Cutting a wave
A clipper is a circuit that cuts off the part of a wave that goes beyond some level, and passes the rest unchanged. The half-wave rectifier of lesson 5 is already one. Its diode in series with the load passes the positive halves and cuts the negative halves off at zero. Turn the diode round, and it passes the negative halves instead. This is a series clipper.
A diode across the output
A shunt clipper puts the diode across the output instead. The input feeds the output through a resistor \(R\). Across the output is the diode in series with a battery \(V_B\), the diode pointing towards the battery's positive terminal. The battery sets the level at which the wave is cut.
Follow one cycle with an ideal diode. While the input is below \(V_B\), the diode's anode is lower than its cathode, so it is off. No current flows through \(R\), and the output equals the input. Once the input rises above \(V_B\), the diode turns on and holds the output at \(V_B\). The extra voltage appears across \(R\). So the positive peaks are cut flat at \(V_B\), and the negative halves pass untouched.
A silicon diode needs 0.7 V across it before it conducts. So the output is cut at \(V_B + 0.7\,\text{V}\) instead. With no battery at all, the level is just the diode's own drop, 0.7 V.
The input of a shunt clipper is \(8\sin\omega t\) volt. The battery is \(2\,\text{V}\), and the diode is silicon. Find where the output is cut, and its greatest and least values. What if the diode and battery are both turned round?
The diode conducts when the output tries to rise above \(V_B + 0.7 = 2.7\,\text{V}\).
While \(8\sin\omega t < 2.7\), the output equals the input. While \(8\sin\omega t > 2.7\), the output is held at 2.7 V.
The output rises to 2.7 V and stays flat there near each positive peak. It goes down to \(-8\,\text{V}\) in the negative halves.
The output runs from \(-8\,\text{V}\) to \(+2.7\,\text{V}\). Turned round, the circuit cuts the negative peaks instead, at \(-2.7\,\text{V}\), and passes the positive halves up to 8 V.
Two such branches can work together. One diode and battery cut the positive peaks, and a second pair, turned the other way, cut the negative peaks. The output is then a sine wave with both tops flattened, almost a square wave. This double clipper keeps a signal between two chosen limits.
A shunt clipper uses a silicon diode and a \(4\,\text{V}\) battery. The input is \(10\sin\omega t\) volt. At what voltage are the positive peaks cut?
A double clipper has two silicon diodes, each in series with a \(3\,\text{V}\) battery, one branch for each sign. The input is \(10\sin\omega t\) volt. Between which values does the output swing?
Lifting a wave
A clamper does something a clipper cannot. It keeps the whole shape of the wave but shifts it up or down, adding a steady DC level. It needs a capacitor \(C\) in series with the signal, and a diode across the output. In a positive clamper the diode's anode is on the common line, marked zero, and its cathode on the output.
A positive clamper with an ideal diode is fed with \(6\sin\omega t\) volt. After the first cycle, between which values does the output swing?
Follow the figure above. In the first positive half-cycle the diode is reverse biased, so nothing happens. As the input goes negative, the output tries to go below zero. The diode then conducts and holds the output at zero, and the capacitor charges. It charges until it holds the full negative peak, 6 V.
After that the input never goes lower than it did, so the diode never conducts again. The capacitor has nowhere to discharge, and it acts like a 6 V battery in series with the input. The output is the input plus 6 V: \(6\sin\omega t + 6\), swinging from 0 to 12 V. The shape is the same, and the peak-to-peak value is still 12 V. Only the level has moved.
With a silicon diode the capacitor charges only to \(6 - 0.7 = 5.3\,\text{V}\). The output then swings from \(-0.7\,\text{V}\) to \(11.3\,\text{V}\). Turn the diode round, cathode on the common line, and the wave is pushed down instead: a negative clamper gives \(-12\) to \(0\,\text{V}\). The clamping holds only if the capacitor loses little charge in a cycle. So the load resistance times \(C\) must be much longer than one period.
A clipper changes the shape of a wave by cutting off the part beyond a chosen level, and leaves the rest unchanged. A clamper keeps the shape and the peak-to-peak value of a wave, and shifts the whole wave up or down by a steady voltage.
A negative clamper with a silicon diode is fed with \(5\sin\omega t\) volt. After the first cycle, what is the lowest value of the output, in volts?
A wave of peak-to-peak value \(12\,\text{V}\) goes into a circuit. The wave that comes out has the same peak-to-peak value, but it runs from \(0\) to \(12\,\text{V}\). Is the circuit a clipper or a clamper? Explain.
It is a clamper. A clipper would cut part of the wave off, so its output would have a smaller peak-to-peak value. Here the whole wave is still there, with the same shape and the same 12 V from bottom to top, only moved up by 6 V. That shift of the DC level, with the shape kept, is what a clamper does.
Every input pin of a chip in a phone carries two tiny diodes. One runs from the pin to the positive supply, and one from the negative supply to the pin. They are a double clipper. A spark of static from a finger can put hundreds of volts on a pin for a moment. The diodes conduct as soon as the pin goes more than a diode drop outside the supply voltages, and hold it there. The delicate circuits inside never see the spark.
The workbook questions on rectifiers, clippers and clampers are placed in the workbook of the next chapter, beside those on the Zener diode. Everything they need has been taught in lessons 4, 5 and 6.
- Forgetting the diode drop in a clipper with a silicon diode.The output is cut at \(V_B + 0.7\,\text{V}\), not at \(V_B\). Use the drop unless the diode is called ideal.
- Thinking a clamper cuts the wave.A clamper keeps the whole wave and its peak-to-peak value; it only shifts the level. A clipper cuts.
- Charging a clamper's capacitor to the peak with a silicon diode.It charges to the peak minus 0.7 V, so the clamped wave reaches 0.7 V past zero.