1、Chapter 13: Introduction to Switched-Capacitor Circuits13.1 General Considerations13.2 Sampling Switches13.3 Switched-Capacitor Amplifiers13.4 Switched-Capacitor Integrator13.5 Switched-Capacitor Common-Mode Feedback2General ConsiderationsFor continuous-time amplifier Fig. (a), Vout/Vin = -R2/R1 ide

2、allyDifficult to implement in CMOS technologyTypically, open-loop output resistance of CMOS op-amps is maximized to maximize AvR2 heavily drops open-loop gain, affecting precision3General ConsiderationsIn equivalent circuit of Fig. (b), we can writeHence,Closed-loop gain is inaccurate compared to wh

3、en Rout = 0General Considerations4To reduce open-loop gain, resistors can be replaced by capacitors Fig. (a)Gain of this circuit is ideally C1/C2To set bias voltage at node X, large feedback resistor can be added Fig. (b)General Considerations5Feedback resistor is not suited to amplify wideband sign

4、als Charge on C2 is lost through RF resulting in “tail”Circuit exhibits high-pass transfer function given byDdd only if General Considerations6RF can be replaced by a switch S2 is turned on to place op amp in unity gain feedback to force VX equal to VB, a suitable common-mode valueWhen S2 turns off,

5、 node X retains the voltage allowing amplification When S2 is on, circuit does not amplify VinGeneral Considerations7In above circuit, S1 and S3 connect left plate of C1 to Vin and ground, S2 for unity-gain feedbackAssume large open-loop gain of op ampFirst phase: S1 and S2 on, S3 off Fig. (a)Genera

6、l Considerations8Here, and C1 samples the input VinSecond phase: At t = t0, S1 and S2 turn off and S3 turns on, pulling node A to ground Fig. (b)VA changes from Vin to 0, therefore Vout must change from zero to Vin0C1/C2 Fig. (c)General Considerations9Circuit devotes some time to sample input, setti

7、ng output to zero and providing no amplification After sampling, for t t0, circuit ignores input voltage, amplifies sampled voltage General Considerations10Switched-capacitor amplifiers operate in two phases: Sampling and AmplificationClock needed in addition to analog input VinMOSFETS as Switches11

8、Sampling circuit consists of a switch and a capacitor Fig. (a) MOS transistor can function as switch Fig. (b) since it can be on while carrying zero current MOSFETS as Switches12CK goes high at t = t0Assume Vin = 0 and capacitor has initial voltage VDDAt t = t0, M1 is in saturation and draws current

9、 As Vout falls, at some point M1 goes into triode regionCH is discharged until Vout reaches zero For Vout 2(VDD - VTH), transistor is an equivalent resistor MOSFETS as Switches13If Vin = +1 V, Vout(t = t0) = +0 V and VDD = +3 VTerminal of M1 connected to CH acts as source, and the transistor turns o

10、n with VGS = +3 V but VDS = +1 VM1 operates in triode region and charges CH until Vout approaches +1 VFor Vout +1 V, M1 exhibits an on-resistance of MOSFETS as Switches14When switch is on Fig. (a), Vout follows VinWhen switch is off Fig. (b), Vout remains constantCircuit “tracks” signal when CK is h

11、igh and “freezes” instantaneous value of Vin across CH when CK goes lowMOSFETS as Switches15Suppose Vin = V0 instead of +1 VM1 is saturated and we have:Solving, As t , Vout VDD - VTH so NMOS cannot pull up to VDD MOSFETS as Switches16Similarly, PMOS transistor fails to operate as a switch if gate is

12、 grounded and drain senses an input voltage of |VTHP| or less On resistance rises rapidly as input and output levels fall to |VTHP| above ground MOSFETS as Switches: Speed Considerations17Measure of speed is the time required for output to go from zero to the maximum input level after switch turns o

13、n Consider output settled within a certain “error band” V around final value If output settles to 0.1% accuracy after tS seconds, then V/Vin0 = 0.1% After t = tS, consider source and drain voltages to be approximately equal MOSFETS as Switches: Speed Considerations18Sampling speed is given by two fa

14、ctors: switch on-resistance and sampling capacitance For higher speed, large aspect ratio and small capacitance are needed On-resistance also depends on input level for both NMOS and PMOS MOSFETS as Switches: Speed Considerations19To allow greater input swings, we can use “complementary” switches, r

