Differentiating chains. High pass filter. Differentiation circuits Differentiation in electronics
We have every right to move on to the consideration of circuits consisting of these elements :) This is what we will do today.
And the first circuit whose operation we will consider is differentiating RC circuit.
Differentiating RC circuit.
From the name of the circuit, in principle, it is already clear what kind of elements are included in its composition - a capacitor and a resistor :) And it looks like this:
The operation of this scheme is based on the fact that current flowing through a capacitor, is directly proportional to the rate of change of voltage applied to it:
The voltages in the circuit are related as follows (according to Kirchhoff’s law):
At the same time, according to Ohm's law we can write:
Let's express it from the first expression and substitute it into the second:
Assuming that (i.e. the rate of change of voltage is low) we get an approximate dependence for the output voltage:
Thus, the circuit fully lives up to its name, because the output voltage is differential input signal.
But another case is also possible, when title="Rendered by QuickLaTeX.com" height="22" width="134" style="vertical-align: -6px;"> (быстрое изменение напряжения). При выполнении этого равенства мы получаем такую ситуацию:!}
That is: .
It can be noted that the condition will be better satisfied for small values of the product, which is called circuit time constant:
Let's figure out the meaning of this characteristic of the circuit :)
The charge and discharge of the capacitor occurs according to the exponential law:
Here is the voltage across the charged capacitor at the initial time. Let's see what the voltage value will be after time:
The voltage across the capacitor will decrease to 37% of the original.
It turns out that this is the time during which the capacitor:
- when charging – will charge up to 63%
- when discharged - discharged by 63% (discharged up to 37%)
Now that we've figured out the time constant of the circuit, let's go back to differentiating RC circuit 🙂
We've covered the theoretical aspects of the circuit's functioning, so let's see how it works in practice. And to do this, let’s try to apply some signal to the input and see what happens at the output. As an example, let’s apply a sequence of rectangular pulses to the input:
And here’s what the oscillogram of the output signal looks like (the second channel is blue):
What do we see here?
Most of the time, the input voltage is constant, which means its differential is 0 (derivative of the constant = 0). This is exactly what we see on the graph, which means the chain performs its differentiating function. What are the reasons for the bursts in the output oscillogram? Everything is simple - when the input signal is “turned on,” the process of charging the capacitor occurs, that is, a charging current passes through the circuit and the output voltage is maximum. And then, as the charging process proceeds, the current decreases exponentially to zero, and along with it the output voltage decreases, because it is equal to . Let's zoom in on the waveform and then we'll get a clear illustration of the charging process:
When the signal is “turned off” at the input of the differentiating circuit, a similar transient process occurs, but it is caused not by charging, but by discharging the capacitor:
In this case, the time constant of the circuit is small, so the circuit differentiates the input signal well. According to our theoretical calculations, the more we increase the time constant, the more similar the output signal will be to the input. Let's check this in practice :)
We will increase the resistance of the resistor, which will lead to an increase:
There’s no need to comment anything here - the result is obvious :) We have confirmed the theoretical calculations by conducting practical experiments, so let’s move on to the next question - to integrating RC circuits.
Let's write down the expressions for calculating the current and voltage of this circuit:
At the same time, we can determine the current from Ohm's Law:
We equate these expressions and get:
Let's integrate the right and left sides of the equality:
As is the case with differentiating RC chain There are two possible cases here:
In order to make sure that the circuit is working, let's apply exactly the same signal to its input as we used when analyzing the operation of the differentiating circuit, that is, a sequence of rectangular pulses. At small values, the output signal will be very similar to the input signal, and at large values of the circuit time constant, at the output we will see a signal approximately equal to the integral of the input. What kind of signal will this be? The sequence of pulses represents sections of equal voltage, and the integral of the constant is a linear function (). Thus, we should see a sawtooth voltage at the output. Let's check the theoretical calculations in practice:
The yellow color here shows the input signal, and the blue color, respectively, shows the output signals at different values of the circuit time constant. As you can see, we got exactly the result we expected to see :)
This is where we end today’s article, but we don’t finish studying electronics, so see you in new articles! 🙂
And together they form an RC circuit, that is, it is a circuit that consists of a capacitor and a resistor. It's simple ;-)
As you remember, a capacitor consists of two plates at some distance from each other.
