Demo 3.3b: Clipper Circuits (Optional, EEN1018)

The demonstrations on diodes and on rectifiers took the output across the load and asked how much of the input got through. A clipper takes the output across the diode instead. While the diode is off, it is an open circuit and the output follows the input exactly. Once the diode turns on, the voltage across it hardly changes, so the output cannot go any further. The waveform is cut off flat at the diode’s threshold, and everything above it appears across the resistor.

That is the whole idea, and the rest of this page is variations on it: turn the diode round, fit two, put a bias source in series with each, replace the diodes and their sources with a pair of Zeners, or shift the input before it reaches the diode. The last tab is a circuit that looks almost the same and does something quite different: a clamper moves the waveform up or down without cutting any of it off.

The notes call the first circuit a clipper. Textbooks often call it a shunt (or parallel) clipper, because the diode sits across the output, and some call it a limiter. The same circuit can have more than one name, so expect all three if you read around the topic. The clamper is not in the notes; it is here because it looks so like a clipper that the two are easily confused.

Why you would want to cut a waveform off

The most common reason is protection. An amplifier input, an ADC pin or a sensitive meter will be damaged by a voltage spike far above its normal range. Two diodes across that input do nothing at all to a normal signal, and they hold a spike to less than a volt. The same circuit with Zeners holds it to a few volts. Microcontroller pins have this protection built in, which is why their data sheets limit the current you may push into a pin rather than only its voltage.

The second reason is to shape a signal. Clip a large sine wave well below its peak and what is left is close to a square wave with the same frequency, which is a quick way to turn a sensor’s analogue output into something a digital input can count.

How to use it

One Diode model selector at the top of every tab chooses how the diodes behave, and it stays set when you change tabs:

Ideal is a perfect switch, with 0 V across it when on. The dc offset example in the notes uses this.
Si, 0.7 V and Ge, 0.3 V are the constant-drop model from the first section of the notes.
Real silicon is the exponential curve from the demonstration on diodes, so the corners are rounded and the clip level moves a little with current.

Every tab draws the input (cyan, dashed) and the output (green) against time, with the clip levels marked. Beside it is the transfer characteristic, vout against vin. Drag Where in the cycle or press Play, and watch which diode lights up in the circuit underneath.

One diode is the first circuit in the notes, with a switch to turn the diode round. Two diodes, and bias fits one each way and lets you add a bias source in series with each. Zener clipper replaces both with a back-to-back Zener pair. A dc offset is the worked example from the notes: V1 = 2 V in series with vin = 5 sin(500πt). Series clipper moves the diode into the signal path. Clamper replaces the resistor with a capacitor.

The input is 250 Hz throughout, which is 500πt as in the notes, so one period is 4 ms and the positive peak is at t = 1 ms.

Clipper Circuits

Diodes that limit a waveform at a chosen level, drawn against time and as a transfer characteristic.

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A diode across the output holds one half of the waveform at its threshold and leaves the other half alone.

The output is taken across the diode, not across the resistor.
While the diode is off it is an open circuit, no current flows, there is no drop across R, and vout follows vin.
Once the diode turns on, the voltage across it stays at its threshold (0.7 V in this model), so vout cannot go any further.
The rest of the input appears across R instead.
Turn the diode round and it is the other half cycle that is clipped, at minus the threshold.

Diode model

Off below 0.7 V, then a constant 0.7 V however much current flows.

The diode

Peak input5.0 V

Series resistor R1 kΩ

With no load, R sets only the current, not the shape.

Where in the cyclet = 1.00 ms

Drag through the cycle, or play it, and watch the diodes.

vin, the input (dashed) vout, the output

transfer characteristic, with vout = vin dotted

vin at the cursor

5.00 V

t = 1.00 ms

vout at the cursor

0.70 V

a diode is on, so vout is held

Across R

4.30 V

current 4.30 mA

Clip levels

+0.70 V / none

top / bottom

Where the output stops

The highest the output reaches is 0.70 V, from an input that reaches 5.00 V.
The constant-drop model gives exactly +0.70 V, the answer the hand analysis gives.
The transfer characteristic says the same thing without time: it follows the dotted vout = vin line, then goes flat where the diode turns on.

