Differential amplifier (Analog Electronic Circuits)
- 1. ECE315 / ECE515 Lecture – 11 Date: 15.09.2016 • MOS Differential Pair • Quantitative Analysis – differential input • Small Signal Analysis
- 2. ECE315 / ECE515 MOS Differential Pair Variation of input CM level regulates the bias currents of M1 and M2 → Undesired!!! Solution?? Current source is ideal: constant current, infinite output impedance To overcome the issues emanating from non-ideal CM level M1 and M2 are perfectly matched (at least in theory!) ensures M1 and M2 in saturation
- 3. ECE315 / ECE515 MOS Differential Pair Qualitative Analysis – differential input • Let us check the effect of Vin1 – Vin2 variation from -∞ to ∞ • Vin1 is much more –ve than Vin2 then: • M1 if OFF and M2 is ON • ID2 = ISS • Vout1 = VDD and Vout2 = VDD – ISSRD • Vin1 is brought closer to Vin2 then: • M1 gradually turns ON and M2 is ON • Draws a fraction of ISS and lowers Vout1 • ID2 decreases and Vout2 rises • Vin1 = Vin2 • Vout1 = Vout2 = VDD- ISSRD/2
- 4. ECE315 / ECE515 MOS Differential Pair • Vin1 becomes more +ve than Vin2 then: • M1 if ON and M2 is ON • M1 carries greater ISS than M2 • Let us check the effect of Vin1 – Vin2 variation from -∞ to ∞ • For sufficiently large Vin1 – Vin2 : • All of the ISS goes through M1 → M2 is OFF • Vout1 = VDD – ISSRD and Vout2 = VDD Qualitative Analysis – differential input
- 5. ECE315 / ECE515 MOS Differential Pair • Plotting Vout1 – Vout2 versus Vin1 – Vin2 The maximum and minimum levels at the output are well defined and is independent of input CM level (Vin,cm) Minimum Slope ↔ Minimum Gain Maximum Slope ↔Maximum Gain The circuit becomes more nonlinear as the input voltage swing increases (i.e., Vin1 – Vin2 increases) ↔ at Vin1 = Vin2, the circuit is said to be in equilibrium Qualitative Analysis – differential input
- 6. ECE315 / ECE515 MOS Differential Pair Qualitative Analysis – common mode input • Now let us consider the common mode behavior of the circuit As mentioned, the tail current source is used to suppress the effect of input CM level variation (Vin,cm) Does this enable us to set any arbitrary level of input CM (Vin,cm) To understand this: • Set Vin1 = Vin2 = Vin,CM • Then vary Vin,CM from 0 to VDD • Also implement ISS with an NFET • Lower bound of Vin,cm: VP should be sufficiently high in order for M3 to act as a current source. • Upper bound of Vin, cm: M1 and M2 need to remain in saturation.
- 7. ECE315 / ECE515 MOS Differential Pair • What happens when Vin,CM = 0? • M1 and M2 will be OFF and M3 can be in triode for high enough Vb • ID1 = ID2 = 0 ← circuit is incapable of amplification • Now suppose Vin,CM becomes more +ve • M1 and M2 will turn ON if Vin,CM exceeds VT • ID1 and ID2 will continue to rise with the increase in Vin,CM • VP will track Vin,CM as M1 and M2 work like a source follower • For high enough Vin,CM, M3 will be in saturation as well • If Vin,CM rises further • M1 and M2 will remain in saturation if: , 1 in CM T out V V V , 2 SS in CM DD D T I V V R V Qualitative Analysis – common mode input
- 8. ECE315 / ECE515 MOS Differential Pair • For M1 and M2 to remain in saturation: 1,2 1,2 GS T DS V V V , 2 SS in CM T DD D I V V V R , 2 SS in CM T DD D I V V V R • The lowest value of Vin,CM is determined by the need to keep the constant current source operational: , 1,2 3 in CM GS GS T V V V V , max ( ) 2 SS in CM T DD D I V V V R , 1,2 3 3 ( ) in CM GS GS T V V V V 1,2 3 , ( ) min , 2 SS GS GS T in CM DD D T DD I V V V V V R V V Qualitative Analysis – common mode input
- 9. ECE315 / ECE515 MOS Differential Pair • Thus, Vin,CM is bounded as: 1,2 3 , ( ) min , 2 SS GS GS T in CM DD D T DD I V V V V V R V V M3=Linear M1=M2 =Off M1=M2 =Off M1=M2 =Off M1=M2 =On M1=M2 =On M1=M2 =On • Summary: M3=Linear M3=Linear Qualitative Analysis – common mode input
- 10. ECE315 / ECE515 MOS Differential Pair • How large can the output voltage swings of a differential pair be? ,max out DD V V ,min , out in CM T V V V The higher the input CM level, the smaller the allowable output swings. Qualitative Analysis – common mode input
