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Chapter Basic MOS Device Physics 0/4/3 Ch Device

MOS Device Structure 0/4/3 Ch Device

NMOS and PMOS with Well 0/4/3 Ch Device 3

MOS Symbols 0/4/3 Ch Device 4

MOS Channel Formation 0/4/3 Ch Device 5

I/ Characteristics 0/4/3 Ch Device 6

I/ Characteristics I = Qd v Qd = WCox(GS TH) Qd(x) = WCox(GS (x) TH) 0/4/3 Ch Device 7

L I/ Characteristics (cont.) ID = WCox[GS (x) TH]v IDdx = WCoxμn[GS (x) TH]d x =0 Given v = μe and E(x) = ID = WCox[GS (x) TH]μn DS = 0 d (x) dx d(x) dx ID = μncox W L [(GS TH)DS DS ] 0/4/3 Ch Device 8

I/ Characteristics (cont.) ID = μncox W L [(GS TH)DS DS ] 0/4/3 Ch Device 9

Operation in Triode eion ID = μncox W L [(GS TH)DS DS ] ID = μncox W L (GS TH)DS, DS << (GS TH) ON = μncox W L (GS TH) 0/4/3 Ch Device 0

Operation in Active (Saturation) eion ID = μncox W L [(GS TH)DS DS ] ' DS = GS TH (Pinch off ) ID = μncox W L (GS TH) 0/4/3 Ch Device

Active eion (cont.) Active eion 0/4/3 Ch Device

Transconductance, m m = ID GS DS cons tan t m = μncox W L ID = μncox W L (GS TH) = ID GS TH 0/4/3 Ch Device 3

Triode and Active eion Transition Active Active 0/4/3 Ch Device 4

Threshold oltae and Body Effect (cont.) TH = TH0 +γ ( ΦF + SB ) ΦF, γ = qεsinsub Cox No Body Effect With Body Effect 0/4/3 Ch Device 5

Channel Lenth Modulation L L ID = μncox W L (GS TH) ( + λds) 0/4/3 Ch Device 6

Channel Lenth Modulation (cont.) ID = μncox m = μncox W L (GS TH) ( + λds) W L (GS TH)( + λds) m = ID GS TH, (unchaned) 0/4/3 Ch Device 7

Subthreshold Conduction GS ID = I 0 exp ζ kt q DS fixed 0/4/3 Ch Device 8

MOS Layout 0/4/3 Ch Device 9

Device Capacitances 0/4/3 Ch Device 0

Layout for Low Capacitance 0/4/3 Ch Device

G-S and G-D Capacitance 0/4/3 Ch Device

MOS Small Sinal Models ro = DS ID = ID / DS = μncox W = L (GS TH) λ λid 0/4/3 Ch Device 3

Bulk Transconductance, mb mb = ID BS = μncox W L (GS TH) TH BS Also, TH BS = TH SB = γ (ΦF + SB) / mb = m γ ΦF +SB = ηm 0/4/3 Ch Device 4

Gate esistance 0/4/3 Ch Device 5

MOS Small Sinal Model with Capacitance 0/4/3 Ch Device 6

C- of NMOS 0/4/3 Ch Device 7

第三章單級放大器 0/4/3 Ch3 Sinle_stae_amp

簡目 3. 基本觀念 3. 共源極組態 3.. 負載電阻之共源極組態 3.. 負載二極體之共源極組態 3..3 負載電流源之共源極組態 3..4 負載三極管之共源極組態 3..5 源極退化之共源極組態 3.3 源極隨耦器 3.4 共閘極組態 3.5 疊接組態 3.5. 摺疊疊接組態 3.6 元件模型的選擇 0/4/3 Ch3 Sinle_stae_amp

放大器之輸入 - 輸出特性一個放大器之輸入 - 輸出特性通常為一非線函數 n y( t) α + α x( t) + α x ( t) + + x ( ) x x x 0 α n t x 的範圍夠小時 y t) α + α x( ) ( 0 t 0/4/3 Ch3 Sinle_stae_amp 3

類比設計八邊形 0/4/3 Ch3 Sinle_stae_amp 4

負載電阻之共源極組態 Line: in-th=out out = DD out <<( in - TH ) out W = DD D μncox ( in TH ) L W out = in TH = DD D μncox ( in TH ) L out = DD D μnc X and Y axis not same scale W [( ) ox in TH out out out 0/4/3 Ch3 Sinle_stae_amp 5 = DD L on on + D DD = W + μncox D ( in TH L ] )

負載電阻之共源極組態 Line: in-th=out X and Y axis not same scale 0/4/3 Ch3 Sinle_stae_amp 6

繪出圖 3.3(a) 中 M 之汲極電流和轉導對於輸入電壓之關係圖解 : 例題 3. m =μ n C ox (W/L) DS m =μ n C ox (W/L)( GS - TH ) W out = in TH = DD D μncox ( in TH ) L 0/4/3 Ch3 Sinle_stae_amp 7

