Switched-Capacitor Circuits
David Johns and Ken Martin University of Toronto ([email protected]) ([email protected])
University of Toronto 1 of 60
© D. Johns, K. Martin, 1997
Basic Building Blocks Opamps • Ideal opamps usually assumed. • Important non-idealities — dc gain: sets the accuracy of charge transfer, hence, transfer-function accuracy. — unity-gain freq, phase margin & slew-rate: sets the max clocking frequency. A general rule is that unity-gain freq should be 5 times (or more) higher than the clock-freq. — dc offset: Can create dc offset at output. Circuit techniques to combat this which also reduce 1/f noise.
University of Toronto 2 of 60
© D. Johns, K. Martin, 1997 Basic Building Blocks Double-Poly Capacitors
metal C1 metal poly1
Cp1 thin oxide bottom plate C1 poly2 Cp2 thick oxide C p1 Cp2
(substrate - ac ground)
cross-section view equivalent circuit • Substantial parasitics with large bottom plate capacitance (20 percent of C1 ) • Also, metal-metal capacitors are used but have even larger parasitic capacitances.
University of Toronto 3 of 60
© D. Johns, K. Martin, 1997
Basic Building Blocks Switches I I Symbol n-channel
v1 v2 v1 v2
I transmission I I gate v1 v p-channel v 2 1 v2 I • Mosfet switches are good switches. — off-resistance near G: range — on-resistance in 100: to 5k: range (depends on transistor sizing) • However, have non-linear parasitic capacitances.
University of Toronto 4 of 60
© D. Johns, K. Martin, 1997 Basic Building Blocks Non-Overlapping Clocks I1 T Von I I1 Voff n – 2 n – 1 n n + 1 tTe delay 1 I fs { --- delay V 2 T on I Voff 2 n – 32e n – 12e n + 12e tTe • Non-overlapping clocks — both clocks are never on at same time • Needed to ensure charge is not inadvertently lost.
• Integer values occur at end of I1 .
• End of I2 is 1/2 off integer value.
University of Toronto 5 of 60
© D. Johns, K. Martin, 1997
Switched-Capacitor Resistor Equivalent I I 1 2 Req V V V1 V2 1 2 C1 T 'QC= V – V every clock period Req = ------1 1 2 C1
Qx = CxVx (1)
• C1 charged to V1 and then V2 during each clk period.
'Q1 = C1 V1 – V2 (2) • Find equivalent average current C V – V I = ------1 1 2 (3) avg T where T is the clk period.
University of Toronto 6 of 60
© D. Johns, K. Martin, 1997 Switched-Capacitor Resistor Equivalent • For equivalent resistor circuit
V1 – V2 Ieq = ------(4) Req • Equating two, we have T 1 Req ==------(5) C1 C1fs • This equivalence is useful when looking at low-freq portion of a SC-circuit. • For higher frequencies, discrete-time analysis is used.
University of Toronto 7 of 60
© D. Johns, K. Martin, 1997
Resistor Equivalence Example • What is the equivalent resistance of a 5pF capacitance sampled at a clock frequency of 100kHz . • Using (5), we have 1 R ==------2M: eq –12 3 510u 100u 10 • Note that a very large equivalent resistance of 2M: can be realized. • Requires only 2 transistors, a clock and a relatively small capacitance. • In a typical CMOS process, such a large resistor would normally require a huge amount of silicon area.
University of Toronto 8 of 60
© D. Johns, K. Martin, 1997 Parasitic-Sensitive Integrator
vc2()nT
I1 I2 C vcx()t 2 I1 vci()t vco()t
vc1()t C1 v ()n = v ()nT vi()n = vci()nT o co
• Start by looking at an integrator which IS affected by parasitic capacitances
• Want to find output voltage at end of I1 in relation to input sampled at end of I1 .
University of Toronto 9 of 60
© D. Johns, K. Martin, 1997
Parasitic-Sensitive Integrator
C2 C2 v ()nT– T ci vci()nT– T e 2 C 1 v ()nT– T C1 co vco()nT– T e 2
I1 on I2 on
• At end of I2
C2vco()nT– T e 2 = C2vco()nT– T – C1vci()nT– T (6)
• But would like to know the output at end of I1
C2vco()nT = C2vco()nT– T e 2 (7) • leading to
C2vco()nT = C2vco()nT– T – C1vci()nT– T (8)
University of Toronto 10 of 60
© D. Johns, K. Martin, 1997 Parasitic-Sensitive Integrator • Modify above to write
C1 vo()n = vo()n – 1 – ------vi()n – 1 (9) C2 and taking z-transform and re-arranging, leads to
Vo()z §·C1 1 Hz(){ ------= –¨¸------(10) Vi()z ©¹C2 z – 1 • Note that gain-coefficient is determined by a ratio of two capacitance values. • Ratios of capacitors can be set VERY accurately on an integrated circuit (within 0.1 percent) • Leads to very accurate transfer-functions.
