EE101: Op Amp circuits (Part 2) - Electrical Engineering

117 Pages·2015·1.21 MB·English

EE101: Op Amp circuits (Part 2)
M. B. Patil
[email protected]
www.ee.iitb.ac.in/~sequel
DepartmentofElectricalEngineering
IndianInstituteofTechnologyBombay
M.B.Patil,IITBombay
Considerabridgecircuitforsensingtemperature,pressure,etc.,withRa=Rb=Rc =R.
Rd =R+∆R varieswiththequantitytobemeasured. Typically,∆R isasmallfractionofR.
Thebridgeconverts∆R toasignalvoltagewhichcanthenbesuitablyamplifiedandusedfor
displayorcontrol.
Assumingthattheamplifierhasalargeinputresistance,
R 1
v1= R+R VCC = 2VCC.
(R+∆R) 1 1+x 1 1
v2= R+(R+∆R)VCC = 2 1+x/2VCC ≈ 2(1+x)(1−x/2)VCC = 2(1+x/2)VCC,
wherex=∆R/R.
Forexample,withVCC =15V,R=1k,∆R=0.01k ,
v1=7.5V,
v2=7.5+0.0375V.
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
M.B.Patil,IITBombay
Thebridgeconverts∆R toasignalvoltagewhichcanthenbesuitablyamplifiedandusedfor
displayorcontrol.
Assumingthattheamplifierhasalargeinputresistance,
R 1
v1= R+R VCC = 2VCC.
(R+∆R) 1 1+x 1 1
v2= R+(R+∆R)VCC = 2 1+x/2VCC ≈ 2(1+x)(1−x/2)VCC = 2(1+x/2)VCC,
wherex=∆R/R.
Forexample,withVCC =15V,R=1k,∆R=0.01k ,
v1=7.5V,
v2=7.5+0.0375V.
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
Considerabridgecircuitforsensingtemperature,pressure,etc.,withRa=Rb=Rc =R.
Rd =R+∆R varieswiththequantitytobemeasured. Typically,∆R isasmallfractionofR.
M.B.Patil,IITBombay
Assumingthattheamplifierhasalargeinputresistance,
R 1
v1= R+R VCC = 2VCC.
(R+∆R) 1 1+x 1 1
v2= R+(R+∆R)VCC = 2 1+x/2VCC ≈ 2(1+x)(1−x/2)VCC = 2(1+x/2)VCC,
wherex=∆R/R.
Forexample,withVCC =15V,R=1k,∆R=0.01k ,
v1=7.5V,
v2=7.5+0.0375V.
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
Considerabridgecircuitforsensingtemperature,pressure,etc.,withRa=Rb=Rc =R.
Rd =R+∆R varieswiththequantitytobemeasured. Typically,∆R isasmallfractionofR.
Thebridgeconverts∆R toasignalvoltagewhichcanthenbesuitablyamplifiedandusedfor
displayorcontrol.
M.B.Patil,IITBombay
R 1
v1= R+R VCC = 2VCC.
(R+∆R) 1 1+x 1 1
v2= R+(R+∆R)VCC = 2 1+x/2VCC ≈ 2(1+x)(1−x/2)VCC = 2(1+x/2)VCC,
wherex=∆R/R.
Forexample,withVCC =15V,R=1k,∆R=0.01k ,
v1=7.5V,
v2=7.5+0.0375V.
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
Considerabridgecircuitforsensingtemperature,pressure,etc.,withRa=Rb=Rc =R.
Rd =R+∆R varieswiththequantitytobemeasured. Typically,∆R isasmallfractionofR.
Thebridgeconverts∆R toasignalvoltagewhichcanthenbesuitablyamplifiedandusedfor
displayorcontrol.
Assumingthattheamplifierhasalargeinputresistance,
M.B.Patil,IITBombay
(R+∆R) 1 1+x 1 1
v2= R+(R+∆R)VCC = 2 1+x/2VCC ≈ 2(1+x)(1−x/2)VCC = 2(1+x/2)VCC,
wherex=∆R/R.
