Rectification and smoothing
| English | Chinese | Pinyin |
|---|---|---|
| diodes | 二极管 | èr jí guǎn |
| capacitor | 电容器 | diàn róng qì |
| rectification | 整流 | zhěng liú |
| smoothing | 平滑 | píng huá |
| ripple | 纹波 | wén bō |
| forward-biased | 正向偏置 | zhèng xiàng piān zhì |
| reverse-biased | 反向偏置 | fǎn xiàng piān zhì |
| bridge rectifier | 桥式整流器 | qiáo shì zhěng liú qì |
Inside a phone charger
- A phone charger takes mains alternating voltage that reverses fifty times a second, and hands the phone a steady $5\ \text{V}$.
- Between those two things sit exactly the components of this lesson: diodes 二极管 to stop the reversing, and a capacitor 电容器 to iron out what is left.
- Rectification 整流 is the conversion of an alternating current into a direct (one-direction) current. That is the marked definition.
- Smoothing 平滑 is then using a capacitor across the load to reduce the ripple of a rectified output.
What a diode does
- A diode conducts only when it is forward-biased 正向偏置: its anode, the flat end of the symbol's triangle, is more positive than its cathode, the bar.
- Otherwise it is reverse-biased 反向偏置 and behaves like an open switch.
- Treat it as ideal: zero resistance one way, infinite the other.
- Conventional current always flows through a diode in the direction the triangle points. Every "complete the circuit" question is that one rule, applied to each diode in turn.
A diode conducts current:
This one-way behaviour is what lets diodes rectify a.c. into d.c.
Match each term to the definition the examiner marks.
Rectified current still varies; what makes it direct is that it never reverses.
Half-wave rectification
- A single diode in series with the load passes only the positive half of each cycle. In the negative half it is reverse-biased and no current flows at all.
- The output is positive half-waves separated by flat zero gaps of equal length. Mean output $= V_0/\pi \approx 0.32\,V_0$.
- Half the input is thrown away, and the output is very uneven.
Humps, then nothing, then humps
Half-wave rectification uses ____ diode.
A single diode passes the positive half-cycles and blocks the negative ones (leaving gaps).
Drawing the half-wave circuit
- Source, one diode in series, load, and $V_{\text{OUT}}$ taken across the load. The triangle points the way the output current must flow.
- If a smoothing capacitor is wanted it goes in parallel with the load, never in series. A capacitor in series would block the d.c. altogether, which is the opposite of the point.
One diode in the line, one capacitor across the load
Where does a smoothing capacitor go?
It must discharge THROUGH the load between peaks, which needs it in parallel. In series it would block the d.c. entirely, which is the opposite of the point.
Full-wave rectification
- A bridge rectifier 桥式整流器 uses four diodes so that the load current runs the same way whichever input terminal is positive.
- The output is a continuous run of positive humps with no gaps, at twice the input frequency. Mean output $= 2V_0/\pi \approx 0.64\,V_0$, double the half-wave value.
- All of the input is used, and what remains is far easier to smooth.
No gaps, and twice as often
Rectifier and smoothing route
Watch alternating input become a smoother direct output.
How many diodes are in a bridge (full-wave) rectifier?
Four diodes, arranged so the load current always flows the same way whichever input terminal is positive.
A full-wave rectified output repeats at twice the frequency of the a.c. input.
Both half-cycles become positive humps, so the output ripples at double the input frequency.
Worked example: explaining the bridge
- Explain how the four diodes produce full-wave rectification. This is the standard four-marker, and it is marked on naming the diodes.
- When terminal P is positive: current leaves P, passes through the diode pointing away from P to the top of the load, flows down through the load, and returns to Q through the diode pointing towards Q. The other two are reverse-biased and carry nothing.
- When Q is positive: the other pair conducts, but they are arranged so the current still enters the load at the top.
- So the current in the load is in the same direction in both half-cycles. Say which pair conducts in each half-cycle and which end of the load is positive.
Two diodes at a time, one direction through the load
Put the explanation of one half-cycle of a bridge rectifier in order.
The four-mark answer is marked on naming the conducting pair in each half-cycle, not on saying the current "goes the same way".
