---
# System prepended metadata

title: CH10. Feedback
tags: [MC2 WebNotes]

---

# CH10. Feedback

:::info
**Disclaimer**
Cases involving BJTs are temporarily omitted, I might work on them someday if I have time. Also, if you spot any error, please contact me via my email: bigbeeismusic@gmail.com
:::

## 10.1 The General Feedback Structure

### 10.1.1 Signal-Flow Diagram

![截圖 2025-01-09 下午5.45.38](https://hackmd.io/_uploads/SyuOiMpUye.png)

:::warning
++Definition++.
**Open-loop gain**: $A=\frac{x_o}{x_i}$
**Feedback factor**: $\beta=\frac{x_f}{x_o}$
$x_i=x_s-x_f$
:::

### 10.1.2 The Closed-Loop Gain

$A_f\equiv\frac{x_o}{x_s}=\frac{A}{1+A\beta}$ ($A\beta$: **loop gain** / $(1+A\beta)$: **amount of feedback**)
When $A\beta\gg 1$, $A_f\simeq\frac{1}{\beta}\Rightarrow A_f|_{\text{ideal}}=\frac{1}{\beta}$
$\Rightarrow A_f=A_f|_{\text{ideal}}\frac{1}{1+(1/A\beta)}$

:::info
++Note++. We can rewrite the expression of signals:
$x_f=\frac{A\beta}{1+A\beta}x_s$, $x_i=\frac{1}{1+A\beta}x_s$
$x_i$ is sometimes referred to as the **error signal**.
:::

### 10.1.3 The Loop Gain

:::info
++Observation++.
- $A\beta>0\Leftrightarrow$ negative feedback.
- $A\beta\nearrow\Rightarrow A_f\rightarrow A_f|_{\text{ideal}}$
:::

**find the loop gain**
![截圖 2025-01-09 下午6.07.43](https://hackmd.io/_uploads/rknGxQT8kx.png)

1. Set $x_s=0$.
2. Break the feedback loop.
3. Apply $x_t$ at where the break is and determine the return signal $x_r$, we have $x_r=-A\beta x_t\Rightarrow A\beta=-\frac{x_r}{x_t}$

### 10.1.4 The Ideal Case of Infinite Open-Loop Gain $A$

![截圖 2025-01-09 下午6.14.12](https://hackmd.io/_uploads/rJuqb7681e.png)

$\Rightarrow x_i=0$, $x_f=x_s$
$\Rightarrow x_o=\frac{1}{\beta}x_s$
$\Rightarrow A_f|_{\text{ideal}}=\frac{x_o}{x_s}$

## 10.2 Some Properties of Negative Feedback

### 10.2.1 Gain Desensitivity

$dA_f=\frac{dA}{(1+A\beta)^2}\Rightarrow\frac{dA_f}{A_f}=\frac{1}{1+A\beta}\frac{dA}{A}$ ($(1+A\beta)$: **desensitivity factor**)

### 10.2.2 Bandwidth Extension

Consider $A(s)=\frac{A_M}{1+s/\omega_H}$
$\Rightarrow A_f(s)=\frac{A(s)}{1+\beta A(s)}=\frac{A_M/(1+s/\omega_H)}{1+\beta[A_M/(1+s/\omega_H)]}=\frac{A_M/(1+A_M\beta)}{1+s/[\omega_H(1+A_M\beta)]}$
$\Rightarrow \omega_{Hf}=\omega_H(1+A_M\beta)$
![截圖 2025-01-09 下午6.28.44](https://hackmd.io/_uploads/Sk9xBQaU1g.png)

### 10.2.3 Reduction in Nonlinear Distortion

![截圖 2025-01-09 下午6.31.31](https://hackmd.io/_uploads/r1mcHm6Lkl.png)
((a): without feedback, (b): with feedback)

## 10.3 The Feedback Voltage Amplifier

### 10.3.1 The Series-Shunt Feedback Topology

![截圖 2025-01-09 下午6.35.36](https://hackmd.io/_uploads/r1_tUXpLJg.png)
(voltage-mixing, voltage-sampling)

### 10.3.2 Examples of Series-Shunt Feedback Amplifiers

![截圖 2025-01-14 下午5.27.39](https://hackmd.io/_uploads/By0fRomPkx.png)

### 10.3.3 Analysis of the Feedback Voltage Amplifier

#### Step 1: Determine $\beta$ and $A_f|_{\text{ideal}}$

![截圖 2025-01-14 下午5.47.06](https://hackmd.io/_uploads/Byx3G2XPJe.png)

First assume that $A=\infty$, find $A_f|_{\text{ideal}}=\frac{1}{\beta}=\frac{V_o}{V_f}=\frac{V_o}{V_s}$.

