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tags: Microelectronic Circuits
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# Chapter 6-Transistor Amplifier
## Basic Principles
### The Voltage-Transfer Characteristic (VTC)
1. MOSFET (NMOS)
$\begin{cases}v_{DS}=V_{DD}-i_DR_D\\i_D=\frac{1}{2}k_n(v_{GS}-V_t)^2\end{cases}\\\Rightarrow v_{DS}=V_{DD}-\frac{1}{2}k_nR_D(v_{GS}-V_t)^2$
At Point B $\begin{cases}v_{GS}=V_{GS}|_B=V_t+\frac{\sqrt{2k_nR_DV_{DD}+1}-1}{k_nR_D}\\v_{DS}=V_{DS}|_B=V_{GS}|_B-V_t=V_{OV}\end{cases}\\\Rightarrow V_{OV}=\frac{\sqrt{2k_nR_DV_{DD}+1}-1}{k_nR_D}$
2. BJT (npn)
1. npn
$v_{CE}=V_{CC}-R_CI_Se^{v_{BE}/V_T}$
2. pnp
$V_{EC}=-V_CC+I_CR_C$
### Obtaining Linear Amplification by Biasing the Transistor
Biasing enables us to obtain almost-linear amplification from the MOSFET and the BJT. We select a dc voltage to obtain operation at a point $Q$ on the segment $AB$ of the VTC. The point $Q$ is knowns the dc operating point or the bias point or the quiescent point.
1. MOSFET (NMOS)
The signal to be amplified, $v_{gs}(t)$, is superimposed on the bias voltage $V_{GS}$.
$v_{GS}(t)=V_{GS}+v_{gs}(t)$
2. BJT (npn)
Supermposing a small-signal $v{be}$ on the dc bias voltage $V_{BE}$ results in
$v_{BE}(t)=V_{BE}+v_{be}(t)$
### The Small-Signal Voltage Gain
1. MOSFET
$\begin{cases}A_v=\frac{dv_{DS}}{dv_{GS}}|_{v_{GS}=V_{GS}}\\v_{DS}=V_{DD}-\frac{1}{2}k_nR_D(v_{GS}-V_t)^2\end{cases}\\\Rightarrow A_v=-k_n(V_{GS}-V_t)R_D=-k_nV_{OV}R_D\\I_D=\frac{1}{2}k_nV_{OV}^2\\\Rightarrow A_v=-\frac{I_DR_D}{V_{OV}/2}=-\frac{V_{DD}-V_{DS}}{V_{OV}/2}$
Since the maximum slope of the VTC occurs at point $B$, the maximum gain magnitude $|A_{vmax}|$ is obtained by biasing the transistor at point $B$.
$|A_{vmax}|=\frac{V_{DD}-V_{DS}|_B}{V_{OV}|_B/2}=\frac{V_{DD}-V_{OV}|_B}{V_{OV}|_B/2}$
2. BJT
$\begin{cases}A_v=\frac{dv_{CE}}{dv_{BE}}|_{v_{BE}=V_{BE}}\\v_{CE}=V_{CC}-R_CI_Se^{v_{BE}/V_T}\end{cases}\\\Rightarrow A_v=-(\frac{I_C}{V_T})R_C=-\frac{V_{CC}-V_{CE}}{V_T}$
The maximum gain is achieved when $V_{CE}$ is at its minimun value of about $0.3V$
$|A_{vmax}|=\frac{V_{CC}-0.3}{V_T}$

The gain is negative, which tells us that the amplifier is inverting; that is, there is a 180° phase shift between the input and the output.
### Determining the VTC by Graphical Analysis (MOSFET)
1. At A Point
2. At C Point
3. Bias Points
Bias point Q1 does not leave sufficient room for positive signal swing at the drain (too close to VDD). Bias point Q2 is too close to the boundary of the triode region and might not allow for sufficient negative signal swing.
