Diode Clipper Circuit - Biased, Unbiased, Positive and Negative

Diode Clipper Circuit - Biased Clippers, Unbiased Clippers, Positive Clippers, Negative Clippers, Shunt Clippers and Double Diode Clipper Circuits:

Clippers are also called amplitude selectors or slicers. A circuit arrangement is used to cut off the part of the signal.  Diode clippers can clip off unwanted parts of the waveform. We have set a reference point. The Signal port lies above or below the reference point clips off. You can think of a half-wave rectifier as the simplest form of clipper (see figure 1).

Figure 1 Simple clipper circuit or rectifier

The basic components of the clipping circuit are a diode and a resistor. In the above circuit, the negative half-cycle was eliminated. You can decide the clipping level of your own choice. To get the desired level of clipping, you have to add a DC source in series with the diode. These are called biased clippers.

Positive Clippers:

It removes the positive part of the input signal.

Negative Clippers:

It removes the negative part of the input signal.

Series Clippers:

In series clipper circuits, the diode is in series with input and output terminals.

Shunt Clippers:

In shunt clippers, the diode is in parallel with input and output terminals.

      

Example 1: Positive Series Biased Clippers:

Figure 2 Positive series clipper

1. Positive series clipper, diode D1 direction shows it is a positive clipper
2. Observe the circuit without sinusoidal input (V1).
3. D1 is reverse biased. Because of biasing voltage polarity. At anode 0V while at cathode +2V
4. Now consider a circuit with sinusoidal input  (V1)
5. During the positive half cycle more positive voltage at cathode +6V and +2V. The diode remains off. No voltage at the output

Vo = 0

6. During the negative half-cycle, the diode is forward biased for a small portion of the input wave. We have to evaluate the time of the diode.
7. A diode is forward biased when the anode is more positive than the cathode. The cathode is at 2V (biasing voltage). To make the diode forward bias, the applied voltage is less than -2V (for the ideal diode) or -2.7V (for the practical diode)
8. Apply KVL on the equivalent circuit (negative half cycle)

V1 - 2 - 0.7 +Vo = 0

Vo = - V1 + 2.7

Vo = -3.3V

Look at the output waveform (Blue). We get a new peak shifted in the upward direction. During the negative half, the cycle diode remains turned off until V1 reaches up to -2.7V.

It is simple, when the diode turns on, the signal appears at the output. And we have evaluated the on-time of the diode.

Example 2: Shunt Biased Negative Clippers :

Figure 3 Shunt clippers/negative clippers

1. The figure above is a negative shunt clipper. The direction of the diode shows it is a negative clipper
2. Have a look at the circuit diagram, output is taken across the diode and  DC voltage source
3. During the positive half cycle, the diode is reversed biased (open circuit). No voltage drop occurs. A positive half cycle appears at the output
4. During the negative half-cycle, voltage half-cycle, a small portion of the input wave appears at the output, rest of the portion clips
5. We get the output until the diode remains off. When does diode forward bias?
6. The diode turns on when the anode is at a higher potential than the cathode
7. The anode is at -3V. To make it forward bias input is less than -3.7V (consider diode drop)
8. Look at the output waveform (blue), it clips after -3.7V

It is simple, when the diode turns off, the input signal appears at the output.

Example 3: Biased Shunt Clippers Clippers | Double Diode Clippers:

Figure 4 Double diode clippers

I would like to explain the double diode clipper with biasing level 3V.

During the positive half cycle, D2 reverse bias D1 forward bias for a small portion of the input waveform.

1. D1 turns on when the anode is at a higher potential than the cathode.
2. Look at D1, the cathode is at +3V. To turn on (forward bias) the D1, applied voltage greater than 3.7V
3. After 3.7V, no voltage appears at the output
4. Look at the output voltage waveform (blue) which clips off after 3.7V

During the negative half-cycle, D1 reverse bias, and D2 forward bias for a small portion of the input waveform.

