By the end of this chapter you should be able to:
- Distinguish wave shaping from rectification, and explain what each circuit family is asked to preserve and to destroy.
- Analyse series and shunt clippers, sketch their transfer characteristics, and place the two resistances that make each one work.
- Compute the clipping level of a biased clipper including the diode drop, and find the conduction angle over which limiting occurs.
- Derive the RC condition \(\tau \gg T\) for a clamper from the exponential droop of the capacitor, and size \(C\) for a stated tilt.
- State the clamping theorem and use it to explain why the clamped level is set by the ratio of forward to reverse resistance.
- Analyse peak and peak-to-peak detectors, and quantify the droop and acquisition time of each.
- Explain the operation of half-wave and full-wave doublers, triplers and quadruplers, and predict their output droop and ripple under load.
Chapters 8 to 10 used the diode for one purpose only: to convert alternating power into direct power. Rectification is a power operation, judged by average values, ripple factors and efficiency, and the waveform that comes out is a by-product of the energy conversion rather than the point of it. This chapter uses exactly the same component for something different. Here the diode is asked to reshape a signal — to remove part of it, to shift the whole of it up or down, or to add successive peaks together — and the figure of merit is the shape of the output, not the power in it.
Three families of circuit do this, and they are close relatives. A clipper (also called a limiter) removes the part of a waveform that lies beyond some chosen level, leaving the rest untouched; it is a series-connected diode or a shunt-connected one, and it changes the extremes of the signal while leaving its DC content largely alone. A clamper (a DC restorer) does the opposite: it changes nothing about the shape and adds a DC offset, so that some chosen part of the waveform always sits at a fixed level. A voltage multiplier is a clamper and a peak detector cascaded, and it produces a DC output several times larger than the peak of the input without any transformer at all.
All three depend on the same two facts about a diode that Chapter 6 established: it conducts above about 0.7 V and blocks below it, and a capacitor charged through a conducting diode holds its charge when the diode blocks. Nearly every error made with these circuits comes from forgetting the 0.7 V or from choosing a time constant that lets the capacitor forget its charge too quickly, and both are treated quantitatively below. The circuits appear in oscilloscope input stages, in television video processing, in mains-fed protection networks and in the high-voltage supplies of photomultipliers and cathode-ray tubes.
1 Wave Shaping Is Not Rectification
It is tempting to look at a shunt clipper and a half-wave rectifier and see the same circuit, because both are a resistor and a diode fed from a source. The difference is in what each is allowed to lose. A rectifier is a power converter: the load resistance is small, the diode current is large, and the four figures of merit derived in Chapter 8 all describe how much of the source's power reaches the load as DC. A wave-shaping circuit is a signal processor: the load is usually the input of the next stage and draws almost nothing, the currents are milliamperes, and nobody computes an efficiency because the point of the circuit is the shape at its output terminals.
That difference in purpose leads to a difference in analysis. For a rectifier we integrate over a cycle and report averages. For a clipper we build a transfer characteristic — a plot of output voltage against input voltage with time eliminated — and read the output waveform off it. The transfer characteristic of any diode wave-shaping network is piecewise linear: straight segments whose slopes are set by resistor ratios and whose break points are set by battery voltages plus or minus 0.7 V. Once the segments are known, the response to any input follows, sinusoidal or not.
A useful way to see the three families is by what each does to the two descriptors of a waveform, its shape and its DC level.
| Circuit | Effect on shape | Effect on DC level | Key component |
|---|---|---|---|
| Rectifier | Destroys one half-cycle | Creates a DC component from nothing | Diode and load |
| Clipper | Truncates beyond a set level | Changes as a consequence | Series resistor |
| Clamper | Preserved exactly | Shifted by a chosen amount | Series capacitor |
| Multiplier | Irrelevant — output is DC | Set to \(n\) times the input peak | Capacitor ladder |
One further point separates the two worlds. A rectifier is fed through a transformer and the diode's forward drop is a small fraction of a large voltage, so 0.7 V is a 2 % correction on a 35 V peak. A clipper often works on signals of a few volts, where 0.7 V is not a correction but a large part of the answer: a clipper intended to limit at 3.0 V actually limits at 3.7 V, an error of 23 %. Every result in this chapter therefore carries the diode drop explicitly.
Note also what a clipper cannot do. Because it removes signal rather than adding it, a clipper is lossy and non-linear; the output of one driven hard is closer to a square wave than to the sinusoid that produced it. Driving a \(\pm 10\) V sinusoid into a symmetrical \(\pm 3.7\) V clipper gives an output of RMS value \(3.39\) V against the \(3.70\) V of a perfect square wave, so the flanks occupy little of the period. That is sometimes the object and sometimes a disaster, and the difference between those intentions decides how the clipping level is chosen.
