Finally a good explenation! Tried to follow few other videos and it was a mess, couldn't undertand almost nothing and had zero success prototyping this circuit on breadboard. I am so happy I've found this video, not I finally managed to make this circuit :)
Круть. Мы такой на городских соревнованиях в 10 классе паяли. Команда из 3 человека. Нужно было спаять, подключить к блоку питания, и для дополнительного контроля к осциллографу. Уложились в 47 секунд.
Thanks! Although voltage is isn’t measurable at only one point. It has to be measured between points. So figuring out what was going on in that diagram because of that was somewhat confusing. Also clearly labeling + and - on caps would be better.
OK I think I'm finally beginning to understand, but in my own imaginary way of vacuum suction of particles and a high pressure buildup. Basically the capacitor being charged, so one side is a vacuum of particles (-) and other side is under highpressure loaded with particles(+) (even though it's the positive that pulls negative, but it's better to think the other way around. So one side of a capacitor you have high pressure build up ,while other side of capacitor (-)is vacuum lack of particles. And eventually capacitor becomes fully charged. Than during that time the other side capacitor begins to charge and when that happens, the first capacitor transistor gate closes and the collector begins to fully conduct electricity. And because the capacitor positive side is connected to collector and it was under high pressure the positive particles flow through the collector (releasing the pressure) and discharges in the negative side of the battery power,while the negative side of the first capacitor that was charged if we remember it became a vacuum (lack of particles), so it cannot suck positive charges from the gate of the other transistor which is connected to negative terminal. Instead the vacuum sucks the positive particles from top resistor that is connected to positive terminal. And it will continue to suck the particles of positive until it becomes completely neutral (or discharged capacitor)... And after it's discharged than ,the direction of the current changes and the capacitor begins to recharge again by turning on the other side transistor l. Something like that.
I have to agree. R1=R4 needs to be less than R2=R3. With R1=R3 < R2=R4 both C1 and C2 will be charging/discharging through the same value resistor, hence s/b at the same rate. Question I have is, do both C1 and C2 have the same potential to charge/discharge, and does that make a difference? Why does one Capacitor have to charge/discharge faster than the other? What would happen if the rate was the same? What would happen if C1 rate was slower than C2, instead of higher when indicated here? By not making the R pairs equal in value, can the duty cycle be changed, so one led is on longer than the other. Can the frequency be changed, just by changing the value of the capacitors, just the resistors, or do both have to be changed?
Maybe just figure the tension on the dielectric pulls the base of the base of the transistor negative Briefly. Until you make a list of of each changing state of each component as they change. List the components across the top of the page and the steps down the left side. The voltage on the bases can be seen going negative Unless it was the tension on the pointer spring of the analog voltmeter. Maybe use a digital voltmeter or create a makeshift one with a special analog to digital converter and a positve negative digital counter
Именно для этого грамотные люди и двигали технический прогресс. Чтобы любой человек, с любой точки планеты, мог открывать для себя новое, интересное, а самое главное, правильную информацию. Спасибо за хорошее изложение материала.
It’s a good tutorial, I have a question for circuit at 08:12 when the Q1 goes to off state, C1 at the moment has potential difference between Q and P points, from my understanding the P point will jump to VCc which is 9v and Q becomes 9.7v? Can you please explain this ?
Thanks, very well explained! Cleared all my doubts, just one thing is bothering me is that if both pair of transistor has same components of same specification then both should operate simultaneously, I know due to some operating or environmental conditions little delay can happen, but thinking technically, this is not digesting.😅
If T2 is in off condition the resistance of the transistor is high and almost all voltage Vcc is assumed to be acting across across it by this fact it will be assumed that voltage across R4 will be zero and no current flows through load resistor R4 capacitor C2 will be discharging not charging as there will be no current through it...thats according to my understanding i wish to be corrected if am wrong????
Non electrolytic capacitor voltage range starting from minimum 230 volt . But our requirement 10 mfd /25 volt. So , select electrolytic capacitor, and its cost 5 rupees only.
I'm an idiot when it comes to electronics, but I really want to learn. Am I understanding correctly by saying the on/off timing of the two separate bulbs is created by the difference in resistance on R1/R3 vs. R2/R4?
It depends, as an example you can use followings R1, R4 : 740 Ohm R2, R3 : 470K Ohm C1, C2 : 10 uf By increasing the capacitor value, the vibrating frequency can be lowered.
Thanks for the video. I am very new to electronics, Can i ask a quesiton? if so, why is current not going to T2 from R2 but only going to C1 when C1 is Discharging can't it go both T2 and C1 at 07:03 ?
T2 is OFF because (at the time that you are making reference ) the voltage at T2 base is less than 0.7V, which is the minimum threshold to turn on T2. Therefore, there is NO current flowing (remember, in order to have current flowing you need to ‘complete’ the circuit). Current does not flow through capacitors. Capacitors charge up to the connected voltage potential, and the electrostatic charges that accumulate in it when voltage is applied, discharge through the connection to the circuit return (there is no ground in this circuit) made when T2 turns ON.
