BJT Small Signal Analysis: Common Emitter Amplifier without Bypass Capacitor

In this video, the Small Signal Analysis of the Common Emitter Amplifiers (without Bypass Capacitor) is Explained.
The following topics are covered in the video:
0:00 Introduction
1:30 Small Signal Analysis of Fixed Bias Circuit with Emitter Resistor (without Bypass Capacitor)
13:08 Small Signal Analysis of Voltage Divider Bias Circuit (without Bypass Capacitor)
14:54 Solved Example (Common Emitter Amplifier with and without Bypass Capacitor)
In this video, the Small Signal Analysis of Common Emitter Amplifier (without bypass capacitor) is explained in detail. And it is explained with two different circuits.
1) Fixed Bias Circuit with Emitter Resistor (without Bypass Capacitor)
2) Voltage Divider Circuit (without Bypass Capacitor)
In Common Emitter Amplifier, without a bypass capacitor, the voltage gain reduces and the input impedance increases.
The Common Emitter Amplifier with a bypass capacitor provides high gain but has relatively low input impedance.
After the Small Signal Analysis, one example is taken in the end to showcase the effect of the bypass capacitor (with and without bypass capacitor) on the Common Emitter Amplifier parameters. (like voltage gain and input impedance.)
The link for the other useful videos related to Small Signal Analysis of BJT:
1) BJT Small Signal Analysis: Common Emitter Fixed Bias and Voltage Divider Bias
• BJT Small Signal Analy...
2) BJT Small-Signal Model:
• BJT - Small Signal Mo...
3) BJT Large Signal Model:
• BJT Large Signal Model...
4) Transistor Biasing: What is Q-point? What is Load Line? Fixed Bias Configuration Explained
• Transistor Biasing: Wh...
This video will helpful to all the students of science and engineering in understanding how to do the small-signal analysis of the Common Emitter Amplifier.
#BJTSmallSignalAnalysis
#BJT
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[@ALLABOUTELECTRONICS]

The timestamps for the different topics covered in the video: 1:30 Small Signal Analysis of Fixed Bias Circuit with Emitter Resistor (without Bypass Capacitor) 13:08 Small Signal Analysis of Voltage Divider Bias Circuit (without Bypass Capacitor) 14:54 Solved Example (Common Emitter Amplifier with and without Bypass Capacitor)

[@uygarbaran8515]

I literally learned a large portion of circuits through you. Thank you for your effort and wonderful teaching style.

[@sampadasuryavanshi4131]

Thank you for such a nice explaination 👍

[@johnbowman9734]

Thank you so much for making this video!!!!!!!!

[@JahanZeb1976]

*Wonderful explanation. I'm also engineering KZbinr.*

[@cloudblue7772]

You are life saver, your video help me a lot 🙏🙏

[@aniljamre1530]

Thanks so much sir 💓💓 क्या धासू कांसेप्ट पढ़ाते हो sir

[@williamchigamba2512]

Good and impressive explanation, so easy to understand

[@tahirbintariq8934]

Sir I can't explain the respect you gain after solving this numerical. It was my assignment and I've been working for it for a long time. Thank you sir

[@rafeedyasir7893]

You are really awesome ❤

[@mayurshah9131]

EXCELLENT

[@priyajitmondal6661]

Thank you so much sir 🙏👍🥺

[@farvezfarook3422]

what will happen to output impedance if the dependant current source has a parallel resistance(early effect), will the emitter resistor be part of the output impedance too?? . I am asking for the case without bypass capacitor ..

[@dhankthapamagar3389]

Best!

[@poojashah6183]

Best👌🏻👌🏻

[@farhanupaul]

While doing the thevenin for output impedence, couldn't we just open the current source without going through the equations?

[@shreyashawaghad760]

THANKU SIR ....🙏

[@alialwafi9229]

Which App you used to make this video and draw the circuit diagrams? answer please

[@prajjwalpandey253]

Nice 👍

[@subratadutta7710]

Sir, please upload a playlist on the negative feedback amplifiers on the topics like voltage-series, voltage-shunt and all others. I couldn't understand the actual circuits from the books or from any other channels. The schematic diagrams are easy to understand but when I think about how the actual circuits look and work I couldn't catch the difference between those circuits.I have understood a lot of concepts clearly from your videos. Please help Sir.

