Alternating currents

A substation's transformers step alternating voltage up or down.

An alternating current 交流电 (a.c.) keeps reversing direction. Mains supply is sinusoidal a.c.: $I$ or $V$ follows a sine wave in time:

(For a purely resistive load the voltage 电压 and current 电流 are in phase, which is the case in this syllabus.)

A steady direct current compared with a sinusoidal alternating current of peak $I_0$ and period $T$

Key terms

• period 周期 $T$ — the time for one full cycle. Unit: s.
• frequency 频率 $f$ — cycles per second; $f = 1/T$. Mains is often $50\ \text{Hz}$ or $60\ \text{Hz}$.
• angular frequency 角频率 $\omega = 2\pi f = 2\pi/T$.
• peak value 峰值 $I_{0}$ or $V_{0}$ — the largest value in a cycle (also called the amplitude).
• peak-to-peak value 峰峰值 $2 I_{0}$ — from $+I_{0}$ to $-I_{0}$. Useful when reading an oscilloscope.

Reading a CRO trace

Same as for any wave (Topic 7), using a cathode-ray oscilloscope 示波器:

• horizontal divisions × time-base 时基 → period $T$, so $f = 1/T$.
• vertical divisions × $y$-gain → peak voltage $V_{0}$ (measure centre to peak, or peak-to-peak then halve).

For a resistive load $R$, the instant power 功率 is $P(t) = I(t)^{2} R$. With $I = I_{0}\sin(\omega t)$:

The power in a resistor pulses; its average is half the peak

This is always positive, with peak $I_{0}^{2} R$ and minimum zero, oscillating at twice the frequency of $I$. The mean of $\sin^{2}(\omega t)$ over a cycle is $\tfrac{1}{2}$, so the average power is

Average a.c. power in a resistor is half the peak power.

The r.m.s. current $I_{\text{r.m.s.}}$ is the steady direct current that would give the same average power in the same resistance 电阻 $R$. From $\langle P \rangle = I_{\text{r.m.s.}}^{2} R = \tfrac{1}{2} I_{0}^{2} R$:

The r.m.s. value is the steady d.c. level ($V_0/\sqrt{2}$) that delivers the same average power as the a.c.

Worked example. An a.c. supply has a peak voltage of $12\ \text{V}$. Find its r.m.s. voltage.

The $\sqrt{2}$ comes from the name root-mean-square 均方根: $I_{\text{r.m.s.}} = \sqrt{\langle I^{2} \rangle}$ and $\langle \sin^{2}\rangle = \tfrac{1}{2}$. (Only the sinusoidal case is needed.)

Why r.m.s. matters

Quoted a.c. values are r.m.s. values. "$230\ \text{V}$ mains" means $V_{\text{r.m.s.}} = 230\ \text{V}$, with peak $V_{0} = 230\sqrt{2} \approx 325\ \text{V}$. Components must be rated for the peak, not the r.m.s. Average power then takes the d.c. form:

Worked example. A heater of resistance $50\ \Omega$ is connected to the $230\ \text{V}$ r.m.s. mains. Find the r.m.s. current and the average power dissipated.

An oscilloscope shows how a voltage varies with time.

Rectification 整流 turns an alternating voltage into a one-direction (d.c.-like) voltage, using diodes 二极管 (which conduct in only one direction).

Half-wave rectification

A single diode in series with the load passes only the positive half of each cycle; in the negative half the diode is reverse-biased 反向偏置 and no current flows. This is half-wave rectification 半波整流.

Output: positive half-waves with flat zero gaps. The mean output is $V_{0}/\pi \approx 0.32 V_{0}$. Drawback: half the input is wasted and the output is very uneven.

In half-wave rectification a single diode passes only the positive half-cycles

Full-wave rectification (bridge rectifier)

A bridge rectifier 桥式整流器 uses four diodes arranged so the current through the load always flows the same way, whichever a.c. terminal is positive — full-wave rectification 全波整流. On each half-cycle a different pair of diodes conducts, but the load always sees the same direction.

A four-diode bridge sends the load current the same way whichever a.c. terminal is positive

Output: a continuous run of positive half-waves (no gaps), at twice the input frequency. The mean output is $2V_{0}/\pi \approx 0.64 V_{0}$ — double the half-wave value. It uses all the input and is smoother and easier to filter.

In full-wave rectification every half-cycle is used, giving a continuous run of positive humps

Drawing the diagrams

• half-wave: a.c. source — single diode — load $R$, in series.
• full-wave bridge: four diodes as the arms of a "diamond"; the a.c. input goes to one pair of opposite corners, the load $R$ across the other pair. The diode directions make the load terminals keep the same polarity for either input polarity.

A rectifier's output is still bumpy. To smooth it, put a capacitor 电容器 $C$ in parallel with the load $R$.

How it works

• on the rising part of each pulse, the capacitor charges up to near the peak.
• on the falling part (and any gap), the diodes are reverse-biased, so the capacitor discharges through the load, keeping current flowing. The voltage falls with time constant 时间常数 $RC$ (Topic 19).
• at the next peak, the capacitor charges again, and the cycle repeats.

The output now sits near the peak with small dips. The size of the dips is the ripple 纹波 (this whole step is called smoothing 平滑).

A capacitor across the load smooths the rectified output, leaving only a small ripple

What reduces the ripple

• larger $C$ → more stored charge → smaller dip between peaks → smaller ripple.
• larger $R$ → smaller load current → slower discharge → smaller ripple.
• higher rectified frequency (full-wave is twice the input) → less time to discharge between peaks → smaller ripple.

In short, a large $RC$ compared with the time between peaks gives a smoother output.

Purpose in summary

The smoothing capacitor reduces the ripple, giving a steadier d.c. voltage suitable for sensitive electronics.

• $I_{rms} = I_0/\sqrt{2}$ for a sinusoidal current; use r.m.s. values for power ($P = I_{rms}^2 R$).
• Mean power in a resistor $= \frac{1}{2}$ of the peak power.
• Explain rectification (diode or bridge) and how a smoothing capacitor reduces the ripple.