You probably remember that its capacity depends on the area of the plates, the distance between them, and also on the substance that is between the plates. Or the formula for a flat capacitor:
Where
Okay, let's get to the point. Let us have a capacitor. What can you do with it? That's right, charge it;-) To do this, take a constant voltage source and apply a charge to the capacitor, thereby charging it:
As a result, our capacitor will charge. One plate will have a positive charge, and the other plate will have a negative charge:
Even if we remove the battery, we will still have a charge on the capacitor for some time.
Charge retention depends on the resistance of the material between the plates. The smaller it is, the faster the capacitor will discharge over time, creating leakage current. Therefore, the worst in terms of charge retention are electrolytic capacitors, or popularly electrolytes:
But what happens if we connect a resistor to the capacitor?
The capacitor will discharge as the circuit becomes closed.
RC circuit time constant
Anyone who knows even a little about electronics understands these processes perfectly. This is all banality. But the fact is that we cannot observe the process of discharging a capacitor just by looking at the circuit. For this we need a signal recording function. Fortunately, I already have a place for this device on my desktop:
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So, the action plan will be this: we will charge the capacitor using the power supply, and then discharge it through a resistor and watch the oscillogram of how the capacitor is discharged. Let's assemble a classic circuit that can be found in any electronics textbook:
at this moment we charge the capacitor
then we switch the toggle switch S to another position and discharge the capacitor, observing the process of discharging the capacitor on an oscilloscope
I think this is all clear. Well, let's start assembling.
We take a breadboard and assemble the circuit. I took a capacitor with a capacity of 100 μF, and a resistor of 1 KiloOhm.
Instead of the S toggle switch, I will manually toss the yellow wire.
Well, that's it, we hook the oscilloscope probe to the resistor
and watch the oscillogram of how the capacitor is discharged.
Those who are reading about RC circuits for the first time, I think, are a little surprised. Logically, the discharge should proceed in a straight line, but here we see a problem. The discharge occurs according to the so-called exponential . Since I do not like algebra and mathematical analysis, I will not give various mathematical calculations. By the way, what is an exponent? Well, an exponential is a graph of the function “e to the power of x”. In short, everyone went to school, you know better ;-)
Since when we close the toggle switch we have an RC circuit, it has such a parameter as RC circuit time constant. The time constant of an RC circuit is denoted by the letter t, in other literature it is denoted by a capital letter T. To make it easier to understand, let's also denote the time constant of an RC circuit by a capital letter T.
So, I think it's worth remembering that the time constant of an RC circuit is equal to the product of the resistance and capacitance ratings and is expressed in seconds, or by the formula:
T=RC
Where T– time constant, Seconds
R– resistance, Ohm
WITH– capacitance, Farads
Let's calculate what the time constant of our circuit is. Since I have a capacitor with a capacity of 100 μF and a resistor of 1 kOhm, the time constant is T = 100 x 10 -6 x 1 x 10 3 = 100 x 10 -3 = 100 milliseconds.
For those who like to count with their eyes, you can plot a level of 37% of the signal amplitude and then approximate it to the time axis. This will be the time constant of the RC circuit. As you can see, our algebraic calculations almost completely agreed with the geometric ones, since the cost of dividing the side of one square in time is 50 milliseconds.
Ideally, the capacitor charges immediately when voltage is applied to it. But in reality there is still some resistance from the legs, but we can still assume that the charge occurs almost instantly. But what happens if you charge a capacitor through a resistor? Let's disassemble the previous scheme and cook up a new one:
starting position
as soon as we close the S key, our capacitor begins to charge from zero to a value of 10 Volts, that is, to the value that we set on the power supply
We observe the oscillogram taken from the capacitor
Did you see anything in common with the previous oscillogram, where we discharged a capacitor to a resistor? Yes, that's right. The charge also proceeds exponentially ;-). Since our radio components are the same, the time constant is also the same. Graphically, it is calculated as 63% of the signal amplitude
As you can see, we got the same 100 milliseconds.