[@ALLABOUTELECTRONICS]

I will try to cover it soon.

[@manasdwivedi5217]

Doubt: Sir, what will happen to the Voltage Gain (Av) ,if we also consider Output Resistance (ro)? Will the gain be -gm*(RC || ro)/ [1+gm*RE]

[@kesavanb3982]

Very useful..can you please explain the small signal analysis of collector to base bias in a another video

[@ALLABOUTELECTRONICS]

Yes, I will also cover it soon.

[@emmetray9703]

How did you calculate capacitor values?

[@andyalvarez5846]

Great video! How can I find the voltage gain when there is a load resistance? Also, how can you solve for the current gain?

[@ALLABOUTELECTRONICS]

The load resistor RL will come in parallel with Rc in the small-signal analysis. In the voltage gain equation, you can replace Rc with (Rc || RL).

[@ALLABOUTELECTRONICS]

Regarding the current gain, Ai = Io / Ii. Io = -Vo / Rc (if RL is not present) and Ii = Vi / Zi (Zi is input impedance) So, current gain = Io / Ii = -(Vo / Rc) / (Vi / Zi) = -Av * Zi / Rc Where Av is the voltage gain.

[@RexxSchneider]

@@ALLABOUTELECTRONICS and writing re = 1/gm, we have current gain = -Av * Zi / Rc = Rc/(re +Re) * β(re + Re) / Rc = β As expected.

[@abhijitkumarmanna2031]

Why input impedance in derivation of ce fixed bias is not used in the numerical ?18:53

[@JK-xm6rz]

Sir how gain and Zo change when we consider ro also

[@daniaazzahra1698]

excuse me sir, just want to ask. i still dont get it, where are those Vn and rn coming from? and also with ic, is it the current that flowing through resistor c?

[@ALLABOUTELECTRONICS]

I recommend you to go through these two previous videos, which will help you in understanding it. The link for the other useful videos related to topic is already provided in the description of the video. Please go through it if required. 1) BJT Small-Signal Model: kzbin.info/www/bejne/n2PXanqKdsllf5I 2) BJT Large Signal Model: kzbin.info/www/bejne/g3aXYqiGqKmrfNE

[@joshbiju7771]

Sir how to find current gain in this case

[@NihalVishwakarma-ol1qc]

Hello honourable Sir the question which I have asked to sister Briti Panday... I am asking to you also... Please don't take my question in another way... I am also like your brother. .. Please tell sir that's question's answer . That which channel I should follow I am confused

[@PrashantKumar-js2gu]

Please make the video for re Model

[@uygarbaran8515]

How are you able to plug in V_in for V_pi? Isn't V_in = V_pi + IeRe?

[@ALLABOUTELECTRONICS]

Would you please mention the timestamp where you are referring to?

[@ndtan9137]

Can i ask what is the Vt that you use find gm in equation gm=ic/vt

[@ALLABOUTELECTRONICS]

Vt is the thermal voltage. It is the voltage produced within the pn junction due to a change in temperature. As you can see from the equation, kT/q, as the temperature increases, the thermal voltage increases. At room temperature, it is around 26 mV.

[@ndtan9137]

Thx for explain ur video make my life easier :)

[@swayambhuray5762]

Sir, why do we consider the thevenin equivalent resistance of the circuit as the output impedance?

[@ALLABOUTELECTRONICS]

The method for finding the output impedance is, consider all the independent sources in the circuit as zero and find the equivalent impedance seen from the output side. Or considering all the independent sources in the circuit as zero, apply test voltage, and find the test current. The ratio of test voltage to the test current gives the output impedance. For finding the Thevenin's equivalent impedance/ resistance across two terminals also, we used to follow the same procedure. Therefore we can say that the output impedance is Thevenin's equivalent impedance of the circuit looking from the output side. I hope it will clear your doubt.

[@swayambhuray5762]

@@ALLABOUTELECTRONICS Sir, what would happen if we don't consider the independent sources in circuit as zero?

