Differential Amplifier Common & Differential Modes

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1 Differential Amplifier Common & Differential Modes Common & Differential Modes BJT Differential Amplifier Diff. Amp Voltage Gain and Input Impedance Small Signal Analysis Differential Mode Small Signal Analysis Common Mode 1

2 Common and Differential Modes Consider a circuit with the two port configuration shown Define (convention):the voltages: ±a V 1 I 2 Z 2 Z 3 V =V 1 V 2 c = V 1 V 2 2 = common mode voltage = differential mode voltage ±a V 2 1. Let V 1 = V 2 = V => = V & = 0 2. Let V 1 = -V 2 = V => = 0 & = 2V 2

3 Replace V 1, V 2 with CM and DM Sources V 1 Simultaneous Equations: ±a I 2 Z 2 Z 3 V 1 V 2 =2 V 1 V 2 = V 2 I 2 ±a ±. - Adding: 2V 1 =2 V 1 = Subtracting: 2V 2 =2 V 2 = Definition = V 1 V 2 2 =V 1 V 2 3

4 Common and Differential Mode Currents Simultaneous Equations: ±. ±. ±. I 2 - Z 2 Define (convention): = I 2 I d = I 2 2 Z 3 I 2 = I 2 =2 I d Adding: 2 = 2 I d = / 2 I d Subtracting: 2 I 2 = 2 I d I 2 = I d 4

5 CM-DM Circuit Description ±. ±. I 2 ±. - 2I d = - I 2 Z 2 Z3 = I 2 1. Voltage sources redefined as common and differential mode quantities. 2. Differential mode currents take blue path. 3. Common mode take red path. = / 2 I d I 2 = I d 5

6 ±. ESE319 Introduction to Microelectronics Analyze Circuits by Superposition Balanced Circuit Differential Mode (V C = 0) I 2 = - V Z1-2I d Z 2 - V Z2 V x Z3 = ±. = = I 2 2 I d = 2 2 Z 2 =0 V x =0 = 2 Z = I 2 = =0 2 V x 2 V x = 2V x Z =0 Z 2 balanced circuit => =Z 2 =Z Z i n dm = 2 I d =2 Z NOTE: If Z 2 = Z, then 0 => V x 0. 6

7 Analyze Circuits by Superposition Balanced Circuit Common Mode ( = 0) = Z 2 Z 3 = Z 2 Z 3 I 2 2I d Z 2 Z3 2I d = I 2 = 2 2 =0 I d=0 ±. Z i n cm = = Z 2 Z 3 balanced circuit => =Z 2 =Z 7

8 Summary In a balanced ( = Z 2 = Z) circuit, differential mode voltages result in differential mode currents and common mode voltages result in common mode currents. The differential mode input impedance of a balanced circuit is: Z i n dm =2 Z The common mode input impedance of a balanced circuit is: Z i n cm = Z 2 Z 3 8

9 BJT Differential Amplifier Bias View I C = m m Q1=Q2 R C1 =R C2 =R C R B1 =R B2 =R B balanced circuit I B I E I E I B Collector bias path inherently common mode. Icm Choose m and R C for approx I E = m 2 ½ V CC drop across R C. I B = m 2 1 I C = m 2 m 2 9

10 Voltage Gain & Input Impedance v c1g-dm i b-dm - vo-dm v c2g-dm i b-cm v c1g-cm - vo-cm v c2g-cm v-dm v-dm ro v-cm - ro - r in-dm Single-ended outputs: v c1g -dm, v c2g-dm Differential output: v o-dm = v c2g-dm - v c1g-dm r in dm = i b dm G dm = v o dm r in-cm Single-ended outputs: v c1g-cm, v c2g-cm Differential output: v o-cm = v c2g-cm - v c1g-cm r in cm = i b cm G cm = v o cm 10

