The timestamps for the different topics covered in the video: 1:00 Why a coupling capacitors are used in the Amplifier Circuit 2:45 Steps to follow for the Small Signal Analysis 4:35 Small Signal Analysis of CE Fixed Bias Circuit 10:40 Small Signal Analysis (with output resistance) 11:44 Small Signal Analysis of CE Voltage Divider Bias Circuit
plz..can you tell me which software you are using as explanation board..? I'm also started one channel I want one software like you are using ....! Thank you
A small increase in the voltage applied to the base will produce an increase in the base current. That will produce an increase in the collector current. The increased collector current will produce a larger voltage across the collector resistor. But one end of that resistor is fixed at the supply voltage, so the voltage at the other end of the resistor must _decrease._ That is the collector of the transistor, and is the output of the stage. That shows that as the input voltage increases, the output voltage decreases. That is why we call it a negative gain.
Sir, I have observed that when there is a talk about approximation of anything, You call it so lightly that It is just for knowledge. Please tell that what exactly we will use during the question-solving side. I hope you will solve it.
Many examples based on small signal analysis are also solved on the second channel. On this channel also, if you go through the playlist page, then you will find the BJT Solved example playlist. If you go through it, you will find many solved examples based on small signal analysis.
As we consider capacitor is shorted for small signal analysis, does that mean small signal analysis is always done for high frequency application purpose?
No, here it is done for mid- band frequencies. But here it is assumed that the value of capacitor is selected such that, at the operating frequency , its reactance is very low and it can be assumed as a short circuit.
Base resistor Rb is between base and ground. So, in AC equivalent circuit it is represented between base and emitter ( as emitter is ground). Similarly, collector resistor is between collector and ground terminal . So, in AC equivalent circuit, it is represented between collector and emitter terminal. I hope, it will clear your doubt.
Sir on the input the decoupling capacitor is placed on before the fixed biasing Resistor Rb, so how will it block the DC bias voltage from Rb given to base of bjt?
It will block the DC voltage of the other BJT amplifier which is cascaded with this. Many times the AC input signal is the output of pre-amplifier (which is also biased with DC voltage). The DC voltage of that amplifier, should not affect the biasing of the BJT amplifier. That's where the decoupling capacitor is very useful for isolating the two stages. I hope, it will clear your doubt.
plz..can you tell me which software you are using as explanation board..? I'm also started one channel I want one software like you are using ....! Thank you
No. The input impedance is the parallel combination of the resistors. It should be obvious that if those resistances increase, the input impedance must increase. Your formula would give a smaller answer if you made the resistances larger, and therefore must be wrong.
In the equivalent circuit, if you see the current source gm*Vπ, it is flowing downwards. Applying KCL at the output, Io = - gm*Vπ. Therefore, Vo = -gm*Vπ*Rc And since Vin = Vπ. Vo / Vin or the voltage gain = -gm*Rc I hope it will clear your doubt.
@@user-fi3cd3ig6l As I mentioned earlier, in the equivalent circuit, if you see the current source gm*Vπ is flowing downwards. Applying KCL at the output, Io = - gm*Vπ. Therefore, Vo = -gm*Vπ*Rc And since Vin = Vπ. Vo / Vin or the voltage gain = -gm*Rc I hope it will clear your doubt.
Not quite. The ac response is idealised so that we assume the changes in quiescent voltages and currents are small compared to their static values. You may have to consider sometimes the ac response of a circuit where the output swing is large, and then you will have to take into account, for example, the change in gm (and hence gain) as the collector current varies.
@@ALLABOUTELECTRONICS Respected i thought this subject is harrd and i am going to fail in the sem exam but u saved my day just now started to study for the xam i have days of solid time to study ...i will get good marks with this tutorial ....need to say something sir i am also teaching to my friends .....THank U SO much Sir..................................
While calculating the output resistance, the input voltage source is considered as zero. And the dependent current source, gmVbe, depends on the input voltage. So, once input voltage and hence vbe is zero, the department current source will also become zero. I hope it will clear your doubt.
