Lecture 330 Low Power Op Amps (3/27/02) Page 330-1

Transcription

1 Lecture 33 Low Power Op Amps (3/27/2) Page 33 LECTURE 33 LOW POWER OP AMPS (READING: AH 39342) Objective The objective of this presentation is:.) Examine op amps that have minimum static power Minimize power dissipation Work at low values of power supply Tradeoff speed for less power Outline Weak inversion Methods of creating an overdrive Examples Summary Lecture 33 Low Power Op Amps (3/27/2) Page 332 Subthreshold Operation Most micropower op amps use transistors in the subthreshold region. Subthreshold characteristics: i D i D Square Law Strong Inversion µa Transition na;;;; na Exponential Weak Inversion v GS =V T v GS V T V T v GS v DS V 2V Fig. 7.4A i D = W L I qv GS DO exp nkt (λv DS) g m = qi D nkt and g ds λi D Operation with channel length = L min also will normally be in weak inversion.

2 Lecture 33 Low Power Op Amps (3/27/2) Page 333 TwoStage, Miller Op Amp Operating in Weak Inversion C c v out M M2 v in VBias C L Fig.7.4 Low frequency response: r A vo = g m2 g o2 r o4 m6 r o2 r r o6 r o7 o4 r o6 r o7 = n 2 n 6 (kt/q)2(λ 2 λ 4 )(λ 6 λ 7 ) (No longer ) I D GB and SR: GB = I D (n kt/q)c and SR = I D5 C = 2 I D C = 2GB n kt q = 2GBn V t Lecture 33 Low Power Op Amps (3/27/2) Page 334 Example 7.4 Gain and GB Calculations for Subthreshold Op Amp. Calculate the gain, GB, and SR of the op amp shown above. The currents are I D5 = 2 na and I D7 = 5 na. The device lengths are µm. Values for n are.5 and 2.5 for pchannel and nchannel transistors respectively. The compensation capacitor is 5 pf. Use Table 3.2 as required. Assume that the temperature is 27 C. If =.5V and =.5V, what is the power dissipation of this op amp? Solution The lowfrequency smallsignal gain is, A v = (.5)(2.5)(.26)2(.4.5)(.4.5) = 43,7 V/V The gain bandwidth is 9 GB = 2.5(.26)(5 2) = 37,69 rps 49. khz The slew rate is SR = (2)(3769)(2.5)(.26) =.4 V/µs The power dissipation is, P diss = 3(.7µA) =2.µW

3 Lecture 33 Low Power Op Amps (3/27/2) Page 335 PushPull Output Op Amp in Weak Inversion First stage gain is, A vo = g m2 g m4 = I D2n 4 V t I D4 n 2 V t = I D2n 4 I D4 n 2 Total gain is, M8 v i2 M M2 A vo = g m(s 6 /S 4 ) (g ds6 g ds7 ) = (S 6 /S 4 ) (λ 6 λ 7 )n V t At room temperature (V t =.259V) and for typical device lengths, gains of 6dB can be obtained. The GB is, GB = g m S 6 C S 4 = g mb C M9 V Bias Fig v out C c Lecture 33 Low Power Op Amps (3/27/2) Page 336 Increasing the Gain of the Previous Op Amp.) Can reduce the currents in and and introduce gain in the current mirrors. M8 2.) Use a cascode output stage (can t use selfbiased cascode, currents are too low). g m g m2 A v = 2 R out = g ds6 g ds g m g m 2n n V t g ds7g ds g m = M9 = I 7 2 λ n 2 I 7 2 λ p 2 2I 7 n n V t 2(n n λ n 2n p λ p 2) I 7 I 7 n n V t n p V t Can easily achieve gains greater than 8dB with power dissipation of less than µw. v i2 V Bias M M2 v i M3 V T 2V ON M M2 M5 M V T 2V ON Fig. 7.43A v out C c

