UMCP ENEE 204 Spring 2000, Lecture 30 1 Lecture #31 April 24 st, 2000 ENEE 204 Sections 0201 –0206 HW#11 due Friday April 28 th

UMCP ENEE 204 Spring 2000, Lecture 30 2 New topic: dependent sources Independent sources Dependent sources ++ ++

UMCP ENEE 204 Spring 2000, Lecture 30 3 There are four types of dependent sources AixAix ixix Current controlled currrent source +vx+vx Gv x Voltage controlled current source ++ RixRix ixix Current controlled voltage source ++ +vx+vx Av x Voltage controlled voltage source

UMCP ENEE 204 Spring 2000, Lecture 30 4 Operation of transistors can be modeled by using dependent sources BJT b e c MOSFET current controlled current source voltage controlled current source d g s

UMCP ENEE 204 Spring 2000, Lecture 30 5 Simple examples ++ +v2+v2 100 v 2 R2R2 R1R1 R3R3 R4R4 ++ vsvs +v2+v2 5i25i2 R2R2 R1R1 R3R3 R 4R 4 ++ vsvs i2i2

UMCP ENEE 204 Spring 2000, Lecture 30 6 For more complex circuits we can use nodal analysis Y3Y3 I s1 I s3 = G 3 v 1 Y2Y2 I s2 = G 2 v 2 Y1Y1 +v2+v2 +v1+v1 v1v1 v3v3 v2v2 v1v1 i 3 =? I s2 = G 2 v 2 = G 2 v 2 I s3 = G 3 v 1 = G 3 v 1 ^ ^ I 3 =Y 3 (v 1 -v 2 ) ^

UMCP ENEE 204 Spring 2000, Lecture 30 7 Some circuits are easier to solve using mesh analysis Z3Z3 V s1 Z2Z2 V s2 = A v 2 V s3 = R I 3 Z1Z1 +v2+v2 +v1+v1 +  ++ ++ I1I1 I2I2 Z4Z4 V s2 = A Z 2 I 1 V s3 = R (I 1 -I 2 )

UMCP ENEE 204 Spring 2000, Lecture 30 8 Another example: a real circuit problem Find v out / v s for  >>1 ReRe RBRB + v out  vsvs rr RCRC ibib   i b BJT

UMCP ENEE 204 Spring 2000, Lecture 30 9 Mesh analysis ReRe RBRB + v out  vsvs rr RCRC   i b I1I1 I2I2 vxvx I 2 =  i b =  I 1 (r  +R e )I 1 + R e  I 1 = v s I 1 = v s /(r  + (  +1)R e ) ibib I 2 =  I 1 =  v s  /(r  + (  +1)R e )

UMCP ENEE 204 Spring 2000, Lecture Calculating the gain of the amplifier ReRe RBRB + v out  vsvs rr RCRC   i b I1I1 I2I2 vxvx ibib I 2 =  v s  /(r  + (  +1)R e ) v out = R C I 2 v out =  v s  R C r  + (  +1)R e

UMCP ENEE 204 Spring 2000, Lecture Thevenin and Norton equivalents can be found for networks containing dependent sources ++ +v2+v2 100 v 2 R2R2 R1R1 R3R3 R4R4 vsvs A B Simple example: + - Z Th V OC A B

UMCP ENEE 204 Spring 2000, Lecture Another simple situation  KVL: iR iR 2 +iR 2  v s =0 i (R 1  99R 2 )  v s i =  v s / (R 1  99R 2 ) + - Z Th V OC A B Turn off sources: +v2+v2 100 v 2 R1R1 R2R2 vsvs A B  + +

UMCP ENEE 204 Spring 2000, Lecture Try again! Find short circuit current.. +v2+v2 100 v 2 R1R1 R2R2 vsvs A B  + + iR 1  v s =0 i  v s /R 1 = I SC i + - Z Th V OC A B

UMCP ENEE 204 Spring 2000, Lecture Another way to solve this problem - use probing current method +v2+v2 100 v 2 R1R1 R2R2 vsvs A B  + + iPiP I 2 =i p I1I1 I2I2 vxvx

UMCP ENEE 204 Spring 2000, Lecture Solving the equations I 2 =i p (R 1 +R 2 ) I 1  R 2 i p = v s +100R 2 (I 1 -i p ) (R 1  99R 2 ) I 1 = v s  99R 2 i p

UMCP ENEE 204 Spring 2000, Lecture Thevenin equivalent B + - Z Th V OC A iPiP +v2+v2 v 2 =V OC - Z Th i p

UMCP ENEE 204 Spring 2000, Lecture An amplifier can be characterized by three parameters Gain Input impedance Output impedance Z in Z out ++ Av in v in ++ v out ++

UMCP ENEE 204 Spring 2000, Lecture Characterizing our amplifier ReRe RBRB + v out  vsvs rr RCRC   i b Z in Z out ++ Av in v in ++

UMCP ENEE 204 Spring 2000, Lecture Mesh analysis ReRe RBRB + v out  vsvs rr RCRC   i b I1I1 I2I2 vxvx I 2 =  i b =  I 1 (r  +R e )I 1 + R e  I 1 = v s I 1 = v s /(r  + (  +1)R e ) ibib I 2 =  I 1 =  v s  /(r  + (  +1)R e )

