Op-Amps Microprocessor Interface Operational Amplifier (Op-Amp) +V cc Very high differential Input 1 gain + V o High input impedance Vd Output Low output impedance Input 2 Provide voltage changes -Vcc Rout~0 R in~inf (amplitude and polarity) Vo GdVd Used in oscillator, filter and instrumentation Gd : differential gain normally Accumulate a very high very large, say 105 gain by multiple stages IC Product OFFSET NULL 1 8 N.C.
-IN 2 7 V+ +IN 3 6 OUTPUT V 4 5 OFFSET NULL + DIP-741 OUTPUT A 1 -IN A 2 +IN A 3
V 4 8 V+ 7 OUTPUT B + -IN B 6 + 5 +IN B Dual op-amp 1458 device Distortion +V =+5V cc + V V +5V o d 0 5V
V = 5V cc The output voltage never excess the DC voltage supply of the Op-Amp Op-Amp Properties (1) Infinite Open Loop gain - The gain without feedback Equal to differential gain Zero common-mode gain Pratically, Gd = 20,000 to 200,000 (2) Infinite Input impedance - Input current ii ~0A T- in high-grade op-amp m-A input current in low-grade opamp (3) Zero Output Impedance - act as perfect internal voltage source No internal resistance Output impedance in series with load Reducing output voltage to the load Practically, Rout ~ 20-100 V1 + V2 i1~0 +
i2~0 Vo Vo Rout Vo' + Vload Rload Rload Vo Rload Rout Frequency-Gain Relation Ideally, signals are amplified from DC to the highest AC (Voltage Gain) frequency Gd Practically, bandwidth is limited 0.707Gd 741 family op-amp have an limit bandwidth of few KHz. Unity Gain frequency f1: the gain at unity Cutoff frequency fc: the gain
drop by 3dB from dc gain Gd GB Product : f1 = Gd fc 20log(0.707)=3dB 1 0 fc f1 (frequency) GainBandwidth Product Example: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV (Voltage Gain) Sol: Since f1 = 10 MHz Gd 0.707Gd ? Hz By using GB production equation f1 = G d fc 10MHz fc = f1 / Gd = 10 MHz / 20 V/mV = 10 106 / 20 103 1 = 500 Hz 0 fc
f1 (frequency) Ideal Op-Amp Applications Analysis Method : Two ideal Op-Amp Properties: (1) The voltage between V+ and V is zero V+ = V (2) The current into both V+ and V termainals is zero For ideal Op-Amp circuit: (1) Write the kirchhoff node equation at the noninverting terminal V+ (2) Write the kirchhoff node eqaution at the inverting terminal V (3) Set V+ = V and solve for the desired closed-loop gain Noninverting Amplifier (1) Kirchhoff node equation at V+ yields, V Vi V + in V o (2) Kirchhoff node equation at V yields, V 0 V Vo 0 Ra Rf (3) Setting V+ = V yields
Vi Vi Vo 0 Ra Rf Rf Vo 1 or Vi Ra Ra Rf v+ v + v- i v a v Rf Ra )vi v+ i + v-
v R f Voltage follower vo vi R2 + v- Noninverting input with voltage divider Rf R2 vo (1 )( )vi Ra R1 R2 v o vo Rf Ra f Noninverting amplifier vo (1 R1 o
R R v+ vi v+ i R 1 R 2 + v- v o R f Less than unity gain vo R2 vi R1 R2 Inverting Amplifier Rf
(1) Kirchhoff node equation at V+ yields, V 0 (2) Kirchhoff node equation at V yields, Vin V_ Vo V 0 Ra Rf (3) Setting V+ = V yields Vo R f Vin Ra Ra V ~ in V o + Notice: The closed-loop gain Vo/Vin is dependent upon the ratio of two resistors, and is independent of the open-loop gain. This is caused by the use of feedback output voltage to subtract from the input voltage. Multiple Inputs (1) Kirchhoff node equation at V+ yields, V 0 Va Vb (2) Kirchhoff node equation at V Vc
yields, V_ Vo Rf V Va V Vb V Vc 0 Ra Rb Rc (3) Setting V+ = V yields c V Va Vb Vc j Vo R f R f j a R j Ra Rb Rc Rf Ra Rb Rc V o + Inverting Integrator Now replace resistors Ra and Rf by complex components Za and Zf, respectively, Zf therefore Vo
Vin Za Supposing 1 The feedback Z f component is a capacitor C, jC i.e., Zf Za V ~ vi (t ) Vi e What happens if Za = 1/jC whereas, Zf = R Inverting differentiator o C R jt where V + in The input component is a resistor R, Za = R 1 Therefore, the vo (tclosed-loop )
vi (tgain )dt (Vo/Vin) become: RC V ~ in V o + Op-Amp Integrator Example: (a) Determine the rate of change of the output voltage. +5V 0 R 100s (b) Draw the output waveform. V i 10 k C 0.01F
V o + Vo(max)=10 V Solution: (a) Rate of change of the output voltage Vo V 5V i t RC (10 k)(0.01 F) 50 mV/s +5V 0 (b) In 100 s, the voltage decrease -5V -10V Vo ( 50 mV/s)(100ss) 5V V i 0 V o Op-Amp Differentiator R 0
to C t1 t2 V i V 0 o + to dV vo i RC dt t1 t2 Slew Rate The maximum possible rate at which an amplifiers output voltage can change, in volts per second, is called its slew rate. FIGURE 10-17 The rate of change of a linear, or ramp, signal is the change in voltage divided by the change in time
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