2 Series and Shunt Clippers
There are exactly two ways to put a diode into a signal path, and they give the two clipper families.
In a series clipper the diode sits between source and load, with the load resistor \(R_L\) to ground. When it conducts, the circuit is a divider of \(r_f\) and \(R_L\) and the output follows the input almost exactly, since \(R_L/(R_L+r_f) = 0.999\) for \(R_L = 10\ \text{k}\Omega\) and \(r_f = 10\ \Omega\); when it blocks, no current flows and the output is zero. The series clipper therefore transmits one polarity and removes the other completely. Its transfer characteristic is two segments meeting at the origin, of slope \(\approx 1\) and \(\approx 0\).
In a shunt clipper the diode sits across the output, with a resistor \(R\) in series with the source. When the diode blocks, it is an open circuit and the output equals the input less the drop across \(R\), which is negligible if the load draws nothing. When it conducts, the diode holds the output at its forward drop and the surplus of the source voltage is dropped across \(R\). The shunt clipper therefore transmits everything below the conduction threshold and limits everything above it, which is what most protection circuits want. Its transfer characteristic is two segments meeting at the threshold, one of slope 1 and one of slope \(r_f/(R+r_f)\).
The two resistances in a shunt clipper are chosen by a single inequality. \(R\) must be large compared with \(r_f\) so that the clipped top is flat, and small compared with \(R_L\) so that the transmitted region is not attenuated. The standard compromise is the geometric mean:
With \(r_f = 10\ \Omega\) and \(R_L = 100\ \text{k}\Omega\) this gives \(R = 1\ \text{k}\Omega\), the value used throughout this chapter. It makes the clipped region flat to one part in a hundred and the transmitted region accurate to one part in a hundred, and no other choice does better on both at once.
The residual slope is worth a number, because students often draw the clipped top as perfectly horizontal and then cannot explain a measurement. With \(R = 1\ \text{k}\Omega\) and \(r_f = 10\ \Omega\), the slope above the break is \(10/1010 = 0.0099\), so a source that overshoots the clipping level by 6.3 V lifts the clipped top by 62 mV. That is small, but it is not zero, and in a limiter protecting a 5 V logic input it is the difference between a specification met and a specification missed.
The choice between the topologies is usually made on impedance grounds. The series clipper presents an open circuit to the source during the removed half-cycle, which is kind to the source but leaves the output floating and vulnerable to pickup unless \(R_L\) is fitted. The shunt clipper presents a low impedance to the source during limiting, so the source must supply the clipping current — here 6.3 mA at the peak. Where the source is a mains-derived transient rather than a signal generator, that current can be amperes, which is why practical protection clippers use transient-voltage-suppressor diodes rather than small-signal ones.
3 Biased and Combination Clippers
An unbiased shunt clipper limits at 0.7 V, which is rarely the level anyone wants. Putting a battery \(V_B\) in series with the diode moves the break point, because the diode cannot conduct until the output node is \(V_B + V_F\) above ground. This is the circuit of Figure 11.1, and its behaviour is summarised by one equation:
The second line is usually simplified to \(v_o \approx V_B + V_F\). Reversing the diode and the battery gives a clipper that limits negative excursions at \(-(V_B + V_F)\), and the two can be combined.
The battery and the diode drop add in a shunt clipper whose diode points towards the battery, because the diode must be forward biased by 0.7 V before the branch conducts. If the branch is reversed so that the battery drives the diode into conduction, the two subtract and the level is \(V_B - V_F = 2.3\) V. Deciding which case applies takes one sentence — which way does the diode point relative to the battery? — and skipping it produces an answer wrong by 1.4 V.
Two biased clippers in parallel across the same node form a combination or double-ended clipper, which limits both extremes. Figure 11.2 shows a \(10\) V peak sinusoid limited at \(+3.7\) V and \(-5.7\) V; the two levels are independent, and the deliberate asymmetry makes the mechanism visible.
A shunt clipper uses \(R = 1\ \text{k}\Omega\), a silicon diode with \(V_F = 0.7\) V and \(r_f = 10\ \Omega\), and a 3 V bias battery. The input is \(v_i = 10\sin\omega t\) at 1 kHz. Find the clipping level, the fraction of the period for which clipping occurs, the peak diode current and the peak diode dissipation.