Hello can you help me in my research Who is talking about star-Delta transformation and its application for electric circuit containing impedances (complex values) similar.
It is the rate of change of collector current as the base-emitter voltage changes. It increases as the collector current increases and is almost independent of the transistor used. It is found to be equal to the collector current divided by 25mV at room temperature.
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The timestamps for the different topics covered in the video:
1:00 Why a coupling capacitors are used in the Amplifier Circuit
2:45 Steps to follow for the Small Signal Analysis
4:35 Small Signal Analysis of CE Fixed Bias Circuit
10:40 Small Signal Analysis (with output resistance)
11:44 Small Signal Analysis of CE Voltage Divider Bias Circuit
plz..can you tell me which software you are using as explanation board..?
I'm also started one channel I want one software like you are using ....!
Thank you
Sir is this hybrid π model please help me...
@@jineshks8271yess
Best Channel for Analog Electronics!
Thank you so much. You just saved me from hours of work.
salute from germany bro..
extremely helpful! Amazing explanation 🙏🙏
Excellent 👌 as usual
thank you sir.
Can you explain more why is negative sign with the Voltage gain?
Thank you
Due to out of phase at output
Dear when output is taken from collector then input signal in out of phase to the output (invert at output)
A small increase in the voltage applied to the base will produce an increase in the base current. That will produce an increase in the collector current. The increased collector current will produce a larger voltage across the collector resistor. But one end of that resistor is fixed at the supply voltage, so the voltage at the other end of the resistor must _decrease._ That is the collector of the transistor, and is the output of the stage.
That shows that as the input voltage increases, the output voltage decreases. That is why we call it a negative gain.
Sir, I have observed that when there is a talk about approximation of anything, You call it so lightly that It is just for knowledge. Please tell that what exactly we will use during the question-solving side.
I hope you will solve it.
Many examples based on small signal analysis are also solved on the second channel. On this channel also, if you go through the playlist page, then you will find the BJT Solved example playlist. If you go through it, you will find many solved examples based on small signal analysis.
Bestest 👌🏻👌🏻
Thank you Sir.
Loved this video
Thank you sir🙏🥺👍❤️
3:27 how this circuit will be fixed bias config? Base voltage is derived from Vcc. So can't it be voltage divider bias configuration sir?
As we consider capacitor is shorted for small signal analysis, does that mean small signal analysis is always done for high frequency application purpose?
No, here it is done for mid- band frequencies. But here it is assumed that the value of capacitor is selected such that, at the operating frequency , its reactance is very low and it can be assumed as a short circuit.
Thank u sir..it helps alot❤️👍
can you explain in which configuration the automatic street light using bc547 and ldr without relay
🙏 Thanks 🙏👍
Sir why can’t we use vout/ iout for output impedance calculations
Sir 6:16 i am confused about the resistors Rb and Rc how u placed them it seems odd
Base resistor Rb is between base and ground. So, in AC equivalent circuit it is represented between base and emitter ( as emitter is ground).
Similarly, collector resistor is between collector and ground terminal . So, in AC equivalent circuit, it is represented between collector and emitter terminal. I hope, it will clear your doubt.
Can you explain the condition at 16:30..? because the resultant of a parallel combination is always less then the least value
Apne Gain Ka Derivation With Early Effect Miss Kar Diya Hai Wahi Zaruri Tha 😪
Can input impedance also be calculated as thevenin equivalent resistance as seen through input terminal?
Not quite sure, but I believe the small impedence resistance is ignored in comparison to the base/input resistance.
Sir on the input the decoupling capacitor is placed on before the fixed biasing Resistor Rb, so how will it block the DC bias voltage from Rb given to base of bjt?
It will block the DC voltage of the other BJT amplifier which is cascaded with this. Many times the AC input signal is the output of pre-amplifier (which is also biased with DC voltage). The DC voltage of that amplifier, should not affect the biasing of the BJT amplifier. That's where the decoupling capacitor is very useful for isolating the two stages.
I hope, it will clear your doubt.
When the voltage source is shorted how is resistance ignored, cant current flow through it??
plz..can you tell me which software you are using as explanation board..?
