Voltage AmpliÞer Frequency Response

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1 Voltage AmpliÞer Frequency Response Chapter 9 multistage voltage ampliþer 5 V M 7B M 7 M 5 R 35 kω M 6B M 6 Q µa X M 3 Q B Q v OUT V s M 1 M 8 M9 V BIAS M 10 Approaches: 1. brute force OCTC -- do for all capacitances in the circuit. identify the largest contributor(s) and calculate the ThŽvenin resistance associated with them Try second approach: identify four nodes in signal path Lecture 3

2 Qualitative Evaluation of Time Constants Input node: no contribution since R S = 0 Drain of M 1 : Òlow impedance nodeó since R in(cb) = 1/ g m Node X: extremely high resistance... could be a big time constant Source of M 3 : Òlow impedance nodeó since R out(cd) = 1/ g m3 Output node: Òlow impedance nodeó since R o(cc) = 1/ g m4 Note: we are not considering capacitances between nodes here... since we are doing an approximate analysis, we will use MillerÕs theorem to Þnd their effective capacitance to ground Lecture 3

3 Small-Signal Model of Node X Account for all of the capacitances from node X to small-signal ground: 1. Base-collector capacitance C µ of Q (since it is connected to diodeconnected transistors M 6B and M 7B that are in turn connected to V DD ). Gate-drain capacitance C gd6 of M 6 (since it is connected to the same node as C µ ) 3. Gate-drain capacitance C gd3 of M 3 (since it is connected directly to ground) Omitted from Section 10.7Õs analysis: parasitic capacitances to substrate -- C cs and C db6 are both present between node X and ground 4. Effective capacitance due to C gs3 connected between the input and the output of the common-drain stage: C eff = C gs3 ( 1 Ð A vcgs3 ) The gain across C gs3 is about 1, but an accurate calculation would include any backgate effect on M 3 if present. Small-signal model with all (non-parasitic) capacitances X β o r o r oc C µ C gd6 C gd3 (1 A v3 )C gs3 Lecture 3

4 Gain-Bandwidth Product Considering only the time constant from node X 1 ω 3dB [( β o r o g m6 r o6 r o7 )]( C µ C gd6 C gd3 ( 1 Ð A vcgs3 )C gs3 ) Approximate low-frequency gain -- from Chapter 9 A vo Ðg m1 ( β o r o g m6 r o6 r o7 ) Gain-bandwidth product A vo ω g m1 3dB C µ C gd6 C gd3 ( 1 Ð A vcgs3 )C gs3 Lecture 3

5 Magnitude Bode Plot Assuming that the second pole is greater than the unity gain frequency, the Bode plot is V out V s g m1 β o r o 1 1 (r oc β o r o ) {C µ C gd6 C gd3 (1A v3 ) C gs3 } ω g log scale m1 {C µ C gd6 C gd3 (1A v3 ) C gs3 } Lecture 3

6 Differential AmpliÞers General structure: two inputs, two outputs V 1 I 1 I V O1 V O V I1 v i1 = 0 V 1 V IN1 V IN V I v i = 0 V V Consider balanced situation: V I1 = V I and 1 = and devices 1 and are in their constant-current modes. = I 1 I = (V / ) Adjust so V O1 = V O are about 0 V (note the dual supplies) Lecture 3

7 Decomposition of Small-Signal Input Voltages Inputs v i1 and v i are not the Ònatural inputsó to understand how the differential ampliþer works... DeÞne two new voltages differential-mode input voltage = v id = v i1 - v i common-mode input voltage = v ic = (1/)(v i1 v i ) Expressing the inputs v i1 and v i in terms of v id and v ic v id v i1 = v ic v id v i = v ic Ð V V v o1 v o v o1 v o 1 1 v i1 v i v id v id v ic V (a) V (b) Lecture 3

8 Example of Signal Decomposition v i1 = v in and v i = 0 V results in both a differential-modeand a common-mode input to the ampliþer V V v o1 v o v o1 v o 1 1 v in v in v in v in V (a) V (b) Goal: determine the response of the differential ampliþer to v id and v ic separately and then reconstruct the response Lecture 3

9 Small-Signal Model of Bipolar Diff. AmpliÞer DeÞne r ob as the internal (source) resistance of the bias current source V v o1 v o Q 1 Q v i1 v i r ob V Small-signal model with purely differential inputs v o1 v o v id r π g m v π1 R C g m v v π rπ π1 v π v id i 1 i v x r ob Lecture 3

10 Small-Signal Model with Purely Differential Inputs The voltage at v x must be zero for purely differential inputs --> it can be considered a small-signal or incremental ground reference Only need to solve half of the circuit, whose outputs are or - v od / v v od od v id r π g m v π1 R v C g m v π rπ π1 v π v id i 1 i v x Solve left half to Þnd v od v id = Ðg m deþne differential-mode gain a dm by a dm v od = = Ðg v m id Lecture 3

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