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6 October University Faculty of Applied Medical Sciences Department of Biomedical equipment and systems BIOELECTRONICS 1 Lec 9: Op Amp Applications By Dr. Eng. Hani Kasban Mahmoud 2017 1 The Unity-gain Amplifier or “Buffer” • This is a special case of the non-inverting amplifier, which is also called a voltage follower, with infinite R1 and zero R2. Hence Av = 1. • It provides an excellent impedance-level transformation while maintaining the signal voltage level. • The “ideal” buffer does not require any input current and can drive any desired load resistance without loss of signal voltage. • Such a buffer is used in many sensor and data acquisition system applications. The Summing Amplifier Since the negative amplifier input is at virtual ground, v v 1 i i 2 i vo 1 R 2 R R 1 2 3 3 Since i-=0, i3= i1 + i2, R R vo 3 v 3 v R 1 R 2 1 2 • Scale factors for the 2 inputs can be independently adjusted by the proper choice of R2 and R1 . • Any number of inputs can be connected to a summing junction through extra resistors. • This circuit can be used as a simple digital-to-analog converter. This will be illustrated in more detail, later. The Difference Amplifier R Since v-= v+ vo 2 (v v ) R 1 2 1 For R2= R1 vo (v1 v2) • This circuit is also called a differential amplifier, since it amplifies the difference between the input signals. v o v- i R v- i R • Rin2 is series combination of R1 2 2 1 2 and R2 because i+ is zero. R R R R 2 v- 2 v • For v2=0, Rin1= R1, as the circuit v- 2 ( v v- ) 1 R 1 R R 1 reduces to an inverting amplifier. 1 1 1 • For general case, i1 is a function R of both v1 and v2. 2 v Also, v R R 2 1 2 Difference Amplifier: Example • • • • Problem: Determine vo Given Data: R1= 10kW, R2 =100kW, v1=5 V, v2=3 V Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0. Analysis: Using dc values, R 100kW A 2 10 dm R 10kW 1 Vo A V V 10(5 3) dm 1 2 Vo 20.0 V Here Adm is called the“differential mode voltage gain” of the difference amplifier. Finite Common-Mode Rejection Ratio (CMRR) A(or Adm) = differential-mode gain Acm = common-mode gain vid = differential-mode input voltage vic = common-mode input voltage v v v v id v v id 1 ic 2 2 ic 2 A real amplifier responds to signal common to both inputs, called the common-mode input voltage (vic). In general, v v vo A (v v ) Acm 1 2 dm 1 2 2 vo A (v ) Acm(v ) ic dm id An ideal amplifier has Acm = 0, but for a real amplifier, Acm v v ic A v ic vo A v dm id dm id CMRR A dm A CMRR dm Acm and CMRR(dB) 20log (CMRR) 10 Finite Common-Mode Rejection Ratio: Example • Problem: Find output voltage error introduced by finite CMRR. • Given Data: Adm= 2500, CMRR = 80 dB, v1 = 5.001 V, v2 = 4.999 V • Assumptions: Op amp is ideal, except for CMRR. Here, a CMRR in dB of 80 dB corresponds to a CMRR of 104. • Analysis: v 5.001V 4.999V id v 5.001V 4.999V 5.000V ic 2 v 5.000 ic V 6.25V vo A v 25000.002 dm id CMRR 104 In the "ideal" case, vo A v 5.00 V dm id 6.255.00 % output error 100% 25% 5.00 The output error introduced by finite CMRR is 25% of the expected ideal output. uA741 CMRR Test: Differential Gain Differential Gain Adm = 5 V/5 mV = 1000 uA741 CMRR Test: Common Mode Gain Common Mode Gain Acm = 160 mV/5 V = .032 CMRR Calculation for uA741 Adm 1000 CMRR 3.125x10 4 Acm .032 CMRR(dB) 20log 10 CMRR 89.9 dB Instrumentation