Golovna Let's take a look at the electrical circuit with a resistor support R capacitor capacity C
performances for the little ones. Let's take a look at the electrical circuit with a resistor supportі capacitor capacity Elementi connected sequentially, therefore, the current flow can be expressed based on the similar voltage to the charge of the capacitor dQ/dt = C(dU/dt) Ohm's law U/R ..
The voltage at the resistor points is significant
U R 
To this end there is a place of jealousy: Let's integrate the remaining virus. The integral of the left side of the equation is more modern U out + Const
. Moved to a permanent warehouse Const
to the right, partly with this very sign. The right side has constant gain R.C. weighted for the integral sign:.
The result showed that the voltage is output The integral of the left side of the equation is more modern U out
directly proportional to the integral of the voltage at the resistor heads, then, and the input stream The right side has constant gain I in Stationary warehouse do not lie below the nominal values of the Lanzug elements.
To ensure direct proportional output voltage from the input integral U in , the proportionality of the input voltage to the input stream is required Non-linear relationship U in /I in at the input lens is selected so that the charge and discharge of the capacitor is generated behind the exponential e-t/τ, which is the most nonlinear at τ
.
t/τ e≥ 1, then, if the value
τ
= Moved to a permanent warehouse t Let's take a look at the electrical circuit with a resistor supportі capacitor capacity.
equal to more Moved to a permanent warehouse Here τ
- an hour to charge or discharge the capacitor is no more than a period. e- steady hour - additional quantities τ
How to take denominations
Lanzug, Koli Moved to a permanent warehouse will be much more then the cob plot is exponential for a short period (just.
) can be added to the linear ones to ensure the necessary proportionality between the input voltage and the flow. The right side has constant gain For a simple Lanzug Stationary warehouse.
Steadily, you should take 1-2 orders of magnitude larger than the period of the variable input signal, so that the main and significant part of the input voltage falls on the resistor pins, ensuring sufficient linear density Moved to a permanent warehouse U in /I in ≈ R
This time there is no voltage The integral of the left side of the equation is more modern with an acceptable loss proportional to the integral of the input Moved to a permanent warehouse The greater the value of the denominations
, the less changeable the warehouse at the output, the more accurate the curve of the function will be. Moved to a permanent warehouse In most cases, a changeable storage integral is not needed for the recovery of such lantzugs, it is only required permanently
Let's take a look at the electrical circuit with a resistor support the same denominations You can choose larger ones to ensure the input support of the offensive cascade. As a butt, the signal from the generator is a positive 1V meander with a period of 2 mS applied to the input of a simple integrating lanc. τ
= Moved to a permanent warehouse with denominations:
At one time there are more than five times more than an hour period, but the visual integration can be done accurately.
The graph shows that the voltage at the level of the stationary warehouse 0.5v will be of tricute form, since the plots, which do not change in hours, for the integral will be a constant (significantly a), and the integral of the constant will be a linear function. ∫adx = ax + Const. a Constant value
calculate the tangent of the slope of the linear function We integrate the sine wave and remove the cosine from the reversal sign..
∫sinxdx = -cosx + Const The integral of the left side of the equation is more modern = 0.
Whose location has a permanent warehouse?
If a tricutaneous signal is applied to the input, the output will be a sinusoidal voltage. The integral of a linear division of a function is a parabola..
In the simplest way
∫xdx = x 2 /2 + Const
The multiplier sign directly denotes the parabola.
The simplest problem is that the output voltage is even less than the input voltage.
Let's take a look at the integrator of the Operational Powerhouse (OP) behind the diagram shown in the little one..
With the assurances of the infinitely great support of the OU and Kirchhoff’s rules, jealousy will be fair: I in = I R = U in / R = - I C .
The voltage at the inputs of an ideal op-amp is equal to zero here, as well as at the capacitor heads The right side has constant gain U C = U out = - U in
Otje, Moved to a permanent warehouse appear, emerging from the struma of the zagalal Lanzug. τ
With element ratings
, if
= 1 Sec, the output variable voltage is equal to the values of the input integral.
Ale, protilenno behind the sign. An ideal integrator-inverter for ideal circuit elements. Differential Lancug RC
Let's take a look at the differentiator from the standpoint of the Operational Supporter. The ideal op-amp here is to ensure the equality of streams .
I R = - I C
following Kirchhoff's rule.
