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Russian
00:00:10
May God’s peace, mercy, and blessings be upon you. In a00:00:12
new lecture of practical course lectures in00:00:14
electronics with engineer Walid Issa, and today00:00:16
we want to complete with each other another application of the00:00:19
diode applications. We took in previous lectures, what is a00:00:22
diode, how does it work, and what are the working conditions in00:00:25
forward and reverse bias? We took circuits or the00:00:27
first application of what It is the half-wave and00:00:30
full-wave calendar. We also took the last lecture,00:00:33
which is the clipper circuits, and we learned how they work,00:00:36
and we made them practical and theoretical, and we also worked on them, or00:00:40
we learned how the applications could be.00:00:42
Ok, today we want to take new circuits or00:00:46
another application from the Dade applications, which are circuits called00:00:49
clamps, and by God, of course, my sections. It's different again.00:00:52
Now we will see it, God willing. What are the circuits of the00:00:54
clamper? What can it be used for and how can I analyze the00:00:56
circuits that follow it? God willing. We want to see it00:00:59
now. The circuits of the enlargement before I know it00:01:05
and let you know. I show you an example of how it00:01:09
works. If I had a perfect signal, the00:01:13
signal is like this. This is a00:01:16
sine signal or a signal, any signal, IC, or any signal.00:01:19
The important thing is that it is AC, square,00:01:22
or triangular, and it is a sinusoid that enters the00:01:27
clutch circuit. This signal appears, shifted to the top. It is a00:01:31
word shifted to the top, and it may00:01:35
be shifted to the bottom. Here we see it as a word shifted00:01:38
to the top, and it comes out. It is the same, meaning it is the AC,00:01:41
it is, and this is it, but what did they all become in the00:01:47
positive half? What is the meaning of this speech? It means that it has an00:01:50
addition of nothing to a second value, which is a DC00:01:56
current or a continuous voltage. The meaning of the speech is that00:02:00
these clamp circuits worked to add a DC voltage00:02:05
or voltage. Continuous current to the right side of the wave that is an00:02:09
alternating current. What happens to me? When00:02:12
the two are combined, they come together, so I get the00:02:14
signal shifted to the top. It's the same story. It could be shifted00:02:17
to the bottom. The same signal00:02:20
entered the clamp circuit and came out. It00:02:23
shifted to the bottom, so it stopped in this way. It00:02:26
all moved. For the negative part, it is fine. If the00:02:30
circle when growing is shifted upward, then the00:02:34
positive clamper is shifted. If it shifts downwards, then the negative00:02:36
clamper smiles. So, now I have seen00:02:40
the two examples. Is it possible that we alone know00:02:42
this word or this circuit? Of course, it is possible that it is a00:02:46
circuit that works to00:02:48
add a DC voltage. This is D. This is the C level to the right side of the00:02:53
alternating wave. Okay. If this voltage is00:02:57
positive, move it up. If this is the DC00:03:12
voltage, this is negative. Then I shift it down. Okay. And of course it has second names. Now, God willing, we want to see them. It is used to add a continuous wave, which is this continuous DC wave. Hesitant,00:03:17
which is this, of course, it could be a triangle, a square,00:03:20
or anything, so what is going up or down? Of00:03:25
course, this continuous wave we call DC00:03:29
Level or DC of State. These names are very famous,00:03:31
and therefore if you have a DC00:03:35
of Six or DC Level, what it means is It is00:03:59
DC. It is a DC part added to an AC part. Of course, it works on a vertical displacement of the reciprocating wave. You can raise it above and you can bring it down exactly. If you remember how the telescope works, you can not shift the wave above and below this exact same thing, as if you were adding DC to it. Of course, if you add it to the positive, it will00:04:03
shift upward, and if you add it to the negative, it will shift downward. This00:04:06
circle may be the first use00:04:09
