INF3410 Fall Book Chapter 3: Basic Current Mirrors and Single-Stage Amplifiers

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1 INF3410 Fall 2013 Amplifiers

2 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 2

3 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 3

4 Simple Current Mirror When will I out = I in, I out I in, I out I in, I out xi in? Amplifiers 4

5 Diode Connected nmos Is the simplified small signal model very useful here? For example when you want to know the voltage when you apply a given current? Amplifiers 5

6 Current Mirror Small Signal Model How do you describe the effect of different W L ratios of the two transistors in terms of the small signal model? Amplifiers 6

7 Current Mirror Small Signal Model, Example 3.1 (1/2) A detour solution (shorter in the book): 2 V ef f = V in V t0 = β I in 2 = 10µm/0.4µm 190µA/V 2 100µA = 205mV V in = 205mV + 570mV = 775mV Amplifiers 7

8 Current Mirror Small Signal Model, Example 3.1 (2/4) Continued detour solution (shorter in the book): I out = 1 2 βv 2 ef f = 10µm/0.4µm 190µA/V2 205mV 2 2 = 100µA Amplifiers 8

9 Current Mirror Small Signal Model, Example 3.1 (3/4) Wrong (!!!) approach: g m = β V ef f = 10µm/0.4µm 190µA/V V = 974µA/V That was still correct but is a wrong conclusion!!! I out = V in g m = 755µA Amplifiers 9

10 Current Mirror Small Signal Model, Example 3.1 (4/4) Continued correct detour solution: r out = = 1 λi out L λli out 0.4µm = 0.16µm/V 100µA = 0.025V/µA = 25kΩ Amplifiers 10

11 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 11

12 Common-Source Amplifier with active load Active: current mirror or FET with fixed gate voltage as load Passive: a resistor to Vdd (normally lower output impedance or excessive voltage across resistor) What are the conditions for this circuit to work as an amplifier? When will it not work? Amplifiers 12

13 Small Signal Model A V = v out v in = g m R 2 = g m (r ds1 r ds2 ) (R in not well motivated in the book: makes only sense with including parasitic capacitances. It is actually attributed to a more reslistic voltage source, i.e. not the transistor.) Amplifiers 13

14 Example 3.2 (1/3) gm1 2 = 2βI bias = µa V 2 12µm 0.5µm I D/bias = 9120 µa V 2 I D r ds1/2 = L = 0.5µm λli D/bias 0.16 µm V 1 I D/bias = 3.125V 1 I D/bias Amplifiers 14

15 Example 3.2 (2/3) A 2 V = 152 = ( g m1 (r ds1 r ds2 ) ) ( 2 = g m1 r ) 2 ds1 2 = 9120 µa ( V 2 I D/bias [3.125V] 2 1 = 22266µA 1 I D/bias I D/bias = 22266µA 15 2 = 99µA 2I D/bias ) 2 Amplifiers 15

16 Example 3.2 (3/3) 2I D V ef f = V in/dc = β = 2 99µA 190 µa V 2 12µm 0.5µm = 208.4mV This V ef f is the DC level (large signal value) of the input! For the V ef f of the pfets one would need to use a different µc ox resp. β! Amplifiers 16

17 Example 3.2 alternative For Common-Source with active load where r ds1 = r ds2 : A V = A i 2 = g m1r ds1 2 V ef f = 1 λ15 = = 208.4mV 0.5µm 0.16 µm V 15 I D = 1 2 βv 2 ef f = 99µA 1 λv ef f Amplifiers 17

18 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 18

19 Output resistance The output resistance of circuit s output is the rate of change of output voltage when forcing an output current. R out = V out I out Consequently, an ideal voltage source has a zero output resistance and an ideal current source has infinite output resistance. Amplifiers 19

20 Output resistance, small signal model R out R out Amplifiers 20

21 Input resistance equivalently the input resistance is the change in voltage seen when forcing a current into a circuit. R in = V in I in The small signal equivalent circuit is a resistance from the input node to ground. If the input is a current signal, R in should be small. If the input is a voltage signal R in should be large. Amplifiers 21