The voltage at the inputs of the op-amp is equal to zero, therefore, the output voltage The right side has constant gain U out = U R = - U in = - U C Based on the similarity of the charge of the capacitor, Ohm's law and the equivalence of the values of the currents in the capacitor and resistors, we write: U out = RI R = - RI C = - RC(dU C /dt) = - RC(dU in /dt)
It's important to know that the voltage is out Moved to a permanent warehouse proportional to the capacitor charge
dU in /dt
how fast the input voltage changes.
The similarity of the linear function will be a constant, the sign of which is determined by the strength of the linear function.
For the simplest differential RC switch made of two elements, the proportional amount of the output voltage is equal to the output voltage at the capacitor heads.
U out = RI R = RI C = RC(dU C /dt)
If we take the values of the RC elements so that the constant hour is 1-2 orders of magnitude less than the last period, then an increase in the input voltage to an increase in the hour between the period can increase the fluidity of the change under the strain of the singing world for sure.
Ideally, the increase should drop to zero.
In this case, the main part of the input voltage falls at the ends of the capacitor, and the output becomes an insignificant part of the input voltage, so it is practically not necessary to calculate such circuits. , the proportionality of the input voltage to the input stream is required Most often, differentiating and integrating RC lances are used to change the pulse in logical and digital devices.
In such cases, RC denominations are insured for the exhibitor -t/RC coming from the last pulse during the period of necessary changes. For example, below on the little one it is shown what is needed for the impulse τ
T i
at the output of the integrating lanyard will increase by an hour 3
. τ
For an hour the capacitor is discharged to 5% of the amplitude value.
At the output of the differential lancer, the amplitude voltage after applying the pulse appears to be mitten, since at the ends of the discharged capacitor it is equal to zero.
Then the charging process follows and the voltage at the resistor ends changes.
In 3 hours (10+)
it will change to 5% of the amplitude value.
Here 5% is a show value.
In practical applications, this threshold is determined by the input parameters of logical elements, which then stagnate.
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If you write down the differential voltage that connects the string and the voltage in this circuit, and then calculate it, then you can find out how much charge and discharge of the capacitor is generated.
I’m not trying to get carried away with fancy mathematics here, I’m just going to marvel at the final result:
If the discharge and charge of the capacitor follows the exponential law, look at the graphs:
What do you think, is there a significant time here?
Remember this value – it is constant to the hour of RC Lantzug and von: τ = R*C.
The graphs, in principle, indicate how much the capacitor is charged/discharged in an hour, so we won’t belabor the point again. Well, before speaking, a rather practical rule is that in an hour, which is equal to the fifth hour of the RC Lancs, the capacitor is charged or discharged by 99%, well, then you can do it again)
What does all this mean and what is the point of capacitors?And everything is simple, on the right is that if a constant voltage is applied to the capacitor, then it will simply charge and that’s it, but if a variable voltage is applied, then everything will start.
The capacitor is either discharged or charged, apparently in a rich stream.
And as a result, a noticeably important connection is removed - a changing stream easily flows through the condenser, but a stationary stream cannot. Therefore, one of the most important purposes of the condenser is to separate the permanent and variable storage space in the lancus. Let's get together, and now let's talk about
differentiating and integrating RC lancers.
Differentiation RC Lanzug. The differential filter is also called a high-pass filter, the diagram of which is presented below: As it stands out from the name, so, more powerfully, it is visible behind the scheme We reject direct impulses.
IntegratingAnd everything is simple, on the right is that if a constant voltage is applied to the capacitor, then it will simply charge and that’s it, but if a variable voltage is applied, then everything will start.
Now the time has come for the integrating lancer.
Also called a low-pass filter.
By analogy, it is difficult to guess that the lance that integrates passes through the constant storage, and the change goes through the capacitor and does not pass to the output.
The diagram looks like this:
If you do a little math and write down the expressions for the voltage and flow, you will find that the output voltage is the integral of the input voltage.
Through the whole process and having lost my name)
Well, they looked at us with great importance, although at first glance they had awkward schemes.
It is important to understand how this all works, and in the end, there is a demand so that, given the most specific tasks, we can immediately develop a unique circuit design solution.
Zagalom, before the Swedish hawk in the upcoming articles, if you have any food, be sure to drink it 😉
And at once the stench is created by the RC-lancug, which is the whole of the lancet, which consists of a capacitor and a resistor.
It's simple ;-) As you remember, the capacitor has two plates on each side, one facing the other. You, melodiously, remember that there is room for it to lie beneath the surface of the plates, the space between them, and also the words that are between the plates.
Or the formula for a flat capacitor:
de
Garazd, closer to the point.