Clipping level. \(V_B + V_F = 3.0 + 0.7 = 3.7\) V.
Conduction angle. The diode turns on when \(10\sin\theta = 3.7\), that is \(\theta_1 = \arcsin(0.37) = 21.72^\circ\), and turns off again at \(180 - 21.72 = 158.28^\circ\). It conducts for \(136.57^\circ\) of every \(360^\circ\), that is 37.94 % of the period, or 379 µs in each 1 ms cycle.
Peak diode current. At the crest the source is 10 V and the output is held at 3.7 V, so the whole 6.3 V surplus appears across \(R\): \(I_{D,\text{peak}} = 6.3/1000 = 6.3\) mA.
Peak dissipation. \(P_D = 3.7 \times 6.3\ \text{mA} = 23.3\) mW at the crest, and much less than that averaged over a cycle. Any small-signal diode is comfortable; a 1N4148 is rated at 500 mW.
Flatness. The clipped top is not perfectly flat: it rises by \(6.3 \times 10/1010 = 62\) mV between the moment of turn-on and the crest. Quoting the output as "3.7 V" is therefore accurate to 1.7 %.
Diodes are not the only way to set a clipping level. A Zener clips at \(V_Z\) in one direction and at \(V_F\) in the other with no battery at all, so two in series back to back limit symmetrically at \(\pm(V_Z+V_F)\) — a pair of 5.1 V devices gives \(\pm 5.8\) V from two components and no supply rail. Chapter 12 develops both the circuit and the device.
4 Clampers and the Time-Constant Condition
A clamper leaves the shape of a waveform completely alone and moves the whole of it up or down, so that one extreme sits at a chosen level. The circuit is a series capacitor followed by a shunt diode — the mirror image of the shunt clipper, in which the series element was a resistor.
Take the negative clamper of Figure 11.3, in which the diode's cathode faces the source. On the first positive half-cycle the diode conducts and the capacitor charges through it with a time constant of only \(r_f C\); by the first crest it holds \(V_m\), left plate positive. After that crest the source falls, the diode reverse biases, and the capacitor is left as a battery in series with the source. The output is therefore
which swings from \(0\) down to \(-2V_m\). Every feature of the input survives — peak-to-peak value, frequency, shape — and only the DC level has moved, by \(-V_m\). Reverse the diode and the same argument clamps the negative peak to zero, giving an output from \(0\) to \(+2V_m\); add a battery in series with the diode and the clamped level becomes \(\pm V_B\) instead of zero.
All of that depends on the capacitor holding its charge from one crest to the next, and it cannot hold it perfectly, because the load resistance \(R_L\) — and the reverse resistance of the diode — discharge it between crests. During the interval when the diode is off the capacitor voltage decays exponentially to \(V_C e^{-t/R_LC}\), so the output tilts by
Taking the worst case \(t = T/2\), the longest interval for which the diode is off in a symmetric waveform, gives the design condition. The tilt or droop is the fractional sag of what should be a flat or a smoothly curved section, and it is the single number that decides whether a clamper is good enough.
Work the numbers for a 1 kHz signal into \(R_L = 100\ \text{k}\Omega\):
| \(C\) | \(\tau = R_LC\) | \(\tau/T\) | Droop over \(T/2\) | Sag on a 10 V clamped peak |
|---|---|---|---|---|
| 0.1 µF | 10 ms | 10 | 4.88 % | 488 mV |
| 0.47 µF | 47 ms | 47 | 1.06 % | 106 mV |
| 1 µF | 100 ms | 100 | 0.499 % | 50 mV |
| 10 µF | 1 s | 1000 | 0.050 % | 5 mV |
The familiar rule of thumb \(\tau \ge 10T\) is the first row, and it costs nearly 5 % droop — acceptable for a television DC restorer, useless for an instrument. Inverting the exponential gives the design equation for a droop \(d\):
For \(d = 1\ \%\), \(T = 1\) ms and \(R_L = 100\ \text{k}\Omega\), the exact form gives \(\tau \ge 49.75\) ms and \(C \ge 0.4975\) µF; specify 0.47 µF and accept 1.06 %, or 1 µF for 0.50 %. Note the frequency dependence: the same 1 % at 50 Hz needs \(\tau \ge 0.995\) s, which into 10 k\(\Omega\) means 100 µF. A clamper designed for audio will not clamp mains-frequency waveforms.