電壓增益? out = DD D μnc ox W L ( in TH ) A v W out = = DμnCox ( in TH ) in L = m D A v = W D μncox I D = L I D μ C n ox W L D I D D fixed for next stae bias 0/4/3 Ch3 Sinle_stae_amp 8

包含輸出電阻之共源極小信號模型 r A O D v = m r O + D 0/4/3 Ch3 Sinle_stae_amp 9

例題 3. 假設圖 3.6 之 M 操作於飽和區, 計算電路之小信號電壓增益解 : 因為 I 之阻抗為無限大, 增益將被 M 之輸出電阻所限制 :A v =- m r O 0/4/3 Ch3 Sinle_stae_amp 0

負載二極體二極體之阻抗? (/ m ) r O / m = X I X = X /r O + m X 0/4/3 Ch3 Sinle_stae_amp

考慮基板效應之負載二極體 =- X bs=- X X ( + ) + = I r m mb X X O I X X =? = = r = I r X O X m + mb + O m + mb m + mb 0/4/3 Ch3 Sinle_stae_amp

0/4/3 Ch3 Sinle_stae_amp 3 負載二極體之共源極組態 +η = + = m m mb m m v A η η μ μ + = + = ) / ( ) / ( ) / ( ) / ( L W L W I L W C I L W C A D ox n D ox n v

負載二極體元件之電路 Cut off Saturation No need to pass (0,0) 0/4/3 Ch3 Sinle_stae_amp 4

0/4/3 Ch3 Sinle_stae_amp 5 PMOS 負載二極體元件 ) / ( TH GS ox m L W C = μ ) / ( ) ( ) / ( TH GS ox p TH GS ox n m m v L W C L W C A = = μ μ

在圖 3.3 電路中,M 偏壓於飽和區, 且其汲極電流為 I, 當電流源 I S = 0.75I 加至電路中,What is Av? 解 : = μ C ( W / L) I = μ C ( W / L) ( ) m p ox D p ox GS THP Now, I D,new =I /4 A vnew, 例題 3.3 When I D =I μ ( W / L) I μ ( W / L) Av = = μ ( W / L) I μ ( W / L) If fixin (W/L), If fixin ( GS - THP ), I I A D, new m n n m p D p m μn( W / L) I μn( W / L) I = = m μ p( W / L) I I D, new μ p ( W / L) 4, new D, new p ox, new GS THP = = = ( W / L) 4 ID μ pcox( W / L) ( GS THP) 4 m μn( W / L) I μn( W / L) I, = = m, new μ p ( W / L) I = 4 μn( W / L) 4 μn( W / L) 4A μ ( / ) p( W / L) = μ, p W L = new, new ID, new μ p( W / L), new 4 0/4/3 Ch3 Sinle_stae_amp 6 vnew μ pcox( W / L), new( GS THP) I = = 4 μ C ( W / L) ( ) I μ C ( W / L) ( ) p ox GS THP D = = I ( W / L) 4 4 μ ( W / L) μ ( W / L) n = = p A v v

負載電流源之共源極組態 A v = - m (r O r O ) 0/4/3 Ch3 Sinle_stae_amp 7

負載三極管之共源極組態 on = μ C p ox ( W / L) ( DD b THP ) 0/4/3 Ch3 Sinle_stae_amp 8

源極退化之共源極組態線性特性 Fiure 3.6 in 增加 I D 及 s 之跨壓亦會增加 0/4/3 Ch3 Sinle_stae_amp 9

G m = 源極退化共源極組態的小信號增益 Av = out / in = ( ID / in) D md = GmD = + ID f GS ( in s ID ) f I D f = = = = ( G ) in GS in in GS in GS G A = / =? m v out in out = DD - I D D = + m m S 假設 I D =f( GS ) S s m m 0/4/3 Ch3 Sinle_stae_amp 0 m S Without r o here! ID = μncox W L (GS TH) ( + λds)

源極退化共源極組態之小信號模型推導 G m G Output short ckt current = I = I out = = m m ( r in mb I out X S I ) + out m O m m m in S + [ + ( m + mb) S] ro S + [ + ( ) ] ms m + mb + S ro 0/4/3 Ch3 Sinle_stae_amp = r out O S mb ( I out S ) I r out O S

共源極元件之汲極電流和轉導值無 / 有退化 m m = ID GS DS cons tant Iout Gm = = W in = μncox ( GS TH) Small sinal L G = + S m m S mr + [ + ( + m O mb ) S ] r O 0/4/3 Ch3 Sinle_stae_amp

源極路徑上所視之電阻 A v = + m m D S = m + D S 0/4/3 Ch3 Sinle_stae_amp 3

繪出圖 3.6 之電路小信號電壓增益 Av 和輸入偏壓電壓 i 之關係圖解 : 例題 3.4 Fiure 3.6 A v = D + m S When s=0 0/4/3 Ch3 Sinle_stae_amp 4