for it in the telescope. It is that it disperses the waves00:04:12
and separates them completely from each other. Does it have a00:04:16
second name? Of course, a second name is possible in some books or00:04:19
some references by naming it. Circuits called DC00:04:23
Restoration Okay, it is DC restoration,00:04:27
and this of course can be famous. This circuit or this00:04:30
name is known in the old television circuits. Quite a few00:04:33
times in circuits, he has the AC signal00:04:37
and wants to convert it to nothing.00:04:40
The clamp00:04:46
and then I call it the DC Restoration circuit, so00:04:49
if you find the DC Restoration circuit, or the00:04:52
Clamp, or the DC of Set Edition, or00:04:56
anything like that, what it means is the same00:04:59
function that we are explaining now, let us see the00:05:02
internal circuit structure here, analyze it, and00:05:05
see an example of it00:05:13
in the militia. We have two very important elements in the00:05:17
clamp circuits, which is the so-called00:05:21
well-known diode and the so-called capacitor. The capacitor is complete with the00:05:24
diode along. From the first time you see them, you understand that it00:05:27
could be a clamper circuit. Guys, an00:05:30
important point. We must not only memorize the diode. Now,00:05:35
how does it work and with the capacitor or We know that this is a00:05:37
lamp circuit, and what is its function? By God, it goes up00:05:40
and down. You have to remember the shape of the circuit as well. I00:05:44
mean, look at the shape of the circuit, how is the capacitor, how is it with00:05:47
the diode, and if you want to dismantle resistors and talk like00:05:50
this, if you have a safe of circuits in its brain, there are00:05:55
many, and therefore, if after that, you can look for any00:05:58
circuit. Even if it is large, you say, “Oh, this part is the00:06:01
Clamper circuit, this part is the Clipper circuit, this00:06:04
part is Multiplayer, this part is enlarged, this00:06:06
part is, for example, Voltage Defender,” and therefore you00:06:10
must have in your brain almost all of these circuits00:06:14
an image of them, so that you increase your experience in analyzing00:06:18
circuits in the future. After that, we will not say that. By00:06:20
God, this is a00:06:21
circuit. In our analysis, this means circuits. This is a00:06:24
Clamper circuit. No, you must roughly know it00:06:27
and what else it uses in the circuit. You should expect because00:06:31
this is almost the first step on your way to00:06:36
designing your own circuits. Okay. I mean, now,00:06:39
these large, complex television circuits. Who00:06:42
designed them? People who understand circuits. And memorize which00:06:47
Clamper circuits are filled by the Clippers, and thus00:06:49
they conjure these circles whenever they encounter a00:06:53
function that intersects with the function of a circle of00:06:56
these circles. Therefore, just this quick point00:06:59
is that you must take a stock and have a stock00:07:03
of the shapes of the circles that you encounter theoretically or00:07:06
practically. Okay. Now we return to the Clamper circuit in00:07:10
Here we have the capacitor and we have the diode. These are the two00:07:12
most important elements in the clamp. In my case, the00:07:15
capacitor is connected in the opposite way to the diode. The00:07:20
waves are here, but the diode will be the negative00:07:23
followed by above and the positive below. And we will see how00:07:26
this wave appeared in this form. Look at the wave with00:07:29
one end, or half positive and half negative,00:07:33
and the peak to peak following it. 20 volts, but it00:07:37
went out, almost all of it shifted, nothing00:07:40
went up. Okay, so we want to know roughly how00:07:45
this happened, because now this step is now the00:07:49
circle of growth, and this will be its function. You00:07:51
may now know it, okay, we know it, but I want to00:07:54
tell you more about how we analyze the circle, if you would like00:07:57
to think about how. This circuit was analyzed and how did this00:08:00
robot appear for this autobot? In analyzing00:08:03
diode or diode circuits with AC00:08:06
exactly, you must take the positive and negative halves of the wave.00:08:10
