Clamper

 The word Clamp means 'to hold things together or fasten two things together'. In electronics it means 'hold voltage level of a wave at a particular value which is different from its normal.

Clamping circuit is used to shifts a signal by a dc voltage. How it is possible to shift a sine wave?

Fig a

This dc level shifting is explained in Fig a. Dc level of  sine wave (leftmost figure) is at zero volt. Sine waves having dc level at +2v and dc level at -2v are followed. A clamping circuit can be used for this purpose. 

Clamper circuits -section I

A simple clamping circuit is given below in fig 1. The circuit consists of a capacitor and a diode. Across the diode output voltage is taking. Here the capacitor act as a battery(Capacitor has a charge storing capability and thus can be act as a cell. At the first instant of switching on the circuit, capacitor is charged to maximum value of input voltage(Vm). It will maintain this voltage in it until the input is switched off). In fig 1, positive side of the diode is at the top side and so leftmost plate  of the capacitor is positive(current direction through the diode is from top to bottom and that in capacitor is from left to right - so left plate is taken as positive. Since right plate is negative, donation to output voltage by the capacitor is negative).

fig 1

To get an idea about how the output is obtained, keep these two things in mind:
1.During positive half cycle of input voltage, diode is forward biased and no input voltage is dropped across diode(so no donation to output voltage by diode). But our capacitor battery is there, and its voltage(Vc) is equal to maximum value of input sine wave(Vm). Then have a look into voltages in the circuit.

 Positive half cycle of the input sine wave is shown with Vm=10v, its polarity is also indicated. Capacitor is charged to 10v with polarity as shown. Since diode is forward biased it is represented with a small forward resistance. Output voltage is taking across points 'a' and 'd'(Vad). Vad = Voltage across 'abcd' =  -Vm+Vin = -10v+ Vin. (Vin is varying from 0 to +10 then to 0). So -10v +Vin will be as follows:

                                                This is how the output in fig 1 is obtained during positive half of input.

2. During negative half cycle of input, diode is in reverse bias and it will behave as an open circuit.

Now have a look into polarities of input voltage and capacitor. Since capacitor charges only during the positive half cycle of input sine wave, polarity is same. But because of negative half cycle of input, polarity of Vin is reversed. Output voltage is taken across points 'a ' and 'd' (Vad). Vad = Voltage acrosss 'abcd' =  -Vm -Vin  =  -10v -Vin. 

This explanation is only for keeping in your mind, don't pour it into your papers. Only one thing you need to do is --- assign a polarity for the capacitor according to forward bias condition of diode. If  -ve polarity is nearer to diode, keep in mind that output will be in the -ve side of Y axis. Keeping circuit of fig 1 in mind, it is easy to find outputs of fig 2 and fig3. Lets see..

fig 2

 Compare outputs of fig 1 and fig 2. If output of fig 1 is slightly shift towards top we will get output of fig 2. This shift is because of the presence of positive cell voltage Vr.

fig 3

Similarly if we shift the output of fig 1 towards bottom, output of fig 3 will be obtained. This is due to negative battery voltage Vr attached to the diode.

Clamper circuits - Section II

It is easy to diegest  the remaining circuits if we relate them to that in section I. Just reverse the direction of diode in fig 1, then look below. Yaa, next circuit.

fig 4

 With the reversal of diode, don't forget to reverse the polarity of capacitor - here current direction through the capacitor is taken from right plate to left plate. A tip to remember - if the capacitor plate nearer to the diode is positive then output wave will be in the positive side of y axis.

Follow the same route of transition from fig 1 to fig 2, for fig 4 to fig 5..

fig 5

 When a  positive bias is introduced into fig 4, we get fig 5 and waveform shifts up.

Then again find an old route, that is from fig 1 to fig 3 and follow it for fig 4 to fig 6.

fig 6

With an extra negative bias in fig 6 compared to fig 4, waveform shifts down.



Easy???

