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BJT stands for Bipolar Junction Transistor. It is a three-terminal electronic device constructed of doped semiconductor material and may be used in amplifying or switching applications.

$r_\pi$ is an input resistance looking into the base with emitter terminal at AC ground. $\Large r_\pi\ = \frac{d V_{BE}}{dI_B} = \frac{V_T}{I_B} = \frac{\beta}{g_m} = (\beta+1)r_e$ On the o …
answered Apr 9 '18 by G36
This diagram should explain everything Any additional questions?
We know that : $$h_{fe}= \frac{β_o}{1+s(Cπ+Cµ)r_π} = \frac{β_o}{\sqrt{1 +(Cπ+Cµ)^2\:{r_π}^2 }}$$ Additional we know thta: $$β_o = g_m r_{\pi}$$ And if we assume that this 50MHz is $1/\omega_B$ …
answered Nov 12 '17 by G36
that now the the situation is much more clearer. And now we need to ask yourself on simple question: What voltage at transistor base (VB) is needed to open transistor? To turn-ON a BJT the voltage at … voltage at the load resistance is equal to: VEE - VE = 12V - 9.894V = 2.1V Because the BJT is in Cut-off region. So, BJT is no longer able to provide any current to the load resistance. …
answered Nov 28 '16 by G36
The task is: provide the highest possible voltage gain while allowing for a maximum signal swing at the collector of ±2 V. Use +10 V and -5 V for power supplies. So you need to find the lar …
answered Aug 24 '18 by G36
I will try to show you a much more practical example for a Common Base amplifier. Also in this example I teat BJT as a Current Controlled Current Source. I also ignore the base current influence …
answered Aug 8 '17 by G36
For this circuit simulate this circuit – Schematic created using CircuitLab You can write these KVL equations: $$V_E = V_{BE1}+I_{B1}*R_F$$ $$V_E = (I_{E2}-I_{B1})*R_E$$ And this give us …
answered May 13 '17 by G36
Yes, we have: 1 - "beta" $\beta = \frac{I_C}{I_B}$ for common-emitter current gain. 2 - "alpha" $\alpha = \frac{I_C}{I_E} =\frac{I_C}{I_B +I_C}= \frac{\beta \cdot I_B}{I_B +\beta \cdot I_B}= \f … answered Jan 2 by G36 At$1 \textrm{kHz}$the$C_E$capacitor reactance is around :$X_C \approx \frac{0.16}{F \cdot C}\approx 1.6\textrm{k}\Omega$So the$Z_E$impedance will be around:$Z_E = R_{E1} + R_{E2} …
answered Apr 22 '18 by G36
part of it. A question about Vce of an NPN BJT in saturation region For this circuit with ideal transistor (current controlled current source CCCS) any base current large than: $\Large I_B … > \frac{\frac{V_{cc}}{R_c}}{\beta }$ will saturate the BJT. But in real life, ideal transistor don't exist. For any real world transistor, the β is not constant. β varies with Ic, Vce, temperature. And …
answered Jun 16 '17 by G36
You must distinguish between DC analysis and small-signal (AC signal only 10mV peak). When we are doing DC analysis we usually treat the BJT as a current controlled current source with fix Vbe … small-signal analysis, we are using the highly linearized BJT's small-signal model. So we can use linear equations and theory. Because do not forget that in real life the is BJT highly nonlinear device and this is why we using linearized models. Because the reality is too complicated for us. …
answered Jun 27 '18 by G36
expression is $g_m = \frac{I_{out}}{V_{in}}$. For example if is $g_m = 1\:Siemens$ any change in the input voltage by $1V$ will change the output current by $1A$ (1 Ampere per Volt). For BJT
answered May 1 '17 by G36
They simply do AC small-signal analysis. So you can skip $V_{BE}$ if you do AC analysis. The $r_e$ resistance "represents" the change in $V_{BE}$. $\Delta V_{BE} = i_e\cdot r_e$ As for …
answered Nov 4 '17 by G36
If we assume $\beta = 200$ the voltage gain is: $$A_v \approx - \frac{R_3}{\frac{R2 + r_\pi}{(\beta+1)}}\approx -8.7\; V/V$$ And you get the wrong result because you don't include R2 influence. …
answered May 22 '17 by G36
Try this circuit (with negative feedback instead of a positive feedback). simulate this circuit – Schematic created using CircuitLab This is all you need to properly DC biased this simple cir …
answered Dec 7 '17 by G36

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