You ask yourself a question: when is this00:08:15
diode in a state of forward bias and when is it in a00:08:18
state of backward bias? So the first question we can00:08:22
ask. When is this David in a state of00:08:25
forward bias, it looks like this? The negative is above and the00:08:29
positive is below. Its meaning is I. In order for it to be in a00:08:31
forward bias, it must come from the positive infinitive here below, and00:08:35
the negative comes above. Perfect as long as the infinitive is00:08:39
from the infinitive. It must be positive and under its meaning.00:08:42
What am I talking about? Regarding the negative half of the vector,00:08:46
this negative half, this, this half, or this00:08:49
half, and so on, this positive half is named, and00:08:52
this negative half, the negative half, condemns the positive00:08:55
below and the negative above. As for the positive half, the positive half00:08:58
condemns me, what is above, and the negative, what is below? Well,00:09:01
its meaning is as long as I talked about the negative half, the00:09:03
negative remains. Above and the positive is below the positive here.00:09:07
Surely the current is red and it goes to the capacitor. Okay. The00:09:12
negative following the source is connected to the negative of the capacitor. Of course00:09:16
we assume that it is any00:09:19
20 volts. Of course from the 20 volt peak then00:09:24
each half is 10 volts. Okay. The negative00:09:28
following the source is now connected to the negative. Who is the capacitor and the00:09:31
positive that follows? The source went through the diode and00:09:34
of course lost 0.7 volts because it will take a charge from it00:09:38
and then the rest will go to the capacitor. So what00:09:41
will happen to the positive on the positive and the negative on the negative? What00:09:44
will happen to the capacitor? Will it charge? It will charge.00:09:47
What will it charge for00:09:50
9.3 volts? Why because it is the positive 10? It will pass. You will lose00:09:54
0.7 of its charge and then it will go to whom. For the capacitor in the00:09:58
mix, now in the negative half of the wave, a00:10:02
charge of 9.3 volts is charged. Guys,00:10:06
what is this charge considered? This is now considered00:10:11
DC. The capacitor carries a constant charge. I mean, it is like a00:10:15
battery. The amount of this battery is00:10:18
9.3 volts, so now it is like here is a continuous voltage source of00:10:26
9.3 volts. Okay, so the capacitor is as if it were a00:10:29
fixed battery. This battery, which is the capacitor,00:10:33
took its charge when it was the negative half. Okay,00:10:38
then we will see what happens in the positive half.00:10:40
Let us see that it is the same circuit. But this time, what00:10:43
do we want to talk about about the positive half? The positive half00:10:46
means what will be the L and will be the David.00:10:51
In a case of back bias, why? Because in the00:10:54
positive half, the positive will be here and the negative will be here. Come on, the00:10:57
negative will come here. He will find that oh, then it00:11:00
will not enter perfectly. Its meaning is this, it will be in H. In a00:11:03
case of back bias, it means this is as if it is off, it will never00:11:06
work. Ok, I know that the capacitor is already00:11:13
charged at 9.3. Of the volts from who from the00:11:17
previous position from the negative half of the wave, I got it00:11:20
charged and ready. I got it. It came charged and ready. The00:11:26
positive followed it here and the negative followed it here. It's the same as the one before. Ok, I got the positive half of the00:11:29
wave. The positive followed it here and the negative followed it00:11:32
here. How much is it 10 volts because it's every half? It is00:11:35
about 10 volts. Well, I have it. I have a battery00:11:39
, capacitor, or continuous voltage source with a value of 9.300:11:45
volts, and a second voltage source with a value of 10 volts. The positive00:11:49
here and the negative here and the positive here and the negative here are00:11:52
like two voltage sources connected in00:11:56
series, as long as they are connected in series and the00:11:58
order is positive, negative, positive. Negative means what do00:12:01