Forexample,withVCC =15V,R=1k,∆R=0.01k ,
v1=7.5V,
v2=7.5+0.0375V.
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
Considerabridgecircuitforsensingtemperature,pressure,etc.,withRa=Rb=Rc =R.
Rd =R+∆R varieswiththequantitytobemeasured. Typically,∆R isasmallfractionofR.
Thebridgeconverts∆R toasignalvoltagewhichcanthenbesuitablyamplifiedandusedfor
displayorcontrol.
Assumingthattheamplifierhasalargeinputresistance,
R 1
v1= R+R VCC = 2VCC.
M.B.Patil,IITBombay
Forexample,withVCC =15V,R=1k,∆R=0.01k ,
v1=7.5V,
v2=7.5+0.0375V.
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
Considerabridgecircuitforsensingtemperature,pressure,etc.,withRa=Rb=Rc =R.
Rd =R+∆R varieswiththequantitytobemeasured. Typically,∆R isasmallfractionofR.
Thebridgeconverts∆R toasignalvoltagewhichcanthenbesuitablyamplifiedandusedfor
displayorcontrol.
Assumingthattheamplifierhasalargeinputresistance,
R 1
v1= R+R VCC = 2VCC.
(R+∆R) 1 1+x 1 1
v2= R+(R+∆R)VCC = 2 1+x/2VCC ≈ 2(1+x)(1−x/2)VCC = 2(1+x/2)VCC,
wherex=∆R/R.
M.B.Patil,IITBombay
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
Considerabridgecircuitforsensingtemperature,pressure,etc.,withRa=Rb=Rc =R.
Rd =R+∆R varieswiththequantitytobemeasured. Typically,∆R isasmallfractionofR.
Thebridgeconverts∆R toasignalvoltagewhichcanthenbesuitablyamplifiedandusedfor
displayorcontrol.
Assumingthattheamplifierhasalargeinputresistance,
R 1
v1= R+R VCC = 2VCC.
(R+∆R) 1 1+x 1 1
v2= R+(R+∆R)VCC = 2 1+x/2VCC ≈ 2(1+x)(1−x/2)VCC = 2(1+x/2)VCC,
wherex=∆R/R.
Forexample,withVCC =15V,R=1k,∆R=0.01k ,
v1=7.5V,
v2=7.5+0.0375V.
M.B.Patil,IITBombay
Theamplifiershouldonlyamplifyv2−v1=0.0375V (sincethatisthesignalarisingfrom∆R).
Definitions:
Givenv1andv2,
1
vc = 2(v1+v2)=common-modevoltage,
vd =(v2−v1)=differential-modevoltage.
v1andv2canberewrittenas,
v1=vc−vd/2, v2=vc+vd/2.
Intheaboveexample,vc ≈7.5V, vd =37.5mV.
Notethatthecommon-modevoltageisquitelargecomparedtothedifferential-modevoltage.
Thisisacommonsituationintransducercircuits.
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
v1=7.5V, v2=7.5+0.0375V.
M.B.Patil,IITBombay
Definitions:
Givenv1andv2,
1
vc = 2(v1+v2)=common-modevoltage,
vd =(v2−v1)=differential-modevoltage.
v1andv2canberewrittenas,
v1=vc−vd/2, v2=vc+vd/2.
Intheaboveexample,vc ≈7.5V, vd =37.5mV.
Notethatthecommon-modevoltageisquitelargecomparedtothedifferential-modevoltage.
Thisisacommonsituationintransducercircuits.
Common-mode and differential-mode voltages
VCC
Ra Rc
v1 v2 Amplifier Vo
Rb Rd
v1=7.5V, v2=7.5+0.0375V.
Theamplifiershouldonlyamplifyv2−v1=0.0375V (sincethatisthesignalarisingfrom∆R).
M.B.Patil,IITBombay