### Small-Signal Operation
1. The MOSFET Case
If we apply the input signal $v_{gs}$.
$v_{GS}=V_{GS}+v_{gs}\\\begin{split}\Rightarrow i_D&=\frac{1}{2}k_n(V_{GS}+v_{gs}-V_t)^2\\&=\frac{1}{2}k_n(V_{GS}-V_t)+k_n(V_{GS}-V_t)v_{gs}+\frac{1}{2}k_nv_{gs}^2\end{split}$
We should keep the input signal small,
$\frac{1}{2}k_nv_{gs}^2<<k_n(V_{GS}-V_t)v_{gs}\\\Rightarrow v_{gs}<<2(V_{GS}-V_t)\\\Rightarrow v_{gs}<<2V_{OV}$
If this small-signal condition is satisfied, we may express $i_D$ as
$\begin{cases}i_D\approx\frac{1}{2}k_n(V_{GS}-V_t)^2+k_n(V_{GS}-V_t)v_{gs}\\i_D\approx I_D+i_d\end{cases},\ where\ i_d=k_n(V_{GS}-V_t)v_{gs}$
1. Transconductance
$g_m\equiv\frac{i_d}{v_{gs}}=k_n(V_{GS}-V_t)=k_nV_{OV}$
Or
$g_m\equiv\frac{\partial i_D}{\partial v_{GS}}|_{v_{GS}=V_{GS}}$
2. The Voltage Gain
Under the small-signal condition, we have
$v_{DS}=V_{DD}-R_D(I_D+i_d)$
We can rewrite it as
$v_{DS}=V_{DS}-i_dR_D$
Thus the signal component of the drain voltage is
$v_{ds}=-i_dR_D=-g_mv_{gs}R_D$
And
$A_v\equiv\frac{v_{ds}}{v_{gs}}=-g_mR_D$
2. The BJT Case
1. The DC Bias Point
$\begin{cases}I_C=I_Se^{V_{BE}/V_T}\\I_E=\frac{I_C}{\alpha}\\I_B=\frac{I_C}{\beta}\\V_{CE}=V_{CC}-I_CR_C\end{cases}$
For active-mode operation, $V_{CE}\geq(V_{BE}-0.4)$.
2. The Collector Current and the Transconductance
1. The Collector Current
If a signal $v_{be}$ is applied, the total instantaneous base-emitter voltage $v_{BE}$ becomes
$v_{BE}=V_{BE}+v_{be}\\\Rightarrow i_C=I_Se^{V_{BE}/V_T}e^{v_{be}/V_T}\\\Rightarrow i_C=I_Ce^{v_{be}/V_T}$
If $v_{be}<<V_T$,
$i_C\approx I_C(1+\frac{v_{be}}{V_T})=I_C+\frac{I_C}{V_T}v_{be}\\\Rightarrow i_c=\frac{I_C}{V_T}v_{be}$
2. The Transconductance
Let $g_m=\frac{I_C}{V_T}$ be the transconductance,
$i_c=g_mv_{be}$
$g_m=\frac{\partial i_C}{\partial v_{BE}}|_{i_C=I_C}$
3. The Base Current and the Input Resistance at the Base
1. The Base Current
$\begin{cases}I_B=\frac{i_C}{\beta}=\frac{I_C}{\beta}+\frac{1}{\beta}\frac{I_C}{V_T}v_{be}\\i_B=I_B+i_b\end{cases}\\\Rightarrow i_b=\frac{1}{\beta}\frac{I_C}{V_T}v_{be}=\frac{g_m}{\beta}v_{be}$
2. The Input Resistance at the Base
$r_\pi\equiv\frac{v_{be}}{i_b}=\frac{\beta}{g_m}=\frac{V_T}{I_B}$
4. The Emitter Current and the Input Resistance at the Emitter
1. The Emitter Current
$i_E=\frac{i_C}{\alpha}=\frac{I_C}{\alpha V_T}v_{be}=\frac{I_E}{V_T}v_{be}$
2. The Input Resistance at the Emitter
$r_e\equiv\frac{v_{be}}{i_e}=\frac{V_T}{I_E}=\frac{\alpha}{g_m}\approx\frac{1}{g_m}$
