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Download Set 6A: Frequency Response (Part A)
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6. Frequency Response Reading: Sedra & Smith: Chapter 1.6, Chapter 3.6 and Chapter 9 (MOS portions), ECE 102, Winter 2011, F. Najmabadi Typical Frequency response of an Amplifier Up to now we have ignored the capacitors. To include the capacitors, we need to solve the circuit in the frequency domain (or use Phasors). o Lower cut-off frequency: fL o Upper cut-off frequency: fH o Band-width: B = fH − fL Classification of amplifiers based on the frequency response AC amplifier (capacitively-coupled) DC amplifier (directly-coupled) fL = 0 Tuned or Band-pass amplifier (High Q) How to find which capacitors contribute to the lower cut-off frequency Consider each capacitor individually. Let f = 0 (capacitor is open circuit): o If vo (or AM) does not change, capacitor does NOT contribute to fL o If vo (or AM) → 0 or reduced substantially, capacitor contributes to fL Example: Cc1 vi = 0 → v o = 0 Contributes to fL CL No change in vo Does NOT contribute to fL How to find which capacitors contribute to the higher cut-off frequency Consider each capacitor individually. Let f → ∞ (capacitor is short circuit): o If vo (or AM) does not change, capacitor does NOT contribute to fH o If vo (or AM) → 0 or reduced substantially, capacitor contributes to fH Example: Cc1 No change in vo Does NOT contribute to fH CL vo = 0 Contributes to fH How to find “mid-frequency” circuit All capacitors that contribute to low-frequency response should be short circuit. All capacitors that contribute to high-frequency response should be open circuit. Example: Cc1 contributes to fL → short circuit CL contributes to fH → open circuit Low-Frequency Response Low-frequency response of a CS amplifier Each capacitors gives a pole. All poles contribute to fL (exact value of fL from simulation) If one pole is at least two octave (factor of 4) higher than others (e.g., fp2 in the above figure), fL is approximately equal to that pole (e.g., fL = fp2 in above) A good approximation for design & hand calculations: fL = fp1 + fp2 + fp3 + … Low-frequency response of a CS amplifier All capacitors contribute to fL (vo is reduced when f → 0 or caps open circuit) Cc1 open: vi = 0 → v o = 0 Cc2 open: vo = 0 Cs open: Gain is reduced substantially (from CS amp. To CS amp. With RS) See S&S pp689-692 for detailed calculations (S&S assumes ro → ∞ and RS → ∞ ) Vo s s s = AM x x x Vsig s + ω p1 s + ω p 2 s + ω p 3 AM = − RG g m (ro || RD || RL ) RG + Rsig ω p1 = 1 1 , ω p3 = Cc1 ( RG + Rsig ) Cc 2 ( RD || ro + RL ) ω p2 ≈ 1 , Cs [ RS || (1 / g m + RD || RL / ro g m )] Finding poles by inspection 1. Set vsig = 0 2. Consider each capacitor separately (assume others are short circuit!), e.g., Cn 3. Find the total resistance seen between the terminals of the capacitor, e.g., Rn (treat ground as a regular “node”). 