---
# System prepended metadata

title: CH9. Frequency Response
tags: [MC2 WebNotes]

---

# CH9. Frequency Response

:::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
:::

:::success
++Recall++. 
Frequency response of MOS/BJT amplifiers:
![截圖 2025-01-07 凌晨2.47.47](https://hackmd.io/_uploads/ByQOrot8ke.png)
Frequency response of DC amplifiers:
![截圖 2025-01-07 凌晨2.49.18](https://hackmd.io/_uploads/ryNpHsKUye.png)

Bandwidth $\text{BW}=f_H-f_L\simeq f_H$
Gain-bandwidth product $\text{GB}=|A_M|BW$
:::

## 9.1 High-Frequency Transistor Models

### 9.1.1 The MOSFET

![截圖 2025-01-07 凌晨2.52.06](https://hackmd.io/_uploads/SyOwLitIkg.png)
- $C_{gs}$, $C_{gd}$: result from the gate-capacitance effect.
- $C_{sb}$, $C_{db}$: depletion capacitances of the pn junctions.

When pinched off (as showned in figure), the gate capacitance is $\frac{2}{3}WLC_{ox}$. In addition, there are other small capacitances due to the overlap of the gate & source/drain ($C_{ov}=WL_{ov}C_{ox}$, $L_{ov}=0.05$ to $0.1L$). Therefore,

$C_{gs}=\frac{2}{3}WLC_{ox}+C_{ov}$, $C_{gd}=C_{ov}$

And, by the result of junction capacitance discussed in Ch1,

$C_{sb}=\frac{C_{sb0}}{\sqrt{1+\frac{V_{SB}}{V_0}}}$, $C_{db}=\frac{C_{db0}}{\sqrt{1+\frac{V_{DB}}{V_0}}}$

:::success
++Recall++. Junction capacitance: $C_j=\frac{C_{j0}}{\sqrt{1+\frac{V_R}{V_0}}}$
:::

#### The High-Frequency MOSFET Model

![截圖 2025-01-07 下午4.41.22](https://hackmd.io/_uploads/S1_RdD5Lkx.png)

**When the source is connected to the body $\Rightarrow$**

![截圖 2025-01-07 下午4.41.33](https://hackmd.io/_uploads/HJNftv9LJe.png)

**When $C_{db}$ can be omitted $\Rightarrow$**

![截圖 2025-01-07 下午4.41.41](https://hackmd.io/_uploads/HJsEYPcI1l.png)

#### The MOSFET Unity-Gain Frequency ($f_T$, aka transition frequency)

![截圖 2025-01-07 下午4.51.43](https://hackmd.io/_uploads/ryr4iP9UJe.png)

:::success
++Recall++. Impedance in s-domain: $Z_C=\frac{1}{sC}$, $Z_L=sL$
:::

$I_i=sC_{gs}V_{gs}+sC_{gd}V_{gs}$
$I_o=g_mV_{gs}-sC_{gd}V_{gs}$
$\Rightarrow\frac{I_o}{I_i}=\frac{g_m-sC_{gd}}{s(C_{gs}+C_{gd})}\Rightarrow \left|\frac{I_o}{I_i}\right|\simeq\frac{g_m}{\omega(C_{gs}+C_{gd})}$
$\Rightarrow \omega_T=\frac{g_m}{C_{gs}+C_{gd}}\Rightarrow f_T=\frac{g_m}{2\pi(C_{gs}+C_{gd})}$

## 9.2 High-Frequency Response of CS and CE Amplifiers

### 9.2.1 Frequency Response of the Low-Pass Single-Time-Constant Circuit

:::success
++Recall++. For low-pass STC:
$\tau =CR$, $\frac{V_o}{V_i}=\frac{1}{1+\frac{s}{\omega_P}}$, $\omega_p=\frac{1}{\tau}$, $\left|\frac{V_o}{V_i}\right|=\frac{1}{\sqrt{1+\left(\frac{\omega}{\omega_P}\right)^2}}$, $\omega_H=\omega_P$
:::

![截圖 2025-01-07 下午5.25.05](https://hackmd.io/_uploads/SJu-Q_qI1l.png)

### 9.2.2 The Common-Source Amplifier

![截圖 2025-01-07 下午5.27.06](https://hackmd.io/_uploads/HJoF7_c81x.png)

**SS circuit**

![截圖 2025-01-07 下午5.28.00](https://hackmd.io/_uploads/rkf-Nd5I1g.png)
![截圖 2025-01-07 下午5.28.12 1](https://hackmd.io/_uploads/ryhkEd5Iyl.png)
($R_L'=r_o\|R_L$)

Midband gain $A_m=\frac{V_o}{V_{sig}}=-g_mR_L'$
If we do analysis now, the transfer function will be **of second order (not very simple)**. Therefore, we aim to find a simplfied circuit.