University of Toronto 11 of 60
© D. Johns, K. Martin, 1997
Typical Waveforms
I1
t I2
t vci()t
t
vcx()t
t
vco()t
t
University of Toronto 12 of 60
© D. Johns, K. Martin, 1997 Low Frequency Behavior • Equation (10) can be re-written as
§·C z–12/ Hz()= – ------1 ------(11) ¨¸12/ –12/ ©¹C2 z – z • To find freq response, recall
ze==jZT cos ZT + jsin ZT (12) ZT ZT z12/ = cos§·------+ jsin§·------(13) ©¹2 ©¹2 ZT ZT z–12/ = cos§·------– jsin§·------(14) ©¹2 ©¹2 ZT ZT cos§·------– jsin§·------jZT §·C1 ©¹2 ©¹2 He()= –¨¸------(15) C ZT ©¹2 j2sin§·------©¹2
University of Toronto 13 of 60
© D. Johns, K. Martin, 1997
Low Frequency Behavior • Above is exact but when ZT « 1 (i.e., at low freq)
jZT §·C1 1 He()# –¨¸------(16) ©¹C2 jZT • Thus, the transfer function is same as a continuous- time integrator having a gain constant of
C1 1 KI # ------(17) C2T which is a function of the integrator capacitor ratio and clock frequency only.
University of Toronto 14 of 60
© D. Johns, K. Martin, 1997 Parasitic Capacitance Effects Cp3 Cp4 C2
I1 I2 I1 vi()n vo()n C1 Cp1
Cp2 • Accounting for parasitic capacitances, we have
§·C1 + Cp1 1 Hz()= –¨¸------(18) ©¹C2 z – 1 • Thus, gain coefficient is not well controlled and partially non-linear (due to Cp1 being non-linear).
University of Toronto 15 of 60
© D. Johns, K. Martin, 1997
Parasitic-Insensitive Integrators
C2
I1 I C 2 1 I v ()t 1 ci vco()t
I2 I1 vi()n = vci()nT vo()n = vco()nT
• By using 2 extra switches, integrator can be made insensitive to parasitic capacitances — more accurate transfer-functions — better linearity (since non-linear capacitances unimportant)
University of Toronto 16 of 60
© D. Johns, K. Martin, 1997 Parasitic-Insensitive Integrators C 2 C2 vci()nT– T vci()nT– T e 2 + - C - 1 C1 vco()nT– T + vco()nT– T e 2
I1 on I2 on
• Same analysis as before except that C1 is switched in polarity before discharging into C2 .
Vo()z §·C1 1 Hz(){ ------= ¨¸------(19) Vi()z ©¹C2 z – 1 • A positive integrator (rather than negative as before)
University of Toronto 17 of 60
© D. Johns, K. Martin, 1997
Parasitic-Insensitive Integrators
Cp3 Cp4
C2 I1 I C 2 1 I1 vi()n vo()n I2 I1
Cp1 Cp2
• Cp3 has little effect since it is connected to virtual gnd
• Cp4 has little effect since it is driven by output
• Cp2 has little effect since it is either connected to virtual gnd or physical gnd.
University of Toronto 18 of 60
© D. Johns, K. Martin, 1997 Parasitic-Insensitive Integrators
• Cp1 is continuously being charged to vi()n and discharged to ground.
• I1 on — the fact that Cp1 is also charged to vi()n – 1 does not affect C1 charge.
• I2 on — Cp1 is discharged through the I2 switch attached to its node and does not affect the charge accumulating on C2 . • While the parasitic capacitances may slow down settling time behavior, they do not affect the discrete- time difference equation
University of Toronto 19 of 60
© D. Johns, K. Martin, 1997
Parasitic-Insensitive Inverting Integrator C2 I1 I1 C1 I1 vi()n vo()n Vi()z I2 I2 Vo()z
C2vco()nT– T e 2 = C2vco()nT– T (20)
C2vco()nT = C2vco()nT– T e 2 – C1vci()nT (21) • Present output depends on present input(delay-free)
Vo()z §·C1 z Hz(){ ------= –¨¸------(22) Vi()z ©¹C2 z – 1 • Delay-free integrator has negative gain while delaying integrator has positive gain.
