Supplies: ripple and the zener window
Introduced| English | Meaning |
|---|---|
| half-wave rectifier | A rectifier producing output from one polarity or half-cycle of an AC input. |
| bridge rectifier/brɪdʒ ˈrektɪfaɪə/ | Four diodes arranged so that both halves of an AC wave come out with the same polarity. |
| peak inverse voltage | The maximum reverse voltage a rectifier diode must survive without breaking down. |
| filter capacitor | A capacitor after a rectifier that smooths the output towards a steady DC voltage. |
| ripple/ˈrɪpl/ | A periodic variation remaining in a nominally steady electrical voltage or current, such as after rectification. |
| zener diode | A diode intended for controlled reverse-breakdown operation, often used with current limiting as a voltage reference or regulator. |
From the wall to 5 V
- The wall socket gives 230 V AC; a circuit board wants a steady DC voltage. Every mains-powered gadget contains the chain that gets from one to the other.
- Transformer, rectifier, smoothing capacitor, regulator: this lesson designs the last three with numbers you can check.
The half-wave rectifier
- One diode in series with the load: the half-wave rectifier 半波整流器 passes the positive half-cycles and blocks the negative ones.
- The output is DC in direction but pulses on and off, with a gap of a whole cycle between peaks.
A half-wave rectifier's output never reverses direction.
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The diode blocks the negative half-cycles, so the load current only ever flows one way. It is DC in direction, though far from steady.
The bridge uses both halves
- The bridge rectifier 桥式整流 uses four diodes. On each half-cycle two of them conduct, steering current through the load the same way.
- Two diodes in the path cost $2 \times 0.7 = 1.4\,\text{V}$ of the peak. Each off diode must withstand a peak inverse voltage 反向峰值电压 of about one secondary peak.
Each bridge diode must survive a peak inverse voltage of about
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The pair that is off sees one full secondary peak, $12\sqrt{2} = 17$ V. Rate the diodes above that.
A bridge rectifies a 9 V RMS secondary. What is the peak output, in volts (0.7 V per diode)?
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$9\sqrt{2} = 12.73$ V, less two diode drops: $12.73 - 1.4 = 11.3$ V.
The capacitor fills the gaps
- A filter capacitor 滤波电容 across the load charges at each peak and supplies the load until the next one. The output sags a little: the ripple 纹波.
- With the load drawing $I$, ripple is $\Delta V \approx \dfrac{I}{fC}$ for half-wave and $\dfrac{I}{2fC}$ for a bridge, which recharges twice per cycle.
Worked example: a bridge supply
- A 12 V RMS, 50 Hz secondary: the peak output is $12\sqrt{2} - 1.4 = 17.0 - 1.4 = 15.6\,\text{V}$.
- At 100 mA with 2200 µF: $\Delta V = \dfrac{0.1}{100 \times 2200 \times 10^{-6}} = 0.455\,\text{V}$, so the output swings between 15.6 and 15.1 V. Half-wave would give 0.91 V.
A bridge on a 12 V RMS, 50 Hz secondary with 2200 µF feeding 100 mA has ripple
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$\Delta V = \dfrac{I}{2fC} = \dfrac{0.1}{100 \times 2200 \times 10^{-6}} = 0.455$ V. Half-wave would double it to 0.91 V.
A bridge at 50 Hz feeds 0.5 A. What capacitance keeps the ripple to 1 V, in µF?
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$C = \dfrac{I}{2f\,\Delta V} = \dfrac{0.5}{100 \times 1} = 5000\,\mu\text{F}$. Choose the next standard size up.
The zener regulator
- A zener diode 齐纳二极管 is built to conduct in reverse at a fixed breakdown voltage. In parallel with the load, it holds the output at that voltage.
- A series resistor drops the surplus. The zener takes whatever current the load does not, but it needs a minimum current to stay in breakdown.
12 V to 5.1 V, load 0 to 10 mA, zener minimum 2 mA. At full load the resistor must pass 12 mA: $R \le \dfrac{12 - 5.1}{12\,\text{mA}} = 575\,\Omega$, so choose 560 Ω. At no load all $\dfrac{6.9}{560} = 12.3\,\text{mA}$ goes through the zener: $5.1 \times 12.3\,\text{mA} = 63\,\text{mW}$ to rate it for.
One resistor, two opposite design limits
$I_Z=\frac{V_{in}-V_Z}{R}-I_L$
For the 12 V to 5.1 V example, 560 Ω leaves 2.32 mA for the zener at a 10 mA load and about 63 mW at no load. With an illustrative 100 mW power rating, compare the lower power-bound resistance with the 575 Ω upper current bound. A smaller resistor can exceed the power rating; a larger one can lose regulation.
A 12 V to 5.1 V zener regulator, load 0–10 mA, minimum zener current 2 mA, allows R up to
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At full load the resistor must carry 10 + 2 = 12 mA: $R \le \dfrac{6.9}{12\,\text{mA}} = 575\,\Omega$. Then check no load: 12.3 mA in the zener, 63 mW.
Three common slips
- $f$ for $2f$. A bridge recharges every half-cycle, so its ripple uses $2f$.
- Forgetting the 1.4 V. Two diodes conduct in a bridge, so the peak output is 1.4 V below the secondary peak.
- Designing a zener at one load. Check the minimum zener current at full load and its power at no load.
Put the stages of a mains DC supply in order, from the socket.
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Step down, rectify, smooth, then regulate. Each stage fixes what the one before leaves behind.
Recap
- A bridge rectifies both half-cycles at a cost of 1.4 V; each diode sees one peak in reverse.
- Ripple $\approx \dfrac{I}{2fC}$ for a bridge: the capacitor bridges the gap between recharges.
- A zener regulator is designed at both ends of the load range.