Let us have a capacitor.
.JPG)
So, the plan of action will be this: we charge the capacitor behind the other life block, and then discharge it on the resistor and watch the oscillogram as the capacitor is discharged.
Let’s take a classic diagram, as it would be for any electronics handyman:
At this moment we are charging the capacitor
then we override the toggle switch S in the other position and discharge the capacitor, preventing the process of discharging the capacitor on the oscilloscope
I think this makes everything clear.
Well, let's get started with the collection.
We take a breadboard and assemble a circuit.
I took the capacitor with a capacity of 100 μF, and the resistor 1 KiloOhm.
Instead of the toggle switch S, I will manually transfer the yellow wire. Well, that's it, touch the oscilloscope probe to the resistor And I can see the oscillogram showing how the capacitor is discharging. Those who are reading about RC-lants first, I think, are a little relieved.
According to logic, the discharge may proceed in a straight line, but here we go wrong. The rank is awarded for such a title exponential
.
Since I don’t like algebra and mathematical analysis, I don’t want to make different mathematical calculations.
Before speaking, what is an exhibitor? Well, the exponent is the graph of the function “e in stage x”. In short, it all started at school, you better know ;-)
Let's take a look at the electrical circuit with a resistor support So, when the toggle switch is closed, we have the highest RC-Lance, then the following parameter is
You can choose larger ones to ensure the input support of the offensive cascade. RC-Lantsyuga
.
The stationary hour RC lancet is designated by the letter t, in other literature it is designated by the great letter T. To make it easier for understanding, let us also signify the stationary hour RC lancet by the great letter T.
Well, I think it’s important to remember that the RC-Lancet is a constant source of nominal values of support and capacity and is expressed in seconds, or by the formula:
Closed position
As soon as we close the key S, our capacitor begins to charge from zero to a value of 10 Volts, to the value that we set on the life block
We observe an oscillogram taken from a capacitor
Didn’t you do anything wrong with the last oscillogram, where did you discharge the capacitor onto the resistor?
So everything is true.
The charge is also going to be an exponential ;-).
Since our radio components are new, the time is also the same.
Graphically, the won is calculated as 63% of the signal amplitude
As you know, we took away the same 100 milliseconds.
Using the RC-Lantzug's formula for the constant hour, it is not important to guess that changing the ratings of the support of the capacitor will cause a change in the constant hour.
Because there is less volume and support, then the shortest hour after hour is constant.
Well, the charge and discharge will be faster.
For example, let's change the capacitance value of the capacitor downward.
Well, we have a capacitor with a nominal value of 100 µF, and we set 10 µF, the resistor is deprived of the same nominal value of 1 kom.
Let’s marvel once again at the charge and discharge graphs.
The axis is charged in this way by our capacitor with a nominal value of 10 µF
And the axle is so discharged
As you see, the slow hour of Lanzug has quickly shortened.
Judging by my calculations, it was T = 10 x 10 -6 x 1000 = 10 x 10 -3 = 10 milliseconds.
Let's check it in a graphical-analytical way, why so?
It will be on the charge and discharge chart directly at a consistent level and approximated for the entire hour.
Set to a new frequency of 1 Hertz and a range of 5 Volts
The yellow oscillogram is a signal from the function generator, which is fed to the input of the lancet, which is integrated at terminals X1, X2, and from the output a red oscillogram is taken, then from terminals X3, X4:
As you may have noticed, the capacitor can be charged and discharged all the time.
Ale scho be, yakscho mi dodamo frequency?
I set the frequency on the generator to 10 Hertz.
I marvel at what came out of us:
The capacitor cannot be charged or discharged as soon as a new direct current pulse arrives.
As a matter of fact, the amplitude of the output signal has decreased greatly, one might say, it has shrunk close to zero.
And the signal at 100 Hertz burned out without depriving anything from the signal, other than low-impact signals
The 1 Kilohertz signal at the output went out without giving anything.
Come on!
Try recharging the capacitor at this frequency :-)
Well, they looked at us with great importance, although at first glance they had awkward schemes.
All the same concerns other signals: sinusoidal and tricutaneous. through the output signal may be close to zero at a frequency of 1 KG or higher.
“And what is it all about, what is it that integrates?”- Power up the vi.
Of course, no! This is the beginning of the cob.
You can choose larger ones to ensure the input support of the offensive cascade. Well, first of all, this circuit turns out to be a voltage sink, and, in another way, a capacitor is a frequency-dependent radio element.
This is based on the frequency.