#### Step 2: Determine the Loop Gain $A\beta$

![截圖 2025-01-14 下午5.55.01](https://hackmd.io/_uploads/B1tYV37vke.png)

Break the feedback loop and apply a test voltage $V_t$ and connect another break of the loop with impedance $Z_t$ (whose value is equal to the impeadance looking into $XX'$), we have $A\beta=-\frac{V_r}{V_t}$

#### Step 3: Determine the Closed-Loop Gain $A_f$

Combine two result, we have $A_f=\frac{A}{1+A\beta}$

## 10.4 Systematic Analysis of Feedback Voltage Amplifiers

### 10.4.1 The Ideal Case

![截圖 2025-01-14 下午6.03.33](https://hackmd.io/_uploads/r19tI2QDyg.png)
(In (a), the $\beta$ circuit does **not** load the $A$ circuit)

$\Rightarrow A_f\equiv\frac{V_o}{V_s}=\frac{A}{1+A\beta}$

**Find $R_{if}$**

$V_{i}=\frac{V_{s}}{1+A\beta}\Rightarrow I_i=\frac{V_i}{R_i}=\frac{V_s}{(1+A\beta)R_i}\Rightarrow R_{if}\equiv\frac{V_s}{I_i}=(1+A\beta)R_i$

**Find $R_{of}$** - Let $V_s=0$, apply $V_x$ at output terminal

$I_x=\frac{V_x-AV_i}{R_o}$, $V_i=-V_f=-\beta V_x\Rightarrow R_{of}=\frac{V_x}{I_x}=\frac{R_o}{1+A\beta}$

### 10.4.2 The Practical Case

![截圖 2025-01-14 晚上10.08.11](https://hackmd.io/_uploads/HJ0AkeEDkx.png)

**Interior Structure**
![截圖 2025-01-14 晚上10.09.04](https://hackmd.io/_uploads/Sk5zee4Pyx.png)
($R_s$: source resistance, $R_L$: load resistance, $R_{11}$&$R_{22}$: resistance due to the loading effect of the feedback network, see the following figure)
![截圖 2025-01-14 晚上10.11.41](https://hackmd.io/_uploads/Sy91Wl4PJx.png)

**Find $\beta$**
![截圖 2025-01-14 晚上10.27.55](https://hackmd.io/_uploads/BJ-2El4DJx.png)

$\beta\equiv\left.\frac{V_f}{V_o}\right|_{I_1=0}$

**Find Loop Gain $A\beta$**
![截圖 2025-01-14 晚上10.30.45](https://hackmd.io/_uploads/ByDQSxED1x.png)

$A=\frac{V_o}{V_i}$
Closed loop gain $\Rightarrow A_f\equiv\frac{V_o}{V_s}=\frac{A}{1+A\beta}$

**Find Input/Output Resistance $R_{if}$/$R_{of}$**
First find $R_i$/$R_o$ (see above figure)
$\Rightarrow R_{if}=(1+A\beta)R_i$, $R_{of}=R_o/(1+A\beta)$

:::info
++Note++. Because the $R_{if}$/$R_{of}$ we calculated above have in fact included $R_s$/$R_L$, we have to subtract them when representing the actual $R_{in}$/$R_{out}$:
$R_{in}=R_{if}-R_{s}$
$R_{out}=1/\left(\frac{1}{R_{of}}-\frac{1}{R_{L}}\right)$
:::