11 BJT Differential Amplifier Small-signal View ic1 ic2 Z 2 V x Z - 3 v-dm ib1 ie1 ie2 ib2 v-dm m current source output impedance =Z 2 = 1 r e Z 3 = 1 r o v-cm Note: v-dm and v-cm are ac small signals 11

12 Balanced circuit => ib1 ESE319 Introduction to Microelectronics Superposition Small-signal Analysis Differential Mode ( = 0) ic1 R C1 =R C2, R B1 =R B2, r e1 =r e2, 1 = 2 i e1 =i e2 =i e,i b1 =i b2 =i b, i c1 =i c2 =i c - v o-dm ic2 ib2 DM Path: 2 =i e1 dmr e1 i e2 dm r e2 2 v-dm ie1 ie1-dm ie2 ie2-dm v-dm = 1 i b1 dm r e1 1 i b2 dm r e2 = 2 1 i b dm r e i b dm = 2 1 r e = r in dm =Z 2 = 1 r e i e1 dm =i e1 i b1 dm =i b1 Z 3 = 1 r o i e2 dm = i e2 i b2 dm = i b2 r i n dm =2 1 r e 12

13 Balanced circuit => ib1 v-dm ESE319 Introduction to Microelectronics Superposition Small-signal Analysis Differential Mode ( = 0) cont. ic1 ie1 ie1-dm i e1 dm =i e1 i b1 dm =i b1 R C1 =R C2, R B1 =R B2, r e1 =r e2, 1 = 2 - v o-dm ic2 ie2 ie2-dm i e2 dm= i e2 i b2 dm = i b2 ib2 i b dm = 2 1 r e v o dm = R C2 R C1 v 2 1 r dm v e 2 1 r dm e DM Differential voltage gain: G dm =G dm2 G dm1 = R C R C 1 r e r e v-dm v o dm =v c2g dm v c1g dm v o dm = R C2 i b dm R C1 i b dm DM Single-ended voltage gains: G dm2 = G dm1 = v cg dm = R C 2 1 r e R C 2 r e 13

14 Balanced circuit => ib1 ESE319 Introduction to Microelectronics Superposition Small-signal Analysis Common Mode ( = 0) ic1 ie1 icm =Z 2 = 1 r e i e1 cm =i e1 R C1 =R C2, R B1 =R B2, r e1 =r e2, 1 = 2-0 V ic2 ie2 icm icm v-cm ib2 Z 3 = 1 r o i e2 cm=i e2 CM path: r e1 & r e2 are in parallel =[r e1 r e2 r o ]i e cm i cm = r e 2 r r o o i e cm = i cm 2 = r e 2 r o 2 r o i b cm = 1 r e 2 r o i b1 cm =i b1 i b2 cm =i b2 r in cm = 2i b cm = 1 r e 2 r o 1 r o 14

15 ib1 ESE319 Introduction to Microelectronics Superposition Small-signal Common Mode Balanced circuit =>R C1 =R C2, R B1 =R B2,r e1 =r e2, 1 = 2 ic1 ie1 ie-cm - 0 V ic2 ie2 ie-cm ie-cm ib2 Analysis v o cm =v c2g cm v c1g cm v o cm = R C2 i b cm R C1 i b cm i b cm = 1 r e 2 r o 1 2 r o v o cm = R C2 R C1 v 1 2 r cm v o 1 2 r cm =0 o v-cm CM Differential voltage gain: G cm = v o cm =0 CM Single-ended voltage gains: G cm2 =G cm1 = v cg cm = R C R C 1 2 r o 2 r o 15

16 Summary Differential mode: r i n dm =2 1 r e Differential DM Voltage gain: G dm = v o dm R C r e Single-Ended DM Voltage gain: Common mode: r i n cm = 1 r e 2 r o 1 r o Differential CM Voltage gain: G cm = v o cm =0 Single-Ended CM Voltage gain: G dm1 = G dm2 = v cg dm R C G cm1 =G cm2 = v cg cm R C 2 r e 2 r o 16

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