Operational amplifiers

# Operational amplifiers

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## Operational amplifiers

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1. Operational amplifiers Building blocks of servos

2. Rf Summing junction If Rin - Vin V1 inverting Vout Iin + non-inverting Op amp Operational amplifiers (Op-amps) Op amp with feedback Want perfect amplifier • Infinite gain • Infinite input impedance • will not load down source • Zero output impedance • will drive anything Have operational amplifiers • Gain ~ 106 • Input impedance ~ 100 M W • Output impedance ~ 100 W Problems • Gain too high • slightest input noise causes max output • Other problems to be discussed later Solutions • Use feedback • Gain depends only on resistance: Rf / Rin • can control precisely V1 = (Vout - Vin) Rin/(Rin+Rf) + Vin Vout = -A V1 Vout = -Vin (Rf / Rin) / [1 + (1+ Rf / Rin)/A] Small gain: Vout = -A Vin Rf / (Rf + Rin) ~ divider Large gain: Vout = -Vin (Rf / Rin) Note: V1 = -Vout /A ~ 0

3. - - + + Non-ideal op-amp Rf Rin Ideal op-amp circuit Vin1 Rf Vout Rin Vin2 Rin Vin1 V1 inverting Vout ~ Rf non-inverting Vin2 Differential amplifier • Op amp output actually depends on voltage difference at two inputs • Vout = - (Vin1 - Vin2) (Rf / Rin) • Insensitivity to common voltage at both inputs = CMRR • Real op amps have problems with unbalanced input impedance Solution: • Add input resistor and pot. • Op amp with built-in resistors = instrumentation amp. Resistors to compensate for non-ideal op-amp

4. - + Integrator Cf If Rin Vin V1 Vout Iin Integrators • Put capacitor in op amp feedback path • Vout = - Vin (Zf / Rin) = - Vin / (2 p jf C Rin) • Similar to low pass filter in high frequency limit • except applies to low frequencies also • can show large gain near dc • Recall V1 ~ 0 forces Iin = -Ifeedback • charge on capacitor is integral of If • since Vout = Q/Cf, Vout is integral of Iin • Result is integrator • integration speed ~ 1 / RinCf Gain response Single-pole rolloff 6 dB/octave = 10 dB/decade RC >60dB log(Vout/Vin) Unity gain at f = 1 / 2 p RC 0 log(f ) Phase response log(f ) 0 degrees Phase shift -90 degrees

5. - + Shunted integrator Rf Cf If Rin Vin V1 Vout Iin Shunted integrator • Limit dc gain • Advantages: • dc input voltage no longer saturates op amp output • prevents servo runaway • Dis-advantages • long term errors not well corrected by servo Gain response Max gain = Rf/Rin at f < 1 / 2 p RfCf log(Vout/Vin) Unity gain at f = 1 / 2 p RinCf 0 log(f ) Phase response log(f ) 0 degrees Phase shift -90 degrees

6. - + Real op amp Gain response Max gain Op amp without feedback • Acts like shunted integrator Stability condition: • unity gain freq. before second pole • otherwise feedback becomes positive • oscillation Single pole 6 dB log(Vout/Vin) Unity gain 0 log(f ) Double pole 12 dB Real op amp Vin Vout Phase response log(f ) 0 degrees Phase shift -90 degrees -180 degrees

7. Gain response - Single pole 6 dB log(Vout/Vin) Min gain + 0 log(f ) Integrator with lead Phase response Cf Rf log(f ) If 0 degrees Phase shift -90 degrees Rin Vin V1 Vout Iin Integrator with lead High frequency gain has minimum value Purpose: • Provides phase lead • Can compensate for pole in servo Alternate circuits for lead • Capacitor at input, inductor in feedback • Overall positive phase • analog to faster than light propagation • output anticipates input

8. Op amp Rf If Rin - - Vin inverting Vout V1 Iin + + non-inverting Summing amplifier Rin1 Rin2 Rin3 Vin3 Vin2 Vin1 Rf Summing junction If Vout Iin V1 Summing amplifier • High gain forces V1 ~ 0 • Feedback current must cancel input current • Generalize to multiple inputs • Vout= -RfeedbackS Iin = -(Rf / Rin) SVin • Works because V1 ~ 0 • Summing amplifier • Op amp input is summing junction • Useful for combining multiple inputs

9. - + Drift-compensated integrator Cf If Rin Vin V1 Iin Vout +V Rc Integrator leakage compensation Ic - V Drift-compensated integrator Real op amps have leakage current • Can saturate integrator • Compensate with dc current to summing junction

10. Trans-impedance amplifier Rf If - inverting Current source Vout V1 Iin + non-inverting Vsource Iin Zsource Trans-impedance amplifiers • Input is current source • model as voltage source with high impedance • Iin = Vsource / Zsource • Vout = -Zfeedback Vsource/ Zsource= -Iin Zfeedback • Trans-impedance amplifier • current in, voltage out • gain expressed in Ohms V1

11. Photo-diode amplifier Cs Shunt capacitor - Rf light + Gain response If Unshunted Vbias Vout Shunted Ipd log(Vout/Iin) Rpd Ipd Cpd log(f ) Photo-diode equiv. circuit Photodiode amplifier • Photodiode like current source but with capacitor • Input capacitor causes op amp gain to diverge at high freq. • Amplifies high freq noise • Oscillation • Solution: • Shunt capacitor in feedback

12. - + Shunt switch on integrator Rs switch Filtered and limited supply rails Cf +15V If Rin V1 Vout Vin Iin -15V Integrator design tips Shunting switch with small value resistor • Discharges capacitor • initialize integrator/ servo • simplify servo testing allow rest of circuit Resistors on power supply rails • Limits current to saturated op amp • prevents burn out Capacitors on supply rails • Reduces noise Note on capacitors • High values • electrolytic, polar • leakage resistance • Use low value in parallel