- 11. ECE315 / ECE515 MOS Differential Pair Quantitative Analysis – differential input P For +ve Vin1 → VGS1 is greater than VGS2→ ID1 will be greater than ID2 For +ve Vin2 → VGS2 is greater than VGS1→ ID2 will be greater than ID1 2 2 1 1 ( ) ( ) out DD D D out DD D D V V I R V V I R 1 1 2 2 ( ) ( ) out DD D D out DD D D V V I R V V I R It is thus apparent that the differential pair respond to differential- mode signals → by providing differential output signal between the two drains
- 12. ECE315 / ECE515 P Differential Pair – Large Signal Analysis Quantitative Analysis – differential input • The idea is to define ID1 and ID2 in terms of input differential signal Vin1 – Vin2 • The circuit doesn’t include connection details considering that these drain current equations do not depend on the external circuitries • Assumptions: M1 and M2 are always in saturation; differential pair is perfectly matched; channel length modulation is not present
- 13. ECE315 / ECE515 Differential Pair – Large Signal Analysis Quantitative Analysis – differential input P 1 1 2 2 P in GS in GS V V V V V 1 2 1 2 in in GS GS V V V V We also know: 2 2 D GS T n ox I V V W C L 2 D GS T n ox I V V W C L Therefore: 1 2 1 2 2 2 D D in in n ox n ox I I V V W W C C L L 2 1 2 1 2 1 2 2 2 in in D D D D n ox V V I I I I W C L Squaring
- 14. ECE315 / ECE515 Differential Pair – Large Signal Analysis Quantitative Analysis – differential input 2 1 2 1 2 1 2 2 2 in in D D D D n ox V V I I I I W C L SS I Squaring 2 1 2 1 2 2 2 in in SS D D n ox V V I I I W C L 2 1 2 1 2 1 2 2 n ox in in SS D D W C V V I I I L 2 4 2 2 1 2 1 2 1 2 1 4 4 n ox in in SS SS n ox in in D D W W C V V I I C V V I I L L 2 4 2 2 2 2 1 2 1 2 1 2 1 2 1 4 n ox in in SS SS n ox in in D D D D W W C V V I I C V V I I I I L L
- 15. ECE315 / ECE515 Differential Pair – Large Signal Analysis Quantitative Analysis – differential input 2 4 2 2 1 2 1 2 1 2 1 4 D D n ox in in SS n ox in in W W I I C V V I C V V L L 2 1 2 1 2 1 2 1 4 2 SS D D n ox in in in in n ox W I I I C V V V V W L C L Observations • ID1 – ID2 falls to zero for Vin1 = Vin2 and |ID1 – ID2| increases with increase in |Vin1 – Vin2| • Therefore, ID1 – ID2 is an odd function of Vin1 – Vin2 • Its important to notice that ID1 and ID2 are even functions of their respective gate-source voltage
- 16. ECE315 / ECE515 Differential Pair – Large Signal Analysis Quantitative Analysis – differential input • Equivalent Gm of M1 and M2 → its effectively the slope of the characteristics 1 2 D D D I I I 1 2 in in in V V V Lets denote: 2 1 4 2 SS D n ox in in n ox W I I C V V W L C L 2 2 4 2 1 2 4 SS in n ox D n ox in SS in n ox I V W C I W L C V L I V W C L For ∆Vin = 0: D m n ox SS in I W G C I V L Furthermore: 1 2 out out D D m in V V R I R G V 1 2 | | out out v n ox SS D in V V W A C I R V L
- 17. ECE315 / ECE515 Differential Pair – Large Signal Analysis Quantitative Analysis – differential input 2 2 4 2 1 2 4 SS in n ox m n ox SS in n ox I V W C W L G C L I V W C L 2 SS in n ox I V W C L Gm falls to zero for 1 in V ∆Vin1 represents the maximum differential signal a differential pair can handle. Beyond |∆Vin1|, only one transistor is ON and therefore draws all of the ISS
- 18. ECE315 / ECE515 Differential Pair – Large Signal Analysis Quantitative Analysis – differential input Reduce ∆Vin1 → by increasing W/L ISS Constant Increase ∆Vin1 → by increasing ISS Linearity Improves W/L Constant Linearity Improves Linearity of a differential pair can be improved by decreasing W/L and/or increasing ISS
- 19. ECE315 / ECE515 Quantitative Analysis – differential input MOS Differential Pair – small signal analysis 1 2 | | out out v n ox SS D in V V W A C I R V L • From large signal analysis we achieved: At equilibrium, this is gm 1 2 | | out out v m D in V V A g R V • We apply small signals to Vin1 and Vin2 and assume M1 and M2 are already operating in saturation. • How to arrive at this result using small signal analysis? • Two techniques • Superposition method • Half-circuit concept