例題 3.5 假設 λ=γ= 0, 計算圖 3.(a) 電路之小信號增益解 : A v = m + D m 0/4/3 Ch3 Sinle_stae_amp 5

0/4/3 Ch3 Sinle_stae_amp 6 共源極組態之輸出電阻 X S X X S mb m X O I I I r = + + + ] ) ( [ 流經 s 之電流為 I X, = - I X s O S mb m O S O mb m O S O mb m S O S mb m out r r r r r r ] ) ( [ ) ( ] ) ( [ ] ) ( [ + + = + + + + + = + + + = X out X I = 流經 r O 之電流為 I X - ( m + mb ) =I X + ( m + mb ) s I X

對應外加汲極電壓之汲極電流變化輸出電阻另一想法加一電壓變量 Δ S m + mb 於輸出節點 Δ =Δ S S + ro m + mb Δ Δ I = S =Δ [ + ( + ) ] r + S m mb S O S Δ = = [ + ( + ) ] r + Exact Solu out m mb S O S 0/4/3 Ch3 ΔISinle_stae_amp 7

0/4/3 Ch3 Sinle_stae_amp 8 + + = + = D S out mb D S out in m D out bs mb m D out ro I ) ( D S out O D S out mb D S out in m O D out S D out O ro out r r r I + + = = 有限輸出阻抗之電壓增益

r o not infinite 有限輸出阻抗之電壓增益 out in = D + S + r O mro D + ( + m mb ) S r O A v mrod[ S+ ro+ ( m+ mb) SrO] = + + r + ( + ) r + r + ( + ) r D S O m mb S O S O m mb S O mro D[ S+ ro+ ( m+ mb) SrO] = S + ro + ( m + mb) SrO D + S + ro + ( m + mb) SrO = Gm( D // out) Iout mro Gm = = in S + [ + ( m + mb) S ] ro Δ = = [ + ( + ) ] r + ΔI out m mb S O S If r o lare, 0/4/3 Ch3 Sinle_stae_amp 9 A v = G m D md = + m S

例題 3.6 計算圖 3.6 電路之電壓增益, 假設 I 0 為一理想電流源解 : A v = r m O + + r + ( + ) r D S O m mb S O D => mro 0/4/3 Ch3 Sinle_stae_amp 30

第三章單級放大器 0/4/3 Ch3 Sinle_stae_amp

源極隨耦器 source follower in,out>>s Gain? in < TH,M 關閉且 out = 0 in > TH 時, W μ nc ox ( in TH out) S = L out 0/4/3 Ch3 Sinle_stae_amp 3

源極隨耦器之增益 W out Av = μ nc ox ( in TH out) S = out in L W TH out μ out ncox ( in TH out) S = L in in TH TH γ out / TH in = = η out = ( Φ F+ SB) = η = in out SB in W μ ( ) ncox in TH out S out = L W in + μncox ( in TH out) S ( + η) L W m = μncox ( in TH out) L S m S Av = + ( m + mb) S + ( + η) S 0/4/3 Ch3 Sinle_stae_ampm 4 = in TH out

源極隨耦器小信號等效電路 A v = m S + ( + η) S 電壓增益與輸入電壓之關係圖 0/4/3 Ch3 Sinle_stae_amp 5

電流源取代電阻之源極隨耦器使用 NMOS 電晶體作為電流源之源極隨耦器 0/4/3 Ch3 Sinle_stae_amp 6

例題 3.7 假設在圖 3.30(a) 之源極隨耦器中,(W/L) = 0/0.5,I = 00μA, TH0 = 0.6,Φ F = 0.7,μ n C ox = 50μA/ 且 γ= 0.4 (a) 計算 in =. 時的 out (b) 如果 I 如圖 3.30(b) 之 M 來實現, 找出使得 M 維持在飽和區之最小 (W/L) 值解 : ( in TH out) = I D μncox( W / L (a) out 0.9 ) TH = TH 0 + γ ϕf + SB ϕf = 0.635 (b) ( GS - TH ) 0.9 ( ) = I D GS TH μncox( W / L) 0/4/3 Ch3 Sinle_stae_amp 7

源極隨耦器之輸出阻抗 = X I X m out = X m mb + X mb = 0 0/4/3 Ch3 Sinle_stae_amp 8

考慮基板效應之源極隨耦器 out = m mb = m + mb 0/4/3 Ch3 Sinle_stae_amp 9

戴維尼等效電路表示法 A v = m + mb mb = When / mb open from the rest, current is zero, so =0 and out=in m + m mb 0/4/3 Ch3 Sinle_stae_amp 0

考慮有效通道長度調變效應 A v = mb mb r O r O r O r O L + L m 0/4/3 Ch3 Sinle_stae_amp

例題 3.8 計算圖 3.35 電路之電壓增益解 : M 源極所視之阻抗為 [/( m + mb )] r O 因此, A v = m m + + mb mb r O r O r O r O + mb mb m 0/4/3 Ch3 Sinle_stae_amp

無基板效應之源極隨耦器基板效應所造成的非線性可被消除, 通常只可在 PMOSs 中實現, 但會產生比 NMOS 更高之輸出阻抗 Not ood since the ood thin of source follower is low output resistance!! 0/4/3 Ch3 Sinle_stae_amp 3