4. The pole associated with that capacitor is f pn = 1 2πRn Cn 5. Lower-cut-off frequency can be found from fL = fp1 + fp2 + fp3 + … * Although we are calculating frequency response in frequency domain, we will use time-domain notation instead of phasor form (i.e., vsig instead of Vsig ) to avoid confusion with the bias values. Example: Low-frequency response of a CS amplifier Examination of circuit shows that ALL capacitors contribute to the low-frequency response. In the following slides with compute poles introduced by each capacitor (compare with the detailed calculations and note that we exactly get the same poles). Then fL = fp1 + fp2 + fp3 Example: Low-frequency response of a CS amplifier f p1 = 1. Consider Cc1 : 1 2π Cc1 ( RG + Rsig ) Terminals of Cc1 ∞ 2. Find resistance between Capacitor terminals Example: Low-frequency response of a CS amplifier f p2 = 1 2π C S [ RS || (1 / g m + RD || RL / ro g m )] 1/ gm + ( RD || RL ) / ro g m 1. Consider CS : 11//ggmm++ (RDD || RLL )/ /rorgo gmm 1/ gm + ( RD || RL ) / ro g m Terminals of CS 2. Find resistance between Capacitor terminals Example: Low-frequency response of a CS amplifier f p3 = 1. Consider Cc2 : 1 2π Cc 2 ( RL + RD || ro ) Terminals of Cc2 2. Find resistance between Capacitor terminals High-Frequency Response Amplifier gain falls off due to the internal capacitive effects of transistors Capacitive Effects in pn Junction Majority Carriers Charge stored is a function of applied voltage. We can define a “small-signal” capacitance, Cj Cj = dQJ dVR VR =VQ In reverse-bias region, analysis show (see S&S pp154-156): C j0 Cj = (1 + VR / V0 ) m V0 : Junction built-in voltage Cj0 : Capacitance at zero reversed-bias voltage. m : grading coefficient (1/2 to 1/3). For forward-bias region: Cj ≈ 2Cj0 Capacitive Effects in pn Junction Minority Carriers Excess minority carriers are stored in p and n sides of the junction. The charge depends on the minority carrier “life-time” (i.e., how long it would take for them to diffuse through the junction and recombine. Gives Diffusion Capacitance, Cd Cd is proportional to current (Cd = 0 for reverse-bias) Cd = τT ⋅ ID VT Small Signal Model for a diode Reverse Bias Cj + Cd C j ≈ 2 ⋅ C j0 Cd = τT ⋅ ID VT Forward Bias Cj = C j0 (1 + VR / V0 ) m Cd = 0 rD Junction capacitances are small and are given in femto-Farad (fF) 1 fF = 10-12 F Capacitive Effects in MOS 1. Capacitance between Gate and channel (Parallel-plate capacitor) 3. Junction capacitance between Source and Body (Reverse-bias junction) 2. Capacitance between Gate & Source and Gate & Drain due to the overlap of gate electrode (Parallel-plate capacitor) 4. Junction capacitance between Drain and Body (Reverse-bias junction) Capacitive Effects in MOS “Parallel-Plate” capacitances (depends on the channel shape) Define: C gate = W ⋅ L ⋅ Cox Cov = W ⋅ Lov ⋅ Cox Triode Saturation 1 ⋅ C gate + Cov 2 1 = ⋅ C gate + Cov 2 C gs = C gd C gs = Cut-off 2 ⋅ C gate + Cov 3 C gd = Cov C gb = C gate Pinched Channel No Channel “Junction” capacitances Csb = Csb 0 (1 + VSB / V0 ) m C gs = C gd = Cov Cdb = Cdb 0 (1 + VDB / V0 ) m Small signal for MOS in high-frequencies For source connected to body Saturation C gs = 2 ⋅ C gate + Cov 3 C gd = Cov Cdb = Cdb 0 (1 + VDB / V0 ) m High-frequency response of a CS amplifier 1) MOS “internal” capacitors are shown “outside” of the transistor to see their impact. 