## 10.5 Other Feedback-Amplifier Types

### 10.5.1 Basic Principles

1. **Sensing**: The feedback network must sample the output signal of interest.
    - If $V_o$ is of interested, the feedback network is connected in parallel (shunt).
    - If $I_o$ is of interested, the feedback network is connected in series.
![截圖 2025-01-14 晚上10.54.20](https://hackmd.io/_uploads/B1G29g4Pyx.png)
2. **Mixing**: 
    - If the input signal to be amplified is a voltage, $V_f$ is connected in series with the input signal source $V_s$.
    - If the signal to be amplified is a current, the feedback current signal $I_f$ is connected in parallel (shunt) with the input signal source $I_s$.
![截圖 2025-01-14 晚上10.56.04](https://hackmd.io/_uploads/r1Omil4wJe.png)
3. **Feedback topology**:
![截圖 2025-01-14 晚上10.59.57](https://hackmd.io/_uploads/rJ6ehlVw1e.png)
4. **Input and output resistance**: The increase or decrease of the input or output resistance depends solely on the type of connection.
    - Series connection always increases the
resistance by the amount of feedback, $(1+A\beta)$.
    - Parallel (shunt) connection always decreases the resistance by the amount of feedback, $(1+A\beta)$.
5. **Dimensions of $A$, $\beta$, $A\beta$, and $A_f$**:
    - Depending on the amplifier type, $A$, $\beta$ and $A_f$ have the dimensions of V/V, A/A, V/A, or A/V. However, 
    - $A\beta$ is always dimensionless.
6. **Determining $\beta$ and $A_f|_{\text{ideal}}$**: To determine $\beta$, set $A=\infty$.
    - For the voltage (series) mixing case, $V_i=0$ and hence $I_i=0$, and $\beta$ is found as the ratio of the open-circuit voltage $V_f$ to the output quantity being sensed.
    - For the case of current (shunt) mixing, $I_i=0$ and hence $V_i=0$, and $\beta$ is found as the ratio of the short-circuit current $I_f$ to the output quantity being sensed.
    - The ideal value of the closed-loop gain is $A_f|_{\text{ideal}}=1/\beta$.
7. **Analysis using the loop gain**: $A_f=\frac{A}{1+A\beta}$.
8. **Find input and output resistance**: described later.

### 10.5.2 The Feedback Transconductance Amplifier (Series-Series)

**Ideal Structure**
![截圖 2025-01-14 晚上11.20.02](https://hackmd.io/_uploads/SJOplW4vkl.png)

**Equivalent Circuit**
![截圖 2025-01-14 晚上11.21.13](https://hackmd.io/_uploads/SyYbZ-4Dke.png)

**Formulas**
$A_f=\frac{I_o}{V_s}=\frac{A}{1+A\beta}$, $A_f|_{\text{ideal}}=\frac{1}{\beta}$
$R_{if}=(1+A\beta)R_i$, $R_{of}=(1+A\beta)R_o$

**General Structure**
![截圖 2025-01-14 晚上11.25.20](https://hackmd.io/_uploads/r1_ZfWNP1l.png)

**Finding the $A$ circuit and $\beta$**
![截圖 2025-01-14 晚上11.27.08](https://hackmd.io/_uploads/H1n1XWEwJl.png)
![截圖 2025-01-14 晚上11.27.17](https://hackmd.io/_uploads/ry31mWEvkl.png)

**Gain, Input, and Output Resistance**
Use formula mentioned to find $A_f|_{\text{ideal}}$, $A_f$, $R_{if}$ and $R_{of}$.
$\Rightarrow R_{in}=R_{if}-R_s$, $R_{out}=R_{of}-R_L$


### 10.5.3 The Feedback Transresistance Amplifier (Shunt-Shunt)

**Ideal Structure**
![截圖 2025-01-14 晚上11.34.43](https://hackmd.io/_uploads/B1EN4-ND1x.png)

**Equivalent Circuit**
![截圖 2025-01-14 晚上11.34.52](https://hackmd.io/_uploads/r1IHVWEvkg.png)

**Formulas**
$A_f=\frac{V_o}{I_s}=\frac{A}{1+A\beta}$, $A_f|_{\text{ideal}}=\frac{1}{\beta}$
$R_{if}=R_i/(1+A\beta)$, $R_{of}=R_o/(1+A\beta)$

