The timestamps for the different topics covered in the video: 0:23 What is Monostable Multivibrator? (A brief introduction) 0:53 Description of the Monostable Multivibrator Circuit 2:28 The operation (working) of the Monostable Multivibrator 7:33 The derivation of the time duration expression of Monostable Multivibrator
Sir thank you for your great lecture. Could you make a video about interface (I2C, SPI..). As a new analog engineer, it is pretty hard to understand these interfaces. Thank you!
Because when a capacitor suddenly discharges off the positive plate, there is a sudden negative voltage off its negative plate. Current draws toward the cathode side of a capacitor during discharge due to what's called displacement current.
Capacitor ke Left pe VCC Right pe Vbe Capacitor ki khud me voltage jispe vo charge hoga = X (X= Vcc-Vbe left to right) Ab Q1 grounds So change happens Capacitor ke Left pe 0 Right pe Vbe Capacitor ki khud me voltage jo held thi pehle se = X (same as before) (X= Vcc-Vbe left to right) Now left to right jao X neeche Higher to lower X neeche 0 gnd se X neeche = -X = -Vcc+Vbe
I dont understand how Q2 will remain in ON state forever bcoz there is capacitor in the circuit. It will start charging through the path Rc1Q2 and then its right plate will develope negative polarity and hence which will also make the base of Q2 neagative and Q2 will be off due to reversed bias. Please correct me if I am wrong?
To whom it may concern We are developing an online programme for the South African Department of Higher Education and would like to include your video in the unit on Electronics. Please also indicate if all your videos will have the same permissions. In order to assist learners who do not have regular access, all the media will be loaded onto a learning management system for download to a device. I would, therefore, like to request permission to download your video from TH-cam to include on the LMS. We will provide full attribution to the video with whatever details you require. I look forward to hearing from you. Many thanks.
why aren't both transistors in saturation mode from the beginning? Doesn't the voltage source immediately have paths to both bases? Why exactly would Q2 start conducting before Q1?
The voltage is applied through a base resistor. The resistor will also play a role in it. Q2 is connected to Vcc via R2. While Q1 is connected to Vcc, via R1 and Rc2. The current in Q2 will be more and hence it will go into saturation first. I hope it will clear your doubt. If you still have any doubt, then let me know here.
@@ALLABOUTELECTRONICS Wow thanks for the quick reply! I thought that might be the case, but was thrown-off by the example given by electronics tutorials ( www.electronics-tutorials.ws/waveforms/monostable.html ), where RT = 6*(R2+R3). Am I missing something?
When Q1 is in saturation - won't capacitor C be reverse biased? That is no good if C is an electrolytic if I understand correct? Must then C be a non-polar capacitor?
So I want to see a TRUE one shot... where a switch is flipped to trigger it and the trigger changes states and stays, so if the switch is state 1 and flipping it sets state 0 and holds it, I want the circuit to do 1 pulse based on R/C and not re-trigger until the switch is manually toggled between state 1 and state 0 again.
Because when the capacitor C suddenly discharges off the positive plate, there is a sudden negative voltage off its negative plate. Current draws toward the cathode side of a capacitor during discharge due to what's called displacement current. This sudden drop into negative voltage on the Base of Q2 is what turns it off.
It was derived using the transient analysis. The expression which I have written at 7:52 is the general expression of capacitor voltage during the transients. I have already discussed about in detail in one of the earlier videos. Please go through this video for more information. th-cam.com/video/KylJ2v1_c-o/w-d-xo.html If you still have any doubt then let me know here.
Since Q2 is ON and operating in saturation, it can be assumed as short circuit. So, one end of R1 is connected to ground terminal. And other end is connected to Q1. That means the base of Q1 is connected to ground via R1. So, we can say that, the voltage at the base of Q1 is zero. (Diode D is reversed biased and will act as a open circuit) I hope, it will clear your doubt.
at the end of the video, (around 12:00) I have shown how to select the values of R1 and Rc. At 3V, just make sure that, transistor operates in the required region. You can tune with R1 and C. You can reduce R1 and increase C. With little tuning, the circuit can work.
The timestamps for the different topics covered in the video:
0:23 What is Monostable Multivibrator? (A brief introduction)
0:53 Description of the Monostable Multivibrator Circuit
2:28 The operation (working) of the Monostable Multivibrator
7:33 The derivation of the time duration expression of Monostable Multivibrator
Please tell me how you derive that equation for Vc(t)
The only video and best help i found on the internet. Thank you so much!
Very nice video man with good detail.
It was really helpful!!!!
Thank you for this calculated time.
Sir thank you for your great lecture.
Could you make a video about interface (I2C, SPI..).
As a new analog engineer, it is pretty hard to understand these interfaces.
Thank you!
I will make the videos on them. (I2C, SPI)
So just to be sure. In this diagram both of the transistors act as on off switches.
Tq so much sirrr 😊😊.....
Well explained
I want to MOSFET series videos from this channel like Op-Amps series
Why the negative voltage of - (Vcc-Vbe) is getting applied to Q2 and how?
