Using Diodes as Clippers or Clampers

Wave-shaping circuits are circuits that usually clip off portions of signal voltages above or below certain levels, or add or restore a DC level to an electrical signal, or transform one waveform to another type of waveform. In this tutorial, we will get to know wave-shaping diode circuits such as the clipper and clamper circuits. We will learn how these circuits modify a signal’s waveform and know their applications. Also for the sake of simplicity of our discussion, we will present and discuss our examples using ideal analysis.

Clippers, also known as ‘diode limiters’ are usually a diode and resistor network that “clips” away a portion of an input signal without distorting the remaining part of the applied input signal.

One of the simplest examples of a clipper circuit is the half-wave rectifier circuit. A rectifier is basically constructed with a diode network circuit and functions by converting an AC signal to a DC signal. Depending on the configuration of the diode, the positive or negative region of the applied signal is “clipped” off. You can learn more about half-wave rectifiers and other configuration of rectifiers in one of our specific tutorials about half-wave and (bridge and center-tapped) full-wave rectifiers.

Clippers are generally categorized into two: series and parallel.The series configuration is where the diode is placed in “series” with the load, while the parallel configuration has the diode placed in a branch ”parallel” to the load.

Now let’s start learning about the clippers! We can start with the series variety. As mentioned above, the series configured clipper circuit is where the diode is placed in series with the load. Below is a picture of a simple clipper circuit. This is a negative series clipper and is also popularly known as the half-wave rectifier.

Now let’s see why this circuit is called a negative series clipper. We’ll apply a sinusoidal signal at the input of the circuit and analyze.

During the positive interval of the input signal, the diode is forward biased and is in the on-state. Therefore, the diode conducts and the positive interval of the input signal passes through the output. On ideal analysis, the voltage drop across the diode is 0V and is represented by a short circuit across its pins when it is forward biased. During this interval, the output is exactly the same as the input, therefore the output voltage is just equal to the positive region of the input signal.

During the negative interval of the input signal, the diode is reversed biased. Therefore the diode is in off-state creating an open circuit. This is represented in the schematic below as an open circuit across the diode. Therefore, there is no output during this interval. The circuit “clipped” off the negative region of the applied input signal. That is why this circuit is called a negative series clipper.

If we reverse the diode compared to the circuit above, a positive series clipper is created.

In this configuration, using the same analysis, this time the diode is reversed biased during the positive interval and is represented by an open circuit across the diode. Therefore, there is no output during this interval.

During the negative interval, the diode is forward biased. The applied signal during the negative interval passes through the output. During this interval, the output is directly parallel to the input, therefore on this configuration, the output voltage is just equal to the negative region of the input signal.

Though we demonstrated this circuit as a half-wave rectifier, it is not limited to sinusoidal input signals only but also to other waveforms such as square, triangular and sawtooth signals.

Now that we have learned about the series configured clipper, let’s move forward and take a look at another variety which is the parallel clipper. Parallel clipper is configured in a way where the diode is placed in a branch parallel to the load. The circuit shown below is the simplest of parallel diode configurations. The analysis of parallel configurations is similar to what we applied to series configurations, as demonstrated in the next example.

Applying an input signal to the circuit, we can start analyzing the circuit. During the positive interval of the applied signal, the diode is forward biased and is in ‘on-state’. The diode will be represented with a short across its terminals. Because of this, the node where the resistor and the diode meet is shorted to the ground. And hence, the output is also shorted to the ground. Therefore, there is no output during this interval.

During the negative interval of the input signal, the diode is reversed biased and is ‘open’ as shown in the figure below. This makes the resistor in the circuit a floating element. Also, making the output directly parallel to the input. Therefore, the output during the negative interval is equal to the negative region of the input signal. Observing the final waveform of the output diagram, the positive interval of the input signal was “clipped off”, therefore this circuit is called a positive parallel clipper.

Connecting the diode in reverse on the circuit, will make it a negative parallel clipper. This time, during the positive interval of the input signal, the diode is reverse biased and is therefore open. During this interval, the output is directly parallel to the input signal. The output is therefore equal to the positive region of the input signal. While during the negative interval of the input signal, the diode is forward biased and is therefore shorted. During this interval, the output voltage is ideally equal to 0 V.

Now we have finished discussing the basic configurations of clipper circuits, we will now proceed to learn about some of its modified configurations. This one is called the ‘biased clipper’. Biased clippers are formed by adding a DC supply to the network. The addition of a DC supply can have a pronounced effect on the analysis of the clipper configuration. The DC supply may aid or work against the input signal depending on how it is positioned. It can be placed in the leg between the input and the output or in the branch parallel to the output.

Like the basic clipper circuits, biased configurations are applied to both series and parallel variations.