I'm also started one channel I want one software like you are using ....!
Thank you
WILL A SALESMAN TELL YOU WHERE HE GETS THE GOODS FROM,,,,,,,,THEN YOU WILL ALSO BECOME A COMPETITOR AND STAND AGAINST HIM
@@vaibhavgiria5997 😂
in the input impedance it is not the parallel combination of Rb and Rpi rather it is 1/parallel combination Rb and Rpi
No. The input impedance is the parallel combination of the resistors. It should be obvious that if those resistances increase, the input impedance must increase. Your formula would give a smaller answer if you made the resistances larger, and therefore must be wrong.
Why you put minus before gm in voltage gain? Vo=-gmVRc
In the equivalent circuit, if you see the current source gm*Vπ, it is flowing downwards.
Applying KCL at the output, Io = - gm*Vπ.
Therefore, Vo = -gm*Vπ*Rc
And since Vin = Vπ.
Vo / Vin or the voltage gain = -gm*Rc
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Thank you for the answer
@@ALLABOUTELECTRONICS here how Io=-gm x vpi? Please answer me sir
@@ALLABOUTELECTRONICS how did minus sign come?
@@user-fi3cd3ig6l As I mentioned earlier, in the equivalent circuit, if you see the current source gm*Vπ is flowing downwards.
Applying KCL at the output, Io = - gm*Vπ.
Therefore, Vo = -gm*Vπ*Rc
And since Vin = Vπ.
Vo / Vin or the voltage gain = -gm*Rc
I hope it will clear your doubt.
Sir, when does ro appear in the circuit?
When the early effect of BJT is also considered then output resistance should also be considered during the analysis.
The Ai of all common emitter circuits is equal to beta?
How to define the direction of io
Hai sir. I have a doubt. Weather AC analysis and small signal analysis are same?
Yes
Not quite. The ac response is idealised so that we assume the changes in quiescent voltages and currents are small compared to their static values.
You may have to consider sometimes the ac response of a circuit where the output swing is large, and then you will have to take into account, for example, the change in gm (and hence gain) as the collector current varies.
Why don't u find Current gain?
Can we use this model for high speed/ high frequency circuit?
At high frequency, other factors need to be considered as well. Will cover the high frequency model soon.
@@ALLABOUTELECTRONICS Respected i thought this subject is harrd and i am going to fail in the sem exam but u saved my day just now started to study for the xam i have days of solid time to study ...i will get good marks with this tutorial ....need to say something sir i am also teaching to my friends .....THank U SO much Sir..................................
Bro please put video about to explain the instrumentation amplifier
I have already made video on it.
Here is the link: ruclips.net/video/pSctPegtZfc/видео.html
are you sure that , you can see the generator as a open circuit? because it's not a indipendent generator...
I think you are talking about the equivalent circuit for finding the output resistance right !!
@@ALLABOUTELECTRONICS yes
LoOkB4U is my second account sir...
While calculating the output resistance, the input voltage source is considered as zero. And the dependent current source, gmVbe, depends on the input voltage. So, once input voltage and hence vbe is zero, the department current source will also become zero.
I hope it will clear your doubt.
Hello can you help me in my research Who is talking about star-Delta transformation and its application for electric circuit containing impedances (complex values) similar.
I have already made videos on star to delta transformation.
You can go through those videos.
@@ALLABOUTELECTRONICS thanks so much
This r0 existence is not cleared sir ...
What is this place tell something
Please go through the video of early effect. It will get clear to you.
Here is the link: ruclips.net/video/BOk501WULQs/видео.html
what is gm in the video, could someone plz explain this to me??
gm is transconductance.
It is the rate of change of collector current as the base-emitter voltage changes. It increases as the collector current increases and is almost independent of the transistor used. It is found to be equal to the collector current divided by 25mV at room temperature.
jo jo yeh like karega, utna property aryan ke naam😆🤣🥰😝🥵🥵🏠🏠🏠
1 like= 1 sq.ft
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When the voltage source is shorted how is resistance ignored, cant current flow through it??