**General Structure**
![截圖 2025-01-14 晚上11.35.58](https://hackmd.io/_uploads/SygZiEZ4DJg.png)


**Finding the $A$ circuit and $\beta$**
![截圖 2025-01-14 晚上11.36.08](https://hackmd.io/_uploads/SJvY4-VPke.png)
![截圖 2025-01-14 晚上11.36.21](https://hackmd.io/_uploads/BJx54ZNDkx.png)

**Gain, Input, and Output Resistance**
Use formula mentioned to find $A_f|_{\text{ideal}}$, $A_f$, $R_{if}$ and $R_{of}$.
$\Rightarrow R_{in}=1/\left(\frac{1}{R_{if}}-\frac{1}{R_s}\right)$, $R_{out}=1/\left(\frac{1}{R_{of}}-\frac{1}{R_L}\right)$

### 10.5.4 The Feedback Current Amplifier (Shunt-Series)

**Ideal Structure**
![截圖 2025-01-14 晚上11.38.44](https://hackmd.io/_uploads/ryHHHZEwJe.png)

**Equivalent Circuit**
![截圖 2025-01-14 晚上11.38.53](https://hackmd.io/_uploads/S1yLSZ4PJl.png)

**Formulas**
$A_f=\frac{I_o}{I_s}=\frac{A}{1+A\beta}$, $A_f|_{\text{ideal}}=\frac{1}{\beta}$
$R_{if}=R_i/(1+A\beta)$, $R_{of}=(1+A\beta)R_o$

**General Structure**
![截圖 2025-01-14 晚上11.39.06](https://hackmd.io/_uploads/HkxUS-EP1l.png)

**Finding the $A$ circuit and $\beta$**
![截圖 2025-01-14 晚上11.39.15](https://hackmd.io/_uploads/rkgLSZEDJe.png)
![截圖 2025-01-14 晚上11.39.27](https://hackmd.io/_uploads/rkkLr-NDkl.png)

**Gain, Input, and Output Resistance**
Use formula mentioned to find $A_f|_{\text{ideal}}$, $A_f$, $R_{if}$ and $R_{of}$.
$\Rightarrow R_{in}=1/\left(\frac{1}{R_{if}}-\frac{1}{R_s}\right)$, $R_{out}=R_{of}-R_L$

## 10.6 Summary of the Feedback-Analysis Method

1. Always begin by finding $A_f|_{\text{ideal}}$.
2. Sketch the two-port feedback circuit and determine its loading effects $R_{11}$ and $R_{22}$.
3. Sketch the A circuit and analyze it to determine $A$, $R_i$, and $R_o$.
4. Calculate $A\beta$ and the amount of feedback $(1+A\beta)$.
5. Use formulas to determine $A_f$, $R_{if}$ and $R_{of}$. The difference between $A_f$ and $A_f|_{\text{ideal}}$ should be approximately $(100/A\beta)\%$.

## 10.7 The Stability Problem

**Transfer function**: $A_f(s)=\frac{A(s)}{1+A(s)\beta(s)}$
Assume that dc gain of the amplifier is $A_0$ and at low frequency, $\beta(s)$ is constant.
Consider $A(j\omega)\beta(j\omega)=|A(j\omega)\beta(j\omega)|e^{j\phi(\omega)}$, if $\phi(\omega)\to 180^{\circ}$, the feedback would be **positive**.
- If $|A(j\omega_{180})\beta(j\omega_{180})|<1$, then $A_f>A$ (the amp is still stable, though).
- If $|A(j\omega_{180})\beta(j\omega_{180})|=1$, then $A_f=\infty\Rightarrow$ **oscillator**.
- If $|A(j\omega_{180})\beta(j\omega_{180})|>1$, then the circuit will oscillate, amplitude will increase until reaching nonlinearity.