What this circuit is for

One diode on its own is mostly a way to understand the idea.
Two of them, one each way, limit both halves, which is the next tab.
That pair is used to stop a voltage spike reaching something that cannot take it, such as an amplifier input.

What this demo does not model

A load on the output. A load and R form a divider, so the unclipped part would be smaller than vin.
The diode's capacitance and recovery time, which matter at radio frequencies but not at 250 Hz.
Temperature: a real silicon diode's drop falls by about 2 mV for each degree it warms.

Walkthrough

Step 1: Clip one half of the waveform

Open One diode with the model on Si, 0.7 V and the diode’s anode to the output.

Drag the cursor through the first half cycle. Until vin reaches 0.7 V the diode is off, no current flows, there is no voltage across R and vout = vin. From 0.7 V on, the diode is forward biased, the voltage across it stays at 0.7 V and the output goes flat. At the peak the readings show the split: 0.70 V across the diode, 4.30 V across R and 4.30 mA flowing.

On the negative half cycle the diode is reverse biased and the output follows the input all the way down to −5 V. Press Cathode to the output and the whole picture turns over: now the bottom is clipped at −0.7 V and the top is untouched.

The transfer characteristic says the same thing without time. It sits on the dotted vout = vin line and goes flat where the diode turns on.

Step 2: Look at a real diode

Switch to Real silicon and look closely at the corner where the output goes flat. It is rounded, because a real diode starts to conduct gradually rather than at one voltage, and its voltage keeps rising slowly with current. With R = 1 kΩ the output peaks at 0.74 V. Drag R down to 100 Ω, which pushes more current through the diode, and it rises to 0.80 V; at 10 kΩ it falls to 0.68 V.

So “0.7 V” is a model. It is a good one for hand analysis and the answers are within a few tens of millivolts, but the real clip level depends on how hard the diode is being driven.

Step 3: Clip both halves, then move the levels

Open Two diodes, and bias and go back to Si, 0.7 V. With no bias, D1 clips the top at +0.7 V and D2 clips the bottom at −0.7 V, so a 10 V peak-to-peak input comes out as 1.4 V peak to peak. This is the input protection circuit.

Now set the bias on D1 to 2 V and the bias on D2 to 1 V. A source in series with a diode adds to its threshold: D1 does not conduct until the output reaches 0.7 + 2 = 2.7 V, and D2 not until it falls to −(0.7 + 1) = −1.7 V. The two levels are set independently, so the output is no longer symmetrical.

Push the bias on D1 to 5 V. The top level is now 5.7 V, above the 5 V peak, so D1 never turns on and the top half of the wave comes through untouched.

Step 4: Replace the bias sources with Zeners

Each bias source in Step 3 has to be a floating dc supply with nothing else connected to it, which is awkward to build. Open Zener clipper.

ZD1 and ZD2 are connected back to back. On the positive half ZD1 is reverse biased and breaks down at its Zener voltage, and the same current flows forwards through ZD2. So the top level is Vz1 plus one forward drop. With the defaults (5.1 V and 3.3 V parts, Si model) the levels are +5.80 V and −4.00 V.

Switch to Ideal and the levels become +5.10 V and −3.30 V, which is the answer you get if you quote the clip level as just Vz. That is the ideal-diode answer. In a real circuit the forward drop is always there, and for a 3.3 V part it is about a sixth of the level. With Real silicon the 3.3 V side reaches −4.14 V, because a 3.3 V Zener has 28 Ω of dynamic impedance (see Demo 3.3) and it is carrying about 4 mA at the peak.

Step 5: Work the dc offset example

Open A dc offset and set the model to Ideal, as the notes do. V1 = 2 V is in series with vin = 5 sin(500πt), and d1 has its anode to ground.

The dotted trace is vb = vin + V1, the voltage after both sources. It is the same sine, moved up by 2 V, so it swings from +7 V to −3 V.

While vb is positive, d1’s cathode is above its anode, so d1 is reverse biased and no current flows. With no current there is no voltage across R, and vout = vb. Note that it is the resistor that has no voltage across it here: the whole of vb is across d1. While vb is negative, d1 is forward biased and an ideal diode holds vout at 0 V, with the rest across R.

At t = 1 ms, vin is at its 5 V peak, so vout = 5 + 2 = 7 V. The output is held at 0 V for 37 per cent of each cycle (about 1.5 ms in every 4 ms) rather than half of it, because the offset has lifted most of the wave above zero.