3. Relationships
$v_{be}=i_br_\pi=i_er_e\\\Rightarrow r_\pi=(i_e/i_b)r_e=(1+\beta)r_e$
5. The Voltage Gain
$\begin{split}v_{CE}&=V_{CC}-i_CR_C\\&=V_{CC}-(I_C+i_c)R_C\\&=(V_{CC}-I_CR_C)-i_cR_C\\&=V_{CE}-i_cR_C\end{split}\\\Rightarrow v_{ce}=-i_cR_C=-g_mv_{be}R_C=(-g_mR_C)v_{be}\\A_v\equiv\frac{v_{ce}}{v_be}=-g_mR_C=-\frac{I_CR_C}{V_T}$
### Biasing
1. The MOSFET Case
1. Biasing by Fixing $V_{GS}$
$I_D=\frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{GS}-V_t)^2$
* ※This is not a good technique, because the use of fixed bias $V_{GS}$ can resulting in a large variability in the value of $I_D$.
2. Biasing by Fixing $V_G$ and Connecting a Resitance in the Source
$\begin{cases}V_G=V_{GS}+R_SI_D\\I_D=\frac{1}{2}k_n(V_{GS}-V_t)^2\end{cases}$
If $I_D$ increase for some reason:
$V_{GS}\downarrow\Rightarrow I_D\downarrow
3. Biasing Using a Drain-to-Gate Feedback Resistor
$V_{GS}=V_{DS}=V_{DD}-R_DI_D\\\Rightarrow V_{DD}=V_{GS}+R_DI_D\\\begin{cases}V_{DD}=V_{GS}+R_DI_D\\I_D=\frac{1}{2}k_n(V_{GS}-V_t)^2\end{cases}$
If $I_D$ increase for some reason (with fixed V_{DD}):
$V_{GS}\downarrow\Rightarrow I_D\downarrow$
2. The BJT Case
1. Two Not Good Arrangement
1. Biasing by Fixing $V_{BE}$
2. Biasing by Fixing $I_B$
2. The Classical Discrete-Circuit Bias Arrangement
$\begin{cases}V_{BB}=\frac{R_2}{R_1+R_2}V_{CC}\\R_B=\frac{R_1R_2}{R_1+R_2}\\I_E=\frac{V_{BB}-V_{BE}}{R_E+R_B/(\beta+1)}\\I_CR_C+V_{CB}=V_{CC}-V_{BB}+I_BR_B\end{cases}$
To make $I_E$ insenstitive to temperature and $\beta$ variation, we design the circuit to satisfy,
$\begin{cases}1.\ V_{BB}>>V_{BE}\\2.\ R_E>>\frac{R_B}{\beta+1}\end{cases}$
1. For a given $V_{CC}$,
$V_{BB}\uparrow\Rightarrow (I_CR_C+V_{CB})\downarrow$
We want the voltage acress $R_C$ to be large in order to obtain high voltage gain. We also want $V_{CE}$ to be large to provide a large signal swing. As a rule of thumb, we design for
$\begin{cases}V_{BB}=\frac{1}{3}V_{CC}\\V_{CE}=\frac{1}{3}V_{CC}\\I_CR_C=\frac{1}{3}V_{CC}\end{cases}$
2. If we choose $R_B$ to be small, $I_B$ will be large, which results in a lowering of the input resistance of the amplifier. So it is much better if we make the current in the divider larger than $I_B$. Typically we select $R_1$ and $R_2$ such that their current is in the range of $I_E$ to $0.1I_E$.
3. $R_E$ as a Negative Feedback Resistance
If $I_E$ increases for some reasons
1. $V_E\uparrow$
2. $I_BR_B$ is small so $V_B$ can be regarded as a constant.