The procedure can be read in the statistics of the capacitor in the lancet of a stationary and changeable strum.
But the bad things won’t end there.
Let's guess what the signal from the warehouse is.
This is nothing more than the sum of a variable signal and a constant voltage.
Taking a closer look at the little ones, it will become clear to you.
So in our case we can say that this signal (lower in the picture) has a constant warehouse in its warehouse, so it has a constant voltage
In order to see the constant storage of this signal, it is enough for us to drive it through our integrating lance.
.jpg)
Let's take a look at the butt.
With the help of our function generator, we add our sinusoid “above the subfield”, so it can be generated like this:
Well, everything is like before, the first input signal is Lanzug, the red one is the output signal.
A simple bipolar sine wave gives us 0 Volt at the output of the RC integrated circuit:
To understand the zero number of signals, I marked them with a square:
Now let me add constant voltage to the sinusoid, or more precisely, constant voltage, since the function generator allows me to do this:
As you can see, as soon as I raised the sine “above the subfield”, at the output of the Lantzug I steadily cut off the voltage at 5 Volts.
I raised the signal at the function generator by 5 Volts;-).
Well, now it’s more fun.
At the same time, I change the distortion of our direct-cut signal, because the distortion is worth nothing more than the time period for the triviality of the impulse, therefore, we change the triviality of the impulses.
Decreased severity of impulses
Greater trepidation of impulses
If you haven’t noticed anything yet, just look at the red oscilloscope and everything will become reasonable.
Summary: with care, we can change the level of the warehouse.
This very principle is inherent in PWM (Pulse Width Modulation).
Let's talk about her in the official statistics.
Differential lancet
Another cute word that comes from mathematics is differential.
The head immediately begins to feel ill due to one of her family’s only children.
Where should I go?
Electronics and mathematics are inseparable friends.
And the axle and the differential lock itself
In the circuit, the resistor and capacitor were simply rearranged
Well, now let’s follow the same steps as we did with the integrating lancet.
For the cob, a low-frequency bipolar meander with a frequency of 1.5 Hertz and a swing of 5 Volts is supplied to the input of the differential lancer.
The yellow signal is the signal from the frequency generator, the red one is from the output of the differential switch:
As you see, the capacitor is about to completely discharge, so we have the axis of the oscillogram.
Let's increase the frequency to 10 Hz
As you know, the capacitor does not allow itself to discharge until a new impulse arrives.
As a result, we can see that the main function of the differential lance is to provide a variable storage signal that can accommodate both variable and permanent storage.
In other words, the vision of a variable flow from the signal, which is the sum of the variable flow and the constant flow.
Why are you so excited?
Let's find out.
Let's take a look at our differential lancet:
If you take a close look at this circuit, then we can improve the same voltage circuit as in the circuit that integrates.
The capacitor is a frequency-deposited radio element.
Then, if we give a signal with a frequency of 0 Hertz (steady current), then the capacitor will immediately charge and then stop passing the current through itself.
1 / 3
Lanzug will be shaved.
If a changeable stream is supplied, then it will also pass through the condenser.
The higher the frequency, the lower the capacitor support.
So, the entire variable signal is dropped on a resistor, from which we immediately pick up the signal.
If we supply a mixing signal, either an alternating stream + a constant stream, then at the output we simply reject the alternating stream.
Whom we talked to you about beforehand.
Why did this happen? The reason is that the capacitor does not allow a constant current to pass through it! Visnovok The reason is that the capacitor does not allow a constant current to pass through it! The integrating angle is also called a low-pass filter (LPF), and the differential one is called a high-pass filter (HPF).
Report about filters. The reason is that the capacitor does not allow a constant current to pass through it! is indicated by the following formula:
U c (t) = U 0 (1 − e − t / R C) .(\displaystyle \,\!U_(c)(t)=U_(0)\left(1-e^(-t/RC)\right).)
In this manner, at a steady pace, this aperiodic process is dear toτ = R C. (\displaystyle \tau = RC.) Integrating lances pass a constant warehouse signal, including high frequencies, then low-pass filters.
Under what circumstances is the thing at rest?
τ (\displaystyle \tau) (\displaystyle \tau = RC.) A differential RC-lanc can be obtained by interchanging the resistor R and the capacitor in the integrated lanc. (\displaystyle \tau = RC.) In this case, the input signal goes to the capacitor, and the output signal is taken from the resistor.
For a constant voltage, the capacitor is open to the lancet, so that the constant storage signal in the differential type lancet will be cut off.