Notice that load current (i.e., current flows through $R_L'$) is $g_mV_{gs}-I_{gd}$. When the frequency is near to $f_H$, we can assume that $I_{gd}$ is still small enough to be omitted. That is,

$V_o\simeq-g_mR_L'V_{gs}$
$\Rightarrow I_{gd}=sC_{gd}(V_{gs}-V_o)=sC_{gd}(1+g_mR_L')V_{gs}$
Define $C_{eq}=C_{gd}(1+g_mR_L')$, the circuit becomes:

![截圖 2025-01-07 下午5.28.39](https://hackmd.io/_uploads/HJnJN_581g.png)

The input circuit is now an STC, thus
$V_{gs}=\frac{1}{1+\frac{s}{\omega_P}}$, $\omega_P=\frac{1}{C_{in}R_{sig}}$ where $C_{in}=C_{gs}+C_{gd}(1+g_mR_L')$
$\Rightarrow \frac{V_{o}}{V_{sig}}=\frac{-g_mR_L'}{1+\frac{s}{\omega_P}}=\frac{A_M}{1+\frac{s}{\omega_H}}$ where $\omega_H=\omega_P$ is the higher 3-dB frequency.

:::info
++Note++. The multiplication effect from $C_{gd}$ to $C_{eq}$ is call the **Miller effect** and $(1+g_mR_L')$ is known as the **Miller multiplier**.
:::

:::info
++Observation++.
- Larger $R_{sig}$ causes lower $f_H$.
- $C_{in}$ is usually dominated by $C_{gd}$ ($\because$ Miller effect). Causes CS Amplifier has lower $f_H$.
- Our approximation applies when the frequency is **not too much higher than $f_H$**.
- The CS amplifier is said to have a **dominant high-frequency pole** with frequency $f_P=f_H$.
:::

### 9.2.3 Frequency Response of the CS Amplifier When $R_{sig}$ Is Low

![截圖 2025-01-07 下午6.09.28](https://hackmd.io/_uploads/HyldaucLkl.png)
($V_{gs}=V_{sig}$, $I_{gd}=sC_{gd}(V_{gs}-V_o)$)

$I_{gd}=g_mV_{gs}+\frac{V_o}{R_L'}+sC_LV_o$
$\Rightarrow\frac{V_o}{V_{sig}}=-g_mR_L'\frac{1-s(C_{gd}/g_m)}{1+s(C_L+C_{gd})R_L'}$
$\Rightarrow f_Z=\frac{g_m}{2\pi C_{gd}}$ (transmission zero frequency), $f_H=\frac{1}{2\pi(C_L+C_{gd})R_L'}$ (pole i.e., 3-dB frequency)
unity-gain frequency $f_t=|A_m|f_H=\frac{g_m}{2\pi(C_L+C_{gd})}$

:::info
++Observation++.
- We can see how pole frequency is derived by zero $V_{sig}$:
![截圖 2025-01-07 下午6.37.15](https://hackmd.io/_uploads/SkflNY9Ikx.png)

- Notice that $\frac{f_Z}{f_H}=(g_mR_L')\left(1+\frac{C_L}{C_{gd}}\right)\gg1$, that's why we ignore the effect of $1-s(C_{gd}/g_m)$ term in $\frac{V_o}{V_{sig}}$ when studying the behavior of the circuit with frequency nears $f_H$.
- Bode plot for the circuit:
![截圖 2025-01-07 下午6.41.13](https://hackmd.io/_uploads/Hyy1BY9LJx.png)
:::

### 9.2.5 Miller's Theorem

![截圖 2025-01-07 下午6.44.21](https://hackmd.io/_uploads/SJIrIYc8Jl.png)
(Assume that $V_2=KV_1$)

To transform the circuits, we have:

$\frac{V_1}{Z_1}=\frac{-KV_1}{Z_2}=\frac{(1-K)V_1}{Z}=I$
$\Rightarrow Z_1=\frac{Z}{1-K}$, $Z_2=\frac{Z}{1-\frac{1}{K}}$ ($(1-K)$: the **Miller multiplier**)

## 9.3 The Method of Open-Circuit Time Constants

### 9.3.1 The High-Frequency Gain Function

$A(s)=A_MF_H(s)=A_M\frac{(1+s/\omega_{Z1})(1+s/\omega_{Z2})\cdots(1+s/\omega_{Zn})}{(1+s/\omega_{P1})(1+s/\omega_{P2})\cdots(1+s/\omega_{Pn})}$

### 9.3.2 Determining the 3-dB Frequency $f_n$

(It is assumed that zeros are either at $\infty$ or very high) Find the lowest pole $\omega_{P1}$ which **dominate** the high-frequency response of the amplifier (if we can do so, the amplifier is said to have a **dominant-pole response**). In such case, 
$F_H(s)\simeq\frac{1}{1+s/\omega_{P1}}$ and $\omega_H\simeq\omega_{P1}$.


For the case where poles&zeros are hard to be specified, i.e.,
$F_H(s)=\frac{1+a_1s+a_2s^2+\cdots a_ns^n}{1+b_1s+b_2s^2+\cdots b_ns^n}$
If a dominant pole exist, we can do approximation:
$F_H(s)\simeq\frac{1}{1+b_1s}$, $\omega_H=\frac{1}{b_1}$.