3. $V_{BE}\downarrow\Rightarrow I_E\downarrow,\ I_C\uparrow$
3. A Two-Power-Supply Version of the Classical Bias Arrangement
$I_E=\frac{V_{EE}-V_{BE}}{R_E+R_B/(\beta+1)}$
To make $I_E$ insenstitive to temperature and $\beta$ variation, we design the circuit to satisfy,
$\begin{cases}1.\ V_{BB}>>V_{BE}\\2.\ R_E>>\frac{R_B}{\beta+1}\end{cases}$
1. In the common-base configuration, $R_B$ can be eliminated.
2. If the input signal is to be coupled to the base, then $R_B$ is needed.
4. Biasing Using a Collector-to-Base Feedback Resistor
$V_{CC}=I_ER_C+I_BR_B+V_{BE}=I_ER_C+\frac{I_E}{\beta+1}R_B+V_{BE}\\\Rightarrow I_E=\frac{V_{CC}-V_{BE}}{R_C+R_B/(\beta+1)}$
To obtain a value of $I_E$ that is insensitive to variabtion of $\beta$, we need to select
$R_B/(\beta+1)<<R_C$
However, the value of $R_B$ determines the allowable negative signal swing at the collector because
$V_{CE}=I_BR_B=I_E\frac{R_B}{\beta+1}$
## Models
### The MOSFET Case (NMOS)
1. The Hybrid $\pi$ Model
In following equivalent circuits, we only consider the alternaing current.
1. The Condition Neglating Early Effect
$A_v\equiv\frac{v_{ds}}{v_{gs}}=-g_mR_D$
2. The Condition Including Early Effect
$\begin{cases}A_v\equiv\frac{v_{ds}}{v_{gs}}=-g_m(R_D||r_o)\\r_o=\frac{|V_A|}{I_D}\\V_A=\frac{1}{\lambda}\\I_D=\frac{1}{2}k_nV_{OV}^2\end{cases}$
If we consider the direct current, the equivalent should be
$g_m=k'_n(W/L)(V_{GS}-V_t)=k'_n(W/L)V_{OV}=\sqrt{2k'_nW/LI_D}$
2. The T Equivalent-Circuit Model
If we consider the direct current, the equivalent should be
### The BJT Case
1. The Hybrid $\pi$ Model
2. The T Model
### Summary
## Basic Configurtions
### The CS and CE Amplifiers
1. The CS Amplifier
1. The CS Amplifier without a Source Resistance
$\begin{cases}R_{in}=\infty\\v_i=v_{sig}\\v_{gs}=v_{i}\\v_o=-g_mv_{gs}R_D\end{cases}$
1. The Open Circuit Voltage Gain $A_{vo}$
$A_{vo}\equiv\frac{v_o}{v_i}=-g_mR_D$
2. The Output Resistance $R_o$
$R_o=R_D$
3. The Overall Voltage Gain
If a load resistance $R_L$ is connected across $R_D$, the voltage gain $A_v$ can be obtained from
$\begin{cases}A_v=A_{vo}\frac{R_L}{R_L+R_o}=-g_m(R_D||R_L)\\G_v\equiv\frac{v_o}{v_{sig}}=-g_m(R_D||R_L)\end{cases}$