Using the formula for the time constant of an RC circuit, it is easy to guess that changing the values of the resistance and capacitor will entail a change in the time constant. Therefore, the smaller the capacitance and resistance, the shorter the time constant. Consequently, charging or discharging will occur faster.
For example, let's change the capacitance value of the capacitor downward. So, we had a capacitor with a nominal value of 100 µF, and we will put 10 µF, leaving a resistor of the same nominal value of 1 kOhm. Let's look again at the charge and discharge graphs.
This is how our 10 µF capacitor is charged
And this is how it discharges
As you can see, the time constant of the circuit has decreased significantly. Judging by my calculations, it became equal to T=10 x 10 -6 x 1000 = 10 x 10 -3 = 10 milliseconds. Let's check in a graphical-analytical way, is this true?
We construct a straight line on the charge or discharge graph at the appropriate level and approximate it to the time axis. It will be easier on the discharge graph ;-)
One side of the square along the time axis is 10 milliseconds (just below the working field it says M:10 ms), so it’s easy to calculate that our time constant is 10 milliseconds ;-). Everything is elementary and simple.
The same can be said about resistance. I leave the capacitance the same, that is, 10 μF, and change the resistor from 1 kOhm to 10 kOhm. Let's see what happened:
According to calculations, the time constant should be T=10 x 10 -6 x 10 x 10 3 = 10 x 10 -2 = 0.1 second or 100 milliseconds. Let's look at it graphically and analytically:
100 milliseconds ;-)
Conclusion: the higher the value of the capacitor and resistor, the greater the time constant, and vice versa, the lower the value of these radioelements, the less the time constant. It's simple ;-)
Okay, I think this is all clear. But where can this principle of charging and discharging a capacitor be applied? It turns out that a use has been found...
Integrating circuit
Actually the scheme itself:
What will happen if we feed it a rectangular signal with different frequencies? The Chinese function generator comes into play:
We set the frequency on it to 1 Hertz and a swing of 5 Volts
The yellow oscillogram is a signal from the function generator, which is fed to the input of the integrating circuit at terminals X1, X2, and from the output we remove the red oscillogram, that is, from terminals X3, X4:
As you may have noticed, the capacitor almost completely has time to charge and discharge.
But what happens if we add frequency? I set the frequency on the generator to 10 Hertz. Let's see what we got:
The capacitor does not have time to charge and discharge before a new rectangular pulse arrives. As we can see, the amplitude of the output signal has dropped very much; one might say, it has shrunk closer to zero.
And a signal of 100 Hertz left nothing of the signal at all, except for subtle waves
A 1 Kilohertz signal at the output did not produce anything at all...
Of course! Try recharging the capacitor with such frequency :-)
The same applies to other signals: sinusoid and triangular. everywhere the output signal is almost zero at a frequency of 1 Kilohertz and above.
“Is that all the integrating circuit can do?” – you ask. Of course not! This was just the beginning.
Let's figure it out... Why did our signal begin to close to zero as the frequency increased and then disappeared altogether?
So, firstly, we get this circuit as a voltage divider, and secondly, the capacitor is a frequency-dependent radio element. Its resistance depends on frequency. You can read about this in the article capacitor in direct and alternating current circuits. Consequently, if we supplied direct current to the input (direct current has a frequency of 0 Hertz), then at the output we would also receive the same direct current of the same value that was driven to the input. In this case, the capacitor doesn’t care. All he can do in this situation is stupidly charge exponentially and that’s it. This is where its fate in the direct current circuit ends and it becomes a dielectric for direct current.
But as soon as an AC signal is applied to the circuit, the capacitor comes into play. Here its resistance already depends on frequency. And the larger it is, the less resistance the capacitor has. Formula for capacitor resistance versus frequency:
Where
X C is the resistance of the capacitor, Ohm
P– constant and equals approximately 3.14
F– frequency, Hertz
WITH– capacitance of the capacitor, Farad
So what is the result? What happens is that the higher the frequency, the lower the resistance of the capacitor. At zero frequency, the resistance of the capacitor ideally becomes equal to infinity (put the frequency 0 Hertz in the formula). And since we have a voltage divider
therefore, less voltage drops across less resistance. As the frequency increases, the resistance of the capacitor decreases greatly and therefore the voltage drop across it becomes almost 0 Volt, which we observed on the oscillogram.
