Biased & Unbiased Diode Clampers, Circuit Diagrams & Working

Applications of Diode - Types of Diode Clampers (Positive and Negative clampers, Biased and Unbiased Clampers) - Circuit Diagrams and Working, Waveforms and Comparison 

Another popular diode-based circuit is the clamping circuit. A clamper circuit or network adds DC level to a given AC signal without any change in signal shape. A simple circuit is shown below, comprise of a capacitor, diode and resistor. Carefully choose the capacitance value. The time constant should be large enough to maintain the shape of the output signal. The time constant can be increased by using larger capacitance values. Throughout this article, I consider a practical diode with a drop of 0.7V.

V1 is the sinusoidal input

VP is the peak voltage

Vo is the output voltage

Outline:
In this article, I explain different types of clamper circuits, like biased clampers and unbiased clampers, positive and negative clampers.

Comparison between biased positive clampers and unbiased positive clampers. 

Comparison between biased negative clampers and unbiased negative clampers.

Positive Clamper Circuit Diagram and Working:

It clamps the incoming signal to an upward direction. See the circuit below (figure 1). The diode arrow points upward. It means the signal is going to shift in an upward direction.

During the negative half-cycle, the diode turns on. As a result, the capacitor charges up to peak value (VP - 0.7V) that is 9.3V. (Considering practical diode)

During the positive half cycle, the diode opens (reverse bias). No path to flow current. To determine the output voltage that appears across the resistor, apply KVL to the loop.

Vo =  V1 + VP

Figure 1 Positive Clampers, capacitance value should high to keep time constant large

Negative Clamper Circuit Diagram and Working:

It clamps the incoming signal to a downward direction. See the circuit below (figure 2). The diode arrow points downward. It means the signal is going to shift in a downward direction. Each point on the sine wave is shifted to a downward direction by an amount of VP.

During the positive half cycle, the diode turns on. As a result, the capacitor charges up to peak value (VP) which is 9.3 V (considering diode drop).

During the negative half-cycle, the diode opens (reverse biased). To determine the output voltage that appears across the resistor, apply KVL to the loop.

Vo = - Vi - VP

Figure 2 Negative Clampers

Positive Biased Clamper Circuit Working:

During the negative half-cycle, the diode is reverse biased for a small part of the input signal. This is because of DC source polarity (V2 = 2V). The anode is more positive than the cathode for a small part of the input waveform (V1 <V2). For higher values of V1 (V1> 2V) the diode turns on and capacitor charges with the polarity shown.

V1 - VC1 - 0.7 - 2 = 0

VC1 = 7.3V

It means the capacitor clamps the output signal to a voltage level of 7.3V.

During the positive half cycle, the diode remains to turn off. The capacitor holds its charge because of the larger capacitance value.

The voltage across the load resistor is

-V1 -VC1+Vo= 0

Vo = 17.3 V

Output voltage clamps and swings from 17.3 V to -2.7 V.

See in next section where I compare both positive clampers Vs positively biased clampers.

Figure 3 Positive Biased Clampers

Negative Biased Clamper Clamper Circuit Working:

During the positive half cycle, the diode remains to turn off for a small part of the input waveform. This is because of the DC source (3V) at the anode. To make the diode forward biased, the anode should more positive than the cathode.

V1 > (3 + 0.7)V …. Turn on condition

Where

V1 = input voltage

0.7V is the forward voltage drop

As the diode turns on, the capacitor charges till the peak value.

-V1 + VC1+0.7+3 = 0

VC1 = 10 - 3.7

VC1 = 6.3 V

During the negative half-cycle, the diode remains off. To determine output voltage, apply KVL to the loop.

V1 + VC1 + Vo = 0

10  + 6.3 + Vo = 0

Vo = -16.3 V

Output voltage clamps and swings from -16.3 V to 3.7 V.

Figure 4 Negative biased clampers

Compare the results:

Compare the unbiased clamper with the biased clamper circuit. What is the effect of adding a DC source in series with the diode?

Positive Clamper (Unbiased) Vs Positive Biased Clamper:

Compare the results obtained from biased and unbiased circuits.

Unbiased clampers give clamped waveforms. The clamping value is equal to the peak input voltage.

For example, if we have a peak input voltage equal to 10V (peak to peak), swings from +10V to -10V.

The unbiased clamper produces an output that is exactly similar to input but has added DC value. The output signal clamped at 9.3V. It swings from +20 to 0V.

Figure 5 Positive clampers Vs positive biased clampers

Biased Clampers also give clamped waveforms. The difference is the clamping level. Capacitor charging describes the clamping level. In the above circuit positive biased clampers, the capacitor charges up to 7.3V.

We have peak input voltage equal to 10V (peak to peak), swings from +10V to -10V.

The biased clamper produces an output that is exactly similar to input but has added DC value. The output signal clamped at 7.3 V. It swings from +17.3 to -2.7V.

Negative Clamper (Unbiased) Vs Negative Biased Clamper:

Compare results obtained from negative clampers and negative biased clampers.

Unbiased Clampers give clamped waveforms. The clamping value is equal to the negative peak of the input current voltage.

We have input voltage equal to 10V (peak to peak), swings from +10V to -10V.

The negative clamper added a DC level. The output signal swings from 0.7V to -19.3V.

Figure 6 Negative Clampers Vs negative biased clampers

Biased Clampers: In the above example of negatively biased clamper, we have a 3V DC voltage source. Due to this source, we have a different clamping level. In this case, the capacitor charges up to 6.3V. 

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