Because when a capacitor suddenly discharges off the positive plate, there is a sudden negative voltage off its negative plate. Current draws toward the cathode side of a capacitor during discharge due to what's called displacement current.
Capacitor ke
Left pe VCC
Right pe Vbe
Capacitor ki khud me voltage jispe vo charge hoga = X
(X= Vcc-Vbe left to right)
Ab Q1 grounds
So change happens
Capacitor ke
Left pe 0
Right pe Vbe
Capacitor ki khud me voltage jo held thi pehle se = X (same as before)
(X= Vcc-Vbe left to right)
Now left to right jao X neeche
Higher to lower X neeche
0 gnd se X neeche
= -X
= -Vcc+Vbe
@@asharawat486 Thank You
Sir what is the use of capacitor and diode
I dont understand how Q2 will remain in ON state forever bcoz there is capacitor in the circuit. It will start charging through the path Rc1Q2 and then its right plate will develope negative polarity and hence which will also make the base of Q2 neagative and Q2 will be off due to reversed bias. Please correct me if I am wrong?
This is collector coupled. Make videos on emitter coupled
To whom it may concern
We are developing an online programme for the South African Department of Higher Education and would like to include your video in the unit on Electronics. Please also indicate if all your videos will have the same permissions.
In order to assist learners who do not have regular access, all the media will be loaded onto a learning management system for download to a device. I would, therefore, like to request permission to download your video from TH-cam to include on the LMS. We will provide full attribution to the video with whatever details you require.
I look forward to hearing from you.
Many thanks.
Any replies?
Can u explain modulation and demodulation using CRO
why aren't both transistors in saturation mode from the beginning? Doesn't the voltage source immediately have paths to both bases? Why exactly would Q2 start conducting before Q1?
The voltage is applied through a base resistor. The resistor will also play a role in it. Q2 is connected to Vcc via R2. While Q1 is connected to Vcc, via R1 and Rc2. The current in Q2 will be more and hence it will go into saturation first.
I hope it will clear your doubt. If you still have any doubt, then let me know here.
@@ALLABOUTELECTRONICS Wow thanks for the quick reply!
I thought that might be the case, but was thrown-off by the example given by electronics tutorials ( www.electronics-tutorials.ws/waveforms/monostable.html ), where RT = 6*(R2+R3). Am I missing something?
When Q1 is in saturation - won't capacitor C be reverse biased? That is no good if C is an electrolytic if I understand correct? Must then C be a non-polar capacitor?
Yes.
Sir please provide a lecture on hybrid parameters
Very soon it will be covered.
Hello, how could I do a PWM control?, using this configuration and with BJT transistors.
So I want to see a TRUE one shot... where a switch is flipped to trigger it and the trigger changes states and stays, so if the switch is state 1 and flipping it sets state 0 and holds it, I want the circuit to do 1 pulse based on R/C and not re-trigger until the switch is manually toggled between state 1 and state 0 again.
My problem is I don't understand English language 😑😑😑
Thank you sir
Sir I have doubts sir
Why we took R2>=10R1 ?
And why we applied _(Vcc_Vbe) to transistor Q2 when trigger is applied ??
Because when the capacitor C suddenly discharges off the positive plate, there is a sudden negative voltage off its negative plate. Current draws toward the cathode side of a capacitor during discharge due to what's called displacement current. This sudden drop into negative voltage on the Base of Q2 is what turns it off.
Sir
Please explain pass transistor used in regulator circuit
And PLL with practical examples
Schmit trigger
About voltage regulator and Schmitt trigger, I have already made a video.
Will make a video on PLL.
how was the capacitor charging expression derived?
It was derived using the transient analysis. The expression which I have written at 7:52 is the general expression of capacitor voltage during the transients.
I have already discussed about in detail in one of the earlier videos.
Please go through this video for more information.
th-cam.com/video/KylJ2v1_c-o/w-d-xo.html
If you still have any doubt then let me know here.
at 3.09 how the voltage at the base of Q1 is zero???
Since Q2 is ON and operating in saturation, it can be assumed as short circuit. So, one end of R1 is connected to ground terminal. And other end is connected to Q1. That means the base of Q1 is connected to ground via R1. So, we can say that, the voltage at the base of Q1 is zero. (Diode D is reversed biased and will act as a open circuit)
I hope, it will clear your doubt.
Sir can you please upload bistable multi vibrator using transistor also
Yes. Soon it will uploaded.
How to calculate values if VCC is 3v...
at the end of the video, (around 12:00) I have shown how to select the values of R1 and Rc. At 3V, just make sure that, transistor operates in the required region. You can tune with R1 and C. You can reduce R1 and increase C. With little tuning, the circuit can work.
Sir pls kya aap iske notes banakar de sakte hain kal tak
Written mai
Write from the subtitles. Then break down to what's important to you.
👌🏻👌🏻
I can make both rising edge and falling edge monostable circuit in Minecraft lol.
Buck converter ke upar video banao. Na
Request karte samay please and thank you to boliye please
Capacitor charging path is wrong
Poor explanatipn 😢