EE140 Homework Solutions Problem Set 6 Fall for a single pole roll-off Dominant pole at output:

Transcription

1 EE40 Homework Solutions Problem Set 6 Fall 2009 ) Single-stage op-amp comparison PMOS-input folded cascode Key results are shown in red. a. for a single pole roll-off Dominant pole at output: Plugging in numbers, (See the solutions for homework 3 if you don t know how to get C db or C gd.) M 2RC and M 3RC both carry 40 ua, so assuming V DS = V dsat for maximum swing, 2 So W n = 9.8 um. To give some margin on V ds > V dsat and for a reasonable value for width, W n = 9.75 um. W p = 2W n = 9.5 um. Neglecting channel length modulation, W n = 0um. Therefore,

2 b. c. Current flowing through each transistor Mtail 80 ua ML/R 40 ua M2L/R 80 ua M2CL/R 40 ua M3CL/R 40 ua M3L/R 40 ua For the total current, what we re interested is the total current flowing out of the supply (or into ground same thing). Flowing out of the supply is the current through Mtail, M3L, and M3R: = 60 ua. Flowing into ground is the current through M2L and M2R: = 60 ua. Therefore, the total current consumption of the folded cascode topology is 60 ua. d. The smallest width we re allowed is um, and in general smaller is better in a bias network, because otherwise we re just wasting power. PMOSes are twice as large as NMOSes for the same V DSAT. The first (left) branch is minimum sized, but the current is multiplied by 5 when copied to the next branch because the PMOS in the second branch is sized /5 the size of the other devices to produce an extra V DSAT across itself, with some margin for error due to body effect or mismatch. The third branch is again minimum-sized so as to not waste current. Its purpose is just to flip the mirror for the fourth branch. The size-limiting component in the fourth branch is the NMOS, which is sized down to drop an extra V DSAT just like the PMOS in the second branch. This bias network consumes 4 ua in the first branch, 0 ua in the second branch, 4 ua in the third branch, and 20 ua in the fourth branch for a total of 38 ua.

3 4 ua um VBP 2 um 0 um VBtail 2 um 0 um um 2.5 um um um VBN2 um 2.5 um um VBN

4 Telescopic Cascode Key results are shown in red. Differences in part a from folded cascode are highlighted. a. for a single pole roll-off Dominant pole at output: Plugging in numbers, Double folded cascode R o Double folded cascode A v M RC and M 2RC both carry 40 ua, so assuming V DS = V dsat for maximum swing, 2 So W n = 9.8 um. To give some margin on V ds > V dsat and for a reasonable value for width, W n = 9.75 um. W p = 2W n = 9.5 um. Neglecting channel length modulation, W n = 0um. Therefore, Half folded cascode ω p. Same ω u as folded cascode.

5 b. Vi Vt +Vdsat = Vi 0.3 V c. The two outer branches are each flowing 40 ua, and the middle branch flows ~40 ua if MCbias is sized as 2 um, /5 of the other devices (excluding the tail). Ideally it would flow 32 ua, but this estimation neglects body effect and leaves no room for mismatch. If the middle branch is resized to flow the minimum possible current while maintaining 2V dsat across it, MCbias can be as small as um and flow 6-20 ua. The total current is approximately 00 ua with this sizing. The extra 20 ua is the cost of improving the input common-mode range. d. The telescopic cascode bias network is shown below. The devices are sized identically to those in the folded cascode bias network, except that in this case there is no fourth branch. The total current in this bias network is 8 ua.

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7 Current Mirror Cascode Op-Amp Key results are shown in red. Differences in part a from folded cascode are highlighted. a. for a single pole roll-off Dominant pole at output: Plugging in numbers, M 2RC and M 3RC both carry 40 ua, so assuming V DS = V dsat for maximum swing, 2 So W n = 9.8 um. To give some margin on V ds > V dsat and for a reasonable value for width, W n = 9.75 um. W p = 2W n = 9.5 um. Neglecting channel length modulation, W n = 0um. Therefore, Half folded cascode ω p. Same ω u as folded cascode. NOTE: This is exactly the same as for the telescopic cascode.

8 b. Output Vdd Current Mirror Cascode Op-Amp Vdd - 2Vdsat = = 2.6 V 2Vdsat = 0.4 V Input Vt + 2Vdsat = 0.9 V Vdd Vdd - Vdsat = 2.8 V (Vt+Vdsat from M2M, -Vt from M) c. This circuit has 40 ua flowing in the left branch, 80 ua flowing in the middle branch (through Mtail), and 40 ua flowing in the right branch for a total of 60 ua. d. The bias network for this circuit is drawn below. It s identical to that used to bias the folded cascode in every way. It consumes 38 ua. e. Having now examined each of the topologies, there is no clear winner overall. Each has its advantages in one area and disadvantages in another. If the target is gain then the folded cascode is probably not the way to go, but if the goal is a large swing then it probably is. The telescopic cascode is the lowest power and lowest complexity solution if those are the key metrics. The current mirror cascode op-amp has similar power consumption and a similar level of complexity to the folded cascode, but it trades off a small amount of swing for the higher gain that the folded cascode lacks. The folded cascode has the best bandwidth, but the gain-bandwidth product of the three topologies is identical, so in many cases this won t matter much.

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10 2) OPA344 a. The input common-mode range goes from 0. V below the bottom rail to within.5 V of the top rail, so it s a fair guess that the input stage is a PMOS-input folded cascode (from the table of electrical characteristics). b. The low-frequency gain is 30 db, or 3.6e6 V/V (from the table of electrical characteristics). c. The unity gain frequency is 2 MHz, and the phase margin is about 73 degrees (from the first plot on page 5). d. The slew rate is.6 V/us (from the table of electrical characteristics),.6 MV per second. The maximum slope of a sin wave occurs at the zero crossing, so we evaluate sin 2 2 cos 2 at t = 0, and find that the maximum slope is 2 volts per second. For A = 0. V, the maximum allowable frequency is ) OPA340 a. The unity gain frequency is 5.5 MHz, and the phase margin is around 73 degrees. b. The typical offset voltage is 50 uv, and the worst case is 500 uv. c. The things to notice here are that the G m has roughly doubled compared to the standard folded cascode, the input common mode range is now rail to rail, and the total current (excluding biasing) is 3*I ref. If the current sources were non-ideal transistors as in the first problem, we would lose some output resistance to those transistors (a factor of three, using the numbers from problem ).