Differential Amplifiers (Ch. 10)

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1 Differential Amplifiers (h. 0) 김영석 충북대학교전자정보대학

2 ontents 0. General onsiderations 0. Bipolar Differential Pair 0.3 MOS Differential Pair 0.4 ascode Differential Amplifiers 0.5 ommon-mode Rejection 0.6 Differential Pair with Actie Load 0-

3 0. General onsiderations Humming Noise in Audio Amplifier from 60Hz Power Line out R 0-3

4 0. General onsiderations How to suppress the Hum? out should not to be referenced to GND Since both node X and Y contain the ripple, their difference will be free of ripple But, wasting current in Q out ( + ) R r A + X Y out r X in Y r A in ( NO r ) 0-4

5 0. General onsiderations To utilize the wasting current in Q Apply the inputs differentially the outputs are 80 out of phase enhancing out when sensed differentially But, ommon-mode (M) problem: increase M > increase gm >increasegain A + X Y out in A X in + Y r r A in 0-5

6 0. General onsiderations To hae the differential-mode gain independent of input M leel Add a tail current 0-6

7 0. Bipolar Differential Pair ommon-mode Response M gain BE BE X Y EE R independent of out AM 0 M M EE M input range M M EE < R (Q in Actie Region) > EE + BE 0-7

8 Differential Pair haracteristics (a) (b) EE 0 Y X R EE X Y 0 EE R EE (a) (b) (a) (b) 0-8

9 Small-Signal Analysis EE EE m EE EE g g Δ Δ Δ + + Δ m Y m X m g R R g R R g Δ + Δ + Δ Δ Δ Δ Δ Δ in m Y X out m Y g R Δ Δ Δ Δ m in out R g A irtual Ground: 0 Δ P 0-9

10 0.. Large Signal Analysis in T 4 max, Δ EE + in in S S T in S T in S exp, / exp / exp T in in T EE exp exp + T in in EE exp + T in in EE out out R tanh 0-0 T

11 Linear/Nonlinear Regions The left column operates in linear region, whereas the right column operates in nonlinear region. 0-

12 0..3 Small-Signal Model 0-

13 Half ircuits out R gm π out in out in R + π out in g, m in g π, g + m R m π g m Since P is grounded, we can treat the differential pair as two E half circuits, with its gain equal to one half circuit s single-ended gain. 0-3

14 Example: Differential Gain out in out in g m r O 0-4

15 Half ircuit Example X 0 A g ( ) m ro ro 3 R 0-5

16 Half ircuit Example A g ( ) m ro ro 3 R 0-6

17 Half ircuit Example A R E R + g m 0-7

18 Half ircuit Example A R RE + g m 0-8

19 0.3 MOS Differential Pair s ommon-mode Response M gain 0 The equilibrium oerdrie oltage: nput M Range X Y DD R D SS Δ ( ) GS TH equil μ μ n SS ox W L > (Min Saturation Region) D G M TH SS DD RD > M TH SS M < DD RD + And, > + GS SS TH 0-9

20 Differential Response (a) (b) (a) (b) (a) (b) H 0 Differential Amplifiers 0-0

21 Small-Signal Response Δ 0 A P g m R D Similar to its bipolar counterpart, the MOS differential pair exhibits the same irtual ground node and small signal gain. 0-

22 0.3. MOS Differential Pair s Large-Signal Response D D D μ n μ n + D ox ox SS W L W L ( ( GS GS TH TH ) ) D D ( ) ( ) W 4 SS μn ox in in in L W μnox L in in in max Δ ( ) GS TH equil D μ W n ox / L 0-

23 ontrast Between MOS and Bipolar Differential Pairs n a MOS differential pair, there exists a finite differential input oltage to completely switch the current from one transistor to the other, whereas, in a bipolar pair that oltage is infinite. And, MOS BJT : : in in in max in max 4 Δ T MOS Bipolar 0-3

24 The effects of Doubling the Tail urrent in in max ( ) GS TH equil D μ n ox W / L SS is doubled SS Δ in,max increases by Δ out,max increases by 0-4

25 The effects of Doubling W/L in in max ( ) GS TH equil D μ n ox W / L W/L is doubled in,max decreases by out,max same 0-5