1. D2 turns on when the anode is at a higher potential than the cathode
2. Look at D2, the anode is at -3V. To turn on (forward bias) the D2, applied voltage less than -3.7V
3. After -3.7V, no voltage appears at the output
4. Look at the output voltage waveform (blue) which clips off after -3.7V

Applications of clipping circuit:

• It is used in wave shaping in such a way that it limits the peak of the input signal. It also removes the unwanted peaks
• It is used in communication systems, signal processing

In the end:

Dear students! You don't need to apply KVL or solve mesh equations. You just need to have some practice. Use any circuit analysis software. Draw different clipper circuit configurations

Analysis output waveforms. After some practice, you can easily draw the output waveforms.

Diode As A Switch

Diode As A Switch:

Have a closer look at diode symbol. It represents an arrow, which shows the direction of current. It allows current to flow from anode to cathode. As you know, it is a voltage controlled two terminal device. It has characteristics of a switch. In one direction it allows current to flow, while in other direction it blocks the current. In other words, during forward biased conditions, diode has ideally zero resistance. It behaves as a closed switch. While during reverse biased it behaves as an open switch. Diodes use in switching applications are also known as 'signal diodes’.

Figure 1 Diode as a switch. Practical versus ideal diodes

Apply voltage positive V at anode. When anode is more positive than anode, it behaves as a closed switch. Input is directly coupled to the output. And hence current flows from positive to negative terminals.

Now apply positive voltage at cathode. See figure 1. It means cathode is at greater potential than anode. In this situation the diode acts as an open switch between input and output terminals. Consequently no current flows through the diode.

• Signal diodes are common in mixing circuits (mixers)
• Logic gates can also be implemented with the help of switching diodes.

Positive Logic AND Gate Using Diodes:

The figure below (figure 2) is two input positive logic AND gate circuit. It is made up of diodes. Here diode is using as a digital switch.

Let's understand the working of circuit. Input A and input B is tied to cathode. While both anodes are connected to VCC through resistor R1. The output VOut depends on both inputs. If either input is zero output is zero.

VA is voltage at input A

VB is voltage at input B

VA < VCC

VB < VCC

Vout = 0 (logic 0)

At this condition, both diodes are forward biased (closed switch) because both input terminals (cathodes) are at  lower potential than anode. The current flows from diodes and hence output is low.

VA > VCC

VB < VCC

Vout = 0

At this condition, diode D1 is reverse biased. While diode D2 is forward biased. The current flows from diode D2 and hence output is low.

VA < VCC

VB > VCC

Vout = 0

At this condition, D2 is reverse biased. While D1 is forward biased because input A (cathode) is at lower potential than anode. The current flows from D1 and hence output is low.

VA > VCC

VB > VCC

Vout = 1 (logic 1)

At this condition, both diodes are reverse biased (opened switch). No current flows from diodes and hence output is high.

A	B	Out
0	0	0
0	1	0
1	0	0
1	1	1

Figure 2: AND gate using diode

Positive Logic OR Gate Using Diodes:

Look at figure 3. It is two input OR gate, implemented from diodes only. A and B are two inputs, where output is taken across resistor R1. Truth table is given as well.

Let's understand the working of circuit.

Look at the diode position. Input A is at its anode. While the cathode is tied to resistor R1, which is grounded. Same for input B.

When anyone input A or B is positive (logic high) the output is high. That is positive logic (voltage) at anode, makes the diode forward bias. Hence input appears at the output.

VOut = VA = VB ...for ideal diode

VOut = VA - 0.7 … for practical diode

Where VA is voltage at input terminal A

Vout is voltage across resistor R1

A	B	Out
0	0	0
0	1	1
1	0	1
1	1	1

Figure 3 Diode OR gate

Cockcroft Walton Voltage Multiplier | Diode Capacitor Multiplier

Diode as a voltage multiplier:

A very sophisticated method of obtaining high voltage with the help of simple circuit is Cockcroft-Walton voltage multipliers.  It consists of diodes and capacitors only. There are many possible ways to design voltage multipliers, with the help of diodes and capacitors. This article is intended to provide basic information about Cockcroft-Walton voltage multipliers.