One more result explains why a real clamper never quite reaches the ideal level. The clamping theorem states that in the steady state the charge gained by the capacitor while the diode conducts must equal the charge lost while it does not, so
where \(A_f\) is the area under the output waveform during the forward (conducting) interval, \(A_r\) the area during the reverse interval, and \(R_f\), \(R_r\) the diode's forward and reverse resistances. With \(R_f = 10\ \Omega\) and \(R_r = 100\ \text{k}\Omega\) the ratio is \(10^{-4}\), so the forward area is ten thousand times smaller than the reverse area — which is why the clamped extreme is pinned so hard to the reference level, and why a leaky diode with a small \(R_r\) lets the whole waveform drift.
5 Peak and Peak-to-Peak Detectors
Remove the load resistor from a half-wave rectifier and what remains is a peak detector: a diode charging a capacitor that has nowhere to discharge. The capacitor charges to within a diode drop of the highest voltage the input has ever reached and stays there, so the output is \(V_m - V_F\). This is the circuit inside every AM demodulator, every analogue peak-reading meter and many sample-and-hold front ends.
Two time constants describe it, and they are wildly different. Acquisition happens through the conducting diode with time constant \(r_f C\); at \(r_f = 20\ \Omega\) and \(C = 1\) µF that is 20 µs, so the capacitor is within 1 % of the peak after 100 µs. Droop happens through whatever is connected to the output, with time constant \(R_LC\) — 1 s for \(R_L = 1\ \text{M}\Omega\). Their ratio, here 50 000, is the figure of merit of a peak detector, and it is why the following stage must have a very high input resistance.
The droop between one peak and the next follows the same exponential as the clamper, this time over a full period because the diode conducts only briefly:
At 1 kHz with \(R_L = 1\ \text{M}\Omega\) and \(C = 1\) µF the droop is 0.100 % per cycle, or 10.0 mV on a 10 V peak — the residual ripple on the detected output. Reduce \(C\) to 0.1 µF, or \(R_L\) to 100 k\(\Omega\), and the droop rises tenfold to 0.995 % and 99.5 mV.
There is a conflict built into that equation. A large \(R_LC\) gives low droop but makes the detector slow to follow a falling envelope, and if the amplitude drops faster than \(R_LC\) allows, the output is wrong — the diagonal clipping familiar from AM detectors. The condition for an AM detector to follow a modulation of depth \(m\) at frequency \(f_m\) is \(R_LC \le \sqrt{1-m^2}/(2\pi f_m m)\), and it bounds \(C\) from above as firmly as the ripple requirement bounds it from below.
Cascade a clamper and a peak detector and something more useful appears. The clamper shifts the waveform so that its negative extreme sits at zero; the peak detector then reads the positive extreme, which is now the full peak-to-peak value. This is the peak-to-peak detector, and its output is
The two diode drops are the only loss, and for a 10 V peak sinusoid the output is \(20 - 1.4 = 18.6\) V. This circuit measures the peak-to-peak value of an arbitrary waveform without knowing anything about its shape, which is exactly what a rectifier-based AC voltmeter cannot do.
Count the components of the peak-to-peak detector: a series capacitor, a diode to ground, a second diode in series, a second capacitor. That is precisely the half-wave voltage doubler of the next section, drawn differently. The only distinction is intent — one is asked to measure a signal, the other to deliver power — and intent decides the component values, not the topology.
Two practical points finish the circuit off. The diode's reverse leakage discharges \(C\) just as the load does, so a detector meant to hold for seconds needs a low-leakage diode. And the source must supply the whole charging current in the short interval when the diode conducts: for \(C = 1\) µF on a 10 V 1 kHz sinusoid that current is of the order of \(C\omega V_m = 62.8\) mA, which many signal sources cannot deliver.
6 Voltage Doublers, Triplers and Quadruplers
A transformer can produce any voltage you like, but a transformer for 10 kV is large, expensive and hazardous to wind. Where the current required is small — a photomultiplier bias supply, a cathode-ray tube final anode, an ion pump, an electrostatic air cleaner — it is cheaper to rectify a modest voltage and then multiply it with capacitors and diodes. Every voltage multiplier is a chain of the clamper-plus-peak-detector pair from the last section.
Trace Figure 11.4 with a 24 V RMS secondary, \(V_m = 33.94\) V. On the negative half-cycle \(D_1\) conducts and charges \(C_1\) to \(V_m\); \(D_2\) is reverse biased and nothing reaches the output. On the following positive half-cycle \(D_1\) blocks, the source rises to \(+V_m\), and the charged \(C_1\) adds its \(V_m\) in series, so the anode of \(D_2\) reaches \(2V_m = 67.88\) V and \(C_2\) charges to that value. Subtracting the two forward drops gives \(2(V_m - V_F) = 66.5\) V. Both capacitors must be rated for \(2V_m\), and both diodes block a peak inverse voltage of \(2V_m = 67.9\) V — not the output voltage, which is the fact that makes long ladders practical.