they add together, so what happened to me? I now have00:12:05
10 volts instead of 10 volts.00:12:07
What will come out here? What happened to me here here here?00:12:10
What is right? The00:12:12
9.3 plus the 10 is perfect, so I will have00:12:16
20 volts, but minus them. What is the 7 volts?00:12:21
Okay, I will have the voltage minus 0.7 of a00:12:25
good volt, and therefore00:12:27
now the battery has worked to raise or not00:12:31
the battery, the capacitor, which has become like the00:12:34
battery, has worked to raise the m from 10 volts00:12:37
to minus 20 minus x of the volts that I lost in00:12:42
this area. Ok, what is this area now?00:12:46
In the negative half, what happened? We said in the00:12:48
negative half of the wave, we said the positive is below and00:12:50
the negative is above, and when the positive came here, it lost,00:12:54
this or that became in a state of forward bias, and00:12:57
I said the voltage or potential difference around the00:13:01
diode, which is in a state of forward bias. We took it from the00:13:03
previous lecture and sat down. It has more than half an00:13:07
hour, and we explain at this point when the00:13:09
diode is forward biased along the voltage00:13:11
around it, what will it be?00:13:26
Negative Q7 Who is the negative half that did not come out00:13:30
but came out instead of -77 and the positive half when it00:13:34
came out came out in place of 20 - 0 exactly, so it all became from00:13:41
E to Han is the same 20 but what, most of it00:13:45
is shifted00:13:46
upward by the amount that is00:13:49
EH, the 10 volts, exactly the displacement of this They are all00:13:53
10 volts, so I got 9.3 volts, because there is00:13:56
a loss of 0.7 out of 100. We hope for this00:14:01
easy method of analysis. Ask yourself a question: What time00:14:05
will be in front of me and what time will it be behind me? And on the basis of it,00:14:08
start with the solution. It is perfect, and you will find that things have become easier00:14:11
for you, God willing. Now I have discovered. How did00:14:14
I have something to shift up and shift down00:14:17
through the diode that charges in the half wave? The00:14:21
negative wave, and what helps the positive half wave00:14:25
to rise? Okay, if I reversed the situation. If00:14:29
I reversed the capacitor, I left the positive here and the negative00:14:31
here and reversed the diode. What would happen? The exact opposite would happen.00:14:34
This is what would go down. Before we see,00:14:39
before we see what the second type of00:14:42
clamp is, let us see what this one is, and what is00:14:45
practical as we see here. I have a00:14:47
clamp circuit in mine. I have a source here. Voltage is 50 Hz,00:14:50
14.6 VRMS. I chose it like this, because00:14:52
this is practical for me and I will not put it in. What you have00:14:56
is a capacitor of 1000 microfarads, and it is a negative diode that00:14:59
follows it to the top, and I have a resistance. I can,00:15:02
of course, remove the resistance, but it is preferable to lower the00:15:05
resistance so that the signal appears00:15:09
clearly and that it is working. Even if I have it, if00:15:11
it is ok, let us turn on the circuit and see what00:15:14
happens with us. Here I am taking input from00:15:17
who is the00:15:18
inside signal and I am taking a signal from who is00:15:22
what is coming out on the outside which is about00:15:24
what is on the diode or what is on the resistance. Let us00:15:26
turn on the circuit and see what comes out with us. We00:15:30
can of course put a recording here so that things can be set and we00:15:34
have stopped the simulation. In order to see00:15:37
exactly, this is the robot. This is what is00:15:41
coming from who is from here. Okay, and this is what is00:15:45
coming out. As we can see,00:15:49
the signal has actually been raised. This is not completely up, and we are00:15:52
using here what is Cinewave. I mean, we00:15:55
explained it on Square and here in Cinewave, so the00:15:57
idea is the same. In the negative half, the capacitor00:16:00
charges, and in the positive half, the capacitor helps the positive00:16:04
half to rise, and thus it becomes00:16:06
as if it raises the entire wave, which is upward, and as00:16:10