University of Toronto 20 of 60
© D. Johns, K. Martin, 1997 Signal-Flow-Graph Analysis
C1 V1()z I I 1 C 2 2 CA V2()z I I 2 1 I1 V ()z I1 I1 o C3 V3()z I2 I2
–1 –C1 1 – z V1()z –1 C2z 1 1 §·------V ()z –1 V ()z 2 ©¹CA 1 – z o
–C3 V3()z
University of Toronto 21 of 60
© D. Johns, K. Martin, 1997
First-Order Filter
Vin()s Vout()s
• Start with an active-RC structure and replace resistors with SC equivalents. • Analyze using discrete-time analysis.
University of Toronto 22 of 60
© D. Johns, K. Martin, 1997 First-Order Filter
I I 1 1 I1 I1 C2 C3
I2 I2 I2 I2
CA I C1 1 Vi()z Vo()z
–C3
–C2 1 1 §·------V ()z Vi()z –1 o ©¹CA 1 – z –1 –C1 1 – z
University of Toronto 23 of 60
© D. Johns, K. Martin, 1997
First-Order Filter –1 –1 CA 1 – z Vo()z = – C3Vo()z – C2Vi()z – C1 1 – z Vi()z (23)
§·C1 §·C2 ¨¸------1 – z–1 + ¨¸------©¹CA ©¹CA –------Vo()z C Hz(){ ------= 1 – z–1 + ------3- (24) V ()z i CA
§·C1 + C2 C1 ¨¸------z – ------©¹CA CA = –------§·C3 ¨¸1 + ------z – 1 ©¹CA
University of Toronto 24 of 60
© D. Johns, K. Martin, 1997 First-Order Filter • The pole of (24) is found by equating the denominator to zero
CA zp = ------(25) CA + C3
• For positive capacitance values, this pole is restricted to the real axis between 0 and 1 — circuit is always stable. • The zero of (24) is found to be given by
C1 zz = ------(26) C1 + C2 • Also restricted to real axis between 0 and 1.
University of Toronto 25 of 60
© D. Johns, K. Martin, 1997
First-Order Filter The dc gain is found by setting z = 1 which results in –C H()1 = ------2 (27) C3 • Note that in a fully-differential implementation, effective negative capacitances for C1 , C2 and C3 can be achieved by simply interchanging the input wires. • In this way, a zero at z = –1 could be realized by setting
C1 = –0.5C2 (28)
University of Toronto 26 of 60
© D. Johns, K. Martin, 1997 First-Order Example • Find the capacitance values needed for a first-order SC-circuit such that its 3dB point is at 10kHz when a clock frequency of 100kHz is used. • It is also desired that the filter have zero gain at 50kHz (i.e. z = –1 ) and the dc gain be unity.
• Assume CA = 10pF . Solution • Making use of the bilinear transform pz=1 – 1 e z + 1 the zero at – is mapped to :f= . • The frequency warping maps the -3dB frequency of 10kHz (or 0.2S rad/sample ) to
University of Toronto 27 of 60
© D. Johns, K. Martin, 1997
First-Order Example 0.2S : ==tan§·------0.3249 (29) ©¹2 • in the continuous-time domain leading to the continuous-time pole, pp , required being
pp = –0.3249 (30)
• This pole is mapped back to zp given by
1 + pp zp ==------0.5095 (31) 1 – pp • Therefore, Hz() is given by kz + 1 Hz()= ------(32) z – 0.5095
University of Toronto 28 of 60
© D. Johns, K. Martin, 1997 First-Order Example • where k is determined by setting the dc gain to one (i.e. H()1 = 1) resulting 0.24525 z + 1 Hz()= ------(33) z – 0.5095 • or equivalently, 0.4814z + 0.4814 Hz()= ------(34) 1.9627z – 1 • Equating these coefficients with those of (24) (and assuming CA = 10pF ) results in
C1 = 4.814pF (35) C2 = –9.628pF (36) C3 = 9.628pF (37)
University of Toronto 29 of 60
© D. Johns, K. Martin, 1997
Switch Sharing
I1 C3
I2 C1 CA
I1 I1 C2 ) Vo()z I2 I2
• Share switches that are always connected to the same potentials.