To determine $b_1$, we can (1)**zero the input signal**, (2)**consider each capacitance $C_i$ one at a time (open-circuit for other capacitors)**, (3)**find $R_i$ by either inspection or by replacing $C_i$ with $V_x$ and calculate $R_i=V_x/I_x$**, then we have
$b_1=\sum_{i=1}^nC_iR_i$

### 9.3.3 Applying the Method of Open-Circuit Time Constants to the CS Amplifier

![截圖 2025-01-07 晚上7.28.16](https://hackmd.io/_uploads/Hy3yx59Uke.png)

left 1 capacitance each time $\Rightarrow$

![截圖 2025-01-07 晚上11.41.51](https://hackmd.io/_uploads/H1mIiaq8kx.png)

(a) **$C_{gs}$ case**: $R_{gs}=R_{sig}$
(b) **$C_{gd}$ case**: $R_{gd}=\frac{V_x}{I_x}=\frac{R_L'(I_x-g_m(-I_xR_{sig}))-(-I_xR_{sig})}{I_x}=R_{sig}(1+g_mR_L')+R_L'$
(c\) **$C_L$ case**: $R_{C_L}=R_L'$
$\Rightarrow \tau_H=C_{gs}R_{gs}+C_{gd}R_{gd}+C_LR_{C_L}=C_{gs}R_{sig}+C_{gd}(R_{sig}(1+g_mR_L')+R_L')+C_LR_L'$
$\Rightarrow f_H=\frac{1}{2\pi\tau_H}$

:::info
++Note++. $\tau_H$ can be rewritten as
$\tau_H=C_{in}R_{sig}+(C_{gd}+C_L)R_L'$
where $C_{in}=C_{gs}+C_{gd}(1+g_mR_L')$.

Therefore, the first term is the effect of input STC discussed in 9.2, while the second term is the effect of $C_L$ and it's dominant.
:::

## 9.4 High-Frequency Response of Common-Gate and Cascode Amplifiers

### 9.4.1 High-Frequency Response of the CG Amplifier

![截圖 2025-01-08 凌晨12.13.03](https://hackmd.io/_uploads/HJQ2z09Iye.png)
(All capacitances have a grounded node $\Rightarrow$ **no Miller effect**. Also, $C_L$ includes the effect of $C_{db}$)

**SS circuit (T-model)**

![截圖 2025-01-08 凌晨12.49.10](https://hackmd.io/_uploads/Sk3foAqUJg.png)
(Note that $C_L$ and $C_{gd}$ are now combined together)

**Analysis w/o $r_o$**

![截圖 2025-01-08 凌晨1.57.15](https://hackmd.io/_uploads/SyeGs1jUkg.png)

Two poles: $f_{P1}=\frac{1}{2\pi C_{gs}\left(R_{sig}\|\frac{1}{g_m}\right)}$ and $f_{P2}=\frac{1}{2\pi(C_L+C_{gd})R_L}$
Usually $f_{P2}<f_{P1}\Rightarrow f_{P2}$ can be dominant.
To find $f_H$, we have
$\tau_{gs}=C_{gs}\left(R_{sig}\|\frac{1}{g_m}\right)$, $\tau_{gd}=(C_L+C_{gd})R_L$
$\Rightarrow \tau_H=C_{gs}\left(R_{sig}\|\frac{1}{g_m}\right)+(C_L+C_{gd})R_L$
$f_H=\frac{1}{2\pi\tau_H}=1/\left(\frac{1}{f_{P1}}+\frac{1}{f_{P2}}\right)$

:::info
++Observation++. Typically both $f_{P1}$ and $f_{P2}$ are much larger than frequncy of dominant input pole in a CS amplifier $\Rightarrow$ preferred.
:::

**Analysis with $r_o$**

(a) $C_{gs}$ case
![截圖 2025-01-08 凌晨2.11.11](https://hackmd.io/_uploads/B1uvRyjLyg.png)

$R_{gs}=R_{sig}\|R_{in}$ where $R_{in}=\frac{r_o+R_L}{1+g_mr_o}\simeq\frac{r_o+R_L}{g_mr_o}$

(b) $(C_L+C_{gd})$ case
![截圖 2025-01-08 凌晨2.12.19](https://hackmd.io/_uploads/SJcCRyjL1g.png)

$R_{gd}=R_L\|R_o$ where $R_o=r_o+R_{sig}+g_mr_oR_{sig}$

Combine (a)&(b) $\Rightarrow \tau_H=C_{gs}(R_{sig}\|\frac{r_o+R_L}{g_mr_o})+(C_L+C_{gd})[R_L\|(r_o+R_{sig}+g_mr_oR_{sig})]$
$\Rightarrow f_H=\frac{1}{2\pi [C_{gs}(R_{sig}\|\frac{r_o+R_L}{g_mr_o})+(C_L+C_{gd})[R_L\|(r_o+R_{sig}+g_mr_oR_{sig})]]}$

### 9.4.2 High-Frequency Response of the MOS Cascode Amplifier

![截圖 2025-01-08 凌晨2.21.45](https://hackmd.io/_uploads/SkD0lxo81x.png)
(Resistance seeing from $D_1$: $R_{d1}=r_{o1}\|R_{in2}=r_{o1}\|\frac{r_{o2}+R_L}{g_mr_{o2}}$)
($C_L$ includes the effect of $C_{db2}$)