Combinational Clipper Circuit

This is a continuation of 'Clipping Circuits'. In a simple clipper circuit either +ve or -ve half cycle of sine wave can be altered at a time. Using a combination clipper circuit, both half cycles can be shaped using a single circuit.

fig a

 Circuit in fig a is a combination of fig 2 and fig 5 in 'Clipping circuits'. Its easy to understand the output waveform by considering +ve half cycle of fig 2 output and -ve half cycle of fig 5 output.

fig b

Circuit in fig b is a combination of fig 2 and fig 6. Here also the output can be easily find out by combining output of fig 2 and fig 6. (In fig 6, bottom side of +ve half cycle of sine wave is removed. In fig 2, topside of +ve half cycle of sine wave is removed. Apply some logic and combine two - in a +ve half cycle of sine wave top and bottom side are removed - that is given in above fig b.Please forget about the -ve side of sine wave - there is no alternation for -ve side in fig 2 and nothing is there in the -ve side of fig 6.)

fig c

Circuit in fig c is obtained from fig 3 and fig 5. By removing both sides of -ve half cycle of sine wave, output of fig c is obtained.

Clipping Circuits

Clipping circuits are used as wave shaping circuits. Usually it is used to shape sine waves by cutting off some of its parts, but it can be used for other waves also.The idea will be clear if we familiar with the simplest member of clipper family. 

An example of a clipper circuit is Half wave rectifier. Circuit, Output voltage ('Vo' vs 't' -obtain using CRO) and Transfer characteristics  (' Vo' vs 'Vin' -obtain using CRO) are shown.

fig 1

Here i take two circuits -one in top and other in the bottom position, both give same output, hence same transfer chara. Though you know very well about the working, am just briefing it - for my convenience. Circuit in the top consists of a resistor(R) and a diode(D). Output takes across D ie. CRO probes are connect across two ends of D. 

For positive half of input sine wave, D is in forward bias - current is flowing without any resistance, so voltage zero(diode ON). //A voltage is developed only across a resistance, if resistance of a component is zero then voltage across it is also zero. But this an ideal situation only - not practical. Every component, even 'diode in forward bias' have a small resistance (ON resistance), therefore a small voltage across it ie. 0.7v //

For negative half cycle of sine wave, D is in reverse bias. No current flow through D. Diode act as 'Open circuit'. So whatever the voltage applied in the input side, is completely appeared across D - in other words, output follows sine wave's(input) negative half cycle. //When there is no current flow through a component, resistance of it is taken as infinity(resistance=voltage/current). Comparing to this infinite resistance, all other resistances in the concerned circuit are negligible(even though it has any value -K ohm/M ohm/..). We know V=RI, since resistance is infinite, voltage is very high(not infinite). Therefore whatever the voltage given as input to the circuit, that full voltage is dropped across this infinite resistance. ie. voltage across an 'open circuit' = input voltage --all other components in the concerned circuit have negligible voltage.// 

   We can easily draw transfer chara using input and output waves. Below given explanation will help you for it.

Transfer characteristics : It means input vs output.When both input and output are sine waves, transfer chara is 

fig a

When Vin increases +vely (in transfer chara, from (0,0) towards right), Vout also increases +vely(in transfer chara, from(0,0) towards top). Result is a straight line with 45 degree slope. Similarly for -ve half cycle also.   

Now let's see how transfer characteristics is obtained in fig 1. Take fig a, as reference. Consider transfer chara of fig 1. For +ve half of Vin, Vout is constant at 0.7v - so transfer characteristics is a straight line at 0.7v(output is constant at 0.7v irrespective of change in input). It is indicated both in output waveform and transfer chara. For -ve half of Vin, fig 1 is same as fig a.

Hope the above explanation is satisfactory. Now move to next clipper circuit.

fig 2

 Explanation for obtaining such an output is as follows: We are adding a +ve bias (+ve voltage) in the basic circuit(fig1) using a battery and it shifts the cut off voltage of diode from 0.7v to (Vr + 0.7v). Vr is the voltage of battery(we can use 5v or 7v or any other, but should be less than Vin). You can easily understand its transfer chara by comparing it with fig a.

Next clipper is obtained by adding a -ve bias to circuit in fig 1. It shifts the cut off voltage to (-Vr + 0.7v).

fig 3

 It is easy to remember the output of fig 2 and fig 3 by relating it to output of fig 1.