22 What is a small signal analysis good for? determine input and output resistances/impedances determine gain (voltage or current), transconductance, transresistance, transadmittance, transimpedance Amplifiers 22

23 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 23

24 Source Follower What are the conditions for this circuit to work? Would you call it an amplifier? Amplifiers 24

25 Small Signal Model Amplifiers 25

26 Simplified Small Signal Model This circuit is used to follow a voltage but with an offset and small output resistance. A V = g m1 ( 1 1 r ds1 r ds2 ) g m1 g s1 g m1 = g m1 + g s1 + 1 r + 1 ds1 r ds2 1 (but actually somewhat smaller) Amplifiers 26

27 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 27

28 Common Gate Amplifier Similar to common-source amplifier but with a less than infinite input resistance. Good for e.g terminating resistance for input from a (e.g. 50Ω) cable or current input (transresistance). Amplifiers 28

29 Small Signal Model Amplifiers 29

30 Simplified Small Signal Model and A V A V is approximately the same as for the comon source amplifier. (v out v s1 )g ds1 + v out G L v s1 (g m1 + g s1 ) = 0 A V = v out v s1 = g m1 + g s1 + g ds1 G L + g ds1 g m1 G L + g ds1 = g m1 (R L r ds1 ) Amplifiers 30

31 Small Signal Model and r in (1/2) i in = v in (g m1 + g s1 ) (v out v in )g ds1 = v in (g m1 + g s1 + g ds1 ) v in(g m1 + g s1 + g ds1 ) G L + g ( ds1 ) = v in (g m1 + g s1 + g ds1 ) g ds1 1 G L + g ds1 G L = v in (g m1 + g s1 + g ds1 ) G L + g ds1 1 = v in (g m1 + g s1 + g ds1 ) 1 + g ds1 G L g ds1 Amplifiers 31

32 Small Signal Model ans r in (2/2) g in = i in = (g m1 + g s1 + g ds1 ) v in g m g ds1 G L r in 1 g m1 (1 + g ds1 G L ) g ds1 G L Amplifiers 32

33 A realistic voltage source at input Since the input resistance is not infinit, one needs to be careful here: if considering a resistance R S between ideal source and common-gate amplifier, e.g. a 50Ω cable or simply a realistic source with r out > 0, A V will appear smaller! That is to say when you first meaure v in without connecting your source to the amplifier, then connect the amp, and then measure v out, you will not see the A V computed before, but A V (still called A V in the book)! v s1 v in = r in r in + R S r in A V = A V r in + R S A V g m1(r L r ds1 ) g 1 + R m1 +g s1 +g ds1 S Amplifiers 1+R L /r ds1 33

34 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 34

35 Source-Degenerated Current Mirror A first attempt to provide higher output resistance (i.e. a better current source). Note: requires higher voltage at output to work at all! Amplifiers 35

36 Small Signal Model Important point: to compute small signal r out the input current can be considered constant and thus the small signal input current source is an open circuit, i.e. v gs = v s. r out = r ds2 [ 1 + RS (g m2 + g s2 + g ds2 ) ] r ds2 [ 1 + RS (g m2 + g s2 ) ] Amplifiers 36

37 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 37

38 Cascode Current Mirror A second attempt to provide higher output resistance (i.e. a better current source). Note: also requires higher voltage at output to work at all but is less dependent on (large signal) current! Amplifiers 38

39 Small Signal r out The analysis is based on the source degenerated current mirror where former R S is replaced by r ds2 r out = r ds4 [ 1 + rds2 (g m4 + g s4 + g ds4 ) ] r ds4 r ds2 g m4 (3.37) V OUT! > 2V ef f + V tn (3.42) Amplifiers 39

40 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 40

41 Cascode Gain Stage Offers large gain for a single stage (depends on quality of current source!) and limits voltage accross the input drive Book Chapter transistor 3: Basic Current (e.g. Mirrors avoiding and Single-Stage short channel effects). Amplifiers 41

42 Small Signal Model The small signal analysis makes use of the previous common source and common gate analysis, simply multiplying the prvious two voltage gains: v out = v s2 v out (3.54) v in v in v s2 g m1 g m2 (r ds1 r in2 )(r ds2 R L ) Amplifiers 42

