These videos are dedicated my beloved students It's not in an professional way of presentation.. I'm just forced to do this for the benefit of you my dear students👍
That's a poor example of CE design from an AC standpoint. The DC analysis shows the following: With a 12V supply and Rc = 2.2K, the collector current is likely to be a few milliamps, so the base current will be β times smaller. The BC107 has a β between 125 and 900, so the base current will be between a few microamps and a few tens of microamps. The current through the base bias resistors (R1 + R2) is 12V / (56K + 10K) = 182μA, which is so much larger that we can ignore the effect of the base current on the bias network. The voltage at the base (Vb) is 12V x 10K / (10K + 56K) = 1.82V. The voltage at the emitter (Ve) is Vb - Vbe = 1.82V - 0.65V = 1.17V. The emitter current is Ve / Re = 1.17V / 470R = 2.49mA. This confirms the supposition that the collector current, which is almost the same as the emitter current, is a few milliamps. The voltage across Rc is 2.49mA x 2.2K = 5.47V, so the voltage at the collector (Vc) is 12V - 5.47V = 6.53V. The voltage across the transistor Vce is Vc - Ve = 6.53V - 1.17V = 5.36V. That's good from a DC standpoint; Vce and voltage across Rc are similar. The problems occur with the AC analysis: The intrinsic emitter resistance (re) is 26mV/Ic = 26mV / 2.49mA = 10.4R. At frequencies where the reactance of the emitter bypass capacitor is small compared to re, the voltage gain of the transistor will be Rc/re = 210. The -3dB frequency for 10.4R and 100μF is 1/2πCR = 150Hz. The input impedance of the stage is the parallel combination of (R1 || R2 || the input impedance of the transistor). Unfortunately, the input impedance of the transistor depends on β, being β x re, which is between 125 x 10.4R = 1.3K and 900 x 10.4R = 9.4K. That is the dominant factor in calculating the input impedance which will be somewhere between 1.1K and 4.5K, depending on the particular sample of BC107 used. That's not good design. Consequently, the -3dB frequency for the input capacitor and the input impedance will lie somewhere between 355Hz and 1.5KHz. Again poor design. Finally, the large signal gain shows significant distortion as seen at 13:25 where the top of the output trace is rounded and the bottom of the trace is spikey for a 6V peak-to-peak signal. This is due to the non-linear nature of the intrinsic emitter resistance (re). At the top of the trace, the collector current is about (5.5V - 3V) / 2.2K = 1.1mA, giving re ≈ 24R and an instantaneous gain of 2.2k/24R = 90. At the bottom of the trace, the collector current is about (5.5V + 3V) / 2.2K = 3.9mA, giving re ≈ 6.7R and an instantaneous gain of 2.2k/6.7R = 330. Bypassing the emitter resistor gives a high gain, but at the cost of low and variable input impedance and severe distortion of large signals.
We can use minimum of 12V DC or 15V DC. A electrolytic capacitor is connected across the Emitter resistor, which is reduces the effect of negative feedback
![ร้อยพ่อพันแม่ - WanMai [Official MV]](http://i.ytimg.com/vi/XY2pNYASEbg/mqdefault.jpg)
These videos are dedicated my beloved students
It's not in an professional way of presentation.. I'm just forced to do this for the benefit of you my dear students👍
Thanks a lot sir
Helped me a lot ❣️💯
Thank-you for your efforts for making us understand so well sir!👍
Thanks a ton for a detailed video and describing the concepts in a crystal clear manner sir.Hoping for more such informative videos from you.😊👍
So nice of you
Thank you soo much sir 🙏🙏🙏🙏
love from up ❤️❤️❤️❤️❤️
Thank you very much sir...i completely understood the experiment
Thank you very much sir it is very helpful for me
Wow venky
Love you sir
That's a poor example of CE design from an AC standpoint. The DC analysis shows the following:
With a 12V supply and Rc = 2.2K, the collector current is likely to be a few milliamps, so the base current will be β times smaller. The BC107 has a β between 125 and 900, so the base current will be between a few microamps and a few tens of microamps. The current through the base bias resistors (R1 + R2) is 12V / (56K + 10K) = 182μA, which is so much larger that we can ignore the effect of the base current on the bias network.
The voltage at the base (Vb) is 12V x 10K / (10K + 56K) = 1.82V.
The voltage at the emitter (Ve) is Vb - Vbe = 1.82V - 0.65V = 1.17V.
The emitter current is Ve / Re = 1.17V / 470R = 2.49mA. This confirms the supposition that the collector current, which is almost the same as the emitter current, is a few milliamps.
The voltage across Rc is 2.49mA x 2.2K = 5.47V, so the voltage at the collector (Vc) is 12V - 5.47V = 6.53V.
The voltage across the transistor Vce is Vc - Ve = 6.53V - 1.17V = 5.36V.
That's good from a DC standpoint; Vce and voltage across Rc are similar.
The problems occur with the AC analysis:
The intrinsic emitter resistance (re) is 26mV/Ic = 26mV / 2.49mA = 10.4R.
At frequencies where the reactance of the emitter bypass capacitor is small compared to re, the voltage gain of the transistor will be Rc/re = 210.
The -3dB frequency for 10.4R and 100μF is 1/2πCR = 150Hz.
The input impedance of the stage is the parallel combination of (R1 || R2 || the input impedance of the transistor). Unfortunately, the input impedance of the transistor depends on β, being β x re, which is between 125 x 10.4R = 1.3K and 900 x 10.4R = 9.4K. That is the dominant factor in calculating the input impedance which will be somewhere between 1.1K and 4.5K, depending on the particular sample of BC107 used. That's not good design.
Consequently, the -3dB frequency for the input capacitor and the input impedance will lie somewhere between 355Hz and 1.5KHz. Again poor design.
Finally, the large signal gain shows significant distortion as seen at 13:25 where the top of the output trace is rounded and the bottom of the trace is spikey for a 6V peak-to-peak signal. This is due to the non-linear nature of the intrinsic emitter resistance (re).
At the top of the trace, the collector current is about (5.5V - 3V) / 2.2K = 1.1mA, giving re ≈ 24R and an instantaneous gain of 2.2k/24R = 90.
At the bottom of the trace, the collector current is about (5.5V + 3V) / 2.2K = 3.9mA, giving re ≈ 6.7R and an instantaneous gain of 2.2k/6.7R = 330.
Bypassing the emitter resistor gives a high gain, but at the cost of low and variable input impedance and severe distortion of large signals.
Thank you for these experiments it was really useful
Wow! Nice video..
I have completed understand this experiment..
If any video available other practical please inform me...
Thank you sure I'll upload maximum video's in future
@@vivekpranichealingwelcome sir
Thanks
Sir,why do we use 6 to 12 volts dc regulated Power supply in negative feedback amplifier circuit...??
We can use minimum of 12V DC or 15V DC. A electrolytic capacitor is connected across the Emitter resistor, which is reduces the effect of negative feedback
Why does the input signal is very low in negative feedback amplifier??
Tqs very much
Sir what is the beta gain of the transistor
Thank you so much
Tq sir
Sir Hindi ya Marathi me batate te aur samaj ata
Thanks sir
Can you give a simplified circuit diagram
Sir please do graph alsoo
Circuit connection banaker dikhaye
Kal exam 💀
Thank you so much