## 10.8 Effect of Feedback on the Amplifier Poles

### 10.8.1 Stability and Pole Location

:::success
++Recall++. (in "Signals and Systems" course) For an LTI system to be stable, the ROC of its transfer function should includes the entire $j\omega$-axis. Furthermore, for a causal and stable system, all poles should lie in the **left half of the s plane**.
:::

![截圖 2025-01-17 下午5.34.51](https://hackmd.io/_uploads/rkHU4jPD1l.png)

### 10.8.2 Poles of the Feedback Amplifier

$\Rightarrow 1+A(s)\beta(s)=0$ (the **characteristic equation**)

:::info
++Note++. From now on, we assume that the open loop amplifier has real poles and no finite zeros $\Rightarrow$ **no pole-zero cancellation**. We also assume that $\beta$ is independent of frequency.
:::

### 10.8.3 Amplifiers with a Single-Pole Response

$A(s)=\frac{A_0}{1+s/\omega_P}\Rightarrow A_f(s)=\frac{A_0/(1+A_0\beta)}{1+s/\omega_P(1+A_0\beta)}$
$\omega_{Pf}=\omega_P(1+A_0\beta)$

![截圖 2025-01-17 下午5.43.31](https://hackmd.io/_uploads/BJX2wsDvyg.png)

:::info
++Observation++. At high frequency,
$A_f(s)\simeq\frac{A_0\omega_P}{s}\simeq A(s)$
:::

The amplifier is said to be **unconditionally stable** because it is stable regardless of the value of $\beta$ (one can observe that by noticing that the maximum phase shift for a single-pole response is $90^{\circ}$ and can never reach $180^{\circ}$).

### 10.8.4 Amplifiers with a Two-Pole Response

$A(s)=\frac{A_0}{(1+s/\omega_{P1})(1+s/\omega_{P2})}$
$1+A(s)\beta=0\Rightarrow s^2+s(\omega_{P1}+\omega_{P2})+(1+A_0\beta)\omega_{P1}\omega_{P2}=0$
$s=-\frac{1}{2}(\omega_{P1}+\omega_{P2})\pm\frac{1}{2}\sqrt{(\omega_{P1}+\omega_{P2})^2-4(1+A_0\beta)\omega_{P1}\omega_{P2}}$

![截圖 2025-01-17 下午5.58.18](https://hackmd.io/_uploads/rJYAFivwkg.png)
(The **root-locus diagram**, the arrows indicates how poles move as the $\beta$ increases.)

The amplifier is also **unconditionally stable**.

:::success
++Recall++. The characteristic equation of a second-order network can be written as $1+\frac{1}{Q}\frac{s}{\omega_0}+\frac{s^2}{\omega_0^2}=0$, where $\omega_0$ is called the **pole frequency** and $Q$ is called the **pole $Q$ factor** (if $Q>0.5$, poles are complex numbers).
![截圖 2025-01-17 下午6.08.50](https://hackmd.io/_uploads/HyNH2oPv1g.png)
:::

**Expressing $Q$ using above parameters**: $Q=\frac{\sqrt{(1+A_0\beta)\omega_{P1}\omega_{P2}}}{\omega_{P1}+\omega_{P2}}$

![截圖 2025-01-17 下午6.12.46](https://hackmd.io/_uploads/BJy4Tjwvyx.png)

### 10.8.5 Amplifiers with Three or More Poles

**Root-locus diagram for an amplifier with 3 poles**

![截圖 2025-01-17 下午6.14.14](https://hackmd.io/_uploads/SyIt6ivDJg.png)

$\Rightarrow$ as $\beta\nearrow$, 2 poles reach the right half plane $\Rightarrow$ **become unstable**.

## 10.9 Stability Study Using Bode Plots

### 10.9.1 Gain and Phase Margins

![截圖 2025-01-17 下午6.19.03](https://hackmd.io/_uploads/By9s0jwvJe.png)

$\omega_{180}>\omega_1\Rightarrow$ **stable**. (Or, if $|\phi(\omega_1)|>180^{\circ}\Rightarrow$ **unstable**)