(i)adding Vr to output of fig1     (iii)output of fig2

(i)adding -Vr to output of fig1     (iii)output of fig3

No need to study theoretical explanation of how to obtain the output for a particular clipper circuit. Such essay questions are not yet asked for theory exams. But short answer questions may be asked and it will be like 'given a transfer chara. draw the corresponding circuit'. To answer such questions, first you draw the output wave form using given transfer chara and then the circuit.

Next look into next set of clipper circuits. Here circuit in fig 4 is taken as the basic circuit and is obtained by reversing the diode in fig 1.

fig 4

In fig 4 cut off is at (-0.7v) and is an example of a half wave rectifier. Now see what happens if a negative bias is added to circuit in fig4.

fig 5

A - ve bias of (-Vr) is added to fig 4 to obtain circuit in fig 5. Then cut off voltage changes from -0.7 to (-0.7-Vr). It is adopted in output waveform and transfer characteristics. Similarly if a +ve bias of Vr is added to circuit in fig4, we will get circuit in fig 6. Cut off voltage is shifted to (-0.7+Vr). //For example, battery voltage Vr =5v. Then cut off voltage is -0.7+5 = +4.3//

fig 6

These are the simple clipping circuits. Once you familiarized them, move to combination clippers

RC Differentiator

High-pass RC circuit as Differentiator:

Adding some ‘salt and pepper’ over high pass RC circuit will give differentiator.
In differentiator output voltage, Vout is proportional to the rate of change of input.

     

Circuit diagram:

As in the case of integrator, when the high pass circuit obey some condition, it act as a differentiator. Shall we see how it is?

Lets write the equation for output voltage.



//output is taken across R and R+1/CS is the total resistance of the circuit. So applying voltage divider rule, Vout = Vin{ R/(Total resistance) } //

Laplace transform s=jw (w=2πf) and RC is the time constant.



Rearranging,
     
 
Take the condition that RCw << 1, then

 

// When RCw << 1, then 1/RCjw become large, so 1+1/RCjw is approximately equal to 1/RCjw. Now the equation become
//

We know jw = S, Laplace transform. So

therefore, Vout = RC S Vin

The above equation is a Laplace equation and we take inverse Laplace, then equation become,



Equation shows that output voltage is proportional to derivative of input voltage. So the circuit represents a differentiator circuit.
OK!! then go through output of the differentiator circuit for different inputs.

                                                                                                           



                                                 
                        
















//In lab hours, select different R and C and check output.Also check output just by changing input frequency(by adjusting front nob of  signal generator) //

 If we make a chicken fry, what is next?. Check its taste, whether salt, chilly etc are ok. Likewise if an output is obtained some people check its different properties and we people study it. So two properties of output wave form.

Rise time (tr): The time taken for the output voltage to rise from 10%  to 90% of final value.
//if peak is 100 v, then how much time the wave will take to rise from 10 v to 90v.//

Delay time (td): The time taken for the output voltage to rise from 0 to 10% of its final value.

Low pass RC - integrator and high pass RC - differentiator are very important in lab exams. It will ask in combination with Rectifiers. For theory exams it is not a usual question. Something bigger is going to happen... 

Come back with some interesting circuits..

RC Integrator

How Low pass RC filter work as an Integrator ?- It is explained here.

Low pass RC circuit is explained in detail here - Low pass RC filter , plz visit. Then move to next course - but have same ingredients -that's Integrators . Oops ...Integration. Is this our AC1 or Maths??? ....

Arranging the table for serving...

In electronics also there are Integrators and differentiators because of the following reasons. Many circuits need signals other than sine wave, like square wave, cosine wave, triangular wave, pulses etc. But the basic signal available as input to any device is our Alternating Current, which is sine wave. Square wave can be produced easily either from sine wave or from a dc voltage with the help of some simple circuits. By integrating and differentiating these two waves(square and sine), it is  possible to produce other waves also. In order to realize integration and differentiation, we are   using integrator and differentiator circuits.

Integrator

An integrator produces an output which is proportional to integral of its input.




Circuit diagram:

AC – AC source that provides the input - Vin
Output – Output voltage
Resistors(R) and Capacitors(C) are connected in series with Vin.