共源極組態和源極隨耦器之疊加組態無源極隨耦器時, X 之最小允許值為 GS - TH 考慮源極隨耦器時, X 必須大於 GS +( GS3 - TH3 ) 0/4/3 Ch3 Sinle_stae_amp 4

負載電阻 A v = ro ro L mb ro ro L + mb m (a) (b) out in out in SF CS L + m L / m L 0/4/3 Ch3 Sinle_stae_amp 5

例題 3.9 (a)c 做為一交流短路, 計算其電壓增益使 M 維持飽和狀態之最大直流輸入信號為何? (b) 為使得一輸入直流位準趨近於 DD, 電路修正為圖 3.39(b), 為確保 M 維持飽和狀態,M,M, M 3 之閘極 - 源極電壓間的關係為何? Level Shiftin 解 : A v =- m [r o r o (/ m )] (a) in 之最大允許直流位準為 DD - GS + TH (b) in = DD X = DD - GS3 DD - GS3 DD - GS + TH 0/4/3 Ch3 Sinle_stae_amp 6

共閘極組態 in > b - TH Cut off in decreases M saturation in decreases out decreases M triode 0/4/3 Ch3 Sinle_stae_amp 7

0/4/3 Ch3 Sinle_stae_amp 8 共閘極組態之輸出 - 輸入特性 TH b D TH in b ox n DD L W C = ) ( μ D TH in b ox n DD out L W C ) ( = μ D in TH TH in b ox n in out L W C = ) ( μ D m TH in b D ox n in out L W C ) ( ) )( ( η η μ + = + = =η = SB TH in TH Enterin triode point At saturation Gain?

例題 3.0 在圖 3.4 中, 電晶體 M 感測到 Δ 並傳送一等比例之電流至一 50Ω 傳輸線傳輸線之另一端連接一 50Ω 電阻如圖 3.4(a), 與一共閘極組態如圖 3.4(b) 假設 λ=γ= 0 (a) 計算二種情況在低頻時的 out / in (b) 將節點 X 之波反射最小化的條件為何? D = 50Ω A v =- m D /( m + mb )= 50Ω D can be lare 0/4/3 Ch3 Sinle_stae_amp 9

共閘極組態之電壓增益 out out out S + in = 0 r O m mb S + in = D D D out S out r O ( m + mb) out in S + in = out D D D ( + ) r + = r r out m mb O D D mb = 0, s 0, ro > ( ) in O + m + mb O S + S + D S + 0/4/3 Ch3 Sinle_stae_amp m 0 out

例題 3. 計算圖 3.44(a) 電路之電壓增益, 如果 λ 0 且 γ 0 r O mb in, eq = ro + mb m Source follower equation in out ( m + mb) ro + = r + ( + ) r + + in O m mb O S S D D eq = ro mb m r r O out ( m + mb) O + mb = D in r [ ( ) ] O O + + m + r mb ro ro + + D mb m mb m ( m) ro = D = m D [( ) r ] Application?? in bias can be very low. m O 0/4/3 Ch3 Sinle_stae_amp m

0/4/3 Ch3 Sinle_stae_amp 共閘極組態之輸入與輸出阻抗 X X mb m X O X D I r I = + + ] ) ( [ ( ) ( ) D O X D in X m mb O m mb O m mb r I r r + = = + + + + + D = 0 時, mb m O O mb m O X X r r r I + + = + + = ) (

負載理想電流源之共閘極組態 in D + ro D = + + ( + ) r ( + ) r + m mb O m mb O m mb Common source with source deeneration = {[ + ( + ) r ] + r } out m mb O S O D 0/4/3 Ch3 Sinle_stae_amp 3

計算負載一電流源之共閘極態電壓增益 [ 圖 3.47(a)] 解 : out in A v = r O ( + ( + m m + mb ) r mb = m mb O ( + ) r + O ) r 例題 3. O S + + S + D D D s no voltae drop 0/4/3 Ch3 Sinle_stae_amp 4

0/4/3 Ch3 Sinle_stae_amp 5 如範例 3.0 所見, 共閘極組態之輸入信號可能為一電流而非電壓信號, 如圖 3.49 所示計算 out /I in 和電路之輸出阻抗, 如果輸入電流源顯示其輸出阻抗值等於 P 解 : P D D P P O mb m O O mb m in out r r r I + + + + + + = ) ( ) ( D O P O mb m out r r } ] ) ( {[ + + + = 例題 3.3 D D S S O mb m O O mb m in out r r r + + + + + + = ) ( ) ( 戴維尼等效電路

疊接組態將一共源極組態和一共閘極組態疊加即為一疊接組態 0/4/3 Ch3 Sinle_stae_amp 6

疊接組態之偏壓條件 0/4/3 Ch3 Sinle_stae_amp 7

疊接組態之輸入 - 輸出特性 x < in TH The circle is a not sure area!? out < b TH in < TH 時,M 和 M 關閉且 out = DD in 增加時, out 下降 in 足夠大時,() X 比 in 少 TH, 強迫 M 進入三極管區 () out 比 b 少 TH, 驅動 M 進入三極管區 0/4/3 Ch3 Sinle_stae_amp 8