we have observed here, I have an easy area.00:16:12
Let us discover it to see, meaning we can play with the wave00:16:16
here more and more. Ha, ha, it has become almost00:16:20
one. Volts, so now it's almost here, Boo S or 800:16:25
volts, what is the connection of the wave and what, and the rest is all00:16:29
going up. I'm exactly as I00:16:33
solved the circuit. I got it through the00:16:36
MultiZim program, and let's see it now00:16:38
in operation, and it's also the circuit with a00:16:40
simple name in mine. Here's what it is.00:16:44
The source is 500 Hz, which is the capacitor, and it is the diode.00:16:47
Note that the diode looked at it exactly once,00:16:50
in the meaning of an area in which the current passes and an area in which the00:16:54
current does not pass. The diode looked at it once in a while, and on one occasion, it does00:16:57
not mean that there is an area in which there is a forward bias and in an00:16:59
area in which there is a back bias, so it appears now when it is forward00:17:02
biased and when it is not biased. Back now,00:17:05
you don't know when the current passes. Let's see what time the00:17:08
current passes. Look here, watch here, and watch here.00:17:11
You will find that the current passes at a time when00:17:15
this00:17:16
small negative region comes exactly now now now00:17:19
now ok. Therefore, when this00:17:23
negative region comes, a current will pass through it and there will be a time of00:17:25
forward bias and the voltage on it will be negative. Q Who and00:17:28
the rest, what will this be, because it will be00:17:32
back space, and therefore what will be the voltage around00:17:36
the capacitor plus the voltage around it, or any signal that00:17:38
you enter? Easy, put the two together and they come00:17:41
out, and in fact, what do I find? This signal is00:17:44
as we expected. Now look at the circuit of the two00:17:46
words. It is in me. I am here. Resistance and diode are00:17:49
fine, but I have the new element, which is a00:17:51
capacitor, and we said the capacitor is the00:17:54
negative terminal, as is clear here in the reference. So00:17:56
this is the negative terminal, and in me this is the00:17:58
positive terminal. Of course, I must make sure that it can withstand the00:18:01
voltage that I have. This load is 25 volts, and I00:18:03
have it here. The maximum is 20 volts, and therefore it can be00:18:05
tolerated without problems. Now, of course, how can I connect00:18:08
this capacitor with ease? I have it, as00:18:10
we said in the circuit. Of course, the cathode is above,00:18:13
and therefore my cathode, that is, connects to the positive part of the00:18:17
capacitor, and therefore the positive part of the capacitor00:18:19
connects here, and the negative part of the capacitor connects to the00:18:22
second part, and therefore This is how I connected it,00:18:24
and this is what happened to me. What is the process of converting to00:18:27
what we are talking about? It will add an amount00:18:30
of DC. Of course, I have an eye for the telescope,00:18:33
and therefore I can see the signal between the end of the transformer00:18:38
or the end of the source. The signal I have here and in00:18:41
my case00:18:42
I also see through what it is. How much I have coming out00:18:47
through the condenser. How much I have here on the David is good00:18:50
or I have it on the R-Load. This is me. Here I am. So we check00:18:54
how much is inside and what I have coming out of this circle.00:18:58
We will see what I have on the00:19:00
[music]00:19:01
scope. I have both. Hey, we are as00:19:04
we noticed that the two are zero and The same00:19:06
thing is fine, the two are identical, but when00:19:10
I start I notice that there are two jobs. The first job is the00:19:14
yellow signal. This is the signal. Who is the00:19:18
signal coming from the source? This is the signal coming from the source. What is the00:19:20
yellow signal? But the blue signal, as00:19:23
we see, is that this signal has become00:19:25
shifted. Vertical by a00:19:28
certain amount, of course, it is as we noticed that in00:19:31
me here, the00:19:33
two have the same zero line, meaning this is the00:19:36
zero line well, but in one unit it rose from the00:19:39