But the good stuff doesn’t end there.
Let's remember what a signal with a constant component is. This is nothing more than the sum of an alternating signal and a constant voltage. By looking at the picture below, everything will become clear to you.
That is, in our case we can say that this signal (below in the picture) contains a constant component, in other words, a constant voltage
In order to isolate the constant component from this signal, we just need to drive it through our integrating circuit. Let's look at all this with an example. Using our function generator, we will raise our sinusoid “above the floor”, that is, we will do it like this:
So, everything is as usual, yellow is the input signal of the circuit, red is the output signal. A simple bipolar sine wave gives us 0 Volts at the output of the RC integrating circuit:
To understand where the zero signal level is, I marked them with a square:
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Now let me add a constant component to the sine wave, or rather a constant voltage, since the function generator allows me to do this:
As you can see, as soon as I raised the sine “above the floor”, at the output of the circuit I received a constant voltage of 5 Volts. It was by 5 Volts that I raised the signal in the function generator ;-). The chain extracted the DC component from the elevated sinusoidal signal without problems. Miracles!
But we still haven’t figured out why the circuit is called integrating? Anyone who studied well at school, in grades 8-9, probably remembers the geometric meaning of the integral - this is nothing more than the area under the curve.
Let's look at a bowl of ice cubes in a two-dimensional plane:
What will happen if all the ice melts and turns into water? That’s right, the water will evenly cover the basin in one plane:
But what will this water level be? That's right - average. This is the average of these ice cube towers. So, the integrating chain does the same thing! Stupidly averages the signal value to one constant level! It can be said to average the area to one constant level.
But the most relish comes when we apply a rectangular signal to the input. Let's do just that. Let's apply a positive square wave to the RC integrating circuit.
As you can see, the constant component of a meander is equal to half its amplitude. I think you would have already guessed it yourself if you imagined a bowl with ice cubes). Or just calculate the area of each pulse and spread it evenly over the oscillogram, like gov... like butter on bread;-)
Well, now comes the fun part. Now I will change the duty cycle of our rectangular signal, since the duty cycle is nothing more than the ratio of the period to the pulse duration, therefore, we will change the duration of the pulses.
Reducing the pulse duration
I increase the duration of the pulses
If no one has noticed anything yet, just take a look at the level of the red waveform and everything will become clear. Conclusion: by controlling the duty cycle, we can change the level of the DC component. This is precisely the principle behind PWM (Pulse Width Modulation). We'll talk about it someday in a separate article.
Differentiation chain
Another dirty word that comes from mathematics is differentiating. The head immediately begins to hurt from just their pronunciation. But where to go? Electronics and mathematics are inseparable friends.
And here is the differential chain itself
In the circuit we only swapped the resistor and capacitor in places
Well, now we will also carry out all the experiments, as we did with the integrating circuit. To begin with, we apply a low-frequency bipolar square wave with a frequency of 1.5 Hertz and a swing of 5 Volts to the input of the differential circuit. The yellow signal is the signal from the frequency generator, the red signal is from the output of the differential chain:
As you can see, the capacitor manages to almost completely discharge, so we got such a beautiful oscillogram.
Let's increase the frequency to 10 Hertz
As you can see, the capacitor does not have time to discharge before a new impulse arrives.
The 100 Hertz signal made the discharge curve even less noticeable.
Well, let's add the frequency to 1 Kilohertz
Whichever is at the input, the same is at the output;-) With such a frequency, the capacitor does not have time to discharge at all, so the tips of the output pulses are smooth and even.
But the good things don’t end there either.
Let me raise the input signal above “sea level”, that is, I will bring it completely to the positive part. Let's see what happens at the output (red signal)
Wow, the red signal remains the same in shape and position, look - there is no constant component in it, like in the yellow signal that we supplied from our function generator.