疊接組態之小信號等效電路 0/4/3 Ch3 Sinle_stae_amp 9

例題 3.4 計算圖 3.54 之電路的電壓增益, 如果 λ= 0 答 : I = D m in A v = P + ( + ) m mb m P D P ( m + mb) 0/4/3 Ch3 Sinle_stae_amp 30 + P

疊接組態之輸出阻抗 = + out [ + ( m + mb) ro ] ro ro 0/4/3 Ch3 Sinle_stae_amp 3

疊接組態的延伸三疊接組態疊接可延伸至三個或更多堆疊元件以達到更高的輸出阻抗, 但需要多餘的電壓頭部空間 0/4/3 Ch3 Sinle_stae_amp 3

負載電流源之疊接組態如果 M 和 M 都操作於飽和區, 則 G m 近似於 m 且 + v out A = + ( m mb) ro ro ( m mb) ro mro 0/4/3 Ch3 Sinle_stae_amp 33

例題 3.5 計算圖 3.57 中電路之正確電壓增益值, 如果 consider λ out ( m + mb) ro + = r + ( + ) r + + in O m mb O S S D m o in O m mb O O O D out ( m + mb) ro + = D = ( m + mb) ro + r r + ( + ) r r + r + D A = r [( + ) r + ] v m O m mb O 0/4/3 Ch3 Sinle_stae_amp 34

疊接與增加元件長度的比較 Question: Fixed I D, 相同的輸入 / 輸出 (ovx) 電壓振幅限制 (b) or (c ) better in terms of A v? λ / L W ID = μncox ( in TH) L W W L mro = μncox ID μncox ID L λid L ID 將 L 放大四倍會使得 m r O 加倍, 而疊接使其 Av 約為 ( m r O ) 0/4/3 Ch3 Sinle_stae_amp 35

負載 PMOS 之疊接組態阻抗值為 [ + ( m3 + mb3) ro 3] ro 4 + ro 3 最大輸出移幅為 DD GS TH ) ( GS TH ) 利用輔助定理 Gm m 得到 ( GS3 TH 3 GS4 TH 4 out = {[ + ( m + mb) ro ] ro + ro } {[ + ( m3 + mb3) ro 3] ro 4 + ro 3} A v [( mro ro ) ( m3ro 3r 4)] m out m O 0/4/3 Ch3 Sinle_stae_amp 36

例題 3.6 二個相同的 NMOS 電晶體作為系統中的固定電流源 [ 圖 3.6(a)], 然而, 由於系統的內部電路特性, X 比 Y 高 Δ I I r D D =Δ / o (a) 計算 I D 和 I D 的差異, 如果 λ 0 I I = D D Δ ( + ) r r m3 mb3 O3 O (b) 將疊接元件加至 M 與 M 並重做 (a) 0/4/3 Ch3 Sinle_stae_amp 37

疊接組態屏蔽特性的消失 ( X 減少驅使 M 進入三極管區 ) x drop and make M triode p follows x to drop in order to make I d =I d fixed possible M also enter triode I s fixed I d fixed W [( )( ) ( ) D = μncox b P TH X P X P L ] 0/4/3 Ch3 Sinle_stae_amp 38

摺疊疊接組態 (a) 簡單摺疊疊接組態 ;(b) 適當偏壓之疊接組態 ;(c) 輸入負載一 NMOS 之疊接組態 0/4/3 Ch3 Sinle_stae_amp 39

摺疊疊接組態之大信號特性 M triode M saturation M cut off b makes all these happen!! = I I out DD D D W D = I μ pcox ( DD in TH ) L I D =I I = in DD TH μ pcox( W / L) I D wants larer than I 0/4/3 Ch3 Sinle_stae_amp 40

高電壓增益之摺疊疊接組態負載疊接之摺疊疊接組態 0/4/3 Ch3 Sinle_stae_amp 4

例題 3.7 計算圖 3.65 之摺疊疊接組態之輸出阻抗, 其中 M 3 運作為一電流源答 : = + out [ + ( m + mb) ro ]( ro ro 3) ro 0/4/3 Ch3 Sinle_stae_amp 4

第四章差動放大器 0/4/3 Ch4 Differential amp

簡目 4. 單端運作和差動運作 4. 基本差動對 4.. 定性分析 4.. 定量分析 4.3 共模響應 4.4 負載 MOS 之差動對 4.5 Gilbert 細胞電路 0/4/3 Ch4 Differential amp

單端信號和差動信號單端信號 : 相對於一個定電壓 ( 通常為接地端 ) 量測而得的信號差動信號 : 二個具有相等但方向相反電壓之節點間所量測而得的信號其平均電壓稱為共模電壓 0/4/3 Ch4 Differential amp 3