Voltage multipliers are used to produce high voltages ( hundreds and thousands of volts). They are commonly used in high voltage low current device like cathode ray tubes, Lasers etc.

Diode capacitor voltage multiplier takes AC as an input, and produce DC output which is multiple (2,4,6) of AC peak voltage.

VO = 2*N*VP    Equation 1

VO = 2*N*VP - Vdrop     Equation 2

Where

N = number of stages,

VP = peak input voltage

Vdrop = diode drop

Equation 2 >> we consider the diode drop.

For example from a three stage multiplier, we get

VO = 2*3*VP

VO = 6*VP

A single stage Cockcroft Walton multiplier consists of 2 diodes and 2 capacitors. Each stage produce an output which is multiple of 2. Hence, the number of diodes and capacitors stages increase as the voltage increases.

For example at input if you have 10V (AC) at input, each stage produce 20 V. For example, we have 3 stages we get 60 V (DC) at output. It means each stage produce 20V. (20+20+20 = 60). And the voltage multiplier circuit contains 6 capacitors and 6 diodes.

VO = 2*N*VP

VO = 2*3*10 = 60V (neglect diode drop)

Basically voltage multiplication depends on charging and discharging of capacitors. Each capacitor charges upto 2VP except  C1 , which charges upto VP. The upper rail of capacitors clamps the signal, while the bottom rail of capacitors smooths the DC.

Diode Capacitor Voltage Doubler | Single Stage Cockcroft Walton Multiplier:

During negative half cycle D1 is forward biased and charges the capacitor C1 to peak voltage Vp, while D2 is reverse biased. During positive half cycle D2 is forward biased. C2 will try to charge up to 2Vp, because C1 and source are in series. See figure below, C2 tries to charge upto 2VP, after several cycles it will reach upto 2VP. The output is taken across capacitor C2.

Figure 1 Single stage multiplier

Look a figure 2. Input 10V (AC)

Output 20V (DC) approximately (we are considering practical diodes)

Voltage at capacitor C1

Figure 2 Output of Single stage multiplier (ideally output is 20 V, practically it is a little bit less because of diode drop as shown in the figure)

Voltage Quadrupler | Two Stage Cockcroft Walton Multiplier:

Figure 3 Two stage voltage multiplier

The voltage quadrupler circuit consists of two stages of Cockcroft-Walton multiplier. The upper rail of capacitors stores and clamps the signal. The output is observed across lower rail of capacitors.

During first negative half cycle D1 and D3 are forward biased. D1 charges the capacitor C1 to peak voltage Vp, while D2 and D4 remain reverse biased. During first positive half cycle D2 and D4 are forward biased. C2 will try to charge up to 2Vp, because C1 and source are in series.

During second negative half cycle, again D1 and D3 are forward biased. C1 already charged upto VP. Apply KVL and calculate the voltage across capacitor C3. It is 2VP. Look at figure 4. During second positive half cycle D2 and D4 are forward biased. Capacitor C4 charges upto 2VP. C2 has already charged upto 2VP. The output is taken across the two capacitors.

Figure 4 Working of voltage quadrupler

Look a figure 5: Waveforms obtained at various points. You can visualize easily. Voltage across C1 is shifted or clamped. Also, voltage across C3 is also clamped. I discussed earlier, upper rail of capacitors clamp or shift the waveforms. While, the lower rail of capacitors produce DC at the output.

Input 10V (AC)

Output 20V (DC) approximately (we are considering practical diodes)

Voltage at capacitor C1

Voltage at capacitor C3

Figure 5 Output of voltage quadrupler