43 Small Signal Model Folded Cascode Similar to the telescopic cascode gain stage but non inverting and with an extra load resistor (current source) diminishing the gain: v out v in = v s2 v in v out v s2 (3.54) g m1 g m2 (r ds1 r in2 R L )(r ds2 R L ) Amplifiers 43

44 Assuming High Quality Current Mirrors v out v in g m1 g m2 (r ds1 r in2 )(r ds2 R L ) 1 2 g2 m r 2 ds Amplifiers 44

45 content Simple Current Mirror Common-Source Amplifier Interrupt: A word on output resistance Common-Drain Amplifier with active load / Source Follower Common-Gate Amplifier with active load Source-Degenerated Current Mirrors Cascode Current Mirrors Cascode Gain Stage Diff Pair and Gain Stage Amplifiers 45

46 Diff Pair Realizes a differential input for most integrated amplifiers: immunity to noise/offsets that affect both inputs (e.g. pick-up noise on twisted-pair cables etc.) Amplifiers 46

47 Small signal model The book uses the T-model. Here comes an alternative deduction based on the normal model. i + i - g v m +x r ds g v m -x r ds v x i +/ = (v +/ v x )g m v x 1 r ds i + i = (v + v )g m (3.69) Amplifiers 47

48 Small signal model with resistive loads on each branch i + i - R L R L v out+ v out- g v m +x r ds g v m -x r ds v x 1 i +/ = v out+/ = (v +/ v x )g m + (v out+/ v x ) 1 R L r ds Amplifiers 48

49 Small signal model with resistive loads on each branch i+ i- R L R L vout+ vout- g m v+x rds g mv-x rds vx ( 1 v out+/ + 1 ) ( = v +/ g m v x g m + 1 ) R L r ds r ds (v out+ v out ) = (v + v )g m 1 R L + 1 r ds Amplifiers (v + v )g m R L 49

50 Differential Gain Stage, Transcunductance Amp The simple analysis: consider that the current mirror faithfully copies the current through the left branch and consider a voltage source at the output, then the current into that voltage source is exactly the previous i out = i + i. With no voltage source, the difference in current has to flow trough the output resistance v out = i out r out. With (3.69) and (3.78) v out = g m v in r out A V g m (r ds2 r ds4 ) which is (3.79). Amplifiers 50

51 More Careful Analysis for r out (1/4) (different from the book) 1/g m3 -g m4v1 r ds4 i 1 i out v out v 1 i out = i 2 i 1 i 2 -g v m1 2 r ds1 -g v m2 2 r ds2 v 2 Amplifiers 51

52 More Careful Analysis for r out (2/4) i 1 = 1 r ds4 v out g m4 v 1 1/g m3 v1 -g m4v1 rds4 i1 i out vout i 2 = 1 (v out v 2 ) g m2 v 2 r ds2 1 v out g m2 v 2 r ds2 i2 -g m1v2 rds1 -g m2v2 v2 rds2 i 2 = (v 2 v 1 ) 1 r ds1 + g m1 v 2 v 1 1 r ds1 + g m1 v 2 Amplifiers 52 i 2 = v 1 g m3

53 More Careful Analysis for r out (3/4) -g m1v2 1/g m3 v1 rds1 -g m4v1 rds4 -g m2v2 v2 rds2 i1 i2 i out vout Using 2. and 3. term on previous page for i 2 : v 2 g m3 g m1 v 1 using 1. and 2. term on previous page for i 2 and g m1 = g m2 v out 1 r ds2 v 1 1 r ds1 + g m3 v 1 + g m3g m2 g m1 v 1 2g m3 v 1 v 1 v out 1 2g m3 r ds2 Amplifiers 53

54 More Careful Analysis for r out (4/4) Finally substituting all into i out = i 2 i 1 using the simplest expression for i 2 (3rd on page (2/4)) 1/g m3 -g m4v1 rds4 -g m1v2 v1 rds1 -g m2v2 rds2 i1 vout i out i2 i out = v out 1 2r ds2 + v out 1 2r ds2 + 1 r ds4 v out v2 i out = v out ( 1 r ds2 + 1 r ds4 ) Amplifiers 54

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