These videos are created using Blender 3.2. Blender is a free and open-source 3D computer graphics software tool set used for creating animated films, visual effects, art, 3D-printed models, motion graphics, interactive 3D applications, virtual reality, and, formerly, video games. You can download it from here : www.blender.org
It works because of the variance in tolerances of the components. If they are so very close in values the oscillation won't start. I did this once by accident. I took my twizzers and lightly shorted the transistor legs to start the oscillation. But that may not be your problem. Just double check the connections. This is a very common circuit.
@@peatmoss4415 i tried it on a breadboard before and it was working just fine but when i solder it on a pcb either both LED turned on or just one. I did it a couple times and got the same results. Maybe a component wasn't functioning properly?
And also the speed can be changed connecting the upper end of R2 And R3 together connecting to a potentiometers middle pin , while one pin of the side of potentiometer is connected to the top power line
the first image of the circuit has the tranisstor wired wrong...they have the gate connected to the emitter...I was sitting here confused about it for a good minute...
when you apply the knee voltage (0.7 V) to the base pin of a transistor, the resistance between the collector and emitter pin becomes 0. That means the voltage of the collector should be equals to the emitter. Normally we consider this happens instantaneously. But actually it takes few microseconds. during this time period the voltage difference between the collector and emitter decrease gradually.
"hello world" program is confusing; it does nothing. So it's not helpful to show, what a program can do. Rather a "two number addition" program can be helpful, i think.
Hello, I have a battery holder where 2 mini 1.5v button batteries belong power a LED. Unfortunately, they are always very quickly empty. I now wanted to feed the LED with 3V power supply instead of the 2 1,5V batteries.. If I put the minus cable to one pole of the battery holder and the other plus cable to the other pole, the LED does not light up. Do I have to make a bridge(connect the 2 poles with each other) between the two poles (plus and minus)? I attached a picture of what I mean: drive.google.com/file/d/16HTlUvZGBan05GhirxAmT-myxfN0AEpY/view?usp=sharing
Finally a good explenation! Tried to follow few other videos and it was a mess, couldn't undertand almost nothing and had zero success prototyping this circuit on breadboard. I am so happy I've found this video, not I finally managed to make this circuit :)
finally i understood this circuit! Thank you!
I hate robots, but that was admittedly a great explanation. And can we thank the heavens for the people that invented transistors. Geniuses.
this is the “hello world” of electronics?? Goddamn
The world's best teacher thanks sir
Круть. Мы такой на городских соревнованиях в 10 классе паяли. Команда из 3 человека. Нужно было спаять, подключить к блоку питания, и для дополнительного контроля к осциллографу. Уложились в 47 секунд.
Its working... And more simple the others multivibrator circuit
Thanks! Although voltage is isn’t measurable at only one point. It has to be measured between points. So figuring out what was going on in that diagram because of that was somewhat confusing. Also clearly labeling + and - on caps would be better.
High voltage is positive.
@@lookupverazhou8599 Heard now he said pos sides of caps are connected to the collector pins of the transistors.
The shaded side of a cap is negative.
@@peatmoss4415 In that diagram the darker portion is the minus half of the caps?
OK I think I'm finally beginning to understand, but in my own imaginary way of vacuum suction of particles and a high pressure buildup. Basically the capacitor being charged, so one side is a vacuum of particles (-) and other side is under highpressure loaded with particles(+) (even though it's the positive that pulls negative, but it's better to think the other way around. So one side of a capacitor you have high pressure build up ,while other side of capacitor (-)is vacuum lack of particles. And eventually capacitor becomes fully charged. Than during that time the other side capacitor begins to charge and when that happens, the first capacitor transistor gate closes and the collector begins to fully conduct electricity. And because the capacitor positive side is connected to collector and it was under high pressure the positive particles flow through the collector (releasing the pressure) and discharges in the negative side of the battery power,while the negative side of the first capacitor that was charged if we remember it became a vacuum (lack of particles), so it cannot suck positive charges from the gate of the other transistor which is connected to negative terminal. Instead the vacuum sucks the positive particles from top resistor that is connected to positive terminal. And it will continue to suck the particles of positive until it becomes completely neutral (or discharged capacitor)... And after it's discharged than ,the direction of the current changes and the capacitor begins to recharge again by turning on the other side transistor l. Something like that.
nice explanation, but @8.52 you write the resistors value wrong on left side, because R1=R4 not R3
I have to agree. R1=R4 needs to be less than R2=R3. With R1=R3 < R2=R4 both C1 and C2 will be charging/discharging through the same value resistor, hence s/b at the same rate. Question I have is, do both C1 and C2 have the same potential to charge/discharge, and does that make a difference? Why does one Capacitor have to charge/discharge faster than the other? What would happen if the rate was the same? What would happen if C1 rate was slower than C2, instead of higher when indicated here? By not making the R pairs equal in value, can the duty cycle be changed, so one led is on longer than the other. Can the frequency be changed, just by changing the value of the capacitors, just the resistors, or do both have to be changed?