$R_{gs1}=R_{sig}$
$R_{gd1}=(1+g_{m1}R_{d1})R_{sig}+R_{d1}=(1+g_{m1}R_{d1})R_{sig}+r_{o1}\|\frac{r_{o2}+R_L}{g_mr_{o2}}$
$R_{db1\|gs2}=R_{d1}$
$R_{L\|gd2}=R_L\|R_o=R_L\|(r_{o2}+r_{o1}+(g_{m2}r_{o2})r_{o1})$
$\Rightarrow \tau_H=C_{gs1}R_{sig}+C_{gd1}[(1+g_{m1}R_{d1})R_{sig}+R_{d1}]+(C_{db1}+C_{gs2})R_{d1}+(C_L+C_{gd2})(R_L\|R_o)$
$\Rightarrow f_H=\frac{1}{2\pi\tau_H}$

#### Design Insight and Trade-Offs

Rewrite $\tau_H$:
$\tau_H=R_{sig}[C_{gs1}+C_{gd1}(1+g_{m1}R_{d1})]+R_{d1}(C_{gd1}+C_{db1}+C_{gs2})+(R_L\|R_o)(C_L+C_{gd2})$
$\Rightarrow$ 3 factors: input node, middle node, output node.

If $R_{sig}$ is small, $\tau_H\simeq(C_L+C_{gd2})(R_L\|R_o)$, 
$f_H\simeq\frac{1}{2\pi(C_L+C_{gd2})(R_L\|R_o)}\Rightarrow$ Same form as an CS amplifier.
However, $(R_L\|R_o)$ is larger that CS Amp's $(R_L\|r_o)$ by the factor of about $A_0=g_mr_o$
While $f_t\simeq\frac{1}{2\pi}\frac{g_m}{C_L+C_{gd}}$ is the same as an CS Amp.

![截圖 2025-01-08 晚上8.10.23](https://hackmd.io/_uploads/ryjBjJnIyl.png)
(Note. to achieve high gain, for cascode case, we select load resistor with resistance $A_0R_L$)

![截圖 2025-01-08 晚上8.10.32](https://hackmd.io/_uploads/SJDFjknLke.png)

#### Conclusion

Depending on the usage, cascode amplifiers can have **either high DC gain or large bandwidth**.

## 9.5 High-Frequency Response of Source and Emitter Followers

### 9.5.1 The Source-Follower Case

![截圖 2025-01-08 晚上8.15.07](https://hackmd.io/_uploads/Hyfo21hL1e.png)

**SS circuit (with body-effect considered)**
![截圖 2025-01-08 晚上8.15.29](https://hackmd.io/_uploads/rkfonJh8Jl.png)

**$V_{bs}=V_{ds}\Rightarrow$ view $g_{mb}V_{bs}$ as a resistor**
![截圖 2025-01-08 晚上8.16.03](https://hackmd.io/_uploads/rJzj3khLkx.png)


#### Obtaining the Transfer Function $V_o(s)/V_{sig}(s)$

$V_{sig}=I_iR_{sig}+V_g=I_iR_{sig}+V_{gs}+V_o$
$I_i=sC_{gd}V_g+sC_{gs}V_{gs}=sC_{gd}(V_{gs}+V_o)+sC_{gs}V_{gs}$
$\Rightarrow V_{sig}=[1+s(C_{gs}+C_{gd})R_{sig}]V_{gs}+[1+sC_{gd}R_{sig}]V_o$
$(g_m+sC_{gs})V_{gs}=\left(\frac{1}{R_L'}+sC_L\right)V_o\Rightarrow V_{gs}=\frac{1}{g_mR_L'}\frac{1+sC_LR_L'}{1+s(C_{gs}/g_m)}V_o$
Combine two expressions, we get $\frac{V_o}{V_{sig}}=A_M\frac{1+\frac{s}{\omega_Z}}{1+b_1s+b_2s^2}$, where
$A_M=\frac{R_L'}{R_L'+1/g_m}=\frac{g_mR_L'}{g_mR_L'+1}$
$\omega_Z=\frac{g_m}{C_{gs}}$
$b_1=\left(C_{gd}+\frac{C_{gs}}{g_mR_L'+1}\right)R_{sig}+\left(\frac{C_{gs}+C_L}{g_mR_L'+1}\right)R_L'$
$b_2=\frac{(C_{gs}+C_{gd})C_L+C_{gs}C_{gd}}{g_mR_L'+1}R_{sig}R_L'$

#### Analysis of the Source-Follower Transfer Function

1. DC gain $=$ midband gain $A_M$.
2. Although there are three capacitance, the transfer function is of second order because they form a continuous loop.
3. Two transmission zero: $s=\infty$($\because C_{gd}$) and $s=-\omega_Z$ (slightly larger than MOSFET's unity gain frequency $\omega_T=\frac{g_m}{C_{gs}+C_{gd}}$, which is very high), therefore **the effect of these zero is small**.
4. If the poles are real, we can find two frequency such that $1+b_1s+b_2s^2=\left(1+\frac{s}{\omega_{P1}}\right)\left(1+\frac{s}{\omega_{P2}}\right)$. If $\omega_{P2}\gg\omega_{P1}$ (at least 4 times larger), a dominant pole exists with frequency $\omega_{P1}$, thus the 3-dB frequency $f_H\simeq f_{P1}\simeq\frac{1}{2\pi b_1}$ ($b_1\simeq\tau_H$).
5. If the poles are real but none is dominant, by finding $\left|\frac{V_o}{V_{sig}}\right|=\frac{A_M}{\sqrt{2}}$ we have $f_H\simeq 1/\sqrt{\frac{1}{f_{P1}^2}+\frac{1}{f_{P2}^2}-\frac{2}{f_Z^2}}$
6. If the poles are complex, then rewrite the expression as $1+b_1s+b_2s^2=1+\frac{1}{Q}\frac{s}{\omega_0}+\frac{s^2}{\omega_0^2}$ where
$\omega_0=\frac{1}{\sqrt{b_2}}=\sqrt{\frac{g_mR_L'+1}{R_{sig}R_L'[(C_{gs}+C_{gd})C_L+C_{gs}C_{gd}]}}$
$Q=\frac{\sqrt{b_2}}{b_1}=\frac{\sqrt{g_mR_L'+1}\sqrt{[(C_{gs}+C_{gd})C_L+C_{gs}C_{gd}]R_{sig}R_L'}}{[C_{gs}+C_{gd}(g_mR_L'+1)]R_{sig}+(C_{gs}+C_L)R_L'}$
When $Q=0.707$ ($45^{\circ}$ on the s-plane), the circuit has **maximally flat response** where $f_{3dB}=f_0$.