共模排斥 (a) 耦合所帶來之信號破壞 ;(b) 藉由差動運作以減少耦合效應 0/4/3 Ch4 Differential amp 4

共模排斥供應電壓雜訊對 (a) 單端電路之效應 ;(b) 差動電路之效應 0/4/3 Ch4 Differential amp 5

差動運作藉由差動運作以減少耦合雜訊 0/4/3 Ch4 Differential amp 6

簡單差動電路 (a) 簡單差動電路 ;(b) 對於輸入共模位準之靈敏度 0/4/3 Ch4 Differential amp 7

基本差動對 out vs. in - in? in in vs. out - out? DD DD - D I SS. 輸出端和輸入的共模位準無關. in = in 時, 其小信號增益為最大值 0/4/3 Ch4 Differential amp 8

電路的共模特性 I D I D vs. in,cm? p vs. in,cm? out out vs. in,cm? P = incm + * on3 m 0/4/3 Ch4 Differential amp 9 on3

差動對的輸入 - 輸出振幅限制 + ( ) GS GS 3 TH 3 in, CM ISS + DD D TH ( in, CM ) TH out DD 0/4/3 Ch4 Differential amp 0

畫出差動對的小信號差動增益與輸入共模位準的關係圖解 : 例題 4. M,M 進入三極管區 W A = = μc I L m D ox D D 0/4/3 Ch4 Differential amp

差動對的定量分析 I I in in = W μ C μ C L I D +I D =I SS D D n ox n ox W 4ISS ID ID = μncox ( in in) ( in in) L W μncox y = x a x L 4I Δ W L in - in = GS - GS Two eqs, two unknowns, I D, I D SS m D in ΔI D W μncoxw / L = μnc W ox in L 4I Av = μncox I SS D SS Δ L in μncoxw / L out - out = D ΔI 0/4/3 Ch4 Differential amp When in - in =0 Δ =

汲極電流和總轉導對輸入電壓之變化 in in 0 = ID IS S ID 0 W W μncox μnco x L L Δ = in I μ C n SS ox W L ( GS TH ), = I μ C n SS ox W L W 4ISS Gm ( in in ) = ID ID = μncox ( in in) ( in in) L W μncox L 0/4/3 Ch4 Differential amp 3

畫出差動對的輸入 - 輸出特性圖, 當元件寬度和尾電流變化時答 : 例題 4. W 4ISS ID ID = μncox ( in in) ( in in) L W μncox L W/L I SS Δ = in I μ C n SS ox W L 0/4/3 Ch4 Differential amp 4

差動對小信號特性分析 (from lare sinal) ΔI Δ D in = μnc out - out = D ΔI ox W L 4I SS μ C W n / 4I SS μ C W n ox ox Δ L / L Δ in in When in - in =0 A = ( ) = μ C I W / L v out in in n ox ss D 0/4/3 Ch4 Differential amp 5

差動對小信號分析方法一 : 重疊原理 S = / m = X D in m + m 0/4/3 Ch4 Differential amp 6

差動對小信號分析方法一 : 重疊原理 / m Source follower T = in T =/ m = Y D in m + m 0/4/3 Ch4 Differential amp 7

= 差動對小信號分析方法一 : 重疊原理 X D in = m + m Y D in m + m 當 m = m = m 時 + D ( X Y ) Due to in in = m m ( X Y ) Due to = md in in 由於對稱之故 ( X Y ) Due to = md in in 執行重疊原理 ( X in Y ) in tot = m D 0/4/3 Ch4 Differential amp 8

例題 4.3 在圖 4.7 之電路中,M 為 M 的二倍寬, 如果 in 和 in 之偏壓值相同時, 計算其小信號增益不能用半電路想法答 : M, M ate bias the same GS = GS I D =I D =I SS /3 μ C / L) I 0/4/3 Ch4 Differential amp 9 D ( X Y ) Due to = in = ( W m n ox SS + / 3 = m = μncox(w / L)I SS / 3 m 4 4 D mw ( ) 4 m(w) Av = = m D = D = D + 3 3 3 3 3 m m mw ( ) = μncox( W / L) ISS/ =.m( w) D= 0.8m( w) D = μ C W L I m m w n ox SS m ( ) ( / ) / in

對稱電路的輔助定理 p will not chane? D : I = G * m D : I = G * m I+ I = IT G *( ) + G *( ) = I = G *( +Δ ) + G *( Δ ) m 0 p m 0 p T m 0 in p' m 0 in p' = p p' 戴維尼等效觀點 0/4/3 Ch4 Differential amp 0

差動對小信號分析方法二 : 半電路 ( X Y ) /( in ) = m D 0/4/3 Ch4 Differential amp

計算圖 4.0(a) 之電路差動增益, 如果 λ 0 答 : 例題 4.4 半電路 - m ( D r O ) 0/4/3 Ch4 Differential amp

任意輸入信號的差動和共模成份 0/4/3 Ch4 Differential amp 3

計算圖 4.0(a) 之電路中, 例題 4.5 如果 in in 且 λ 0 時, 計算其 X 和 Y ( in in) X = m( D ro) ( in in ) Y = m( D ro) out independent of in,cm 0/4/3 Ch4 Differential amp 4

