Differential Amplifier Design

Transcription

1 Differential Amplifier Design Design with ideal current source bias. Differential and common mode gain results Add finite output resistance to current source. Replace ideal current source with current mirror. 1

2 I C ESE319 Introduction to Microelectronics Quick Amplifier Design I C By inspection (for Q1 & Q2): I E = I 2 =5mA. - v o + Neglecting I B : I C =I E =5 ma. v-dm/2 I E I E v-dm/2 V Rc =5V V CG =V CC V Rc =7V V EG = V BE = 0.7V V CE =V CG V EG =7.7V ideal current source v-cm Single-ended DM-ode voltage gains w.r.t. v-dm/2 and v-cm (for Q1 side): G dm1/2 = v c1g dm v i =v dm /2 R C R E = 10 G cm1 = v c1g cm v cm R C 2r o =0 2

3 Differential- Mode Gain Results v A =v dm /2 v B =v c1g dm Scope output B at collector of Q1, i.e.. v c1g dm Input voltage v dm /2 : 0.14 V peak arg (0 o ) at f = 1 khz. v c1g dm Output voltage : 1.33 V peak arg (180 o ). Measured gain: G dm1/2 = v c1g dm v i =v dm /2 = 1.33V 0.14 V = 9.5 3

4 v A =v cm ESE319 Introduction to Microelectronics Common Mode Results v B =v c1g cm Scope output B at collector of Q1, i.e.. v c1g cm v cm Input voltage : 0.14 V peak arg (0 o ) at f = 1 khz. Since I is an ideal current source r o = => G cm1 =0. No common mode signal current (i cm = 0 => i e-cm = 0) v c1g cm =0V peak CMRR 20 log = 4

5 Common Mode Results - II v A =v b1g cm v B =v e1g cm Comparing base and emitter voltages to ground for Q1, i.e. v b1g cm, v e1g cm. v b1g cm =v e1g cm =1.4V peak v be cm =0V peak Since i e-cm = 0, we expect v be cm =0V peak, all base voltage appears at the emitter. 5

6 Common Mode Results - Add r o to Model insert finite r o to model non-ideal current source G dm1/2 = v c1g dm v i =v dm /2 R C R E = 10 G cm1 R C 2 r o = =

7 Common Mode Results - Add Finite r o v A =v cm v b =v c1g cm r o =100 k Scope output B at collector of Q1, i.e.. v c1g cm v cm Input voltage : 1.4 V peak arg (0 o ) at f = 1 khz. CMRR 20log dB v c1g cm Output voltage : V peak arg (180 o ). G cm = G dm1/2 = 9.5 (unchanged) 7

8 I C1 I C3 2 ESE319 Introduction to Microelectronics Simulation with Current Mirror I C2 I C3 2 V C1G =11.5V I REF 1 ma Matched 2N2222 BJTs (Q1, Q2, Q3 and Q4). I C3 1 ma I C1 =I C2 0.5 ma I C3 1mA v-cm NOTE: - The zero-to-peak swing at the collector now only 0.5 V! 8

9 Simulation with Current Mirror - II v A =v cm v b =v c1g cm Scope output B at collector of Q1, i.e.. v c1g cm v cm Input voltage : 1.4 V peak arg (0 o ) at f = 1 khz. Output voltage v c1g cm : 7 mv peak arg(180 o ). G cm = khz common mode results almost exactly same as those for r o =100 k model. 9

10 Simulation with Current Mirror - III I C1 5 ma V C1G =7V I C2 5 ma I REF 10 ma Matched 2N2222 BJTs (Q1, Q2, Q3 and Q4). I C3 10 ma I C1 =I C2 5mA I C3 10 ma v-cm I REF = V CC V BE4 V EE R ref 10 ma NOTE: - The zero-to-peak swing at the collector now increased to 5 V! 10

11 Simulation with Current Mirror - III v A =v cm v b =v c1g cm Input voltage : 1.4 V peak arg (0 o ) at f = 1 khz. G cm = v cm Output voltage v c1g cm : 60 mv peak arg (180 o ). G cm Common mode output now about 10X its previous value with 0.5 ma. collector current. Why? Early effect! r o = V A I C 10k 10 times the current means 1/10 the value of r o! 11

