83-4 Analog addition and subtraction The addition circuit input -voltages R, 2 output-voltage R1 at node a 3 R 3 0。. 十 R R3 R (addition-circuit) U1+2 2 3 3 Let R1R3 and The result we have found for two R2/RIk, we have inputs may be extended to any U2=-k(U1+U3) number of inputs. R2,R2、,R2 if: k +U3+U4 RI R3
§3-4 Analog addition and subtraction The addition circuit 2 2 3 3 1 1 0, R R R a − = + = 2 2 3 3 1 1 R R R at node a a a a − = − + − : Let R1=R3 and R2 /R1= k, we have : 1 ( ) k = 2 = − 1 +3 if The result we have found for two inputs may be extended to any number of inputs. ( ) 4 4 2 3 3 2 1 1 2 2 = − + + + R R R R R R output voltage input voltages − − − − − − 2 1 3 , ) R R R R ( 3 3 2 1 1 2 2 = − + ( ) 2 = −k 1 +3
As a example, if it is required to design a circuit to realize U,=-30,-20 R, KI 3 R 2 Ry 十 let:R=60k we=have 20k (addition-circuit) R R 30k 3 D,十 R I 3
and R k R k we have 30 20 3 1 = = − 3 2 3 2 1 2 = = R R and R R let : R2 = 60k ) R R R R ( 3 3 2 1 1 2 2 = − + 2 = −31 − 23 As a example, if it is required to design a circuit to realize
The differential amplifier circuit The circuit may be analyzed by superposition. R By superposition R R R U21+D2y= y +(1+32) rI Ra+R, R2 +1 R RI R RI B IfR Ru,RhR2. 02=(U3-U If we require that D2=203-3U1 3+1 when, R2=RI 3→ 2 R R b b letr=30k we have R=R =R=lOk
The differential amplifier circuit By superposition: 3 1 2 1 1 2 2 21 2 (1 ) a b b y R R R R R R R + = + = − + + If Ra =R1 , Rb =R2 . ( ) 3 1 1 2 2 = − R R 2 1 1 3 1 3 1 2 = = + + = → b a b a R R or R R R R 1 1 2 3 1 2 1 1 R R R R R R b a − + + = let R2 = 30k we have R1 = Ra = Rb = 10k ( ) when,R R 2 3 1 2 1 → = − = If we require that 2 = 23 − 31 The circuit may be analyzed by superposition