![截圖 2025-01-08 晚上11.02.59](https://hackmd.io/_uploads/B1B6XG2I1e.png)
![截圖 2025-01-08 晚上11.03.07](https://hackmd.io/_uploads/SJHpQf2Iye.png)

## 9.6 High-Frequency Response of Differential Amplifiers

### 9.6.1 Analysis of the Resistively Loaded MOS Amplifier

![截圖 2025-01-08 晚上11.13.47](https://hackmd.io/_uploads/SJVn8znI1e.png)

**$\Rightarrow$ differential & CM half-circuits**

![截圖 2025-01-08 晚上11.16.19](https://hackmd.io/_uploads/S1fQwzh8yl.png)
(Only $C_{SS}$ is considered, because it is dominant. Also, since differential half-circuit is basically an CS amp which has been discussed before, we won't do analysis here.)

:::success
++Recall++. (In Ch8.) $A_{cm}=-\left(\frac{R_D}{2R_{SS}}\right)\frac{\Delta R_D}{R_D}$
:::

$A_{cm}(s)=-\left(\frac{R_D}{2Z_{SS}}\right)\frac{\Delta R_D}{R_D}=-\frac{R_D}{2}\frac{\Delta R_D}{R_D}\left(\frac{1}{R_{SS}}+sC_{SS}\right)=-\left(\frac{R_D}{2R_{SS}}\right)\frac{\Delta R_D}{R_D}\left(1+sC_{SS}R_{SS}\right)$
$\Rightarrow \omega_Z=\frac{1}{C_{SS}R_{SS}}$, $f_Z=\frac{1}{2\pi C_{SS}R_{SS}}$

![截圖 2025-01-08 晚上11.30.33](https://hackmd.io/_uploads/ryUiqfn8Jx.png)
![截圖 2025-01-08 晚上11.30.45](https://hackmd.io/_uploads/HkUs5M2Lye.png)
![截圖 2025-01-08 晚上11.30.55](https://hackmd.io/_uploads/rkUs9zhIJl.png)

:::info
++Observation++. (Trade-off between DC voltage $\leftrightarrow \text{CMRR}$)
If we want small $V_{DD}\Rightarrow$ small $V_{DS}\Rightarrow$ small $V_{OV}\Rightarrow$ (given $I$) large $W/L\Rightarrow$ large $C_{SS}\Rightarrow$ lower $f_Z\Rightarrow$ lower $\text{CMRR}$

High $\text{CMRR}$ at high-frequency is needed in differential pair to surpress noise. For instance, consider the following two-stage amplifier.
![截圖 2025-01-08 晚上11.39.18](https://hackmd.io/_uploads/SJTBhf281l.png)
:::

### 9.6.2 Analysis of the Current-Mirror-Loaded MOS Amplifier

![截圖 2025-01-08 晚上11.46.09](https://hackmd.io/_uploads/HkRRpGh8yg.png)
($C_m$: total capacitance at the input node of the current mirror ($C_m=C_{gd1}+C_{db1}+C_{db3}+C_{gs3}+C_{gs4}$); $C_L$: total capacitance at the output node of the amplifier ($C_L=C_{gd2}+C_{db2}+C_{gd4}+C_{db4}+C_x$, where $C_x$ is the input capacitance of a subsequent stage))

#### The Dominant Pole

**Equivalent output circuit**
![截圖 2025-01-08 晚上11.54.16](https://hackmd.io/_uploads/SJg61QnUke.png)

:::success
++Recall++. (In Ch8.) $G_m=g_{m1,2}$, $R_o=r_{o2}\|r_{o4}$
:::

$V_o=G_mV_{id}(\frac{1}{R_o}+sC_L)^{-1}\Rightarrow \frac{V_o}{V_{id}}=\frac{G_mR_o}{1+sC_LR_o}$
$\Rightarrow f_{P1}=\frac{1}{2\pi C_LR_o}$ (much lower than other poles/zeros $\Rightarrow$ **dominant**)