The full-wave doubler rearranges the same four components so that each of two capacitors is charged to \(V_m\) on opposite half-cycles, with the output taken across the two in series. It gives the same \(2V_m\) but its ripple appears at twice the line frequency, and because the two capacitors discharge in antiphase the ripple magnitude is halved as well. Neither terminal of a full-wave doubler can be common with the source, however, so the half-wave version survives wherever a shared ground matters.
Extending the ladder gives the tripler (\(3V_m = 101.8\) V from the same secondary), the quadrupler (\(4V_m = 135.8\) V) and the general Cockcroft–Walton multiplier: each added diode-and-capacitor pair lifts the output by another \(V_m\), ideally. Ideally is doing a lot of work there, because the output impedance rises very steeply with the number of stages. For \(N\) stages, each of capacitance \(C\), driven at frequency \(f\) and delivering load current \(I\), the droop below the ideal \(2NV_m\) is
The \(N^3\) term dominates beyond two or three stages, so doubling the number of stages costs eight times the droop. This is why high-voltage multipliers are driven at tens of kilohertz: \(\Delta V\) falls in proportion to \(f\), so 50 Hz to 20 kHz buys a factor of 400 and lets the capacitors shrink by the same factor.
A multiplier is fed from a 24 V RMS 50 Hz secondary (\(V_m = 33.94\) V) and every capacitor is 22 µF. The load draws 1 mA. Compare the doubler (\(N=1\)) with the quadrupler (\(N=2\)).
The scaling voltage is \(I/(fC) = 0.001/(50 \times 22\times10^{-6}) = 0.9091\) V.
Doubler. \(N = 1\), so the bracket is \(2/3 + 1/2 - 1/6 = 1.000\) and \(\Delta V = 0.909\) V. The ideal output is \(2V_m = 67.88\) V, so the loaded output is 66.97 V, a droop of 1.34 %.
Quadrupler. \(N = 2\), so the bracket is \(16/3 + 2 - 1/3 = 7.000\) and \(\Delta V = 6.364\) V. The ideal output is \(4V_m = 135.76\) V, so the loaded output is 129.40 V, a droop of 4.69 %. Seven times the droop for twice the output.
Source impedance. Dividing droop by load current, the doubler behaves as an ideal 67.88 V source behind 909 \(\Omega\) and the quadrupler as 135.76 V behind 6.36 k\(\Omega\); a six-stage unit at the same \(f\) and \(C\) has a bracket of 22.0, an output impedance of 20 k\(\Omega\) and nearly 10 % droop. Specify the output at the actual load, never at zero load, and if the droop is unacceptable raise \(f\) before raising \(C\) — the capacitors are the expensive, bulky, high-voltage-rated part.
Two further constraints govern a real multiplier. The string as a whole stands off the full output to ground even though no single component does, so the physical layout and the insulation must be designed for the output voltage. And the surge current at switch-on, when every capacitor is discharged, is limited only by the source impedance and the diode resistance, so a series current-limiting resistor is not optional. Chapter 12 turns to the opposite problem — not making a large voltage from a small one, but holding a voltage steady while everything around it moves.
7 Choosing and Building the Circuit
The three families answer three different questions, and the first step in any design is to decide which is being asked. "The signal must never exceed 5 V at this pin" is a clipper. "The sync tips of this video signal must always sit at the same level whatever the picture content" is a clamper. "I need 2 kV at 50 µA from a 12 V supply" is a multiplier driven from a switching oscillator. Once the family is chosen, the values follow from three inequalities that have appeared separately above.
- Clipper: \(r_f \ll R \ll R_L\), best satisfied by \(R = \sqrt{r_fR_L}\). The clipping level is \(V_B \pm V_F\), with the sign decided by which way the diode faces relative to the battery.
- Clamper: \(r_fC \ll T \ll R_LC\). The left-hand inequality lets the capacitor charge within one conducting interval; the right-hand one keeps the droop small. A ratio of \(R_L/r_f\) of \(10^4\) leaves plenty of room for both.
- Multiplier: \(I/(fC)\) must be small compared with \(V_m\), and the number of stages must be kept low because the droop grows as \(N^3\).