2) All MOS capacitors contribute to fH (vo is reduced when f → ∞ or caps short circuit) 3) For f → ∞ , all coupling (Cc1 and Cc2 ) and by-pass capacitors are short circuit Cgd short: Input is connected to output Gain is reduced to 1 For f → ∞ Cgs short: vi = 0 → v o = 0 Cdb short: vo = 0 High-frequency response of a CS amplifier In General: 1) One internal capacitors shorts input to the ground (Cgs here) 2) One internal capacitors shorts output to the ground (Cdb here) 3) One internal capacitors shorts input to output (Cgd here) Cgd appears in parallel to Cdb See S&S pp712-714 for detailed calculations (S&S assumes Cdb → 0) Vo 1 = AM x Vsig 1 + s / ω p1 AM = − RG g m (ro || RD || RL ) RG + Rsig ω p1 = 1 Cin ( RG || Rsig ) , ω p2 = (C gd 1 , + Cdb )(ro || RD || RL ) Cin = C gs + C gd (1 + g m ro || RD || RL ) Cgd appears in parallel to Cgs (with a much larger value) High-frequency-relevant capacitors High-frequency-relevant capacitors appear between o o o input & ground, output & ground, and input & output. Capacitors that are connected between input & output provide feedback. In the case of CS amplifier, we saw that they appeared in the transfer function as capacitors in parallel to input & ground and output & ground capacitors. We can use Miller’s Theorem to replace capacitors connected between input & output and simplify the analysis. Miller’s Theorem Consider an amplifier with a gain A with an impedance Z attached between input and output V1 and V2 “feel” the impedance of Z only through I1 and I2 We can replace Z with any circuit as long as a current I1 flows out of V1 and a current I2 flows out of V2. V2 = A ⋅ V1 I1 = V1 − V2 (1 − A) ⋅ V1 = Z Z I2 = V2 − V1 ( A − 1)V1 ( A − 1)V2 = = Z Z Z⋅A I1 = V1 V = 1, Z /(1 − A) Z1 I2 = V2 V Z = 2 , Z2 = Z ⋅ A /( A − 1) Z 2 1−1/ A Z1 = Z (1 − A) Miller’s Theorem If an impedance Z is attached between input and output an amplifier with a gain A , it can be replaced with two impedances between input & ground and output & ground Other parts of the circuit V2 = A ⋅ V1 V2 = A ⋅ V1 Z Z1 = 1− A Z2 = Z 1− 1 A Example of Miller’s Theorem: Inverting amplifier Solution using Miller’s theorem: Recall from ECE 100, if A0 is large Z 1− A Rf Rf ≈ Rf1 = 1 + A0 A0 Z1 = Rf vo =− vi R1 vo = A0 ⋅ (v p − vn ) = − A0 ⋅ vn Z2 = Rf 2 = Z 1−1/ A Rf 1 + 1 / A0 ≈ Rf Rf 1 vn = vi R1 + R f 1 − A0 R f 1 − A0 ( R f / A0 ) − Rf − Rf vo v = − A0 n = = = ≈ vi vi R1 + R f 1 R1 + ( R f / A0 ) R1 + ( R f / A0 ) R1 Finding fH by inspection 1. Set vsig = 0 2. Use Miller’s Theorem to replace capacitor between input & output with two capacitors at the input and output. 3. Consider each capacitor separately (assume others are open circuit!), e.g., Cn 4. Find the total resistance seen between the terminals of the capacitor, e.g., Rn (treat ground as a regular “node”). 5. Compute the f pn = 1 2πRn Cn 6. Upper cut-off frequency can be fund from: 1 1 1 1 = + + + ... fH f p1 f p 2 f p 3 Caution: Method in previous slide is called the time-constant approximation to fH (see S&S page 724). 1 fH = (S&S Eq. 9.73) 2π Σ Rn Cn n Since f pn 1 = , the above formula give 2πRn Cn 1 1 1 1 = 2π Σ Rn Cn = + + + ... n fH f p1 f p 2 f p 3 This is the correct formula to find fH However, S&S gives a different formula in page 722 (contradicting formulas of pp724). Ignore this formula (S&S Eq. 9.68) 1 = fH 1 1 1 + + + ... 