4. If the circuit includes a $r_o$, than we can simply replace $R_D$ with $(R_D||r_o)$ in the formula.
2. The CS Amplifier with a Source Resistance
$\begin{cases}R_{in}=\infty\\v_{gs}=v_i\frac{1/g_m}{1/g_m+R_s}=\frac{v_i}{1+g_mR_s}\end{cases}$
1. The Open Circuit Voltage Gain $A_{vo}$
$A_{vo}\equiv\frac{v_o}{v_i}=-\frac{g_mR_D}{1+g_mR_s}$
2. The Overall Voltage Gain $G_v$
With a load resistance $R_L$ connected at the output,
$A_v=-\frac{g_m(R_D||R_L)}{1+g_mR_s}$
Because $v_i=v_{sig}$,
$G_v=A_v=-\frac{g_m(R_D||R_L)}{1+g_mR_s}$
* ※Including the resistance $R_s$ reduces the voltage gain by the factor $(1+g_mR_s)$, where $R_s$ is called the source-degeneration resistance. This can be helpful with the following condition: If with $v_{sig}$ and hence $v_{i}$ kept constant, the drain current increases for some reason, the source current will increase, resulting in an increased voltage drop across $R_s$. Thus the source voltage rises, and the gate-to-source voltage decreases, and this will help the circuit be stable.
2. The CE Amplifier
1. The CE Amplifier without a Emitter Resistance
$\begin{cases}R_{in}=r_\pi\\v_i=\frac{r_\pi}{r_\pi+R_{sig}}v_{sig}\\v_\pi=v_i\\v_o=-g_mv_\pi R_C\end{cases}$
1. The Open Circuit Voltage Gain $A_{vo}$
$A_{vo}\equiv\frac{v_o}{v_i}=-g_mR_C$
2. The Output Resistance $R_o$
$R_o=R_C$
3. The Overall Voltage Gain $G_v$
With a load resistance $R_L$ connected across $R_C$,
$\begin{cases}A_v=-g_m(R_C||R_L)\\G_v\equiv\frac{v_o}{v_{sig}}=\frac{v_i}{v_{sig}}\frac{v_o}{v_i}=-\frac{r_\pi}{r_\pi+R_{sig}}g_m(R_C||R_L)\end{cases}$
4. If the circuit includes a $r_o$, than we can simply replace $R_C$ with $(R_C||r_o)$ in the formula.
2. The CE Amplifier with a Emitter Resistance
1. The Input Resistance $R_{in}$
$\begin{cases}R_{in}\equiv\frac{v_i}{i_b}\\i_b=(1-\alpha)i_e=\frac{i_e}{1+\beta}\\i_e\frac{v_i}{r_e+R_e}\end{cases}\\\Rightarrow R_{in}=(1+\beta)(r_e+R_e)$
2. The Open Circuit Voltage Gain $A_{vo}$
$\begin{cases}v_o=-i_cR_C=-\alpha i_eR_C\\i_e=\frac{v_i}{r_e+R_e}\end{cases}\\\begin{split}\Rightarrow A_{vo}&=-\alpha\frac{R_C}{r_e+R_e}\\&=-\frac{\alpha}{r_e}\frac{R_C}{1+R_e/r_e}\\&=-\frac{g_mR_C}{1+R_e/r_e}\\&\approx-\frac{g_mR_C}{1+g_mR_e}\end{split}$
3. The Output Resistance $R_o$
$R_o=R_C$
4. The Overall Voltage Gain $G_v$
If a load resistance $R_L$ is connected at the amplifier output
$\begin{split}A_v&=A_{vo}\frac{R_L}{R_L+R_o}\\&=-\alpha\frac{R_C}{r_e+R_e}\frac{R_L}{R_L+R_C}\\&=-\alpha\frac{R_C||R_L}{r_e+R_e}\end{split}$
And the overall voltage gain is
$\begin{cases}G_v=\frac{R_{in}}{R_{in}+R_{sig}}\times-\alpha\frac{R_C||R_L}{r_e+R_e}\\R_{in}=(\beta+1)(r_e+R_e)\end{cases}\\\Rightarrow G_v=-\beta\frac{R_C||R_L}{R_{sig}+(\beta+1)(r_e+R_e)}$
* ※The Existence of the Emitte Degeneration Resistance $R_e$:
1. The input reistance $R_{in} is increased by the factor $1+(g_mR_e)$
2. Th evoltage gain from base to collector, $A_v$, is reduced by the factor $(1+g_mR_e)$
3. The additional term $(\beta+1)R_e$ inte denominator cause the less dependence on the value of $\beta.
4. $\frac{v_\pi}{v_i}=frac{r_e}{r_e+R_e}\approx\frac{1}{1+g_mR_e}$
Thus for the same $v_\pi$, the input signal $v_i$, can be greater by the factor $(1+g_mR_e)$
5. The hig-frequency response is significantly improved.
### The CG and CB Amplifiers
1. The CG Amplifiers
1. The Input Resistance $R_{in}$
$R_{in}=\frac{1}{g_m}$
2. The Voltage Gain $A_{vo}$
$\begin{cases}v_o=-iR_D\\i=-\frac{v_i}{1/g_m}\end{cases}\\\Rightarrow A_{vo}\equiv\frac{v_o}{v_i}=g_mR_D$
3. The Output Resistace $R_o$
$R_o=R_D$
4. The Overall Voltage Gain $G_v$
With a resistance $R_L$ connected at the output
$\frac{v_i}{v_{sig}}=\frac{R_{in}}{R_{in}+R_{sig}}=\frac{1/g_m}{1/g_m+R_{sig}}\\\Rightarrow G_v=\frac{1/g_m}{1/g_m+R_{sig}}[g_m(R_D||R_L)]\\\Rightarrow G_v=\frac{R_D||R_L}{R_{sig}+1/g_m}$
2. The CB Amplifiers
1. The Input Resistance $R_{in}$
$R_{in}=r_e=\frac{\alpha}{g_m}\approx\frac{1}{g_m}$
2. The Voltage Gain $A_{vo}$
$A_{vo}=\frac{\alpha}{r_e}R_C=g_mR_C$
3. The Output Resistace $R_o$
$R_o=R_C$
4. The Overall Voltage Gain $G_v$
$G_v\equiv\frac{v_o}{v_{sig}}=\alpha\frac{R_C||R_L}{R_{sig}+r_e}$
* ※In practice, we don't use CB and CG amplifiers, because of the very low input resistance $(1/g_m)$.
### The Source and Emitter Followers
1. The Need for Voltage Buffers
Connecting the source to the load directly results in severe attenuation of the signal.
2. Circuits
1. The Source Follower
1. The Input Resistance $R_{in}$
$R_{in}=\infty$
2. The Gain with a Load Resistance $R_L$ $A_v$
$A_v\equiv\frac{v_o}{v_i}=\frac{R_L}{R_L+1/g_m}$
3. The Open Circuit Voltage Gain $A_{vo}$
Setting $R_L=\infty$,
$A_{vo}=1$
4. The Output Resistance $R_o$
$R_o=1/g_m$
5. The Overall Voltage Gain $G_v$
Because the infinite $R_{in}$, $v_i=v_{sig}$, and
$G_v=A_v=\frac{R_L}{R_L+1/g_m}$
6. If the circuit includes a $r_o$, than we can simply replace $R_L$ with $(R_L||r_o)$ in the formula.
2. The Emitter Follower
1. The Input Resistance $R_{in}$
$\begin{cases}R_{in}=\frac{v_i}{i_b}\\i_e=\frac{v_i}{r_e+R_L}\end{cases}\\\Rightarrow R_{in}=(\beta+1)(r_e+R_L)$
2. The Gain with a Load Resistance $R_L$ $A_v$
$A_v\equiv\frac{v_o}{v_i}=\frac{R_L}{R_L+r_e}$
3. The Open Circuit Voltage Gain $A_{vo}$
Setting $R_L=\infty$,
$A_{vo}=1$
4. The Output Resistance $R_o$
$R_o=r_e$
5. The Overall Voltage Gain $G_v$
$\frac{v_i}{v_{sig}}=\frac{R_{in}}{R_{in}+R_{sig}}=\frac{(\beta+1)(r_e+R_L)}{(\beta+1)(r_e+R_L)+R_{sig}}\\\Rightarrow G_v\equiv\frac{v_o}{v_{sig}}=\frac{v_i}{v_{sig}}\times A_v=\frac{(\beta+1)R_L}{(\beta+1)R_L+(\beta+1)r_e+R_{sig}}$