26 Small-Signal Analysis of MOS Differential Pair W 4 SS μ ( ) ( ) in D D n ox in in in L μnox μn ox W L 4 W L W L SS ( ) μ ( ) g ( ) in in μ n ox W L n When the input differential signal is small compared to 4 SS /μ n ox (W/L), the output differential current is linearly proportional to it, and small-signal signal model can be applied. ox SS in in m in in 0-6

27 irtual Ground and Half ircuit Δ P 0 A g Applying the same analysis as the bipolar case, we will arrie at the same conclusion that node P will not moe for small input signals and the concept of half circuit can be used to calculate the gain. m R 0-7

28 MOS Differential Pair Half ircuit Example λ 0 A g m ro 3 ro gm3 0-8

29 MOS Differential Pair Half ircuit Example λ 0 A g g m m3 0-9

30 MOS Differential Pair Half ircuit Example λ 0 A R SS R DD + g m 0-30

31 0.4 Bipolar ascode Differential Pair [ r + ( r r )( g r )] A g m O 3 O π 3 + m3 O 3 0-3

32 Bipolar Telescopic ascode A g [ g r ( r r )] [ g r ( r r )] π 3 m5 O5 O7 5 m m3 O3 O π 0-3

33 Example: Bipolar Telescopic Parasitic Resistance R r r g r R [ ] op O O m m O m O op R r r r g g A r r g r R ) ( π π 0-33

34 MOS ascode Differential Pair A g r g r m O3 m3 O 0-34

35 MOS Telescopic ascode g [ ( g r r ) ( g r r ) ] A g m 5 O 5 7 m m 3 O 3 O O 0-35

36 Example: MOS Telescopic Parasitic Resistance ) ( )] ( [ O m O op m O m O O op r g r R g A R r g r R r R

37 0.5 M Rejection Effect of Finite Tail mpedance Δ Δ out, M in, M R EE R / + / g m f the tail current source is not ideal, then when a input M oltage is applied, the currents in Q and Q and hence output M oltage will change. 0-37

38 nput M Noise with deal Tail urrent 0-38

39 nput M Noise with Non-ideal Tail urrent 0-39

40 M to DM onersion, A M-DM Δout ΔRD Δ / g + R M m EE f finite tail impedance and asymmetry are both present, then the differential output signal will contain a portion of input common-mode mode signal. 0-40

41 Example: A M-DM A M DM g m + ΔR [ r + ( + g r )( R r )] O 3 ( m 3 O 3 π 3 0-4

42 MRR MRR A A DM M DM MRR defines the ratio of wanted amplified differential input signal to unwanted conerted input common-mode noise that appears at the output. 0-4

43 0.6 Diff Pair with Actie Load Differential to Single-ended onersion Many circuits require a differential to single-ended conersion, howeer, the aboe topology is not ery good. 0-43

44 Supply Noise orruption The most critical drawback of this topology is supply noise corruption, since no common-mode cancellation mechanism exists. Also, we lose half of the signal. 0-44

45 Better Alternatie This circuit topology performs differential to single-ended conersion with no loss of gain. 0-45

46 Actie Load With current mirror used as the load, the signal current produced by the Q can be replicated onto Q 4. This type of load is different from the conentional static load and is known as an actie load. 0-46

47 Differential Pair with Actie Load The input differential pair decreases the current drawn from R L by Δ and the actie load pushes an extra Δ into R L by current mirror action; these effects enhance each other. 0-47

48 Actie Load s. Static Load The load on the left responds to the input signal and enhances the single-ended output, whereas the load on the right does not. 0-48

49 MOS Differential Pair with Actie Load Similar to its bipolar counterpart, MOS differential pair can also use actie load to enhance its single-ended output. 0-49

50 Asymmetric Differential Pair Because of the astly different resistance magnitude at the drains of M and M, the oltage swings at these two nodes are different and therefore node P cannot be iewed as a irtual ground. 0-50

51 Theenin Equialent of the nput Pair g r ( ) R The The mn ron on ( in in 0-5

52 Simplified Differential Pair with Actie Load out in in g mn ( ron rop ) 0-5

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