Three practical matters are easy to overlook on paper and impossible to overlook on the bench. The first is source impedance. Every circuit here assumes an ideal source, yet a shunt clipper draws real current — 6.3 mA in Worked Example 11.1. From a sensor with a 10 k\(\Omega\) output impedance that current would need 63 V to drive it; what happens instead is that the source collapses and the clipping level is not what the analysis predicted. Buffer the source, or fold its impedance into \(R\).
The second is diode capacitance. A reverse-biased diode is a capacitor of a few picofarads (Chapter 13 exploits the effect deliberately), and in a clipper it forms a divider with \(R\) that lets fast edges straight through the "off" diode: with \(R = 1\ \text{k}\Omega\) and \(C_j = 4\) pF the feedthrough corner is 40 MHz, so a clipper that limits a 1 kHz sinusoid perfectly passes a 5 ns spike almost untouched.
The third is reverse recovery. In a multiplier driven at 20 kHz, an ordinary rectifier diode with a 30 µs recovery time never turns off at all and the multiplier simply does not work; fast or Schottky diodes are mandatory above a few kilohertz. This is the commonest reason a multiplier that simulates correctly fails on the bench, and it is the subject of the first half of Chapter 13.
A series resistor with a shunt diode is a clipper: it changes the shape and leaves the DC content to look after itself. A series capacitor with a shunt diode is a clamper: it preserves the shape exactly and moves the DC level. Add a second diode and capacitor to a clamper and it becomes a peak-to-peak detector; call the same circuit by a different name and it is a voltage doubler. Every circuit in this chapter is one of those two cells, or a chain of them.
8 Summary and Key Results
| Circuit | Governing relation | Worked value |
|---|---|---|
| Series clipper | Transmits one polarity; \(v_o = v_i R_L/(R_L+r_f)\) | 0.999 \(v_i\) for \(R_L = 10\) k\(\Omega\), \(r_f = 10\ \Omega\) |
| Shunt clipper resistor | \(R = \sqrt{r_fR_L}\) | 1 k\(\Omega\) for \(r_f = 10\ \Omega\), \(R_L = 100\) k\(\Omega\) |
| Biased clipping level | \(V_B + V_F\) (diode towards battery) | 3.0 + 0.7 = 3.7 V, not 3.0 V |
| Flatness of clipped top | slope \(= r_f/(R+r_f)\) | 0.0099; top rises 62 mV over a 6.3 V overdrive |
| Clipping interval | \(180^\circ - 2\arcsin(V_{\text{clip}}/V_m)\) | 136.6° = 37.9 % of the period at \(V_m\) = 10 V |
| Clamper output (negative) | \(v_o = v_i - V_m\) | 0 to \(-2V_m\); shape and \(V_{pp}\) unchanged |
| Clamper droop | \(d = 1 - e^{-T/2R_LC}\) | 4.88 % at \(\tau = 10T\); 0.50 % at \(\tau = 100T\) |
| Clamper capacitor | \(C \ge T/(2R_Ld)\) | 0.4975 µF for 1 % at 1 kHz into 100 k\(\Omega\) |
| Clamping theorem | \(A_f/A_r = R_f/R_r\) | \(10^{-4}\) for \(R_f = 10\ \Omega\), \(R_r = 100\) k\(\Omega\) |
| Peak detector droop | \(\Delta V/V \approx 1/(fR_LC)\) | 0.100 %/cycle at 1 kHz, 1 M\(\Omega\), 1 µF |
| Peak-to-peak detector | \(V_o = V_{pp} - 2V_F\) | 18.6 V from a 10 V peak sinusoid |
| Voltage doubler | \(V_o = 2(V_m - V_F)\), PIV \(= 2V_m\) | 66.5 V from 24 V RMS; PIV 67.9 V |
| Multiplier droop | \(\Delta V = (I/fC)(2N^3/3 + N^2/2 - N/6)\) | 0.909 V for \(N=1\); 6.364 V for \(N=2\) at 1 mA, 22 µF, 50 Hz |
| Multiplier output impedance | \(\Delta V/I\) | 909 \(\Omega\) doubler, 6.36 k\(\Omega\) quadrupler, 20 k\(\Omega\) at \(N=3\) |
9 Common Mistakes
A shunt clipper with a 3 V battery is routinely quoted as clipping at 3 V. It clips at 3.7 V, because the diode branch cannot conduct until the diode itself is forward biased. The 0.7 V is 23 % of the intended level, and in a 5 V logic protection circuit it is the difference between clamping safely below the absolute maximum rating and exceeding it. Trace the branch, write down the loop equation \(v_o = V_B + V_F\) or \(v_o = V_B - V_F\) according to which way the diode faces, and only then substitute numbers. The same error in a clamper puts the clamped crest at +0.7 V instead of 0 V, which matters wherever the clamped level is a reference.