2 2 2 f p1 f p 2 f p 3 Applying Miller’s Theorem to Capacitors V2 = A ⋅ V1 Z1 = Z 1− A Z2 = Z 1− 1 A 1 j ωC Z Z1 = ⇒ C1 = (1 − A)C 1− A Z Z1 = ⇒ C2 = (1 − 1 / A)C 1−1/ A Z= Example: High-frequency response of a CS amplifier o Circuit includes CL which is often used to set the “dominate pole”. o The first step is to identify which capacitors are relevant to highfrequency response (Cgs ,Cdb , Cgd , and CL ). The other capacitors, Cc1 and Cc2 are relevant to low-frequency response. o At high frequency, Cc1 and Cc2 will be short. Example: High-frequency response of a CS amplifier Use Miller’s Theorem to replace capacitor between input & output (Cgd ) with two capacitors at the input and output. A= vd = − g m (ro || RD || RL ) ≡ − g m RL′ vg C gd ,i = C gd (1 − A) = C gd (1 + g m R′) ≈ g m R′C gd C gd ,o = C gd (1 − 1 / A) = C gd (1 + 1 / g m RL′ ) ≈ C gd * Assuming gmR’L >> 1 Cin = C gs + C gd ,i C L′ = Cdb + C gd ,o + C L Example: High-frequency response of a CS amplifier f p1 = 1 2π Cin ( RG || Rsig ) 1. Consider Cin : ∞ Terminals of Cin 2. Find resistance between Capacitor terminals Example: High-frequency response of a CS amplifier f p2 = 1 2π C L′ (ro || RD || RL ) 1. Consider C’L : Terminals of C’L 2. Find resistance between Capacitor terminals High-frequency response of a CS amplifier AM = − RG g m (ro || RD || RL ) RG + Rsig 1/f p1 = 2π Cin ( RG || Rsig ), 1/f p 2 = 2π C L′ (ro || RD || RL ) Cin = C gs + C gd (1 + g m ro || RD || RL ) C L′ = C gd + Cdb + C L 1/f H = 1/f p1 + 1/f p 2 Miller’s Theorem vs Miller’s Approximation For Miller Theorem to work, ratio of V2/V1 (amplifier gain) should be independent of feedback impedance Z. This was correct for OpAmp example where the gain of the chip, A0 , remains constant when Rf is attached (output resistance of the chip is small). However, the capacitor that connect the input and output changes the frequency response of the amplifier (i.e., its gain) and so we cannot “strictly” apply Miller’s Theorem. In our analysis, we used mid-band gain of the transistor and ignored changes in the frequency response due to the feedback capacitor. This is called “Miller’s Approximation.” Miller’s Approximation only gives “approximate” values of the poles and the higher cut-off frequency. More importantly, Miller’s Approximation “misses” the zero introduced by the feedback resistor (which can cause “unstable” operation). Example: High-frequency response of a CG amplifier C L′ = C gd + C L (+Cdb ) o Cdb is ignored in the above. Including body effect, one sees Cdb actually appears between drain and ground (parallel to CL in the above circuit) and is “absorbed” in CL. o Note that Cgd is also between the drain and the ground and is in parallel to CL. Example: High-frequency response of a CG amplifier f p1 = 2π C gs [ Rsig 1 || (ro +RD || RL ) / g m ro ] Terminals of Cgs 1. Consider Cgs : ro +RD || RL g m ro ro +RD || RL g m ro ≈ ro +RD || RL g m ro 2. Find resistance between Capacitor terminals Example: High-frequency response of a CG amplifier f p2 = 1 2π C L′ [ RD || RL || ro (1 + g m Rsig )] 1. Consider C’L : ≈ ro (1 +g m Rsig ) ro (1 +g m Rsig ) 2. Find resistance between Capacitor terminals High-frequency response of a CG amplifier AM = + Ri g m (ro || RD || RL ) Ri + Rsig Ri = (ro +RD || RL ) / g m ro 1/f p1 = 2π C gs ( Rsig ||Ri ) , 1/f p 2 = 2π C L′ [ RD || RL || ro (1 + g m Rsig )] C L′ = C gd + C L (+Cdb ) 1/f H = 1/f p1 + 1/f p 2