I can even output the yellow signal to the negative region, but at the output we will still get the variable component of the signal without any hassle:
And in general, even if the signal has a small negative constant component, we will still get a variable component at the output:
The same applies to any other signals:
As a result of experiments, we see that the main function of a differential circuit is to separate the variable component from a signal that contains both variable and constant components. In other words, it is the separation of alternating current from a signal that consists of the sum of alternating current and direct current.
Why is this happening? Let's figure it out. Consider our differential circuit:
If we look closely at this circuit, we can see the same voltage divider as in the integrating circuit. A capacitor is a frequency-dependent radio element. So, if we apply a signal with a frequency of 0 Hertz (direct current), then our capacitor will stupidly charge and then completely stop passing current through itself. The chain will be broken. But if we supply alternating current, then it will also begin to pass through the capacitor. The higher the frequency, the lower the resistance of the capacitor. Consequently, the entire alternating signal will drop across the resistor, from which we are just removing the signal.
But if we supply a mixed signal, that is, alternating current + direct current, then at the output we will simply get alternating current. We have already seen this from experience. Why did this happen? Yes, because the capacitor does not allow direct current to pass through itself!
Conclusion
The integrating circuit is also called a low-pass filter (LPF), and the differentiating circuit is also called a high-pass filter (HPF). More details about filters. To make them more accurately, you need to carry out a calculation for the frequency you need. RC circuits are used wherever it is necessary to isolate a direct component (PWM), an alternating component (interstage connection of amplifiers), isolate the front of a signal, make a delay, etc. As you delve deeper into electronics, you will often encounter them.
In pulsed devices, the master generator often produces rectangular pulses of a certain duration and amplitude, which are intended to represent numbers and control elements of computing devices, information processing devices, etc. However, for the correct functioning of various elements, in the general case, pulses of a very specific shape other than rectangular are required , having a given duration and amplitude. As a result, there is a need to pre-convert the pulses of the master oscillator. The nature of the transformation may vary. Thus, it may be necessary to change the amplitude or polarity, duration of the master pulses, or delay them in time.
Conversions are mainly carried out using linear circuits - four-terminal networks, which can be passive and active. In the circuits under consideration, passive quadripoles do not contain power supplies; active ones use the energy of internal or external power supplies. With the help of linear circuits, transformations such as differentiation, integration, shortening of pulses, changes in amplitude and polarity, and delay of pulses in time are carried out. The operations of differentiation, integration and shortening of pulses are performed by differentiating, integrating and shortening circuits, respectively. The amplitude and polarity of the pulse can be changed using a pulse transformer, and it can be delayed in time using a delay line.
Integrating circuit. In Fig. 19.5 shows a diagram of the simplest circuit (passive two-terminal network), with which you can perform the operation of integrating the input electrical signal applied to terminals 1-1 | , if the output signal is removed from the 2-2" terminals.
Let's create a circuit equation for instantaneous values of currents and voltages according to Kirchhoff's second law:
It follows that the circuit current will change according to the law
If you choose a time constant large enough, then the second term in the last equation can be neglected, then i(t) = uin(t)/R.
The voltage across the capacitor (at 2-2" terminals) will be equal to
(19.1)
From (19.1) it is clear that the circuit shown in Fig. 19.5, performs the operation of integrating the input voltage and multiplying it by a proportionality coefficient equal to the inverse value of the circuit time constant:
The timing diagram of the output voltage of the integrating circuit when a sequence of rectangular pulses is applied to the input is shown in Fig. 19.6.
Differentiation chain. Using a circuit, the diagram of which is shown in Fig. 19.7 (passive four-port network), you can perform the operation of differentiating the input electrical signal supplied to terminals 1-1", if the output signal is removed from terminals 2-2". Let's create a circuit equation for instantaneous values of current and voltage according to Kirchhoff's second law:
If the resistance R is small and the term i(t)R can be neglected, then the current in the circuit and the output voltage of the circuit removed from R
(19.2)
Analyzing (19.2), we can see that with the help of the circuit under consideration, the operations of differentiating the input voltage and multiplying it by a proportionality coefficient equal to the time constant τ = RC are performed. The shape of the output voltage of the differentiating circuit when a series of rectangular pulses is applied to the input is shown in Fig. 19.8. In this case, theoretically, the output voltage should be alternating pulses of infinitely large amplitude and short (close to zero) duration.
