# 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