$G_v\equiv\frac{v_o}{v_{sig}}=\frac{R_L}{R_L+r_e+R_{sig}/(\beta+1)}$
6. If the circuit includes a $r_o$, than we can simply replace $R_L$ with $(R_L||r_o)$ in the formula.
* ※Thévenin Representation of the Output of the Emitter Follower
## Discrete-Circuit Amplifiers
### A Common-Source (CS) Amplifier
$G_v=-\frac{R_{in}}{R_{in}+R_{sig}}g_m(R_D||R_L||r_o)$
### A Common-Emitter (CE) Amplifier
1. A Common-Emitter (CE) Amplifier without an Emitter Resistance $R_e$
$G_v=-\frac{R_{in}}{R_{in}+R_{sig}}g_m(R_C||R_L||r_o)$
2. A Common-Emitter (CE) Amplifier with an Emitter Resistance $R_e$
$\begin{cases}R_{in}=R_{B1}||R_{B2}||(\beta+1)(r_e+R_e)=R_{B1}||R_{B2}||[r_\pi+(\beta+1)R_e]\\G_v=-\frac{R_{in}}{R_{in}+R_{sig}}\times\alpha\frac{Total\ resistance\ in\ collector}{Total\ resistance\ in\ emitter}=-\alpha\frac{R_{in}}{R_{in}+R_{sig}}\frac{R_C||R_L}{r_e+R_e}\end{cases}$
### A Common-Base (CB) Amplifier
1. $R_{in}$
$R_{in}=R_e||r_e\approx r_e\approx\frac{1}{g_m} (R_e>>r_e)$
2. $v_{in}$
$v_{in}=v_{sig}\frac{R_{in}}{R_{in}+R_{sig}}$
3. $v_{o}$
$v_{o}=-\alpha i_e(R_C||R_L)=-g_mv_\pi(R_C||R_L)=g_mv_i(R_C||R_L)$
4. $A_v$
$A_v\equiv\frac{v_o}{v_i}=g_m(R_C||R_L)$
5. $G_v$
$G_v=\frac{v_o}{v_{sig}}=\frac{v_o}{v_i}\frac{v_i}{v_{sig}}=g_m(R_C||R_L)\frac{R_{in}}{R_{in}+R_{sig}}$
### An Emitter Follower
1. $R_{in}$
$R_{in}=R_B||R_{ib}$
2. $R_{ib}$
$R_{ib}=(\beta+1)[r_e+(R_E||r_o||R_L)]$
3. $v_i$
$v_i=v_{sig}\frac{R_{in}}{R_{in}+R_{sig}}$
4. $v_o$
$v_o=v_i\frac{R_E||r_o||R_L}{r_e+(R_E||r_o||R_L)}$
5. $A_v$
$A_v\equiv\frac{v_o}{v_i}=\frac{R_E||r_o||R_L}{r_e+(R_E||r_o||R_L)}$
6. $G_v$
$G_v\equiv\frac{v_{o}}{v_{sig]}}=\frac{R_{in}}{R_{in}+R_{sig}}\frac{R_E||r_o||R_L}{r_e+(R_E||r_o||R_L)}$
7. $R_{out}$
$R_{out}=r_o||R_E||[r_e+\frac{R_B||R_{sig}}{\beta+1}]$
## The Amplifier Frenquency Response
1. The Bandwidth
$BW=f_H-f_L$
since usually $f_L<<f_H$
$\Rightarrow BW\equiv f_H$
2. The Gain-Bandwidth Product
$GB=|A_M|BW$
## Exercises
### P.359 Example 6.1 (b)

**Reference**
NCKUDPS 1101_Electronics(2)
Microelectronic Circuits 8/e by ADEL.S SEDRA