However, due to the difference in the properties of the real and ideal differentiating circuits, as well as the finite steepness of the pulse front, the output receives pulses whose amplitude is less than the amplitude of the input signal, and their duration is determined as t and = (3 ÷ 4) τ = (3 ÷ 4) RC.
In general, the shape of the output voltage depends on the ratio of the input signal pulse duration t and the time constant of the differentiating circuit τ. At moment t 1, the input voltage is applied to resistor R, since the voltage across the capacitor cannot change abruptly. Then the voltage across the capacitor increases exponentially, and the voltage across the resistor R, i.e., the output voltage, decreases exponentially and becomes equal to zero at the moment t 2 when charging of the capacitor is complete. At small values of τ, the duration of the output voltage is short. When the voltage u BX (t) becomes zero, the capacitor begins to discharge through the resistor R. Thus, a pulse of reverse polarity is formed.
P 
Active integrating and differentiating chains have the following disadvantages: both mathematical operations are implemented approximately, with known errors. It is necessary to introduce corrective elements, which, in turn, greatly reduce the amplitude of the output pulse, i.e., without intermediate amplification of the signals, n-fold differentiation and integration are practically impossible.
These disadvantages are not characteristic of active differentiating and integrating devices. One possible way to implement these devices is to use operational amplifiers (see Chapter 18).
Active differentiator. The circuit diagram of such a device using an operational amplifier is shown in Fig. 19.9. Capacitor C is connected to input 1, and resistor R oc is connected to the feedback circuit. Since the input resistance is extremely high (R in -> ∞), the input current flows around the circuit along the path indicated by the dotted line. On the other hand, the voltage and input amplifier in this connection are very small, since K u -> ∞, therefore the potential of point B of the circuit is practically equal to zero. Therefore, the input current
(19.3)
The output current i(t) is at the same time the charging current of capacitor C: dq= Сdu BX (t), from where
(19.4)
Equating the left sides of equations (19.3) and (19.4), we can write - and out (t)/R oc = С du in (t)/dt, from where
(19.5)
Thus, the output voltage of the operational amplifier is the product of the time derivative of the input voltage multiplied by the time constant τ = R OS C.
A 
active integrating device. The circuit of an integrating device based on an operational amplifier, shown in Fig. 19.10, differs from the differentiating device in Fig. 19.9 only in that capacitor C and resistor R oc (in Fig. 19.10 -R 1) have swapped places. As before, R input -> ∞ and voltage gain K u -> ∞. Consequently, in the device, capacitor C is charged with current i(t) =u BX (t)/R 1 . Since the voltage on the capacitor is almost equal to the output voltage (φ B = 0), and the operational amplifier changes the phase of the input signal at the output by an angle π, we have
(19.6)
Thus, the output voltage of an active integrating device is the product of a certain integral of the input voltage over time by a factor of 1/τ.
In the differentiating circuit (Fig. 11.2, a) the time constant should be small compared to the duration of the pulses. This circuit is used in cases where pulses of relatively long duration need to be converted into short trigger pulses with a steep edge. The circuit maintains the steep edge of the pulse in the same polarity and essentially behaves as a high-pass filter, attenuating the low-frequency components of the pulse and passing the high-frequency components of the pulse.