Where are the values of the components stated? How does anyone build this without that information? But the description of the operation is good.
Very good explanation. I think this type of circuit is also called Flip Flop circuit.
Thank you very much, it is the best explanation I ever had
Maybe just figure the tension on the dielectric pulls the base of the base of the transistor negative Briefly. Until you make a list of of each changing state of each component as they change. List the components across the top of the page and the steps down the left side. The voltage on the bases can be seen going negative Unless it was the tension on the pointer spring of the analog voltmeter. Maybe use a digital voltmeter or create a makeshift one with a special analog to digital converter and a positve negative digital counter
Sehr schön animiert und auch super erklärt. Daumen hoch !👍🙋😀
Именно для этого грамотные люди и двигали технический прогресс. Чтобы любой человек, с любой точки планеты, мог открывать для себя новое, интересное, а самое главное, правильную информацию. Спасибо за хорошее изложение материала.
Thank you, the explanation is obvious and really helpful
Unless I'm mistaken, the crossing should not be drawn as a node.
Changing the capacitance value will change the blink frequency
Subscribed. This type of tutorial video I need to understand and learn. Want to try build this too.
Awesome, thank you!
This trend of putting repetitive tin music in the background is really a shame. Especially when you are explaining something
I am following your videos respected professor
Glad to hear that
@@Profmad sir please make van de graph generator in front of camera , if possible please
I have a question at 9:04. Is your formula right? I would expect R1=R4
you are right, wrong notation
This is very helpful 👍.
Glad it was helpful!
It’s a good tutorial, I have a question for circuit at 08:12 when the Q1 goes to off state, C1 at the moment has potential difference between Q and P points, from my understanding the P point will jump to VCc which is 9v and Q becomes 9.7v? Can you please explain this ?
Nice explanation
is the middle section interconnected ? I mean is current also flowing from Q to T1 ? or those are different cables
well explained
Thanks, very well explained! Cleared all my doubts, just one thing is bothering me is that if both pair of transistor has same components of same specification then both should operate simultaneously, I know due to some operating or environmental conditions little delay can happen, but thinking technically, this is not digesting.😅
You will never find 2 resistors for example that have exactly the same resistance. Most you buy are for example 1%.
@@jinsoku1911 okay, understood, thanks 👍
If T2 is in off condition the resistance of the transistor is high and almost all voltage Vcc is assumed to be acting across across it by this fact it will be assumed that voltage across R4 will be zero and no current flows through load resistor R4 capacitor C2 will be discharging not charging as there will be no current through it...thats according to my understanding i wish to be corrected if am wrong????
I don’t get it, if the polarity changes on the capacitors, then why does everyone use electrolytic capacitors for this circuit and everything works?
Non electrolytic capacitor voltage range starting from minimum 230 volt .
But our requirement 10 mfd /25 volt.
So , select electrolytic capacitor, and its cost 5 rupees only.
May I ask the value of the materials that you used ?
I'm an idiot when it comes to electronics, but I really want to learn. Am I understanding correctly by saying the on/off timing of the two separate bulbs is created by the difference in resistance on R1/R3 vs. R2/R4?
how can that polarized capacitor be charged in oppositite direction please help
Wow thanks you Sir
Most welcome
guys is it safe to connect a capacitor in reverse polarity the one that he used in the video is a polarized capacitor
very nice video. can you make a tutorial on how to make this video Please!!
It was created using Blender 3.2. Ill create a tutorial for that also. 🙂
@@Profmad Thank you very much.
Amazing video and animations are very good thank you for that
Glad you like them!
4km
Generator square signal which is 0 to +5volt DC
Great vid, but please stop the annoying and disctracting background music next time. thanks.
Sir, how much the value of capasitor and resistor that used?
It depends, as an example you can use followings
R1, R4 : 740 Ohm
R2, R3 : 470K Ohm
C1, C2 : 10 uf
By increasing the capacitor value, the vibrating frequency can be lowered.
@@Profmad but you said at 8:52 that R1= R3
@@c2225😂😂😂
Thanks for the video. I am very new to electronics, Can i ask a quesiton? if so, why is current not going to T2 from R2 but only going to C1 when C1 is Discharging can't it go both T2 and C1 at 07:03 ?