Low pass RC circuit itself is acting as an integrator provided it obey one condition. Lets check that condition.


Equation for output voltage          
                                       1/CS –reactance offered by C in terms of Laplace transform

               

Laplace transform s=jw and w=2πf (w-omega)

                  
if we are adopting a condition as wRC >> 1, then , in denominator 1+jwRC is nearly equal to jwRC. Then equation become

                 

Re arranging,

                        Here 1/RC is taken as a constant


Taking inverse Laplace transform,

                

//Those who are not familiar with Laplace-- while converting integral into Laplace 1/s is adding with the equation. During inverse Laplace, 1/s indicates an integration.//


Above equation shows that Vout is proportional to integral of input voltage. So if we have a low pass RC circuit which obeys the condition wRC >> 1, it will act as integrator.


//The product RC is known as time constant(T). With the value of RC there will be changes in output waveform.//


In order to achieve a good integration, the following conditions must be satisfied:

• The time constant RC of the circuit should be very large as compared to the time period of the input signal.
• The value of R should be 10 or more times larger than Xc.

Output for different input waves:


1. For Step input:

 2. For square wave input


Three output waveforms are shown. These are with increase in values of time constant ‘T’.

Applications:
Use to perform integration in certain electronic circuits.
Use to convert square wave into triangular wave.

High pass RC circuit

We can hear so many sounds in nature. These sounds may have different frequencies. How we can separate a particular type of sound(waves) from others?..We can do it by using a filter circuit. A filter circuit helps us to 'filter out' that waves having particular frequencies. Many circuits are used as filters. Here, explain the circuit diagram and working of a high pass RC circuit(high pass filter). Frequency response  is also given. This is only a stepping stone to differentiator. 

Passive High pass filter

A first order high pass RC circuit contains one pair of Resistor(R) and Capacitor(C). It act as a simple high pass filter - it allows high frequency (input)signals to appear at the output side while attenuating low frequency signals.

Circuit diagram:

             Input is an AC source, Output is taken across resistor.Contains one pair of R & C.

Working:

 Voltage across R is taken as output voltage. Let’s see what happened to this voltage at low and high frequencies.For refreshments of basic concept see Refreshments.

For high frequency input: 

Xc is very low(Xc inversely proportional to frequency). So Vc is also low. For very high frequencies, magnitude of Vc is very less and so we can say that Vc << V_R, where V_R is the output voltage. In conclusion, for high frequency input, output is higher.

For low frequency input:
Xc is very high.Vc is also high. Therefore for lower frequencies V_R<< Vc. So output voltage V_R is much less for low frequency input.

If you are comfortable with 'working', directly go to 'Summary'. 
Feel something odd..   just look below.

When  working with filter circuit , we are keeping the amplitude of input AC constant(here 6v) and varying its frequency.
First case: we give a high frequency input(fig a), say 10MHz. For this suppose Xc = 10. Then the 6v input is divided as Vc : V_R =  Xc : R  = 1:2. V_R = 4v. Then Vc= ?? (2v).Vc <V_R .
Second case: increase input frequency as 50MHz. Suppose in this case Xc =5. But R doesn’t depend on frequency, so fixed value R=20 . Then the 6v input is divided as Vc : V_R = Xc : R  = 5:20 = 1:4 . V_R = 4.8v. Then Vc = 1.2v. 
This example shows that as frequency increases, Vc is also increases. So at higher frequencies it is taken as Vc>>V_R .

Then check for a low frequency input(fig b), say 10KHz. Suppose Xc = 40. Find V_R. Then you will get, why V_R << Vc for low frequencies?.

Summarising : 
for high frequency input signal   -----   output voltage is larger
for low frequency input signal    -----   output voltage is lesser

Then we can call it as a “high pass”.

Equation for output voltage

 Vout = V_R =             where Vin indicates amplitude of input signal

Frequency response of first order High pass filter:

A frequency response gives you an idea about 'the amplitude of  wave at different frequencies'. So in X-axis, we plot 'frequency' and in Y-axis, Gain(Gain= amplitude of output wave/amplitude of input wave).

Frequency response of High pass filter