When voltage is applied to a capacitor, the current flowing through it is proportional to the derivative of the voltage applied to the capacitor e s:
(11.4)
At a small time constant, the resistance of the resistor is significantly greater than the reactance of the capacitor. Therefore, the output voltage, equal to the voltage drop across the resistor, is approximately expressed by the formula
(11.5)
In Fig. 11.2,6 and V the pulse shapes at the input and output of the differentiating circuit are shown, respectively. From the initial moment of the pulse action and throughout its entire duration, a constant voltage is applied to the input of the circuit. If the capacitor Ci was not charged when the input pulse was applied, then at the first moment a large current will flow through the capacitor, as well as through the resistor R1. Thus, a large voltage drop immediately appears across the resistor, due to which the pulse front rises very quickly at the output (Fig. 11.2, c). As a capacitor charges, the current flowing through it decreases at a rate dependent on the time constant of the circuit. With a small time constant, the capacitor quickly charges and current stops flowing through the circuit. Thus, when the capacitor is fully charged, the voltage across the resistor R 1 drops to zero level. At the end of the pulse, the input voltage drops to zero and the capacitor begins to discharge. The discharge current of the capacitor has the opposite direction compared to the charge current, therefore, the direction of the current through the resistor is also opposite to the charge current. Therefore, a negative voltage surge will now appear at the output.
Rice. 11.2. Differentiating circuit(s) and input pulse shape (b) and exit (c) chains.
In practice, pulses are usually applied to the input of the differentiating circuit. If sinusoidal oscillations are applied to the input of the differentiating circuit, then their shape will not change, but the phase of the output oscillation will shift and the amplitude of these oscillations will decrease by amounts depending on the frequency of the input signal. Another type of differentiating circuit can be obtained if C 1 is replaced by a resistor and R 1 by inductance. In such a chain, the factor determining the quality of differentiation is also the time constant. As in an integrating circuit, the ohmic resistance of the inductor degrades the circuit's performance. Therefore, such a chain is used quite rarely.
An RC circuit can change the shape of complex signals so that the output shape is completely different from the input. The amount of distortion is determined by the time constant of the RC circuit. The type of distortion is determined by the output component connected in parallel with the output. If a resistor is connected parallel to the output, then the circuit is called differentiating. used in synchronization circuits to obtain narrow pulses from rectangular, as well as for receiving switching pulses and marks. If a capacitor is connected parallel to the output, then the circuit is called integrating. used in signal conditioning circuits in radio, television, radar and computers.
The picture shows differentiating chain.
Recall that complex signals consist of a fundamental frequency and a large number of harmonics. When a complex signal enters a differentiating circuit, it affects each frequency differently. The ratio of capacitance (X s) to R is different for each harmonic. This causes each harmonic to be shifted in phase and reduced in amplitude to varying degrees. As a result, the original signal shape is distorted. The figure shows what happens to a square wave signal that passes through a differentiating circuit.
Similar to differentiating, except that a capacitor is connected parallel to the output.
The figure shows how the shape of a rectangular signal changes after passing through an integrating circuit.
Another type of circuit that changes the waveform is signal limiter. The figure shows the waveform at the limiter input: the negative part of the input signal is cut off.
A clipping circuit can be used to clip peaks of an applied signal, to produce a square wave from a sine wave, to remove positive or negative portions of a signal, or to maintain the amplitude of an input signal at a constant level. The diode is forward biased and conducts current during the positive half cycle of the input signal. During the negative half-cycle of the input signal, the diode is reverse biased and does not conduct current. The circuit is essentially a half-wave rectifier.
Using the offset voltage you can adjust the amount of the signal being cut. The parallel clipper can be offset to change the clipping level of the signal. If it is necessary to limit the signal on both the positive and negative sides, two biased diodes are used in parallel with the output. This allows you to obtain an output signal with an amplitude that does not exceed a predetermined positive and negative level. With this conversion, the output signal takes on a shape close to rectangular. Hence this circuit is called a square wave generator. The figure shows another limiter circuit that limits the signal on both the positive and negative sides using two zener diodes.
The output signal is limited on both sides by the stabilization voltages of the zener diodes. Between these limits, no zener diode conducts and the input signal passes to the output.
Sometimes it is desirable to change the DC sample level of an AC signal. The DC reference level is the level against which the AC signal is measured. A clamp can be used to clamp the high or low value of a signal at a given constant voltage. Unlike a signal limiter, a clamp does not change the waveform. Diode clamp called the constant component reducing agent.
This circuit is commonly used in radar, television, telecommunications and computers. In the circuit shown, a square wave signal is applied to the input. The purpose of the circuit is to limit the maximum value of the signal to 0 volts without changing the waveform.