共模響應 / m A v, CM = out in, CM = /( D m / ) + SS 0/4/3 Ch4 Differential amp 5

例題圖 4.6 之電路使用一電阻而不用電流源來定義一 ma 之尾電流, 假設 4.6 (W/L), = 5/0.5,μ n C ox = 50μA/, TH = 0.6,λ=γ= 0, DD = 3 (a)( SS ) 維持於 0.5 之輸入共模信號為何? in, CM = GS + 0.5 =.73 (b) 計算差動增益為 5 時的 D 值 = 5 D = 3.6kΩ (c) 如果輸入共模位準比 (a) 中所求得的值高 50m 時,M, M 距 triode 多遠?.73 +50m.4 96.8m 0/4/3 Ch4 Differential amp 6 m D μ ncoxw 0.5mA = ID = ( GS TH) L GS =.3 = μ C ( W / L) I m n ox D = 3.5 m*3.6k = 3.58 =.4.4 (.73 0.6) = 0.9 away from triode D / ΔX, Y = Δin CM = 50m.94 = 96. 8m + /( ), = I * X DD D D SS.4 96.8 m (.73+ 50m 0.6) = 0.43 away from triode m SS = 500Ω

共模 - 差動轉換特性雜訊 Δ in, CM Δ X = Δin, CM D + SS m Δ Y = Δ in, CM ( D +ΔD) + SS m 0/4/3 Ch4 Differential amp 7

電晶體不匹配所造成的非對稱性 I D m in, CM m = Itotal = + [( / / ) + ] + SS m m m m m m I I [( / / ) + ] + in, CM m m X Y = ( D D) D = [ ( )] D SS m m m m = = in, CM m m m m [ ( )] D [ Din, CM] [ + ] m + m ( m + m) SS + SS m + m A CM DM = Δ m D ( + ) + m m SS Δ m = m m 0/4/3 Ch4 Differential amp 8

例題 4.7 兩個差動對如圖疊加起來, 電晶體 M 3 和 M 4 遭遇到 m 之不匹配為 Δ m, 在節點 P 之總寄生電容為 C P, 否則電路則為對稱有多少供應雜訊將會在輸出端以差動成份出現? 假設 λ=γ=0 答 : A CM DM = ΔmD ( + ) + m m SS A CM DM = ΔmD + ( m3 + m4) ω 0/4/3 Ch4 Differential amp 9 C P

共模排斥比 CM = A A 只考慮 m 不匹配時 m SS m DM CM DM ( ) = ( = ) D SS D X Y Due to in in in T ' + // SS + SS ( // SS = T ') + m m m m m D = SS + + + // + SS D in in SS SS m m m ( + ) ( = SS m m m SS m SS m m m D in D + ( m + m) SS + SS m + ( m + m) SS ( SS m m + m ) = D + ( + ) m m SS in + ) in SS SS ( + ) ( ) = SS m m m X Y Due to in D in + ( m + m) SS 0/4/3 Ch4 Differential amp 30

A 共模排斥比只考慮 m 不匹配時 CM DM ΔmD = ( + ) + m m SS CM = A A CM DM ( SS m m + m) in ( SS m m + m ) in D X Y + ( m + m) SS ADM = = m 0/4/3 Ch4 Differential amp m SS Δm 3 DM in in in in SS m m m in SS m m m in D + ( m + m) SS ( + ) ( + ) = + m m SS in in ( SS m m + m) ( SS m m + m ) + ( m + m) SS = D m + m + 4m mss = + ( + ) D CM = in = - in, md Smith = D Δm SS m 其中 m =( m + m )/ + + 4 m m m m SS Δ m (+ )

負載 MOS 之差動對 (a) 負載二極體之差動對 A = ( r r v mn mn mp mp μn( W / L) μ ( W / L) ON N OP ) (b) 負載電流源之差動對 A = r r ) mn ( ON OP p P 0/4/3 Ch4 Differential amp 3 v

加入電流源以增加電壓增益 A = = = v m m3 μncox( W / L) ( GS TH) μ C ( W / L) p ox 3 GS3 TH3 I ( ) I GS 3 TH 3 GS TH 3 = = I ( ) 0.I GS 3 TH 3 GS TH GS 3 TH 3 5 ( ) GS TH If fix out bias, 差動增益大約為沒有負載 PMOS 電流源的五倍 0/4/3 Ch4 Differential amp 33

利用疊接以增加電壓增益 A v [( m3ro 3rO ) ( m5ro 5r 7)] m O 0/4/3 Ch4 Differential amp 34

可變增益放大器 (a) 簡單可變增益放大器 ;(b) 提供可變增益的二種組態 0/4/3 Ch4 Differential amp 35

Gilbert 單元 0/4/3 Ch4 Differential amp 36

例題 4.8 解釋為何 Gilbert 單元可以操作為一類比電壓相乘器解 : cont = cont - cont out = in.f( cont ) 將 f( cont ) 以泰勒展開式展開且只留下其一次項,α cont out =α in cont 0/4/3 Ch4 Differential amp 37