University of Toronto 30 of 60
© D. Johns, K. Martin, 1997 Fully-Differential Filters • Most modern SC filters are fully-differential • Difference between two voltages represents signal (also balanced around a common-mode voltage). • Common-mode noise is rejected. • Even order distortion terms cancel
2 3 4 v = k v ++++k v k v k v p1 1 1 2 1 3 1 4 1 } v nonlinear 1 element 3 5 + v = 2k v +++2k v 2k v diff 1 1 3 1 5 1 } - –v nonlinear 1 element 2 3 4 v = – k v +++k v – k v k v n1 1 1 2 1 3 1 4 1 }
University of Toronto 31 of 60
© D. Johns, K. Martin, 1997
Fully-Differential Filters C1 C 3 I1
I2 C I2 A I2 + C2 I1 I1 + V ()z V ()z i I o I1 1 - - C2 C I2 I2 A I2
C 3 I1
C1
University of Toronto 32 of 60
© D. Johns, K. Martin, 1997 Fully-Differential Filters • Negative continuous-time input — equivalent to a negative C1 C1 C 3 I1
I2 C I2 A I2 + C2 I1 I1 + V ()z V ()z i I o I1 1 - - C2 C I2 I2 A I2
C 3 I1
C1
University of Toronto 33 of 60
© D. Johns, K. Martin, 1997
Fully-Differential Filters • Note that fully-differential version is essentially two copies of single-ended version, however ... area penalty not twice. • Only one opamp needed (though common-mode circuit also needed) • Input and output signal swings have been doubled so that same dynamic range can be achieved with half capacitor sizes (from kTe C analysis) • Switches can be reduced in size since small caps used. • However, there is more wiring in fully-differ version but better noise and distortion performance.
University of Toronto 34 of 60
© D. Johns, K. Martin, 1997 Low-Q Biquad Filter 2 Vout()s k2s ++k1sko Ha()s { ------= –------(38) Vin()s 2 Zo 2 s ++§·------s Z ©¹Q o
1 e Zo Q e Zo C = 1 A Vc1()s CB = 1 Z e k o o –1 e Zo Vin()s Vout()s 1 e k1
k2
University of Toronto 35 of 60
© D. Johns, K. Martin, 1997
Low-Q Biquad Filter
I1 I1 K4C1
I2 I2 I1 I1 K6C2 I I C 2 2 1 C V1()z 2 I1 K C I1 1 1 I1 I2 Vi()z K5C2 I1 Vo()z I2 I2 I2 I1
I1 K C I1 2 2
I2 I2
K3C2
University of Toronto 36 of 60
© D. Johns, K. Martin, 1997 Low-Q Biquad Filter
–K4
–K6 V1()z –K 1 K z–1 1 V ()z 1 ------5 ------V ()z i 1 – z–1 1 – z–1 o
–K2 –1 –1K3 – z
2 Vo()z K2 + K3 z ++ K1K5 –2K2 – K3 zK3 Hz(){ ------= –------(39) 2 Vi()z 1 + K6 z ++ K4K5 –2K6 – z 1
University of Toronto 37 of 60
© D. Johns, K. Martin, 1997
Low-Q Biquad Filter Design 2 a z ++a za Hz()= –------2 1 0 (40) 2 b2z ++b1z 1 • we can equate the individual coefficients of “z” in (39) and (40), resulting in:
K3 = a0 (41) K2 = a2 – a0 (42) K1K5 = a0 ++a1 a2 (43) K6 = b2 – 1 (44) K4K5 = b1 ++b2 1 (45) • A degree of freedom is available here in setting internal V1()z output
University of Toronto 38 of 60
© D. Johns, K. Martin, 1997 Low-Q Biquad Filter Design • Can do proper dynamic range scaling • Or let the time-constants of 2 integrators be equal by
K4 ==K5 b1 ++b2 1 (46) Low-Q Biquad Capacitance Ratio • Comparing resistor circuit to SC circuit, we have
K4 ||K5 ZoT (47)
ZoT K | ------(48) 6 Q • However, the sampling-rate, 1 e T , is typically much larger that the approximated pole-frequency, Zo ,
ZoT « 1 (49)
University of Toronto 39 of 60
© D. Johns, K. Martin, 1997
Low-Q Biquad Capacitance Ratio • Thus, the largest capacitors determining pole positions are the integrating capacitors, C1 and C2 .