15、equiring complementary clocks Fig. (a)Equivalent on-resistance shows following behavior Fig. (b), revealing much less variation MOSFETS as Switches: Speed Considerations20For high speed signals, NMOS and PMOS switches must turn off simultaneously to avoid ambiguity in sampled value If NMOS turns off

16、 t seconds before PMOS, output tends to track input for the remaining t seconds, causing distortionFor moderate precision, circuit below is used to provide complementary clocks MOSFETS as Switches: Precision Considerations21Speed trades with precision Channel Charge Injection:For MOSFET to be on, a

17、channel must exist at the oxide-silicon interface Assuming Vin Vout, total charge in the inversion layer is When switch turns off, Qch exits through the source and drain terminals (“channel charge injection”)MOSFETS as Switches: Precision Considerations22Charge injected to the left is absorbed by in

18、put source, creating no errorCharge injected to the right deposited on CH, introducing error in voltage stored on capacitor For half of Qch injected onto CH, error (negative pedestal) equalsMOSFETS as Switches: Precision Considerations23If all of the charge is deposited on CH,Since we assume Qch is

19、a linear function of Vin, circuit exhibits only gain error and dc offsetMOSFETS as Switches: Precision Considerations24Clock Feedthrough:MOS switch couples clock transitions through CGD or CGSSampled output voltage has error due to this give byCov is the overlap capacitance per unit widthError V is

20、independent of input level, manifests as constant offset in the input/output characteristic MOSFETS as Switches: Precision Considerations25kT/C Noise:Resistor charging a capacitor gives a total RMS noise voltage of On resistance of switch introduces thermal noise at output which is stored on the cap

21、acitor when switch turns offRMS voltage of sampled noise is still approximately equal to Charge Injection Cancellation26Charge injected by main transistor removed by a dummy transistor M2M2 is driven by so that after M1 turns off and M2 turns on, channel charge deposited by M1 on CH is absorbed by M

22、2 to create a channel If W2 = 0.5W1, then charge injected by M1, q1 is equal to that absorbed by M2 27Charge Injection CancellationIf W2 = 0.5W1 and L2 = L1, effect of clock feedthrough is suppressedTotal change in Vout is zero because28Charge Injection CancellationIncorporate both PMOS and NMOS dev

23、ices so that opposite charge packets injected cancel each otherFor q1 to cancel q2, we must have Cancellation occurs for only one input levelClock feedthrough is not completely suppressed since CGD of NFETs is not equal to that PFETs29Charge Injection CancellationCharge injection appears as a common

24、-mode disturbance, may be countered by differential operationq1 = q2 only if Vin1 = Vin2, thus overall error is not suppressed for differential signalsRemoves constant offset and nonlinear component 30Unity-Gain Sampler/ BufferFor discrete-time applications, unity-gain amplifier Fig. (a) requires a

25、sampling circuit Fig. (b)Accuracy limited by input-dependent charge injected by S1 onto CH31Unity-Gain Sampler/ BufferConsider the topology shown in Fig. (a)In sampling mode, S1 and S2 are on, S3 is off yielding circuit in Fig. (b)Thus, Vout = VX 0, and the voltage across CH tracks VinAt t = t0, whe

26、n Vin = V0, S1 and S2 turn off and S3 turns on, yielding circuit of Fig. (c) amplification modeOp amp requires node X is still a virtual ground, Vout rises to approximately V0 “frozen” for processing by subsequent stages32Unity-Gain Sampler/ BufferS2 turns off slightly before S1 during transition fr

27、om sampling mode to amplification modeCharge injected by S2, q2 is input-independent and constant, producing only an offsetAfter S2 turns off, total charge at node X stays constant and charge injected by S1 does not affect output voltage 33Unity-Gain Sampler/ BufferInput-independent charge injected

28、by S2 can be cancelled by differential operation as shown Charge injected by S2 and S2 appears as common-mode disturbance at nodes X and YCharge injection mismatch between S2 and S2 resolved by adding another switch Seq that turns off slightly after S2 and S2, equalizing the charge at nodes X and Y3

29、4Unity-Gain Sampler/ BufferPrecision Considerations:Assume op-amp has a finite input capacitance Cin and calculate output voltage when circuit goes from sampling to amplification modeIt can be shown from the above fig. thatCircuit suffers from gain error of approximately 35Unity-Gain Sampler/ Buffer