Gilbert 單元 Another way of connection (a)gilbert 單元利用下端之差動對感測輸入電壓 ;(b) 對非常大之正 out 的信號路徑 ;(c) 對非常大之負 out 的信號路徑 0/4/3 Ch4 Differential amp 38

第五章被動與主動電流鏡 0/4/3 Ch5 Current Mirrors

簡目 5. 基本電流鏡 5. 疊接電流鏡 5.3 主動電流鏡 5.3. 大信號分析 5.3. 小信號分析 5.3.3 共模特性 0/4/3 Ch5 Current Mirrors

電流源的應用 0/4/3 Ch5 Current Mirrors 3

電阻分壓電流源假定 M 位於飽和區, I out μnc ox W L + DD TH 0/4/3 Ch5 Current Mirrors 4

電流源的設計利用一參考電流來產生許多不同的電流源 0/4/3 Ch5 Current Mirrors 5

0/4/3 Ch5 Current Mirrors 6 基本電流鏡 (a) 提供一反函數之連接二極體元件 ;(b) 基本電流鏡 ) ( ) ( DS TH GS ox n D L W C I λ μ + = ) ( ) ( DS TH GS ox n D L W C I λ μ + = ) / ( ) / ( DS DS D D L W L W I I λ λ + + = 考慮通道長度調變效應

偏壓 - 差動放大器之電流源 0/4/3 Ch5 Current Mirrors 7

例題 5. 計算圖 5.8 中電路之小信號電壓增益解 : m L (W/L) 3 /(W/L) 0/4/3 Ch5 Current Mirrors 8

疊接電流鏡 Shieldin M Let Y = X possible O + TH Y =? X N TH 節點 P 之最小允許電壓值為 = = + GS0 GS TH ( GS0 TH ) + ( GS TH ) + TH Smith 0/4/3 Ch5 Current Mirrors 9

例題 5.3 在圖 5.0 中, 繪出 X 和 Y 對於 I EF 之關係圖如果 I EF 需要 0.5 才能做為一電流源時, I EF 其最大值為何? 解 : self study 0/4/3 Ch5 Current Mirrors 0

例題 5.4 在圖 5.(a) 中, 假定所有電晶體都相同, 繪出 I X 和 B 對於 X 之關係圖, 且 X 由一個很大的值開始降低.3 0.5 Tri 0.3 < A TH B A TH M:Tri M3:Sat M3:Tri M:Sat.3 0.5 X < N TH3 T =0.7 O =0.3 0.3 Tri 0/4/3 Ch5 Current Mirrors

低電壓修正疊接組態 () O + TH ( = ) b TH X GS O + TH O Smith ( = ) GS TH A b GS + GS + ( GS TH ) b GS TH O TH + makes minimum o 0/4/3 Ch5 Current Mirrors

()How to create = +? b t O t + O t t + O t GS 7 TH 7 t + O t + O Body effect of t creates t difficult to cancel completely! 0/4/3 Ch5 Current Mirrors 3

()How to create = +? Smith b t O o t + O GSMs THMs W I = μcox ( O) L W = μcox ( O) 4L W5 = μcox ( O 5) L W W =, = 4L = + 5 O 5 O GS 5 t O 0/4/3 Ch5 Current Mirrors 4 t + O t DS!= DS introduces substantial mismatch

主動電流鏡處理信號之電流鏡 I ΔI out out = I in = ΔI in 0/4/3 Ch5 Current Mirrors 5

主動電流鏡差動對 (a) 結合 M 和 M 汲極電流之觀念 ;(b)(a) 中觀念之實現 0/4/3 Ch5 Current Mirrors 6

大信號分析 (a) 負載一主動電流鏡與實際電流源之差動對 ; (b) 大信號之輸入 - 輸出關係圖 0/4/3 Ch5 Current Mirrors 7

例題 5.5 假設完美的對稱性, 當 DD 由 0 變至 3 時, 繪出圖 5.(a) 中電路之輸出電壓, 假設 DD = 3 且所有元件都操作於飽和區對稱性 out = F 答 : 0/4/3 Ch5 Current Mirrors 8

非對稱特性 Loadin is not symmetric!! A? 主動電流鏡差動對中的非對稱電壓振幅 A = G = r r ) Smith v m out m, ( O O4 0/4/3 Ch5 Current Mirrors 9

共模特性 A CM? A CM + m, m, SS m3,4 A cm Smith m3 SS 假設 / m3,4 <<r O3,4, 且忽略 r O, / 之影響 0/4/3 Ch5 Current Mirrors 0

CM? 共模特性 CM Ad r r A = cm ( ) [ ] m o o4 m3 SS Smith 假設 / m3,4 <<r O3,4, 且忽略 r O, / 之影響 CM = = A A DM CM m, = ( + ( r O, m, r O3,4 SS ) ) m3,4 m3,4 ( r ( + O, m, r O3,4 m, ) SS ) 0/4/3 Ch5 Current Mirrors

不匹配之差動對 A CM? A CM ( ) r / m m O3 m m3 + ( + ) m m SS Oriinal term A CM Smith m3 SS 0/4/3 Ch5 Current Mirrors