#### Dependence of $G_m$ on Frequency

$G_m=\frac{I_o}{V_{id}}$, $I_o=I_{d2}+I_{d4}$
$\Rightarrow I_{d2}=I_{d1}=g_m\frac{V_{id}}{2}$, $I_{d4}=-g_{m4}V_{g3}$
$\Rightarrow V_{g3}=-I_{d1}\frac{1}{\frac{1}{R_m}+sC_m}$ where $R_m=r_{o1}\|\frac{1}{g_{m3}}\|r_{o3}\simeq\frac{1}{g_{m3}}$
$\Rightarrow V_{g3}=-\frac{g_m(V_{id}/2)}{g_{m3}+sC_m}\Rightarrow I_{d4}=\frac{g_{m4}g_m(V_{id}/2)}{g_{m3}+sC_m}=\frac{g_mV_{id}/2}{1+s\frac{C_m}{g_{m3}}}$ ($g_{m3}=g_{m4}$)
$\Rightarrow I_o=g_m\frac{V_{id}}{2}+\frac{g_mV_{id}/2}{1+s\frac{C_m}{g_{m3}}}=g_mV_{id}\frac{1+s\frac{C_m}{2g_{m3}}}{1+s\frac{C_m}{g_{m3}}}$
$\Rightarrow G_m=g_m\frac{1+s\frac{C_m}{2g_{m3}}}{1+s\frac{C_m}{g_{m3}}}\Rightarrow f_{P2}=\frac{g_m}{2\pi C_m}$, $f_Z=\frac{2g_m}{2\pi C_m}$

:::info
++Note++. Usually $C_m\simeq C_{gs2}+C_{gs4}=2C_{gs}$, we have $f_{P2}\simeq\frac{g_m}{2\pi 2C_{gs}}=\frac{f_T}{2}$ and $f_Z\simeq f_T$, which are very high.
:::

![截圖 2025-01-09 凌晨12.40.13](https://hackmd.io/_uploads/Bk7YcX2I1e.png)

To sum up, $\frac{V_o}{V_{id}}=(g_mR_o)\left(\frac{1+s\frac{C_m}{2g_{m3}}}{1+s\frac{C_m}{g_{m3}}}\right)\left(\frac{1}{1+sC_LR_o}\right)$

## 9.7 Other Wideband Amplifier Configurations

### 9.7.1 Obtaining Wideband Amplification by Source or Emitter Degeneration

![截圖 2025-01-09 凌晨12.56.34](https://hackmd.io/_uploads/B1Ic0X3I1g.png)

**Equivalent output circuit**
![截圖 2025-01-09 凌晨12.56.49](https://hackmd.io/_uploads/Sya3RQhIJx.png)

:::success
++Recall++. Source degeneration resistor $R_s$ in an CS amplifier introduces negative feedback and reduce the overall gain by the factor of $1/(1+g_mR_s)$
:::

$G_m\simeq\frac{g_m}{1+g_mR_s}$, $R_o\simeq r_o(1+g_mR_s)$
$\Rightarrow A_M=-G_m(R_o\|R_L)=-G_mR_L'$

**Find $C_{in}$**
![截圖 2025-01-09 凌晨1.04.42](https://hackmd.io/_uploads/B1-Ug42UJx.png)

$C_{in}=C_{gd}(1+G_mR_L')\simeq C_{gd}|A_M|$
$\Rightarrow f_H\simeq\frac{1}{2\pi R_{sig}C_{gd}|A_M|}$
$\Rightarrow GB\equiv |A_M|f_H=\frac{1}{2\pi R_{sig}C_{gd}}$

:::info
++Observation++. The above result is merely an approximation as we didn't consider the effect of other capacitors:
![截圖 2025-01-09 凌晨1.11.44](https://hackmd.io/_uploads/r1NZzNhLkg.png)


**Finding $\tau_H$ using the method of open-circuit time constant**
![截圖 2025-01-09 凌晨1.12.07](https://hackmd.io/_uploads/S1NWz42Iye.png)
($R_L'=R_o\|R_L=[r_o(1+g_mR_s)]\|R_L$)

$R_{gd}=R_{sig}(1+G_mR_L')+R_L'$
$R_{C_L}=R_L'$
$R_{gs}\simeq\frac{R_{sig}+R_s+R_{sig}R_s/(r_o+R_L)}{1+g_mR_s\left(\frac{r_o}{r_o+R_L}\right)}$
$\Rightarrow\tau_H=C_{gd}R_{gd}+C_LR_{C_L}+C_{gs}R_{gs}$
$\Rightarrow f_H=\frac{1}{2\pi\tau_H}$
:::

### 9.7.2 Increasing $f_H$ by Buffering the Input Signal Source

![截圖 2025-01-09 凌晨1.32.55](https://hackmd.io/_uploads/SJ_-DEhUyl.png)

We add a source follower before the CS amp, thus increases the frequency of the pole that arises at the amplifier input because the **Miller multiplier** is reduced ($R_{sig}$ is replaced with the buffer's output resistance $\frac{1}{g_m}\|r_o$, which is smaller)

### 9.7.3 Increasing $f_H$ by Eliminating the Miller Effect Using a CG or a CB Configuration with an Input Buffer

![截圖 2025-01-12 凌晨1.05.55](https://hackmd.io/_uploads/HyaWSQgDyg.png)

CG amps have no **Miller effect**, but have small input resistance. We enhance the performance by adding a buffer.