UMCP ENEE 204 Spring 2000, Lecture Calculating the gain of the amplifier ReRe RBRB + v out  vsvs rr RCRC   i b I1I1 I2I2 vxvx ibib I 2 =  v s  /(r  + (  +1)R e ) v out = R C I 2 v out =  v s  R C r  + (  +1)R e

UMCP ENEE 204 Spring 2000, Lecture Finding the input impedance ReRe RBRB + v out  vSvS rr RCRC   i b ibib iSiS i S = v S /R B +i B = v S /R B + v s /(r  + (  +1)R e ) Z in = v S /i S = R B (r  + (  +1)R e )  > R B Y in = i S /v S = 1/R B + 1 /(r  + (  +1)R e )

UMCP ENEE 204 Spring 2000, Lecture Finding the output impedance ReRe RBRB + v out  vSvS rr RCRC   i b Z out = R C Turn off sources: v S = 0 i b = v s /(r  + (  +1)R e ) = 0 ibib   i b = 0

UMCP ENEE 204 Spring 2000, Lecture Characterizing the amplifier ReRe RBRB + v out  vSvS rr RCRC   i b ibib Z in Z out ++ Av in v in ++ Z in = R B Z out = R C for large 

UMCP ENEE 204 Spring 2000, Lecture Performance of an amplifier in a system can be calculated from the three parameters Z in Z out ++ A v in v in ++ RSRS vSvS RLRL out V Sin S R Z Z V  A Lout L R Z R   v out ++ Efficient power transfer requieres impedance matching

UMCP ENEE 204 Spring 2000, Lecture Operational amplifier is an ideal differential amplifier ++ v1v1 v2v2 0  v out = A (v 2 -v 1 ) + Infinite input impedance Zero output impedance Very high (infinite) gain

UMCP ENEE 204 Spring 2000, Lecture Circuit model for an operational amplifier ++ v1v1 v2v2 0  v out = A (v 2 -v 1 ) + ++ Av in v1v1 v2v2 + v in  Typical A= 10 4  > 10 7

UMCP ENEE 204 Spring 2000, Lecture Properties of a good voltage amplifier Z in Z out ++ Av in v in ++ High gain High input impedance RSRS vSvS Low output impedance RLRL A

UMCP ENEE 204 Spring 2000, Lecture Inverting configuration for operational amplifier ++ v1v1 v2v2 + v out = A (v 2 -v 1 )  R1R1 R2R2 ++ vsvs

UMCP ENEE 204 Spring 2000, Lecture Operational amplifier with large, finite A R2R2 ++ v1v1 v2v2 + V out = A (v 2 -v 1 )  R1R1 ++ VsVs i1i1 i2i2  V S +i 1 R 1 + i 2 R 2 + A (v 2 -v 1 ) = 0 i 1 = i 2 due to infinite input impedance  V S +i 1 R 1 + i 1 R 2 + A (  V S + i 1 R 1 ) = 0 i 1 (R 1 (1+A) + R 2 ) = V S (1+A) i 1 = V S (A+1) / ( R 1 (A+1) + R 2 ) i 1  > V S / R 1 v 1 =V S  i 1 R 1 v 1 -> 0 V out =  i 2 R 2 V out   i 1 R 2  V S R 2 / R 1

UMCP ENEE 204 Spring 2000, Lecture Virtual short ++ v1v1 v2v2 + V out = A (v 2 -v 1 )  R1R1 ++ vsvs i1i1 i2i2 In general: 0v vA 12  as i 1 = V S / R 1

UMCP ENEE 204 Spring 2000, Lecture Calculation of gain R2R2 ++ v1v1 v2v2 + V out  R1R1 ++ vsvs i1i1 i2i2 i 1 = i 2 i 2 R 2 +  V out = 0 V out =  i 2 R 2 =  i 1 R 2 i 1 = V S / R 1 V out =  V S R 2 / R 1 What does negative gain mean?

UMCP ENEE 204 Spring 2000, Lecture Procedure for analysis of operational amplifier circuit 1. Find the potential at the + input terminal 2. Using virtual short equate the potentials at + and  terminals 3. Find the current flowing into the feedback resistor 4. Calculate the output voltage

UMCP ENEE 204 Spring 2000, Lecture Example ++ 10 k  ++ vsvs 1M  + V out =  100 V S 

UMCP ENEE 204 Spring 2000, Lecture The load has no effect on the gain of ideal operational amplifier ++ 10 k  ++ vsvs 1M  + V out =  100 V S  50  Output impedance is zero

UMCP ENEE 204 Spring 2000, Lecture The resistor at the + input has no effect on the gain of ideal operational amplifier ++ 10 k  ++ vsvs 1M  + V out =  100 V S  50  100  input impedance is infinite

UMCP ENEE 204 Spring 2000, Lecture The resistor between the input terminals has no effect on the gain of ideal operational amplifier ++ 10 k  ++ vsvs 1M  + V out =  100 V S  50  100  Virtual short is in parallel with the resistor

UMCP ENEE 204 Spring 2000, Lecture Non-inverting configuration operational amplifier ++ v1v1 v2v2 + V out  R1R1 R2R2 ++ vsvs i1i1 i2i2 i 1 =  V S /R 1 i 2 = i 1 =  V S /R 1  V S + i 2 R 2 +V out = 0 v 2 = V S v 1 = v 2

UMCP ENEE 204 Spring 2000, Lecture Voltage follower - buffer amplifier ++ + V out  ++ vsvs  V S +V out = 0 v 2 = V S v 1 = v 2 What is the use of an amplifier with unit gain?