"Make \(\tau\) at least ten times the period" is widely quoted and gives 4.88 % droop, computed exactly from \(1 - e^{-T/2\tau}\). That is fine for a television DC restorer and unacceptable for an instrument front end, where the tilt would be read as signal. Work backwards from the tolerable droop instead: \(C \ge T/(2R_Ld)\). Note also that the requirement scales with period, so a clamper designed at 1 kHz has twenty times too little capacitance at 50 Hz. A clamper is a low-frequency circuit before it is anything else.
The label "voltage quadrupler" suggests \(4V_m\) at the output, and that value is correct only at zero load current. The droop is \((I/fC)(2N^3/3 + N^2/2 - N/6)\), which for the quadrupler of Worked Example 11.2 is 6.36 V at just 1 mA — an output impedance of 6.36 k\(\Omega\). Because the coefficient grows as \(N^3\), a six-stage unit has an output impedance twenty times higher than a two-stage one. Multipliers are for microamperes and milliamperes; anything that draws real current needs a transformer, or a much higher drive frequency, or both.
10 Chapter Review
1. A shunt clipper uses \(R = 2.2\ \text{k}\Omega\), a silicon diode (\(V_F = 0.7\) V, \(r_f = 12\ \Omega\)) and a 4 V bias battery arranged so that the diode conducts on positive peaks. The input is \(15\sin\omega t\). Find the clipping level, the conduction angle, the peak diode current, the peak diode dissipation and the flatness of the clipped top.
The diode conducts only when the output node exceeds the battery voltage plus its own forward drop, so the clipping level is \(V_B + V_F = 4.0 + 0.7 = 4.7\) V. Conduction begins when \(15\sin\theta = 4.7\), that is \(\theta_1 = \arcsin(0.3133) = 18.26^\circ\), and ends at \(180 - 18.26 = 161.74^\circ\), so the diode conducts for \(143.5^\circ\), which is 39.9 % of the period. At the crest the source stands at 15 V while the output is held at 4.7 V, so the surplus 10.3 V appears across \(R\) and the peak diode current is \(10.3/2200 = 4.68\) mA. The peak dissipation in the diode is \(4.7 \times 4.68\ \text{mA} = 22.0\) mW, well within any small-signal device. The clipped top is not flat: its slope is \(r_f/(R+r_f) = 12/2212 = 0.00543\), so over the 10.3 V overdrive the top rises by \(10.3 \times 0.00543 = 55.9\) mV, and the output at the crest is really 4.756 V rather than 4.700 V. Note that the negative half-cycle passes through completely unchanged, reaching \(-15\) V, because this clipper limits one polarity only.
2. Derive the condition on \(R_LC\) for a clamper, and design a clamper for a 2 kHz square wave driving a 47 k\(\Omega\) load with a tilt of no more than 0.5 %.
While the diode conducts, the capacitor charges to the peak with the short time constant \(r_fC\). While it blocks, the only path is through the load, so the capacitor voltage decays as \(V_Ce^{-t/R_LC}\) and the output tilts by a fraction \(1 - e^{-t/R_LC}\). The worst case is the longest non-conducting interval, which for a symmetric waveform is \(T/2\), so the design condition is \(d = 1 - e^{-T/2R_LC}\), or inverted, \(R_LC \ge -(T/2)/\ln(1-d)\). For small \(d\) this reduces to \(C \ge T/(2R_Ld)\). Now the numbers: \(T = 1/2000 = 500\) µs, so \(T/2 = 250\) µs. With \(d = 0.005\), \(\ln(0.995) = -0.005013\), so \(R_LC \ge 250\times10^{-6}/0.005013 = 49.88\) ms. With \(R_L = 47\) k\(\Omega\), \(C \ge 49.88\times10^{-3}/47\times10^{3} = 1.061\) µF. Specify the next standard value up, 1.5 µF, which gives \(\tau = 70.5\) ms and an actual tilt of \(1 - e^{-250\mu/70.5m} = 0.354\) %. Check the other inequality too: with \(r_f = 10\ \Omega\) the charging time constant is 15 µs, which is 3 % of the period — comfortably fast enough to acquire the clamp within one conducting interval, but not so fast that the charging current is unreasonable.