$f_{P1}=\frac{1}{2\pi (\frac{C_{gs}}{2}+C_{gd})R_{sig}}$, $f_{P2}=\frac{1}{2\pi C_{gd}R_D}$

## 9.8 Low-Frequency Response of Discrete-Circuit CS and CE Amplifiers

### 9.8.1 Frequency Response of the High-Pass Single-Time-Constant Circuit

:::success
++Recall++. For high-pass STC:
$\tau =CR$, $\frac{V_o}{V_i}=\frac{s}{s+\omega_P}$, $\omega_p=\frac{1}{\tau}$, $\left|\frac{V_o}{V_i}\right|=\frac{1}{\sqrt{1+\left(\frac{\omega_P}{\omega}\right)^2}}$
:::

![截圖 2025-01-09 凌晨1.49.31](https://hackmd.io/_uploads/BJB0qV2IJg.png)

### 9.8.2 The CS Amplifier

![截圖 2025-01-09 凌晨1.51.08](https://hackmd.io/_uploads/S177oE3Lyg.png)

**SS circuit**
![截圖 2025-01-09 凌晨1.54.09](https://hackmd.io/_uploads/S1b13E38kl.png)
($R_G=R_{G1}\|R_{G2}$, $Z_S=R_S\|C_S$)

$\frac{V_o}{V_{sig}}=\frac{V_g}{V_{sig}}\times\frac{I_d}{V_g}\times\frac{V_o}{I_d}=(1)\times(2)\times(3)$

$(1)$
$V_g=V_{sig}\frac{R_G}{R_G+\frac{1}{sC_{C1}}+R_{sig}}\Rightarrow\frac{V_g}{V_{sig}}=\frac{R_G}{R_G+R_{sig}}\frac{s}{s+\frac{1}{C_{C1}(R_G+R_{sig})}}$
$\Rightarrow \omega_{P1}=1/C_{C1}(R_G+R_{sig})$, (there's also a zero $s=0$)

$(2)$
$I_d=I_s=\frac{V_g}{\frac{1}{g_m}+Z_S}=g_mV_g\frac{Y_S}{g_m+Y_S}$, $Y_S=\frac{1}{Z_S}=\frac{1}{R_S}+sC_S$
$\Rightarrow \frac{I_d}{V_g}=g_m\frac{s+\frac{1}{C_SR_S}}{s+\frac{g_m+1/R_S}{C_S}}$
$\omega_{P2}=\frac{g_m+1/R_S}{C_S}$, $\omega_Z=\frac{1}{C_SR_S}$ (Note that $\omega_{P2}\gg\omega_Z$)
![截圖 2025-01-09 凌晨2.12.06](https://hackmd.io/_uploads/BJMMer3IJe.png)


$(3)$
$I_o=-I_d\frac{R_D}{R_D+\frac{1}{sC_{C2}}+R_L}\Rightarrow \frac{V_o}{I_d}=-\frac{R_DR_L}{R_D+R_L}\frac{s}{s+\frac{1}{C_{C2}(R_D+R_L)}}$
$\Rightarrow \omega_{P3}=\frac{1}{C_{C2}(R_D+R_L)}$, (there's also a zero $s=0$)

Finally,
$\frac{V_o}{V_{sig}}=-\frac{R_G}{R_G+R_{sig}}g_m(R_D\|R_L)\left(\frac{s}{s+\omega_{P1}}\right)\left(\frac{s+\omega_{Z}}{s+\omega_{P2}}\right)\left(\frac{s}{s+\omega_{P3}}\right)$
$\Rightarrow A_M=-\frac{R_G}{R_G+R_{sig}}g_m(R_D\|R_L)$

#### Determining the 3-dB Frequency $f_L$

![截圖 2025-01-09 凌晨2.22.29](https://hackmd.io/_uploads/S1KuzH2IJx.png)


If the highest-frequency pole (in this case, $f_{P2}$) is higher than the nearest pole/zero (in this case, $f_{P3}$) by at least a factor of 4, then we say that a **dominant pole** exists, and $f_L\simeq f_{P2}$. Otherwise, we can use approximation $f_L\simeq\sqrt{f_{P1}^2+f_{P2}^2+f_{P3}^2-2f_Z^2}$

#### Determining the Pole and Zero Frequencies by Inspection

**Zeros**: Find frequency such that $V_o=0$. In this circuit, it is the case that the signal is DC or $Z_S$ happens to be $\infty$.

**Poles**: Set $V_{sig}=0$ and determine the time constant of 3 separated circuit:

![截圖 2025-01-09 凌晨3.06.27](https://hackmd.io/_uploads/HJAphB28kx.png)

:::info
++Note++. This method is feasible is due to the fact that 3 capacitors in the circuit do not interact and each one is responsible for exactly one time constant.
:::

#### Selecting Values for the Coupling and Bypass Capacitors

Select $C_S$ to provide the highest frequency pole, then decide $C_{C1}$ and $C_{C2}$ to make other pole frequency be much lower.