12 Simulation with Current Mirror - Bode Plots (I C = 5 ma) r o 10k f =1 khz 20 log = 27.3 db 1.4 f =9.7 MHz = 24.2dB (+ 3dB) f B 5 ma current: G cm-db + 3 db frequency f B 0.5 ma current: G cm-db + 3 db frequency theory simulation r o 100k f B 0.5mA r o 5mA = 10k f B 5mA r o 0.5mA 100k = MHz = f B 5mA 9.7 MHz =0.2 f B 0.5mA (I C = 0.5 ma) f =1 khz 20 log = 46 db 1.4 f =1.9 MHz = 43.2 db (+ 3dB) 12

13 Simulations with Parasitic Caps I REF =1 ma 2 pf Results with 2 pf capacitance added from collector-to-base of mirror transistor in the I C1 = I C2 = 0.5 ma amplifier emitter return path. f B f =1 khz 20 log = 46 db 1.4 f =645 khz = 43.2 db This drops the amplifier G cm-db + 3 db frequency from 1.9 MHz to about 645 khz! 13

14 Simulations with Parasitic Caps - cont. 2 pf f B f =1 khz 20 log = 46 db 1.4 f =334 khz = 43.2 db Results with 2 pf capacitance added from base-to-emitter of mirror transistor. This drops the amplifier G cm-db +3 db frequency from 1.9 MHz to about 334 khz! RECALL: 2 pf is about the capacitance between 2 rows of Protoboard pins! 14

15 Simulations with Parasitic Caps - cont. 2 pf 2 pf f B f =1 khz 20 log = 46 db 1.4 f =234 khz = 43.2 db Drops amplifier G cm-db break frequency from 1.9 MHz to about 234 khz! 15

16 Simulate the 5 ma Design with 2 pf Parasitics I C1 5 ma I REF 10 ma 2 pf 2 pf f B f =1 khz 27 db 3dB common mode bandwidth with 2 pf base-emitter and base collector capacitances. About 10X the bandwidth as the 0.5 ma design with same low frequency CM gain as the 5 ma design for DM gain of -10. f =2.5 MHz = 24.2dB Drops amplifier G cm-db break frequency from 9.5 MHz to about 2.5 MHz! f B 2.5 MHz 234 khz I C1 =5mA I C1 =0.5mA f =1 khz 27 db 46 db 16

17 Observations 1). For best common mode rejection use small collector currents i.e. increase r o. 2). For best bandwidth use large collector currents, i.e. decrease r o. 3). Minimize parasitic capacitance around mirror transistor to increase common mode rejection bandwidth. 4). Since no differential mode current flows through the mirror transistor (Q3, i.e. r o ), it should have no effect on differential mode performance. 5). Observations 1) and 2) force a trade-off in selecting the bias current. 17

18 Try Redesign for Reasonable Differential I C1 0.5 ma Mode Voltage Swing & large r o I C3 =1 ma r o 100 k I C2 0.5 ma I REF 1 ma Can we beat the r o trade-off? IDEA: 1. Reduce I REF to increase r o. V C1G =7V I C3 1mA v-cm G dm1/ 2 R C R E = 10 G cm1 R C 2r o r o V A I REF 2. Increase R C1 to increase V RC1. R C1 = V RC1 I REF /2 3. Increase R E1 to desired G dm1/2. R E1 = R C G cm1 R C1 2r o = V RC1 I REF /2 RESULT: No help! G dm1/ 2 I REF 2V A = V RC1 V A 18

19 Redesign (I REF = 1 ma) Bode Plot 1. Increasing r o by 10X decreases the CM gain by 10X. r o =10 k 100 k f B Simulation with 2 2 pf caps. 1. f B = 288 khz. 2. Low frequency CM gain -26 db. 1 f B 2 r o C o 2. Increasing R C by increased the CM gain by 10X. 3. Nothing gained! f =1 khz 26 db About same as I REF = 10 ma design! G cm-db break frequency R C =1 k 10 k f B 288 khz About same as I REF = 1 ma design! where ( f B 234 khz) 19

20 v 1 v 2 v 1 v 2 v i v i 2v i v-cm = 0 v 1 v 2 v 1 v 2 v i v i /2 v i /2 v i /2 20

21 v 1 v 2 v 1 v 2 v i v i 2v i v-cm = 0 v 1 v 2 =2v i v 1 v 2 v 1 v 2 v i v i /2 v i /2 v 1 =v i /2 v i / 2 v 2 =v i / 2 v i /2 v i /2 21