zero line more, why because this is a Clamper operation,00:19:42
which is adding DC, so it did the same as if it was added to00:19:46
me, it is in me. The square of two, no, the whole00:19:49
signal is four squares, as if in reality I00:19:52
have two squares added to the top, which is a square, which is two, and each00:19:54
square I have 10 volts, and therefore in reality, as if00:19:57
the piston00:19:58
added 20 volts to me on the signal, but 20 volts,00:20:02
that is, DC, added 10 to me, and also 10 is now00:20:06
20. Volts, so it became as if it was being carried on00:20:09
DC 20 volts, but as we noticed here, I00:20:12
have a part here for the thief straying below in the negative,00:20:15
so we can enlarge the signal a little. I mean, let us00:20:19
cancel the zero so that00:20:21
we can see that it is the one sitting by enlarging the part so that I can00:20:25
actually see almost a volt, because this is 100:20:29
volt now, and it has not reached a full volt.00:20:32
It is approximately 7 volts, which is the amount of00:20:34
what is the occupancy of the diode in the circuit. The00:20:38
previous circuit was called “What is the positive00:20:40
clamper” because it was shifted to the top? Let us see00:20:42
the negative clamper, but this time the explanation is00:20:44
quick because now we understand the method. What time will00:20:47
this be? Forward bias, when the positive is00:20:50
above and the negative is below its meaning, the negative is below its meaning,00:20:53
in the positive half of the negative here, and the positive here,00:20:56
at the time it will charge the capacitor because it is the positive on the00:20:59
positive and the negative. What will happen here on the negative?00:21:02
It will charge the capacitor by how much is the value of this00:21:05
upper peak minus the 7 of what it loses on the00:21:09
diode. Which is approximately if it is 2 volts here,00:21:11
but if we say 20 volts it will be here 9.3 in the00:21:15
negative half of the wave. In the negative half00:21:18
of the wave there will be the positive here. The positive will be00:21:21
opposite the cathode and therefore this will be something like it will00:21:23
not work and therefore it will happen to me here. Here it will be00:21:27
negative00:21:29
9.3 here it will be Negative why because it is the negative of the00:21:32
capacitor and here also in the negative half there will be00:21:34
negative Vs 9.3 and negative 10 so I have negative00:21:39
20 approximately00:21:41
with 3 Syrians negative 20 Na with 7 negative00:21:45
19.3 volts and this is what will come here and will be00:21:49
here but in the case of forward bias how much00:21:52
will it be here in the case of forward bias00:21:54
There will be 07 around Aldad and this. What is00:21:57
this area above? Let us also00:21:59
check this with semolina. The same circle also with00:22:03
semolina. What did we do on the heart of what is in the00:22:07
direction of the David? If I had the condenser00:22:10
polarized, I would also do what on its heart. I mean,00:22:13
if I had the positive here. And the negative here is what I do00:22:15
on its heart, and therefore I have00:22:17
exactly the opposite, so I have this00:22:20
new circuit, and this is called the negative clipper,00:22:22
and we will see how it will turn out. I have what is the shift, but00:22:25
this time what is down,00:22:29
so I have the red signal, which is the00:22:31
input signal, and the orange signal, this is the signal,00:22:34
which is the outside signal that we are in. We saw it, and in fact, exactly00:22:37
as we expected, all the signs were shifted to00:22:39
the bottom, but here in a very small part, what00:22:43
will it be approximately?00:22:55
We want to see it in a00:22:56
circuit after a little while, but why should we pass00:23:00
carefully the practical as we can see?00:23:03
This is the practical circuit in which we have the diode. All that00:23:05
we did not do is that we turned it in the direction of the00:23:07
diode. If you compare it to the previous video and00:23:09
we turned it in the direction of the capacitor, because this is a polarized capacitor, and00:23:12
therefore it became In me, here the diode is00:23:15
inverted and the capacitor is inverted according to the circuit of the negative00:23:18