3. Explain what the clamping theorem says and why it matters in practice.
In the steady state a series capacitor must gain exactly as much charge during the interval when the diode conducts as it loses during the interval when the diode does not, otherwise its voltage would drift from cycle to cycle. Charge is the integral of current, and current is voltage divided by resistance, so the theorem states \(A_f/R_f = A_r/R_r\), that is \(A_f/A_r = R_f/R_r\), where \(A_f\) is the area under the output waveform during forward conduction and \(A_r\) the area during the reverse interval. It explains why the clamped level is pinned so hard: with \(R_f = 10\ \Omega\) and \(R_r = 100\) k\(\Omega\) the ratio is \(10^{-4}\), so the waveform is allowed to cross the reference level by only a ten-thousandth of the area it spends on the other side — a very sharp clamp. It also predicts the failure mode: if the diode leaks, \(R_r\) falls, the permitted forward area rises in proportion, and the clamped extreme sits further away from the reference than it should. A clamper whose reference level drifts with temperature is almost always suffering from diode leakage, which roughly doubles every 10 °C, rather than from a wrong capacitor. The theorem also holds for biased clampers, with the reference level taking the place of zero.
4. A half-wave voltage doubler is fed from a 40 V RMS 50 Hz secondary with \(C_1 = C_2 = 47\) µF, using silicon diodes. Find the no-load output, the PIV of each diode, the loaded output at 2 mA, and the equivalent output resistance. What changes if the drive frequency is raised to 20 kHz?
The peak of the secondary is \(V_m = 40\sqrt2 = 56.57\) V. Ideally the doubler produces \(2V_m = 113.1\) V; subtracting the two forward drops gives \(2(V_m - 0.7) = 111.7\) V at no load. Each diode blocks a peak inverse voltage of \(2V_m = 113.1\) V, and both capacitors must be rated for at least that, so 160 V parts are the sensible choice. Under load, the droop for a single-stage (\(N = 1\)) multiplier is \(\Delta V = I/(fC)\times(2/3 + 1/2 - 1/6) = I/(fC)\). At \(I = 2\) mA, \(f = 50\) Hz and \(C = 47\) µF, \(I/(fC) = 0.002/(50\times47\times10^{-6}) = 0.851\) V, so the loaded output is \(111.7 - 0.85 = 110.8\) V, a droop of 0.76 %. The equivalent output resistance is \(\Delta V/I = 0.851/0.002 = 426\ \Omega\), which is simply \(1/(fC)\). Raising the drive to 20 kHz divides that by 400: the output resistance falls to 1.06 \(\Omega\) and the droop to 2.1 mV. Equivalently, the same droop can be had with capacitors 400 times smaller, 0.12 µF rather than 47 µF, which is the whole reason high-voltage multipliers are driven from a switching oscillator. The one new requirement is that the diodes must recover in far less than the 50 µs period, so ordinary rectifiers with 30 µs recovery times will not do — fast-recovery or Schottky devices are needed.
5. A 10 V peak, 1 kHz sinusoid is applied to (a) a negative clamper, (b) a peak detector, and (c) a peak-to-peak detector, each with a 1 M\(\Omega\) load and a 1 µF capacitor, using silicon diodes. Describe the output of each and compute its principal imperfection.
(a) The negative clamper leaves the shape and the 20 V peak-to-peak value untouched and shifts the whole waveform down so that the crests sit at \(+V_F = +0.7\) V and the troughs at \(-19.3\) V; the new average is \(-9.3\) V instead of zero. Its imperfection is tilt: with \(\tau = R_LC = 1\) s and \(T/2 = 500\) µs, the droop is \(1 - e^{-500\mu/1} = 0.050\) %, that is 10 mV on the 20 V swing — negligible here, because \(\tau\) is a thousand periods. (b) The peak detector produces a nearly constant DC output of \(V_m - V_F = 9.3\) V. Its imperfection is ripple caused by discharge between peaks: \(\Delta V/V = 1 - e^{-T/R_LC} = 1 - e^{-0.001} = 0.100\) % per cycle, that is 9.3 mV of sawtooth ripple on the 9.3 V output. Its acquisition time is set by the much shorter constant \(r_fC\), about 20 µs, so it reaches its final value within one cycle of switch-on. (c) The peak-to-peak detector is the clamper of (a) followed by the peak detector of (b), so its output is \(V_{pp} - 2V_F = 20 - 1.4 = 18.6\) V of DC, with roughly twice the ripple of the simple peak detector because two capacitors now contribute droop. All three outputs are DC or DC-shifted; none of them contains the frequency information of the input except through the ripple, which is why an oscilloscope and not a meter is the instrument for checking these circuits.