- 20. ECE315 / ECE515 MOS Differential Pair – small signal analysis • Method-I: Superposition technique – the idea is to see the effect of Vin1 and Vin2 on the output and then combine to get the differential small signal voltage gain • First set, Vin2 = 0 • Then let us calculate VX/Vin1 This is open for small signal analysis CS-stage Simplified Circuit Input impedance of M2 Provides degeneration resistance to CS-stage of M1 m2 m1 + 1 g 1 g 1 1 1 2 1 1 X D in m m V R V g g 1 1 2 1 1 X D in m m V R V g g
- 21. ECE315 / ECE515 This is open for small signal analysis MOS Differential Pair – small signal analysis • Superposition technique • Now calculate VY/Vin1 Replace M1 by its Thevenin Equivalent Circuit 1 i T n V V = T R m1 = 1 g CG-Stage 1 2 1 1 1 Y D in m m V R V g g Simplified Circuit • combine the expressions to calculate small signal voltage only due to Vin1 1 _ _ 1 1 2 | 2 1 1 in X Y due to V D in m m V V R V g g
- 22. ECE315 / ECE515 MOS Differential Pair – small signal analysis For matched transistors: 1 _ _ 1 | in X Y due to V m D in V V g R V • Similarly: 2 _ _ 2 | in X Y due to V m D in V V g R V • Superposition gives: 2 1 X Y total v m D in in V V A g R V V • The magnitude of differential gain is gmRD regardless of how the inputs are applied • The gain will be halved if single ended output is considered • Half Circuit Approach • If a fully symmetric differential pair senses differential inputs (i.e, the two inputs change by equal and opposite amounts from the equilibrium condition), then the concept of half circuit can be applied. Change this by ∆VT Change this by -∆VT RT1 = RT2 Potential at node P will remain unchanged Node is said to be ac-grounded
- 23. ECE315 / ECE515 MOS Differential Pair – small signal analysis • Half Circuit Approach Ac grounding of node P leads to We can write: 1 X m D in V g R V 1 Y m D in V g R V Vin1 and –Vin1 are the change in input voltage at each side Therefore the differential output can be expressed as: 1 2 X Y in m D V V V g R Thus the small signal voltage given is: 1 2 X Y v m D in V V A g R V
- 24. ECE315 / ECE515 MOS Differential Pair – small signal analysis • How does the gain of a differential amplifier compare with a CS stage? • For a given total bias current ISS, the value of equivalent gm of a differential pair is 1 2 times that of gm of a single transistor biased at the ISS with the same dimensions. Thus the total gain is proportionally less. • Equivalently, for given device dimensions and load impedance, a differential pair achieves the same gain as a CS stage at the cost of twice the bias current. • What is the advantage of differential stage then? • Definitely the noise suppression capability. Right?
- 25. ECE315 / ECE515 in1 V in1 -V 1 M 2 M X V Y V P • Effect of r0 on the gain • How is gain affected if channel length modulation is considered? → the circuit is still symmetric → the voltage at node P will be zero No current through RSS RSS plays no role in differential gain Finite output resistance of current source MOS Differential Pair – small signal analysis
- 26. ECE315 / ECE515 • The virtual ground on the source allows division of two identical CS amplifiers: → differential half circuits 1 X m in D o V g V R r 1 Y m in D o V g V R r 1 2 X Y m in D o V V g V R r MOS Differential Pair – small signal analysis 1 in V 1 in V 1 M 2 M 2 SS I X V Y V 1 2 X Y v m D o in V V A g R r V
- 27. ECE315 / ECE515 Small signal analysis – asymmetric inputs Transform the inputs as Simplify Differential Inputs Common Mode Inputs Simplified Circuit
- 28. ECE315 / ECE515 Circuit for Differential Mode Circuit for Common Mode Small signal analysis – asymmetric inputs 1 2 X Y m D o in in V V g R r V V If the circuit is fully symmetric and ISS is ideal current source, then M1 and M2 draws half of ISS and is independent of Vin,CM. The VX and VY experience no change as Vin,CM varies. In essence, the circuit simply amplifies the difference between Vin1 and Vin2 while eliminating the effect of Vin,CM.