EE 435 Lecture 13. Two-Stage Op Amp Design

Transcription

1 EE 435 Lecture 13 Two-Stae Op Amp Desin

2 Review ascades of three or more amplifier staes are seldom used to build a feedback amplifier because of challenes associated with compensatin the amplifier for stability when feedback is applied Two-stae amplifiers are widely used but compensation is critical for use in feedback applications Internal and output compensation are the two compensation strateies used for two-stae amplifiers The pole separation of the two major poles of an open loop amplifier must be lare and enerally satisfies the relationship 4β A > k > β 0TOT A 0TOT There are many different ways to realize a -stae op amp over 18,000 were identified in the last lecture

3 Review from Last Time Two-stae Architectural hoices ascode-ascade Two-Stae Op Amp

4 Review from Last Time Example Solution V DD V X4 V X5 V IN V IN V OUT V X3

5 First ommercial Operational Amplifier K-W Op Amp by Philbrickk,

6 Inventor of the Two-Stae Op Amp Robert Widlar Many say he started the field of analo I desin, considered a brilliant enineer Widlar bean his career at Fairchild semiconductor, where he desined a couple of pioneerin op amps. By 1966, the commercial success of his desins became apparent, and Widlar asked for a raise. He was turned down, and jumped ship to the fledlin National Semiconductor. At National he continued to turn out amazin desins, and was able to retire just before his 30th birthday in (from posted www site)

7 Inventor of the internally-compensated Op Amp Dave Fullaar (from posted www site) Desined the first internally-compensate op amp, the 741 Fullaar was 6 years old when this was desined (introduced?) Introduced in 1968 Larest sellin interated circuit ever Still in hih-volume production even thouh over 40 years old Fullaar later started the linear desin activities at Intersil ofounder (catalyst) of Maxim

8 Analysis of Internally ompensated Two- Stae Op Amps onsider sinle-ended input-output (differential analysis only slihtly different) an t et everythin but can et most of the small-sinal results Since internally compensated, must have p 1 <<p

9 Analysis of Internally ompensated Two- Stae Op Amps For p 1 << p A ( ) 0 A s s s p p 1 BW p 1

10 Analysis of Internally ompensated Two- Stae Op Amps V 3 MP1 V 3 op1 V 4 MP V 4 op V OUT V IN MF1 V 1 of1 MF V V 1 of V L

11 Analysis of Internally ompensated Two- Stae Op Amps A V0 of1 mf1 + op1 of mf + op p ( + ) of L op p 1 ( + ) of1 op1 BW GB p 1 mf1 mf ( of + op )

12 Analysis of Externally ompensated Two- Stae Op Amps an t et everythin but can et most of the small-sinal results

13 Analysis of Externally ompensated Two- Stae Op Amps V 3 MP1 V 3 op1 V 4 MP V 4 op V OUT V IN MF1 V 1 of1 MF V V 1 1 of V

14 Analysis of Externally ompensated Two- Stae Op Amps A V0 of1 mf1 + op1 of mf + op p ( + ) of op p 1 ( + ) of1 1 op1 BW GB p mf1 mf ( of1 + op1 )

15 onsider Aain the Internally ompensated Two-Stae Op Amp Recall very crude compensation requirements: where p k p 1 Thus, very approximately, p 3β A0TOT p 3β of1 4β A > k > β mf1 mf of + op + op1 of + op L of1 + op1 3β 0TOT A 0TOT Since the pole ratio needs to be very lare, ets very lare! 1 ( + ) of mf1 mf op L

16 Miller apacitance - Review 1 1EQ EQ If V -AV 1 1EQ then for A lare 1 A ( + A) A 1 1 EQ + Thus, a lare effective capacitance can be created with a much smaller capacitor if a capacitor brides two nodes with a lare invertin ain!!

17 Miller apacitance - Review 1 1EQ EQ 1EQ If V -AV 1 then for A lare 1 A ( + A) A 1 1 EQ + If A chanes with frequency, 1EQ and EQ are no loner pure capacitors More useful for ivin a concept than for accurate actual analysis because of frequency dependence of A

18 Miller apacitance - Review The Basic oncept from capacitance multiplication [ ] ( ) I X Vx-(-AV X) s VXs 1+A thus V 1 ( ) Z X IN IX s 1+A So, if A is constant, input looks like a capacitor of value ( ) EQ 1+A

19 Miller apacitance - Review VX 1 Z IN IX s 1+A ( ) A s +1 p If A 0 ( s ) s +1+A 0 p G INs ( 1+A) s s +1 p Does not behave as a capacitor for ω > p

20 Internal Miller-ompensated Two-Stae Op Amp Standard ompensation Miller ompensation ompensation capacitance reduced by approximately the ain of the second stae! Since the ain of the second stae is not constant, however, a new analysis is needed

21 Analysis of Internally Miller-ompensated Two-Stae Op Amps

22 Analysis of Internally Miller-ompensated Two-Stae Op Amps p 1 ( of1 + op1) ( ) of + op p + 1 of1 op1 mf A V0 of1 mf1 + op1 BW of p 1 mf + op

23 Analysis of Internally Miller-ompensated Two-Stae Op Amps To find the hih-frequency pole p, the circuit has chaned Note the F block is now diode connected at hih frequencies

24 Analysis of Internally Miller-ompensated Two-Stae Op Amps p ( + ) of L op p mf L A V0 of1 mf1 + op1 of mf + op BW + of op ( + ) of1 op1 mf GB mf1 mf ( of + op ) GB mf1 Has the GB decreased? No, because the decreased by the same factor!

25 Basic Two-Stae Op Amp o One of the most widely used op amp architectures o Essentially just a cascade of two common-source staes o ompensation apacitor used to et wide pole separation o Two poles in amplifier o No universally accepted stratey for desinin this seeminly simple amplifier Pole spread βa A 01 0 makes unacceptably lare

26 Basic Two-Stae Op Amp (with Miller ompensation) o Reduces by approximately A 0 o Pole spread makes size of manaeable βa 01 A 0

27 Basic Two-Stae Miller ompensated Op Amp By inspection A o m1 o + p 1 o1 o o5 m5 + o5 o6 m5 + o6 p GB Will also et these results from a more complete (and time consumin) analysis m5 L m1

28 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent V d V d Norton Equivalent One-Port Two-Port

29 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent V d V d V d V d

30 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent V d V d I X V I X VX + V V m3 + o m1 V d ( o + o4 ) + m m4v4 d ( ) 0 4 X ( + ) X o 04 + X m 1+ Vd m1 m m3 + m4 o + o3 ( ) o + 04 mvd I V + }

31 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent V d V d X X ( o + 04 ) mvd I V + Since M 1 and M are matched as are M 3 and M 4 md m1 od

32 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent oo o mo m5

33 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent

34 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent V d V d

35 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent Solvin we obtain: V OUT This simplifies to: V V OUT ( s + sl + oo) + mov sv V ( s + od ) + mdvd s OUT V d OUT V s V d L + s s } md( mo s ) [ mo + ( ( oo + od ) + Lod )] + oood md L + ( s ) mo s mo + oo od

36 Small Sinal Analysis of Basic Two-Stae Op Amp Differential Small Sinal Equivalent Summary: where ( ) A s s md L + ( s ) m5 s m5 + oo od md m1 m + od o o4 + oo o5 o6

37 Small Sinal Analysis of Two-Stae Miller- ompensated Op Amp ( ) A s s Note this is of the form: md L + ( s ) m5 s ( ) A 1 A s 0 m5 + s +1 z s s p p 1 This has two neative real-axis poles and one positive real-axis zero oo od 1 1

38 End of Lecture 13