Let’s start analyzing a biased series clipper. Observe that a DC supply is added in series with the diode. In this example, the DC supply is placed where its negative terminal is connected to the anode of the diode while its positive terminal is connected to the positive terminal of the input signal.

Notice that the DC supply reverse biases the diode. The DC supply ‘Vdc’ in this configuration acts as an additional threshold voltage. If you recall from previous tutorials about diodes, the threshold voltage is the required voltage applied across the diode for it to be forward biased and conduct. Ideally, threshold voltage is equal to 0V, but in practical situations this would measure around ~0.7V depending on the semiconductor used in the diode. Again for simplicity, let’s just use the ideal threshold voltage of a diode which is 0V.

As mentioned, the DC supply acts as a threshold voltage adding the diode’s own threshold voltage (0V), meaning that the input voltage ‘Vi’ required to forward bias the diode must be greater than than the DC supply voltage ‘Vdc’.

Let us say that the DC supply added is 5 Vdc, and we have an input sinusoidal signal with peak voltage of 20 V.

As the input signal transitions from 0V - 5V, the diode is ‘off’, as we have established in our discussion earlier, the input signal must be greater than the DC supply presented in this circuit for the diode to be in ‘on-state’.

Here, the diode will only conduct during positive intervals, only when the input signal increases to a voltage greater than the DC supply placed in the circuit which is 5Vdc. During this condition, the diode is ‘on’ and is represented by a short across its terminals in the schematic circuit.

Using Kirchhoff's Voltage Law, or KVL, we can determine the peak voltage of the output.

Therefore, peak voltage of the output during the positive interval is:

During the negative interval, the diode will remain open. Therefore, the output in this interval is 0V. The final output waveform would look like this in the image below. Notice that the output waveform did not start from the origin of the plane diagram but only when the DC supply was surpassed by the input signal.

Now let’s take a look at a biased - parallel clipper.

Here, a DC supply is added to the line which is parallel to the output. In this example, the DC supply is placed where its negative terminal is connected to the ground of the circuit, while its positive terminal is connected to the anode of the diode.

This time we are using a triangular wave as an example input to our circuit. Looking at the circuit , during the positive interval, the diode is reverse biased by the input signal. But also take note that it is being forward biased by the DC supply on a certain time when the input is still less than the DC supply. Therefore, as the input signal is increasing from 0V to 4V, the diode is ‘on’ through the DC supply. During this time, the output is directly parallel to the 4Vdc supply. So the output voltage, Vo, during this portion of the positive interval is equal to 4V.

When the input signal becomes greater than 4V, the diode will now be in ‘off state’. During this time, the off-state diode creates an open circuit, making the voltage output directly parallel to the input signal. As you can observe in the diagram, the output is equal to the input during this duration. This output will continue until such time that the input decreases to 4V and below again as it cycles. When the input signal falls to 4V and below, the diode will be in off-state again. This will reconnect the DC supply to the circuit making the output directly parallel to the 4V DC supply.

During the negative interval, the diode is forward biased by the 4V DC supply. Therefore, the diode is in the ‘on-state’, making the output directly parallel to the 4V DC supply for the whole duration of this interval.

And that’s it for the clippers, just take note that different configurations of the clipper circuit may require a different analysis to tackle the circuit. Clipper circuits can be modified in many varieties and may be configured to clip off a positive and a negative section as determined by the magnitude of the DC supplies added.

Clippers are usually used as a protection circuit or as a distortion circuit in audio applications.

Previously, we have investigated a number of diode configurations that clipped off a portion of the applied signal without changing the remaining part of the waveform known as the clipper circuit. On this section, will examine another diode based circuit that shifts the applied signal to a different level. This circuit is called the “Clamper”.

Basically, a clamper is a network of a diode, resistor, and a capacitor. It is configured in a way such that the capacitor is connected directly from input to output with a resistive element in parallel with the output signal. The diode is also in parallel with the output signal as shown in the schematic diagram.

In constructing a clamper circuit, the resistor and capacitor of the network must be chosen such that the time constant, determined by t RC, is long enough to ensure that the voltage across the capacitor does not discharge significantly during the interval the diode is nonconducting. In our discussion, we will assume that the capacitor fully charges and discharges in five time constants for all practical purposes.

Now let’s start learning about clampers! Here we have a basic clamper with a square wave input signal.

To analyze the circuit, we must first determine on which interval the diode conducts. In this case, the diode will go into the on-state during the positive interval. During this time, the diode will be represented as a short across its terminals. Since we have a square wave input signal, we will represent it with a DC input voltage in our analysis.

The output during this interval is 0V, since the diode effectively shorts the resistor where the output voltage is measured. Also, take note that during this interval the capacitor will charge up quickly to a value closely equal to input signal ‘V’ since the time constant will be very small due to small effective resistance of the shorted line. Take note of the polarity of the capacitor as it is fully charged.