• If Q 1 , the smallest capacitors are K4C1 and K5C2 resulting in an approximate capacitance spread of
1 e ZoT . • If Q ! 1 , then from (48) the smallest capacitor would be K6C2 resulting in an approximate capacitance
spread of Q e ZoT — can be quite large for Q » 1
University of Toronto 40 of 60
© D. Johns, K. Martin, 1997 High-Q Biquad Filter • Use a high-Q biquad filter circuit when Q » 1 • Q-damping done with a cap around both integrators • Active-RC prototype filter
1 e Zo 1 e Q C = 1 1 C2 = 1 Zo e ko –1 e Z Vin()s o Vout()s k1 e Zo
k2
University of Toronto 41 of 60
© D. Johns, K. Martin, 1997
High-Q Biquad Filter
I1 K C I1 4 1
I2 I2 K6C1 C 1 V ()z 1 C2 I1 K C I1 1 1 I I2 V ()z 1 K5C I1 i 2 Vo()z I2 I2 I2 I1
K C 2 1
K3C2
• Q-damping now performed by K6C1
University of Toronto 42 of 60
© D. Johns, K. Martin, 1997 High-Q Biquad Filter
• Input K1C1 : major path for lowpass
• Input K2C1 : major path for band-pass filters
• Input K3C2 : major path for high-pass filters • General transfer-function is: 2 Vo()z K3z ++ K1K5 + K2K5 – 2K3 zK 3 – K2K5 Hz(){ ------= –------(50) 2 Vi()z z ++ K4K5 + K5K6 – 2 z 1 – K5K6 • If matched to the following general form 2 a z ++a za Hz()= –------2 1 0 (51) 2 z ++b1zb0
University of Toronto 43 of 60
© D. Johns, K. Martin, 1997
High-Q Biquad Filter K1K5 = a0 ++a1 a2 (52)
K2K5 = a2 – a0 (53)
K3 = a2 (54)
K4K5 = 1 ++b0 b1 (55)
K5K6 = 1 – b0 (56) • And, as in lowpass case, can set
K4 ==K5 1 ++b0 b1 (57)
• As before, K4 and K5 approx ZoT « 1 but K6 # 1 e Q
University of Toronto 44 of 60
© D. Johns, K. Martin, 1997 Charge Injection • To reduce charge injection (thereby improving distortion) , turn off certain switches first.
I1 C3 Q6 I Q5 2 C1
CA I1 I1a C2 I1 Vi()z Q1 Q4 Vo()z I2 Q2Q3 I2a
• Advance I1a and I2a so that only their charge injection affect circuit (result is a dc offset)
University of Toronto 45 of 60
© D. Johns, K. Martin, 1997
Charge Injection
• Note: I2a connected to ground while I1a connected to virtual ground, therefore ... — can use single n-channel transistors — charge injection NOT signal dependent
QCH ==–WLCoxVeff –WLCox VGS – Vt (58)
• Charge related to VGS and Vt and Vt related to substrate-source voltage.
• Source of Q3 and Q4 remains at 0 volts — amount of charge injected by Q3 Q4 is not signal dependent and can be considered as a dc offset.
University of Toronto 46 of 60
© D. Johns, K. Martin, 1997 Charge Injection Example • Estimate dc offset due to channel-charge injection when C1 = 0 and C2 ==CA 10C3 =10pF .
• Assume switches Q3 Q4 have Vtn = 0.8V , W = 30Pm , –3 2 L = 0.8Pm, Cox = 1.9u 10 pF e Pm , and power supplies are r2.5V .
• Channel-charge of Q3 Q4 (when on) is
QCH3 ==QCH4 –0.8 30 0.0019 2.5– 0.8 (59) –3 = –1077.5 u pC • dc feedback keeps virtual opamp input at zero volts.
University of Toronto 47 of 60
© D. Johns, K. Martin, 1997
Charge Injection Example
• Charge transfer into C3 given by Q = –C v (60) C3 3 out
• We estimate half channel-charges of Q3 , Q4 are injected to the virtual ground leading to 1 --- Q + Q = Q (61) 2 CH3 CH4 C3 which leads to –3 77.5u 10 pC v ==------7 8 m V (62) out 1pF • dc offset affected by the capacitor sizes, switch sizes and power supply voltage.