[@RexxSchneider]

@@swayambhuray5762 The output impedance is the ratio of a small voltage applied to the _output_ to the small change in current that it produces (by definition). If we tried to do that while an input signal was being applied, it would take a lot more work to either measure or calculate that small change in current. So we choose to measure or calculate the output impedance in the absence of an input signal. In other words set the input source to zero. To a first order approximation, it can't affect the result.

[@inioluwaobisakin465]

sir why is vo negative? is because of the direction of the current

[@RexxSchneider]

Yes, it is an ac current flowing from the collector to ac ground. Because the ac ground is fixed (at the supply rail), Vo will decrease as Ic increases, and vice-versa.

[@studywithbriti9058]

Sir,voltage regulation using opamp k simulation kaise kia video provide kijie...project kr rh h...plz Sir...hlp us.

[@ALLABOUTELECTRONICS]

I have already provided simulation link in that video. (Op-amp as voltage regulator) You can go through that simulation link. The simulation was carried out in Multisim Live.

[@studywithbriti9058]

@@ALLABOUTELECTRONICS work toh kregi na?both hardware n simulation? Aapke varose yeh project kr rh h... Plz agar hlp chahiye hog toh ap hlp krn... Time bht kom h,aur hume submit v krn h ... Thanks ... Future aspect k bare m plz plz kch bolie.

[@ALLABOUTELECTRONICS]

In simulation and the actual result there will be some difference. But more or less, it will be close to simulated results. If required any help, I will try to help you.

[@studywithbriti9058]

@@ALLABOUTELECTRONICS thanks.

[@studywithbriti9058]

@@ALLABOUTELECTRONICS Voltage regulator mai ap kya opamp ko use kr rhe hai as an comparator?

[@RexxSchneider]

All of the analysis is accurate, but the formulae for gain and input impedance are not presented in their most useful forms. It is far more convenient to consider 1/gm to be an intrinsic emitter resistance (re), that is equal to 26mV/Ic at room temperature, then express the formulae in terms of re and β, noting that rπ = β/gm = β.re Av = -gm.Rc / (1 + gm.Re) = -Rc / (1/gm + Re) = -Rc / (re + Re) Zin = R1 || R2 || Zb = R1 || R2 || β.re(1 + Re/re) = β(re + Re) The gain is now the ratio of collector resistance to total emitter resistance, and the input impedance is beta times the total emitter resistance, which makes sense. Introducing a bypass capacitor in parallel with Re is now easy to accommodate as it can be seen that the total emitter resistance is just re at frequencies where the capacitor's reactance is significantly smaller than re.

[@sakshamjain2217]

why dont you consider Rc which is on output side for finding input impedance , but considering r(pi) for finding output impedance which is on input side?? and also like for output imp can we say that, input imp is the thev eq seen from input side, if no then why??

[@ALLABOUTELECTRONICS]

Here while finding the output impedance, the R(pi) was considered (during KVL), just to showcase that, V(Pi) is 0.

[@sakshamjain2217]

@@ALLABOUTELECTRONICS then why Rc is not considered for finding input imp??

[@ALLABOUTELECTRONICS]

@@sakshamjain2217 From the smal signal model, the input impedance in this case is the RB || Zb. Zb is the input impedance seen through the base terminal. And that is equal to Vb / ib. So, to find that, we need Ib and Vb. So, as you can see in the video, the expression of Vb and Ib were used to find Zb. And in that, we do not require Rc.

[@aagmanarora2556]

5:06 how is rpi=B/gm?

[@abhijitkumarmanna2031]

ib = V pi /r pi ic / beta=v pi / r pi ic=gm * v pi gm * v pi / beta = V pi / r pi r pi = beta/gm

[@varulven5143]

I didn't get it either

[@nikolaynetsov6529]

How did you find VT?

[@ALLABOUTELECTRONICS]

VT is thermal voltage. It is equal to kT/q. k is Boltzmann constant. q- charge T - temperature.

[@gibsonchileshe5471]

does VT depend on temperature and is the value of VT usually given in the question