### 9.8.3 The Method of Short-Circuit Time Constants

To determine $f_L$, we can (1)**zero the input signal**, (2)**consider each capacitance $C_i$ one at a time (short-circuit for other capacitors)**, (3)**find $R_i$ by either inspection or by replacing $C_i$ with $V_x$ and calculate $R_i=V_x/I_x$**, then we have
$f_L=\frac{1}{2\pi}\sum_{i=1}^n\frac{1}{C_iR_i}$

## Summary

- There are 4 internal capacitance in MOSFET: $C_{gs}$, $C_{gd}$ and $C_{sb}$, $C_{db}$. When modeling the behavior at high frequency, at least $C_{gs}$, $C_{gd}$ should be considered.
- The MOSFET unity-gain frequency (i.e., the frequency s.t. $|I_o/I_i|=1$) is $f_T=\frac{g_m}{2\pi(C_{gs}+C_{gd})}$, which is a figure of merit for the high-frequency operation of the transistor.
- The gain-bandwidth product $GB=A_Mf_H$ is a figure of merit for the amplifier. For amplifier with dominant pole $f_H$, the unity-gain frequency is $f_t=GB$.
- For CS amplifiers, 3-dB frequency $f_H=\frac{1}{2\pi C_{in}R_{sig}}$ where $C_{in}=C_{gs}+C_{gd}(1+g_mR_L')$ is hugely amplified by the **Miller effect**.
    - However, when $R_{sig}$ is small, the pole arised from the input circuit isn't dominant anymore, the 3-dB frequency is mostly affected by the load capacitance. Thus, $f_H=\frac{1}{2\pi(C_L+C_{gd})R_L'}$ and $f_t=\frac{g_m}{2\pi(C_L+C_{gd})}$
- For CG amplifiers, $\tau_H=C_{gs}\left(R_{sig}\|\frac{1}{g_m}\right)+(C_L+C_{gd})R_L$ isn't affected by the Miller effect, hence can provide design with higher bandwidth.
- For cascode amplifiers, $\tau_H=R_{sig}[C_{gs1}+C_{gd1}(1+g_{m1}R_{d1})]+R_{d1}(C_{gd1}+C_{db1}+C_{gs2})+(R_L\|R_o)(C_L+C_{gd2})$
    - When $R_{sig}$ is small, $\tau_H\simeq(C_L+C_{gd2})(R_L\|R_o)$. Comparing to the CS, the 3-dB frequency is divided by a factor of $A_0=g_mr_o$, while the midband-gain is multiplied by $A_0$ (tradeoff between bandwidth and gain).
![截圖 2025-01-08 晚上8.10.23](https://hackmd.io/_uploads/rkSBaIhLJe.png)
- For source followers, $\frac{V_o}{V_{sig}}=A_M\frac{1+\frac{s}{\omega_Z}}{1+b_1s+b_2s^2}$, where
$A_M=\frac{R_L'}{R_L'+1/g_m}=\frac{g_mR_L'}{g_mR_L'+1}$
$\omega_Z=\frac{g_m}{C_{gs}}$
$b_1=\left(C_{gd}+\frac{C_{gs}}{g_mR_L'+1}\right)R_{sig}+\left(\frac{C_{gs}+C_L}{g_mR_L'+1}\right)R_L'$
$b_2=\frac{(C_{gs}+C_{gd})C_L+C_{gs}C_{gd}}{g_mR_L'+1}R_{sig}R_L'$
which exhibits wide bandwidths.
- For diffential pairs, theres a zero transmission frequency $f_Z=\frac{1}{2\pi C_{SS}R_{SS}}$, causing the $\text{CMRR}$ falls at the relatively low frequency.
    - If the DP is current-mirror-loaded, the transfer function is then $\frac{V_o}{V_{id}}=(g_mR_o)\left(\frac{1+s\frac{C_m}{2g_{m3}}}{1+s\frac{C_m}{g_{m3}}}\right)\left(\frac{1}{1+sC_LR_o}\right)$, where the short-circuit current gain $G_m=g_m\frac{1+s\frac{C_m}{2g_{m3}}}{1+s\frac{C_m}{g_{m3}}}$ is reduced to half of $g_m$ at high frequency, and the "$\left(\frac{1}{1+sC_LR_o}\right)$" is due to the load capacitance.
- For CS amplifiers with a source degeneration resistor $R_S$, the 3-dB frequency is $f_H\simeq\frac{1}{2\pi R_{sig}C_{gd}|A_M|}$. As $R_S$ increases, $|A_M|$ decreases and the bandwidth is increased.
- For discrete-circuit CS amplifiers at low frequency, $\frac{V_o}{V_{sig}}=A_M\left(\frac{s}{s+\omega_{P1}}\right)\left(\frac{s+\omega_{Z}}{s+\omega_{P2}}\right)\left(\frac{s}{s+\omega_{P3}}\right)$ where
$A_M=-\frac{R_G}{R_G+R_{sig}}g_m(R_D\|R_L)$
$\omega_{P1}=\frac{1}{C_{C1}(R_G+R_{sig})}$
$\omega_{P2}=\frac{g_m+1/R_S}{C_S}$
$\omega_{P3}=\frac{1}{C_{C2}(R_D+R_L)}$
$\omega_Z=\frac{1}{C_SR_S}$
If we can the determine the dominant pole, then $f_L$ is found.

## Practice

- 9.1, 9.2
- 9.13, 9.16, 9.17, 9.20, 9.21
- 9.38, 9.39, 9.41
- 9.50, 9.51, 9.53
- 9.64, 9.67
- 9.70, 9.71, 9.72
- 9.79, 9.80, 9.82
- 9.93, 9.94