30、Speed Considerations:In sampling mode, circuit appears as in Fig. (a)Use equivalent circuit of Fig. (b) to find time constant in sampling modeTotal resistance in series with CH is Ron1 and the resistance between X and ground, RX36Unity-Gain Sampler/ BufferSince typically and ,Time constant in sampli

31、ng mode is thusConsider circuit as it enters amplification modeCircuit must begin with Vout 0 and eventually produce Vout V0For relatively small Cin, voltages across CL and CH do not change instantaneously so that VX = -V0 at the beginning of amplification37Unity-Gain Sampler/ BufferRepresent charge

32、 on CH by a voltage source VS that goes from zero to V0 at t = t0, while CH carries no charge itselfThe transfer function Vout(s)/Vin(s) can be obtained asThis response is characterized by a time constant independent of op-amp output resistance38Noninverting AmplifierIn non-inverting amplifier of Fi

33、g. (a), in sampling mode, S1 and S2 are on while S3 is off, creating a virtual ground at X and allowing voltage across C1 to track Vin Fig. (b)39Noninverting AmplifierAt the end of sampling mode, S2 turns off first, injecting a constant charge q2 onto node X, after which S1 turns off and S3 turns on

34、 Fig. (c)Since VP goes from Vin0 to 0, output voltage changes from 0 to approximately Vin0(C1/C2), providing a gain of C1/C2Called a “noninverting amplifier” since output polarity is the same as Vin0 and the gain can be greater than unity40Noninverting AmplifierNoninverting amplifier avoids input-de

35、pending charge injection by turning off S2 before S1After S2 is off, total charge at node X remains constant, making the circuit insensitive to charge injection of S1 or charge “absorption” of S341Noninverting AmplifierCharge injected by S1, q1 changes voltage at node P by VP = q1/C1 and output volt

36、age by -q1C1/C2After S3 turns on, VP becomes zero so overall change in VP is 0 Vin0 = -Vin0, producing overall change in output of Vin0(-C1/C2) = Vin0C1/C2VP goes from V0 to 0 with a perturbation due to S1Since output is measure after node P is connected to ground, charge injected by S1 does not aff

37、ect final output42Noninverting AmplifierPrecision Considerations:Calculate actual gain if op amp has finite open-loop gain of Av1 and input capacitance CinIt can be shown that Amplifier suffers from a gain error ofGain error increases with the nominal gain C1/C2 43Noninverting AmplifierSpeed Conside

38、rations:Consider equivalent circuit in amplification mode Fig. (a)It can be shown for a large GmR0 that This gives a time constant of 44Precision Multiply-by-Two CircuitTopology shown in Fig. (a) provides a nominal gain of two while achieving higher speed and lower gain errorIncorporates two equal c

39、apacitors C1 = C2 = CIn sampling mode Fig. (b), node X is a virtual ground, allowing voltage across C1 and C2 to track Vin45Precision Multiply-by-Two CircuitDuring transition to amplification mode Fig. (c), S3 turns off first, placing C1 around op-amp and left plate of C2 is grounded At the moment S

40、3 turns off, total charge on C1 and C2 equals 2Vin0C and since voltage across C2 approaches zero in amplification mode, final voltage across C1 and hence output are approximately 2Vin0(c)46Switched-Capacitor Integrator Output of a continuous-time integrator can be expressed asIn Fig. (a), resistor R

41、 carries a current of (VA VB)/RIn circuit of Fig. (b), CS is alternately connected to nodes A and B at a clock rate fCKAverage current flowing from A to B is the charge moved in one clock period Can be viewed as a resistor of value 47Switched-Capacitor Integrator Fig. (a) shows discrete-time integra

42、torIn every clock cycle, C1 absorbs a charge equal to C1Vin when S1 is on and deposits it on C2 when S2 is onIf Vin is constant, output changes by VinC1/C2 every clock cycle Fig. (b)Final value of Vout after clock cycle can be written asSwitched-Capacitor Integrator 48Input-dependent charge injectio

43、n of S1 introduces nonlinearity in output voltage Nonlinear capacitance at node P resulting from source/drain junctions of S1 and S2 leads to a nonlinear charge-to-voltage conversion when C1 is switched to XCharge stored on the total junction capacitance, Cj is not equal to Vin0Cj, but rather equal to 49Switched-Capacitor Integrator Circuit of Fig. (a) resolves the issues in the simple integratorIn sampling mode Fig. (b), S1 and S3 are on, S2 and S4 are off, allowing voltage across C1 to track Vin while op amp and C2 hold previous valueIn the transition to integration mode, S3 turns