### 10.9.2 Effect of Phase Margin on Closed-Loop Response

$A(j\omega_1)\beta=1\times e^{-j\theta}$ where $\theta=180^{\circ}-\text{phase margin}$.
$\Rightarrow A_f(j\omega_1)=\frac{A(j\omega_1)}{1+A(j\omega_1)\beta}=\frac{(1/\beta)e^{-j\theta}}{1+e^{-j\theta}}$
$\Rightarrow |A_f(j\omega_1)|=\frac{1/\beta}{|1+e^{-j\theta}|}$
Assume that $A_f(0)\simeq\frac{1}{\beta}\Rightarrow\frac{|A_f(j\omega_1)|}{A_f(0)}\simeq\frac{1}{|1+e^{-j\theta}|}$
For example, if phase margin is $45^{\circ}$ ($\theta=135^{\circ}$), then $\frac{|A_f(j\omega_1)|}{A_f(0)}=1.3$.

### 10.9.3 An Alternative Approach for Investigating Stability

$20\log|A(j\omega)|-20\log\frac{1}{\beta}=20\log|A\beta|$

![截圖 2025-01-17 晚上11.12.16](https://hackmd.io/_uploads/B1_DXluDyx.png)

$A=\frac{10^5}{(1+jf/10^5)(1+jf/10^6)(1+jf/10^7)}$
$\phi=-[\tan^{-1}(f/10^5)+\tan^{-1}(f/10^6)+\tan^{-1}(f/10^7)]$
When $\phi=-180^{\circ}$, $|A|=60\text{dB}\Rightarrow \beta_{cr}=10^{-3}$

(a)$\beta=5.623\times10^{-5}$, **stable** $\Rightarrow f_1=5.6\times10^5\text{Hz}$
(b)**unstable**

:::info
++Observation++.
Since the $180^{\circ}$-phase point always occurs on the $-40\text{dB}/\text{decade}$ segment of the Bode plot for $|A|$, the closed-loop amplifier will be stable if the $20\log( 1/\beta)$ line intersects the $20\log|A|$ curve at a point on the $-20\text{dB}/\text{decade}$
segment. Following this rule **ensures that a phase margin of at least $45^{\circ}$** is obtained.

The general rule states that at the intersection of $20\log[1/|\beta(j\omega)|]$ and $20 \log|A(j\omega)|$ the difference of slopes (called the **rate of closure**) should not exceed $20\text{dB}/\text{decade}$.
:::

## 10.10 Frequency Compensation

### 10.10.1 Theory

$\Rightarrow$ introduce a new pole $f_D$ with low frequency $\Rightarrow A'(s)$ intersects with $20\log(1/\beta)$ with a slope difference of $20\text{dB}/\text{decade}$.
![截圖 2025-01-17 晚上11.32.14](https://hackmd.io/_uploads/ryLE_euP1x.png)

:::info
++Note++. Disadvantage: open-loop gain has been drastically reduced.
:::

### 10.10.2 Implementation

![截圖 2025-01-17 晚上11.46.00](https://hackmd.io/_uploads/ByQ8oxOPye.png)

$f_{P1}=\frac{1}{2\pi C_xR_x}$
Adding compensating capacitor $C_C$, the pole becomes $f_D'=\frac{1}{2\pi(C_x+C_C)R_x}$

:::info
++Note++. Disadvantage: the required $C_C$ is usually quite large.
:::

### 10.10.3 Miller Compensation and Pole Splitting

:::success
++Recall++. Miller effect:
![截圖 2025-01-07 下午6.44.21](https://hackmd.io/_uploads/HkS5pl_vJg.png)
:::

![截圖 2025-01-17 晚上11.54.01](https://hackmd.io/_uploads/H1nm6lOPkx.png)
($C_1$ includes the miller component of $C_\mu$, $C_2$ includes $C_L$)