Now drag V1. The clip level stays at 0 V; what moves is how much of the waveform reaches it. On the transfer characteristic, the corner moves sideways, to vin = −V1. Compare that with Step 3, where a bias source in series with the diode moved the corner up or down.

Step 6: Put the diode in series

Open Series clipper. Now the diode is in the signal path and the output is across R. The output is zero until the input passes the diode’s threshold, and after that it is the input less one drop: 4.30 V at the peak with the Si model.

This is the half-wave rectifier from Demo 3.2. The shunt clipper keeps the part of the waveform below the level; the series clipper keeps the part above it.

Step 7: Shift the waveform instead of cutting it

Open Clamper, which starts from switch-on with the capacitor empty and runs for three cycles.

For the first half cycle the output follows the input. On the first negative peak the diode conducts and charges the capacitor to 4.3 V (the 5 V peak less the 0.7 V drop). After that the capacitor holds that charge and adds it to the input, so the output swings from −0.70 V to +9.30 V: the same 10 V peak to peak, moved up by 4.3 V. Nothing has been cut off. The output comes back down to −0.7 V once per cycle, but with no load the diode never needs to conduct again.

Now choose RC = 0.3T, which connects a load that discharges the capacitor between peaks. The flat top sags to about 5.2 V and the waveform is no longer the shape of the input. A clamper needs RC to be much longer than the period of the signal; a clipper has nothing to charge, so the question does not come up.

There is no transfer characteristic on this tab, and that is the difference between the two circuits. A clipper’s output depends only on the input at that instant. A clamper’s output depends on what the input did earlier, and the clamper only gives this output after the first cycle.

What this demonstration does not model

No load is connected to any clipper output. A load would form a divider with R, so the unclipped part of the waveform would be smaller than vin, and there would be a voltage across R even while the diodes are off. The bias sources and the signal source have no internal resistance. The diodes have no capacitance or recovery time, which matters at radio frequencies and not at 250 Hz. The clamper charges its capacitor instantly, which needs a source with no resistance, and it uses the constant-drop model for that reason.

🎓Tutor: predicting a clipper’s output

Predict vout for each of the questions in the notes, then check it against the circuit and the waveform.

Interactive Tutor

Predicting a Clipper's Output

Predict vout at each step, then see the circuit and the waveform.

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Check your understanding

Quiz

Select 0/1

A 5 V peak sine wave drives a resistor and a silicon diode, with the output taken across the diode and the diode's anode to the output. Using the 0.7 V model, what range does the output cover?

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Quiz

Select 0/1

Two 6.2 V Zener diodes are connected back to back across the output of a clipper, each with a 0.7 V forward drop. The input is a 12 V peak sine wave. What are the clip levels?

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Quiz

Select 0/1

In the dc offset example, V1 is turned round so that it subtracts 2 V instead of adding it: vb = 5 sin(500πt) − 2 V, with an ideal diode, anode to ground. What is the highest value vout reaches?

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Concept Match

Match each circuit to what it does to the waveform

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Quiz

Select 0/3

Which of these change the level at which a shunt clipper limits the output? Select all that apply.

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Quiz

Select 0/1

A clamper's output sags between the peaks, so the flat parts are no longer flat. Which change fixes it?

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Wrap-up

Five things to take away.

A shunt clipper takes its output across the diode. While the diode is off, it is an open circuit and vout = vin. Once it is on, its voltage stays at the threshold and the excess appears across R.
The clip level is set by the diode branch: one threshold for a plain diode, threshold plus bias with a series source, and Vz plus one forward drop for back-to-back Zeners. The Zener level is exactly Vz only for an ideal diode.
A dc source in series with the input moves the waveform, not the clip level. In the worked example vb = vin + 2 V swings from +7 V to −3 V, d1 clips it at 0 V, and vout = 7 V at t = 1 ms.
The transfer characteristic shows the whole circuit in one line: it follows vout = vin and goes flat at each clip level. A clamper has no transfer characteristic, because its output depends on what came before.
The constant-drop model is the one to use on paper. A real diode’s corner is rounded and its level moves by tens of millivolts with current, which is visible with the Real silicon model but does not change any design decision on this page.