08 s is much larger than 80 ms
• The capacitor will not discharge completely between the positive cycles of the half-wave rectified signal
• Therefore, the 800 µF capacitor would be suitable for the smoothing task 