Completing a bridge
- Given a bridge with diodes missing, use one rule: both diodes joined to the positive output terminal point towards it, and both joined to the negative output terminal point away from it.
- A single diode the wrong way round either short-circuits the supply on one half-cycle or passes nothing at all.
- Check your answer by tracing one half-cycle right round the loop before you move on.
In a bridge rectifier, both diodes joined to the positive output terminal point towards it.
And both joined to the negative terminal point away from it. One diode reversed either short-circuits the supply on a half-cycle or passes nothing.
Smoothing with a capacitor
- Put a capacitor $C$ in parallel with the load $R$. On the rising part of each pulse it charges up to near the peak.
- On the falling part, and through any gap, the diodes are reverse-biased, so the capacitor discharges through the load and keeps the current flowing. That decay has time constant $RC$.
- At the next peak it charges again. The output now sits near the peak with small dips, and the size of the dips is the ripple 纹波.
Never down to zero, never above the peak
To smooth a rectified output, you connect a capacitor:
In parallel, it charges to the peak and then discharges through the load between peaks, holding the voltage up.
Worked example: how big is the ripple
- A half-wave rectifier fed from a $50\ \text{Hz}$ supply has a $470\ \mu\text{F}$ capacitor across a $1.2\ \text{k}\Omega$ load. Estimate the fractional fall between peaks, and say what a bridge would change.
- Half-wave gives one peak per cycle, so the peaks are $T = 1/50 = 20\ \text{ms}$ apart.
- Time constant: $RC = (1.2\times10^3)(470\times10^{-6}) = 0.56\ \text{s}$.
- Discharge: $V = V_0 e^{-t/RC} = V_0 e^{-0.020/0.56} = 0.965\,V_0$, a fall of about $3.5\%$.
- With a bridge the peaks are only $10\ \text{ms}$ apart, so the fall halves to about $1.8\%$. Because $t \ll RC$, the ripple is roughly $V_0 t/(RC)$: proportional to the time between peaks, and inversely proportional to both $R$ and $C$.
A full-wave rectifier from a 50 Hz supply feeds a 1.2 kilo-ohm load with a 470 microfarad capacitor. How many milliseconds pass between output peaks?
Full-wave output runs at twice the input frequency, so 100 peaks per second and 10 ms between them. Half-wave would give 20 ms and about twice the ripple.
The mean output of a half-wave rectifier is $V_0$ divided by ____.
Half-wave gives 0.32 V0; a bridge uses both halves and doubles it to 2V0/pi = 0.64 V0. The bridge also doubles the ripple frequency, which is why the same capacitor smooths it better.
What reduces the ripple, and how to sketch it
- Larger $C$: more stored charge, so a smaller dip between peaks.
- Larger $R$: a smaller load current, so a slower discharge. Note this is the load resistance, so a heavier load (smaller $R$) makes the ripple worse.
- Full-wave instead of half-wave: less time to discharge between peaks.
- To sketch it: draw the humps faintly, then a curve that touches each peak, sags gently, and turns sharply up where the next hump meets it. It never reaches zero and never rises above the peak.
Bigger $RC$ hugs the peak line
Select all the changes that reduce the ripple.
A bigger $RC$ (or less time between peaks) makes the capacitor discharge less between peaks — smaller ripple. A smaller capacitor does the opposite.
Marks that slip away
- The smoothing capacitor goes across the load, in parallel. In series it blocks the d.c. entirely.
- In the bridge explanation, name which pair of diodes conducts in each half-cycle. "The current goes the same way" alone is one mark of four.
- Full-wave output is at twice the input frequency, not the same frequency.
- A smaller load resistance means a larger ripple, because the capacitor discharges faster.
- Half-wave has flat zero gaps; full-wave has none. Draw the gaps the same width as the humps.
You've got it
- rectification converts alternating current into current in one direction; a diode conducts only when forward-biased, in the direction its triangle points
- half-wave uses one diode and passes half of each cycle, mean $V_0/\pi$; a bridge of four diodes passes both halves, mean $2V_0/\pi$, at twice the frequency
- a capacitor in parallel with the load smooths by discharging through the load between peaks, with time constant $RC$
- the ripple shrinks with larger $C$, larger load resistance, and full-wave rather than half-wave rectification