Amplifier R vo 4 (va v ) b R 3 va iR i(2R ) iR v 2 1 2 b v v i 1 2 2R 1 R R vo 4 1 2 (v v ) R R 1 2 3 1 NOTE Combines 2 non-inverting amplifiers with the difference amplifier to provide higher gain and higher input resistance. Ideal input resistance is infinite because input current to both op amps is zero. The CMRR is determined only by Op Amp 3. Instrumentation Amplifier: Example • Problem: Determine Vo • Given Data: R1 = 15 kW, R2 = 150 kW, R3 = 15 kW,R4 = 30 kW V1 = 2.5 V, V2 = 2.25 V • Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0. • Analysis: Using dc values, R R 30kW 150kW 1 22 A 4 1 2 dm R R 15kW 15kW 3 1 Vo A (V V )22(2.5 2.25)5.50V dm 1 2 The Active Low-pass Filter Use a phasor approach to gain analysis of this inverting amplifier. Let s = jw. v˜o ( jw) Z2( jw ) Z jw R Av 1 1 v˜( jw) Z ( jw ) 1 1 R R 2 jwC 2 Z ( jw) 2 1 jwCR 1 R 2 2 jwC R R e j 1 Av 2 2 R (1 jwCR ) R (1 jw ) 1 2 1 wc wc 2f c 1 f c 1 RC 2R C 2 2 fc is called the high frequency “cutoff” of the low-pass filter. Active Low-pass Filter (continued) j • At frequencies below fc (fH in the figure), the amplifier is an inverting amplifier with gain set by the ratio of resistors R2 and R1 . • At frequencies above fc, the amplifier response “rolls off” at -20dB/decade. • Notice that cutoff frequency and gain can be independently set. j 1(w /w )] R R R j[ tan e e 2 2 c Av 2 e 1(w /w ) R (1 jw ) 2 2 jtan 1 c w w wc phase R 12 e R 1 magnitude 1 1 w w c c Active Low-pass Filter: Example • Problem: Design an active low-pass filter • Given Data: Av= 40 dB, Rin= 5 kW, fH = 2 kHz • Assumptions: Ideal op amp, specified gain represents the desired lowfrequency gain. • Analysis: A 1040dB/ 20dB 100 v Input resistance is controlled by R1 and voltage gain is set by R2 / R1. The cutoff frequency is then set by C. R R R 5kW Av 2 R 100R 500kW and 1 in 2 1 R 1 1 1 C 159pF 2f R 2 (2kHz)(500kW) H 2 The closest standard capacitor value of 160 pF lowers cutoff frequency to 1.99 kHz. Low-pass Filter Example PSpice Simulation Output Voltage Amplitude in dB Output Voltage Amplitude in Volts (V) and Phase in Degrees (d) Cascaded Amplifiers • Connecting several amplifiers in cascade (output of one stage connected to the input of the next) can meet design specifications not met by a single amplifier. • Each amplifer stage is built using an op amp with parameters A, Rid, Ro, called open loop parameters, that describe the op amp with no external elements. • Av, Rin, Rout are closed loop parameters that can be used to describe each closed-loop op amp stage with its feedback network, as well as the overall composite (cascaded) amplifier. Two-port Model for a 3-stage Cascade Amplifier • Each amplifier in the 3-stage cascaded amplifier is replaced by its 2-port model. R R inB inC A vo A vs A vC vA R R vBR R outA inB outB inC vo Av A A A SinceRout= 0 vA vB vC vs Rin= RinA and Rout= RoutC = 0 A Problem: Voltage Follower Closed Loop Gain Error due to A and CMRR The ideal gain for the voltage follower is unity. The gain error here is: A 1 CMRR GE1 Av 1 1 A1 2(CMRR) v v v s o ic 2 v vs vo id vs vo vo A vs vo 2(CMRR) 1 vo 2(CMRR) Av vs 1 1 A1 2(CMRR) A 1 Since, both A and CMRR are normally >>1, 1 1 GE A CMRR Since A ~ 106 and CMRR ~ 104 at low to moderate frequency, the gain error is quite small and is, in fact, usually negligible. Questions????? 24