I think that is because T2 is in an off state and its emitter is not conducting, but I'm as green as you :)
T2 is OFF because (at the time that you are making reference ) the voltage at T2 base is less than 0.7V, which is the minimum threshold to turn on T2. Therefore, there is NO current flowing (remember, in order to have current flowing you need to ‘complete’ the circuit). Current does not flow through capacitors. Capacitors charge up to the connected voltage potential, and the electrostatic charges that accumulate in it when voltage is applied, discharge through the connection to the circuit return (there is no ground in this circuit) made when T2 turns ON.
Sir, your animation is very simple. May i know how you do the animation for your videos .Thanks.
These videos are created using Blender 3.2. Blender is a free and open-source 3D computer graphics software tool set used for creating animated films, visual effects, art, 3D-printed models, motion graphics, interactive 3D applications, virtual reality, and, formerly, video games. You can download it from here : www.blender.org
@@Profmad thank you sir for your valuable information.
What
Da ae
Anyone know 3:27 why do the voltage in P and S decrease?
Yeah I don't understand either, since when does a collector voltage continuously decrease with just a resistor and diode in series with it?
I tried doing this on a pcb but both LED turned on at the same time. Can you explain why it isn't blinking?
It works because of the variance in tolerances of the components. If they are so very close in values the oscillation won't start. I did this once by accident. I took my twizzers and lightly shorted the transistor legs to start the oscillation. But that may not be your problem. Just double check the connections. This is a very common circuit.
@@peatmoss4415 i tried it on a breadboard before and it was working just fine but when i solder it on a pcb either both LED turned on or just one. I did it a couple times and got the same results. Maybe a component wasn't functioning properly?
Plc could you explain how to work SMPS
Hope this video will help you : ua-cam.com/video/F2dCS5qOE8A/v-deo.html
"R1=R3 R2=R4" Are they wrong? Sould be "R1=R4" and "R2=R3"
Yes !
Capacitor I used are 10uF/25v
R2=R3 ==> 2.2K ohm
R1=R4 ==> 220 ohm
Transistor ==> BC547
And also the speed can be changed connecting the upper end of R2 And R3 together connecting to a potentiometers middle pin , while one pin of the side of potentiometer is connected to the top power line
Challenge for those don't now electronic used this circuit to blind 3 led in different time
Circuit operation can not be displayed clearly due to subtitles.
Check now.
You need not increase expenses when your income rises or increases.Jos.
❤ESCARGENCY ❤😮❤
You write that R1=R3 and R2=R4. I think that's a mistake. I think R1=R4 and R2=R3 and R2 and R3 are much higher than R1 and R4. Is't correct?
What does Edith maen
Music plus robotic voice does my head in, sorry but what about a bit of down to earth chatter.
What transistor numbers can be used for this 9V circuit? Thanks.
You can use any general purpose npn transistor like c1815 or c945.
Imagine 3 LEDs in a triangle but instead of leds you make electromagnets turn on and off
ua-cam.com/video/FLk2U3ITdDw/v-deo.htmlfeature=shared
how is the capacitor positively charged on both sides????
5:43 anyone can explained me,why Q side become -0,3V?
so the Q side has to keep being 0.3V lower than the right side.what is it mean?
sir can you plz send the transcript of the video plzzzz
it will help my project
The transcript is in the video description. Under the "...more"
the first image of the circuit has the tranisstor wired wrong...they have the gate connected to the emitter...I was sitting here confused about it for a good minute...
can anyone explain to me, why and what happen in 02:14 .. why voltage gradually decreasing and finally become 0
when you apply the knee voltage (0.7 V) to the base pin of a transistor, the resistance between the collector and emitter pin becomes 0. That means the voltage of the collector should be equals to the emitter. Normally we consider this happens instantaneously. But actually it takes few microseconds. during this time period the voltage difference between the collector and emitter decrease gradually.
@@Profmad thanks a lot
Components plss😊
how is q discharging & charging at the same time...
What is this simulator name?
Blender 3.5
"hello world" program is confusing; it does nothing. So it's not helpful to show, what a program can do. Rather a "two number addition" program can be helpful, i think.
Sir ur name plz
Tranislate to somali
I don't understand, I've been understanding this circuit for years, and I still don't understand this fucking simple circuit😂
OR?? could simply buy Flashing leds .
Nawww too easy.
Hindi yaar bol
Hindi is our mother language plzzz teach us in hindi
Haaa blkul Shahi 😂🎉❤
Hello,
I have a battery holder where 2 mini 1.5v button batteries belong power a LED.
Unfortunately, they are always very quickly empty. I now wanted to feed the LED with 3V power supply instead of the 2 1,5V batteries.. If I put the minus cable to one pole of the battery holder and the other plus cable to the other pole, the LED does not light up. Do I have to make a bridge(connect the 2 poles with each other) between the two poles (plus and minus)?
I attached a picture of what I mean:
drive.google.com/file/d/16HTlUvZGBan05GhirxAmT-myxfN0AEpY/view?usp=sharing