Alternating current circuits: rectification, smoothing and RLC

What the exam asks

• Cambridge: half-wave and full-wave rectification waveforms; how one diode, or a bridge of four diodes, rectifies; how a smoothing capacitor works, including the effect of $C$ and of the load resistance; “mean power = half the peak power”.
• Eduqas/WJEC option: the phase of the current relative to the pd for R, L and C; $X_L = \omega L$ and $X_C = \dfrac{1}{\omega C}$; phasor diagrams; impedance and phase angle; resonance and the Q factor; power; a resonance experiment.

Core ideas

Rectification (Cambridge)

A diode conducts only when it is forward biased.

In a bridge rectifier, whichever supply terminal is positive, the current passes through the load in the same direction. Real silicon diodes drop about 0.7 V each, so a bridge loses about 1.4 V from the peak. Cambridge questions usually say “ideal diodes” so you can ignore this drop.

Smoothing with a capacitor

A capacitor connected in parallel with the load:

1. Charges through the conducting diode(s) while the rectified pd rises, up to (almost) the peak value $V_0$.
2. When the rectified pd falls below the capacitor’s pd, the diodes become reverse biased. The capacitor then discharges through the load, keeping a current in it.
3. The pd across the load falls exponentially, $V = V_0e^{-t/RC}$, until the next pulse rises above it and recharges the capacitor.

The small rise and fall that remains is the ripple. For good smoothing, the time constant $RC$ must be much longer than the time between pulses (20 ms for half-wave, 10 ms for full-wave at 50 Hz). For small ripple,

where $t$ is the time between pulses. A larger capacitance or a larger load resistance (a smaller load current) reduces the ripple. Full-wave rectification halves it for the same $RC$, because the capacitor is topped up twice as often.

Phase in resistors, inductors and capacitors (Eduqas/WJEC)

An ideal inductor or capacitor stores energy for a quarter-cycle and returns it in the next quarter-cycle, so its mean power is zero. In any series RCL circuit, the power is dissipated only in the resistance. For each component, $V_{\text{rms}} = I_{\text{rms}}X$.

Phasors and impedance

In a series circuit the current is the same in every component, so draw it as the reference phasor. Then $V_R$ lies along the current, $V_L$ points $90^\circ$ ahead and points behind. and are in antiphase, so they partly cancel:

If $X_L > X_C$ the circuit is inductive and the current lags the supply pd. If $X_C > X_L$ it is capacitive and the current leads. Drop the missing term for an RC or RL circuit. The rms pds across the separate components can add up to than the supply pd, because they are not in phase.

Resonance and the Q factor

At the resonant frequency $X_L = X_C$, so

At resonance the reactances cancel and $Z = R$ (the minimum). The current is a maximum, $\dfrac{V}{R}$, and in phase with the supply. $V_L$ and are equal and opposite, and each can be many times the supply pd. The measures how sharp the resonance is:

Increasing $R$ lowers the peak current, broadens the curve and reduces $Q$; $f_0$ does not change. This is how a radio tuning circuit selects one station: a high-Q circuit responds strongly to only a narrow band of frequencies.

Worked examples

Exam technique

• Waveform sketches (Cambridge): keep the input period, show zero output between half-wave pulses, and show every half-cycle positive for full-wave. With smoothing, draw a steep rise to each peak, then a gentle exponential fall that meets the next pulse before its peak.
• “Explain the effect of the capacitance” needs the chain: larger $C$ → larger $RC$ → slower discharge → smaller ripple.
• Bridge diodes: trace the path from the positive supply terminal through one diode, through the load, then through the diagonally opposite diode.
• Phasor calculations: write down $X_L$ and first (method marks), then , and the pds. Always say whether the current leads or lags.

Common mistakes

Quick recap

• Half-wave: one diode, and the output is zero for half of each cycle. Full-wave bridge: four diodes, two conducting in each half-cycle, with the current through the load always in the same direction.
• A smoothing capacitor charges to the peak, then discharges through the load when the diodes are reverse biased; the ripple is about $V_0\dfrac{t}{RC}$.
• A larger $C$, a larger or full-wave rectification all reduce the ripple.