clamp, and therefore I have the same00:23:21
circuit, but I reversed the direction of the DAD00:23:23
and the direction of the capacitor. Let us see the output. Now00:23:25
it is the output in me. The one that is the00:23:28
yellow signal is the same as the input signal that is coming from The00:23:31
amount of trans is 14.600:23:34
volts RMS, and I have the external signal.00:23:36
Notice that it is shifted downwards in the same00:23:40
way as the previous circuit was analyzed. Now we00:23:44
have confirmed whether the two circuits are shifted to the bottom or00:23:47
shifted to the top. Now the answer to the question00:23:49
is that it is possible for me to shift it up more or00:23:52
Going down further, we took this circuit,00:23:55
and we took this circuit, this circuit. Now00:23:57
we assume that it is 5 volts as the source, so I get00:24:01
4.3 volts here. It is fine, and if I am the one who is this00:24:05
negative clipper, if I00:24:08
put in, if I reversed the value of the capacitor or reversed the polarity of the00:24:11
capacitor and reversed the direction of the diode, then I get00:24:15
what is the opposite? It is the 4.3 meter00:24:18
of course, 0.7 taken across the diode. Now we will see00:24:22
what happens to the signal inside it that is coming00:24:26
from the source. What is the signal? Do00:24:28
you see that the green one is the signal that is coming from the00:24:30
source in the negative clamper that is the00:24:33
circle? I have an offset that is not below what is the00:24:37
signal. The red one is good, and I have it here.00:25:01
What can I do, quite simply, with00:25:04
this diode? Do you see that I add this diode to it or00:25:08
put it in series with a constant voltage source? It could00:25:12
be a battery. It could be another capacitor charged with a00:25:15
specific charge. So, here we put a00:25:18
5-volt battery. The meaning of the words is that instead of it00:25:23
being an upward shift, only 4.3 volts will be an00:25:27
upward shift. How much is 4.3 and we add 5 volts to it, so it becomes 9.3.00:25:32
See, this is the result. 9.3 volts up. This is00:25:37
9.3 volts up. Therefore, it all became where it was. It was00:25:41
perfect. I can remind you of this thing. I told you00:25:45
that the telescope would be the signal. For example,00:25:48
here I go with a shift up and a shift down. It is00:25:51
actually an application of one of the applications.00:25:53
When you shift00:25:57
up and shift down, you actually control this00:25:59
voltage. It could be five, four, three, or 10.00:26:03
You shift it up or down depending on00:26:06
what you want. Therefore, you control the00:26:09
amount of voltage that is here. You can take00:26:12
this voltage through voltage. The diverter is also perfect, or it is00:26:15
possible through an operation amplifier, but this is a00:26:17
little advanced. I can control this voltage, so I can control the00:26:21
displacement up and down00:26:24
as I do. We hope that the lecture will be enjoyable00:26:27
and that you will benefit from it. We took what is an analysis of the00:26:29
Clamper rotor and what is its benefit, and God willing,00:26:32
you will benefit from it. I hope you share with us via Like00:26:34
and Subscribe and share with us on the site00:26:38
and support us as well, because during this00:26:41
previous period we faced problems on the site. Praise be to00:26:44
God, we have overcome them, but it is possible that you support us by00:26:48
visiting the site, that is, seeing some things.00:26:52
Peace, mercy, and blessings of God be upon you.Description:
للدعم على موقع patreon :: https://www.patreon.com/WalidIssa في هذه الحلقة سنتناول تطبيق اخر من تطبيقات الدايود وهي دوائر Clamper والتي تعمل على اضافة جهد DC الى موجة مترددة فتزيحها عموديا ... سنتناول الموضوع بتحليل تفصيلي وكيف تفكر لتتوقع الخرج واثبات هذا التحليل ببرنامج المحاكاة وبالتجربة العملية ايضا ... يمكنك الاشتراك في الموقع والاستفادة من كل دوراته المجانية والمقالات والدروس العملية في مجال الهندسة الكهربية والالكترونية .... عبر زيارة روابطنا :: رابط تطبيق موقعنا :: http://survey-smiles.com/ رابط الموقع :: www.eletorial.com رابط الصفحة :: www.facebook.com/eletorial شارك معنا ... م. وليد عيسى
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