4 Lecture 33 Low Power Op Amps (3/27/2) Page 337 Increasing the Output Current for Weak Inversion Operation A significant disadvantage of the weak inversion is that very small currents are available to drive output capacitance so the slew rate becomes very small. Dynamically biased differential amplifier input stage: M2 M8 M9 M2 i i vi2 i i 2 i 2 i 2 vi M M2 M22 A(i 2 i ) I A(i i 2 ) 5 M24 M28 V M29 Bias M25 M26 M27 Fig Note that the sinking current for M and M2 is I sink = A(i 2 i ) A(i i 2 ) where (i 2 i ) and (i i 2 ) are only positive or zero. If v i >v i2, then i 2 >i and the sinking current is increased by A(i 2 i ). If v i2 >v i, then i >i 2 and the sinking current is increased by A(i i 2 ). M23 Lecture 33 Low Power Op Amps (3/27/2) Page 338 Dynamically Biased Differential Amplifier Continued How much output current is available from this circuit if there is no current gain from the input to output stage? Assume transistors M8 through M2 are equal to and and that transistors M22 through M27 are all equal. W 28 Let L 28 = A W 26 L 26 and W 29 L 29 = A W 27 L 27 The output current available can be found by assuming that v in = v i v i2 >. i i 2 = A(i 2 i ) The ratio of i 2 to i can be expressed as i 2 v in i = exp nv t Defining the output current as i OUT = b(i 2 i ) and combining the above two equations gives, v in b exp nv t v in i OUT = (A) (A)exp v i in OUT = when A = 2.6 and nv t = nv t where b corresponds to any current gain through current mirrors ( and M8).

5 Lecture 33 Low Power Op Amps (3/27/2) Page 339 Overdrive of the Dynamically Biased Differential Amplifier The enhanced output current is accomplished by the use of positive feedback (M28M2M9M28). The loop gain is, g m28 g m9 LG = g m4 g m26 = A g m9 g m4 = A Note that as the output current increases, the transistors leave the weak inversion region and the above analysis is no longer valid. I OUT 2 A = 2 A =.5 A = A =.3 A = 2 v IN nv t Fig Lecture 33 Low Power Op Amps (3/27/2) Page 33 Increasing the Output Current for Strong Inversion Operation An interesting technique is to bias the output transistor of a current mirror in the active region and then during large overdrive cause the output transistor to become saturated causing a significant current gain. Illustration: 53µA i i 2 M M2 V ds2 Current µa.v ds (sat) i 2 for W 2 /L 2 = 5.3(W /L ) i V ds (sat) Volts Fig. 7.46

6 Lecture 33 Low Power Op Amps (3/27/2) Page 33 Example 7.42 Current Mirror with M2 operating in the Active Region Assume that M2 has a voltage across the drainsource of.v ds (sat). Design the W 2 /L 2 ratio so that I = I 2 = µa if W /L =. Find the value of I 2 if M2 is saturated. Solution Using the parameters of Table 3.2, we find that the saturation voltage of M2 is 2I V ds (sat) = K N (W 2 /L 2 ) = 2 =.4264V Now using the active equation of M2, we set I 2 = µa and solve for W 2 /L 2. µa = K N (W 2 /L 2 )[V ds (sat) V ds2.5v ds2 2] = µa/v2(w 2 /L 2 )[ ]V2 =.883x6(W 2 /L 2 ) Thus, =.883(W 2 /L 2 ) W 2 L 2 = 53.2 Now if M2 should become saturated, the value of the output current of the mirror with µa input would be 53µA or a boosting of 5.3 times I. Lecture 33 Low Power Op Amps (3/27/2) Page 332 Implementation of the Current Mirror Boosting Concept i M8 M7 M M9 M8 i M3 M2 2 v i M M2 v i2 M22 ki M29 ki 2 v o i i 2 M27 M28 i i 2 v o2 ki 2 ki M5 M23 V Bias M25 i 2 i M26 M24 M6 M M2 M9 k = overdrive factor of the current mirror M2 Fig.7.47

7 Lecture 33 Low Power Op Amps (3/27/2) Page 333 A Better Way to Achieve the Current Mirror Boosting It was found that when the current mirror boosting idea illustrated on the previous slide was used that when the current increased through the cascode device (M6) that V GS6 increased limiting the increase of V DS2. This can be overcome by the following circuit. i in I B i in I B ki in 5/ / / / M M2 2/ Fig. 7.47A Lecture 33 Low Power Op Amps (3/27/2) Page 334 SUMMARY Operation of transistors is generally in weak inversion Boosting techniques are needed to get output sourcing and sinking currents that are larger than that available during quiescent operation Be careful about using circuits at weak inversion, i.e. the selfbiased cascode will cause the resistor to be too large