Once the interval switches to the negative, represented by an inverted DC input supply in our diagram, the diode will be reversed biased. During this time, the diode will be open and the resistor is not shorted anymore.

As mentioned earlier, we assume that the time constant of the capacitor and the resistor is long enough for the capacitor to hold its charge. This assumption will be enough for the capacitor to hold its charge during the whole negative interval. Given this, if we determine the output voltage across resistor R, using KVL, we get:

Notice that the negative interval of the output is twice as much of the original signal applied at the input. In other words, the negative interval of the input signal, -V, was shifted to another level, -2V, when it was applied to the clamper circuit.

Though we can say that the input signal was shifted when passing through the circuit, observe that the total swing is the same (2V). This is a property that applies to all clamper circuits.

Just like the clipper circuit, a bias DC supply can also be added to a clamper circuit to modify its output. Let’s take a look at the next example.

Here a 5V DC supply is added to the clamper circuit. Its negative terminal is connected to the ground while its positive terminal is connected to the anode of the diode. Looking at the input signal, the diode will be forward biased during the negative interval. As we have learned earlier, we will start analyzing the circuit on the interval on which the diode goes into on-state, from t₁ to t₂.

During the negative interval, the diode will conduct and the diode will be represented as a short in the schematic diagram.

Resistor R is now directly parallel to the 5V DC supply, hence the output voltage during this interval is equal to the 5V. Also, during the interval, capacitor C charges through the input signal and the DC supply. Note the polarity of the charge of the capacitor. Using KVL, we can determine how much charge the capacitor stored.

The capacitor will therefore charge up to 25 V during the negative interval.

When the input signal switches into the positive interval, the diode will go to the off-state since it will be reverse biased by the input. During this interval the diode will be open, leaving the 5V DC supply with no effect on the circuit.

Checking if the capacitor will hold its charge during the positive interval, let's compute the period and interval time of the input signal and compare it to the discharge time of the capacitor. Given that the input signal frequency is 1000Hz, we can determine the period, 1/f, is equal to 1ms, meaning each interval is equal to 0.5ms. Computing for the discharge time, which is equal to five time constants, 5τ = 5RC.

Therefore we can say that the capacitor will hold its charge during this interval since the discharge time, 50ms is sufficiently large compared to the interval time 0.5ms.

Using KVL, we can determine the voltage output during this interval:

Therefore, the output voltage during the positive interval is equal to 35V. Observe the output’s waveform and see that the total voltage swing is still the same with the input’s.

Materials Needed

• Signal Generator
• 2 Channel Oscilloscope
• UA741 Opamp
• 1N4148 x 2
• 10uF x 2
• 100k resistor x 2
• 10k resistor
• 9V DC supply

This circuit is a simple distortion circuit commonly used as guitar effects is based on clipping diodes. Usually this circuit is made of an amplifier with variable gain as the first stage, but for our demo purposes,we modified it to buffer (unity gain).

1. Gather the listed material and construct the circuit above.
2. The circuit is composed of a buffer circuit in the first stage and a network of diodes in the second stage. Let us call the output of the first stage as output A and the output of the second stage as output B.
3. Input a sinusoidal signal to the input of the circuit using the signal generator.
4. Set the signal generator to 1kHz and a starting voltage 0.1V or the minimum voltage of the signal generator you have.
5. Connect the one channel of the scope to the input of the circuit while the other channel to the output A of the circuit. Set the potentiometer to maximum output.
6. Slowly increase the input voltage to slightly over 1V. Observe the waveform in your scope. Observing the waveform from output A, you can see that there is no clipping.
7. Reset the input to the minimum voltage of your signal generator. Then this time connect the 2nd channel of your scope to output B.
8. Again, slowly increase the input voltage to slightly over 1V. Observe the waveform in your scope. Observing the waveform from output B, you can see now that there is clipping occuring.
9. Checking the datasheet of diode 1N4148 , you can see that it has a forward voltage around 0.6V to 1V, let's assume that we are using the 916B package which has a maximum forward voltage of 0.73V. During the positive interval, D2 will be forward biased as the input signal goes over 0.73V, while D1 is off. When D2 is already forward biased, it clips or limits the output to its forward voltage, 0.73V, since the output is directly parallel to D2.
10. During the negative interval, D1 will be forward biased as the input signal goes over 0.73V, while D2 is off. When D1 is already forward biased, it clips or limits the output to its forward voltage 0.73V, since the output is directly parallel to D1.

That ends the demo for our clipper circuit and I hope you learned something and enjoyed!

So far we have learned two wave-shaping circuits which are mainly based on the diode, the Clipper and the Clamper. We have also discussed some of their variations and how to analyze them. We learned the construction of a series clipper, the parallel clipper as well as the basic clamper circuit. We also studied how a DC bias can affect the output of the circuit and can modify how much level is clipped or shifted.