University of Toronto 48 of 60
© D. Johns, K. Martin, 1997 SC Gain Circuits — Parallel RC R2 R2 e K C1 KC v 1 in vout()t = –Kvin()t
I1 C2 I I 1 1 I KC2 2
I2 I2 C1 KC1 vin()n vout()n = –Kvin()n
University of Toronto 49 of 60
© D. Johns, K. Martin, 1997
SC Gain Circuits — Resettable Gain
I C2 2
I2 I1 I1 C1 §·C1 vin()n v ()n = – ------v ()n I out ¨¸in 2 ©¹C2
vout
Voff time I2 I1 I2 I1 }
University of Toronto 50 of 60
© D. Johns, K. Martin, 1997 SC Gain Circuits Parallel RC Gain Circuit • circuit amplifies 1/f noise as well as opamp offset voltage Resettable Gain Circuit • performs offset cancellation • also highpass filters 1/f noise of opamp • However, requires a high slew-rate from opamp V off- + VC C - 2+ 2 C V 2 vout()n V v ()n + C1- - off+ in V C C off 1 + §·C 1 + Voff 1 V - = –¨¸------vin()n off- ©¹C2
University of Toronto 51 of 60
© D. Johns, K. Martin, 1997
SC Gain Circuits — Capacitive-Reset • Eliminate slew problem and still cancel offset by coupling opamp’s output to invertering input
I1 C3 I2 I2
C2 I1 I2 I1 C1 v ()n §·C1 in v ()n = –¨¸------v ()n I2 out C in I1 ©¹2 C4
• C4 is optional de-glitching capacitor
University of Toronto 52 of 60
© D. Johns, K. Martin, 1997 SC Gain Circuits — Capacitive-Reset V +-off C2 v ()n – 1 out- + C V 3 - off+ vout()n – 1 + Voff C1 + Voff - during reset
C2 during valid output C1 vin()n vout()n + C1 V = –§·------v ()n off C3 ©¹C in - 2
University of Toronto 53 of 60
© D. Johns, K. Martin, 1997
SC Gain Circuits — Differential Cap-Reset I1a C3 I2 I2a C 2 I I 1 + 1 C v 1 §·C1 + - in v = ¨¸------v – v out C in in I2 ©¹2 I2 C - C1 2 I1 vin I1 I 2a C3 I2
I1a • Accepts differential inputs and partially cancels switch clock-feedthrough
University of Toronto 54 of 60
© D. Johns, K. Martin, 1997 Correlated Double-Sampling (CDS) • Preceeding SC gain amp is an example of CDS • Minimizes errors due to opamp offset and 1/f noise • When CDS used, opamps should have low thermal noise (often use n-channel input transistors) • Often use CDS in only a few stages — input stage for oversampling converter — some stages in a filter (where low-freq gain high)
• Basic approach: — Calibration phase: store input offset voltage — Operation phase: error subtracted from signal
University of Toronto 55 of 60
© D. Johns, K. Martin, 1997
Better High-Freq CDS Amplifier I1 C2 I 1 I2 C1 I1 v in vout I2 I I2 2 I2 Cc1 Cc2
I1 I1 I1
• I2 — C1' C2' used but include errors
• I1 — C1 C2 used but here no offset errors
University of Toronto 56 of 60
© D. Johns, K. Martin, 1997 CDS Integrator C2
I2 I1 I I 2 1 I C1 1 vin vout I I I C 1 2 1 c2
• I1 — sample opamp offset on C2'
• I2 — C2' placed in series with opamp to reduce error • Offset errors reduced by opamp gain • Can also apply this technique to gain amps
University of Toronto 57 of 60
© D. Johns, K. Martin, 1997
SC Amplitude Modulator
• Square wave modulate by r1 (i.e. Vout = rVin )
I1 C3 I2 I 2 I1 IB C2 I IA I1 2 C1 vin vout I IB A
Ica IA = II2 Ica + 1 Ica IB = II1 Ica + 2 Ica • Makes use of cap-reset gain circuit.
• Ica is the modulating signal
University of Toronto 58 of 60
© D. Johns, K. Martin, 1997 SC Full-Wave Rectifier
v Square Wave v in Modulator out
Ica
• Use square wave modulator and comparator to make • For proper operation, comparator output should changes synchronously with the sampling instances.
University of Toronto 59 of 60
© D. Johns, K. Martin, 1997
SC Peak Detector
I VDD
vin comp Q opamp 2 vout v vout in Q C 1 CH H
• Left circuit can be fast but less accurate • Right circuit is more accurate due to feedback but slower due to need for compensation (circuit might also slew so opamp’s output should be clamped)
University of Toronto 60 of 60
© D. Johns, K. Martin, 1997