Lesson 3
Adding a Capacitor: Some Smoothing, but Not Enough
A capacitor resists sudden voltage changes, smoothing the output through charge/discharge cycles. But a capacitor alone is not enough.
Fundamental property of capacitors: voltage across them cannot change instantaneously (V = Q/C, charge cannot change instantly). This can 'oppose' the square-wave transitions from the switch.
Place a capacitor in parallel with the load: when the switch is ON, the source charges the capacitor and powers the load simultaneously; when OFF, the capacitor discharges to keep powering the load.
As the switch toggles rapidly, the capacitor continuously charges and discharges. The output fluctuates within a range instead of jumping between 0 and 12V. Smoothing depends on the RC time constant vs. switching frequency.
But there's a problem: when the switch closes, the source charges the capacitor directly with very large current spikes (limited only by switch Ron). Also, achieving low ripple requires impractically large capacitors.
The capacitor helps, but alone it's not elegant enough. We need another energy storage element — the inductor.
Switch + Capacitor + Resistor: Charge/Discharge Smoothing
Added C=100μF. Output is somewhat smoother, but ripple is still significant.
After reading this section, run the simulation and observe the waveforms. To explore further, open the example in a standalone page.
Key Takeaways
- Capacitors resist sudden voltage changes, smoothing output via charge/discharge
- Larger RC time constant relative to switching period means better smoothing
- Downside: large current spikes when switch closes, causing EMI issues
- A capacitor alone cannot elegantly achieve low-ripple output
Watch Items
- Compare with previous lesson: output is no longer a square wave but a sawtooth charge/discharge waveform
- Observe the output ripple — better than a square wave, but far from smooth DC
- Think: what happens if you increase capacitance? (Try changing the C parameter)