If there is no compensating capacitor $C_f$,
$\Rightarrow f_{P1}=\frac{1}{2\pi C_1R_1}$, $f_{P2}=\frac{1}{2\pi C_2R_2}$
When $C_f$ is added:
$\Rightarrow\frac{V_o}{I_i}=\frac{(sC_f-g_m)R_1R_2}{1+s[C_1R_1+C_2R_2+C_f(g_mR_1R_2+R_1+R_2)]+s^2[C_1C_2+C_f(C_1+C_2)]R_1R_2}$
Rewrite the denominator $D(s)$, we have ($\omega_{P1}' \ll \omega_{P2}'$)
$D(s)=\left(1+\frac{s}{\omega_{P1}'}\right)\left(1+\frac{s}{\omega_{P2}'}\right)=1+s\left(\frac{1}{\omega_{P1}'}+\frac{1}{\omega_{P2}'}\right)+\frac{s^2}{\omega_{P1}'\omega_{P2}'}\simeq1+\frac{s}{\omega_{P1}'}+\frac{s^2}{\omega_{P1}'\omega_{P2}'}$
$\Rightarrow \omega_{P1}'=\frac{1}{C_1R_1+C_2R_2+C_f(g_mR_1R_2+R_1+R_2)}\simeq\frac{1}{g_mR_2C_fR_1}$
$\Rightarrow\omega_{P2}'\simeq\frac{g_mC_f}{C_1C_2+C_f(C_1+C_2)}$

:::info
++Observation++. As $C_f$ increases, $\omega_{P1}'$ is reduced and $\omega_{P2}'$ is increased (**pole splitting**).

Also, because of the Miller effect, $C_f$ is increased by $g_mR_2$ and hence can be smaller than that of $C_C$ in 10.10.2.
:::

## Summary

- The key parameter in a feedback amplifier is the **loop gain** $A\beta$ (and the amount of feedback $(1+A\beta)$).
    - The closed-loop gain can be expressed as $A_f=\frac{A}{1+A\beta}$.
    - The ideal value of $A_f$ is $A_f|_{\text{ideal}}=\frac{1}{\beta}$, happens when the open loop gain $A$ is large.
    - $A\beta$ can be determined by breaking the feedback loop (set $x_s=0\Rightarrow A\beta=-\frac{x_r}{x_t}$).
    - $\beta$ can be determined finding $\beta=\frac{x_s}{x_o}$ under the condition $x_i=0$ (i.e., $A=\infty$).
- The properties of four types of feedback topologies are described in section [10.5.1](https://hackmd.io/fWlf9XhPQxykpzsLwgKfkA#1051-Basic-Principles).
    - A series connection at the input or output of a feedback amplifier increases the resistance by $(1+A\beta)$.
    - A parallel (shunt) connection reduces the resistance by $(1+A\beta)$.
- Method of analysing a feedback amplifier:
    - **Begin by finding $A_f|_{\text{ideal}}$**.
    - Sketch the two-port feedback circuit and determine its loading effects $R_{11}$ and $R_{22}$. 
    - Sketch the A circuit and analyze it to determine $A$, $R_i$, and $R_o$. 
    - Calculate $A\beta$ and the amount of feedback $(1+A\beta)$. 
    - Use formulas to determine $A_f$, $R_{if}$ and $R_{of}$.
- Stability (frequency response of the feedback amplifier)
    - The poles are determined by the characteristic equation $1+A(s)\beta(s)=0$.
    - For the amplifier to be stable, **all poles should be in the left half of the s plane**.
    - Also, if at the frequency for which the phase shift of $A\beta$ is $180^{\circ}$ ($\omega_{180}$), we have $|A\beta|<1$, then the amplifier is stable. (The amount by which it is less than $1$, in $\text{dB}$, is the **gain margin**.)
    - Alternatively, if at the frequency for which $|A\beta|=1$ (named $\omega_1$), the phase angle is less than $180^{\circ}$, then the amplifier is stable. (The difference is called the **phase margin**.)
    - We can also check the stability by analyzing the Bode plot of $|A|$ and the curve of $20\log(1/|\beta|)$. Stability is guaranteed if two lines intersect with a difference in slope no greater than $20\text{dB}/\text{decade}$.
- Frequency compensation is the process where we modify the frequency of the pole for the given $\beta$ in order to make the amplifier stable.

## Practice

- 10.1, 10.2, 10.3
- 10.10, 10.11, 10.12, 10.13, 10.14, 10.15, 10.16
- 10.24, 10.25, 10.26
- 10.32, 10.33
- 10.43, 10.50, 10.61, 10.71
- 10.81
- 10.83
- 10.89, 10.90
- 10.95, 10.96