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Op-Amps

Op-Amps. Microprocessor Interface. Operational Amplifier (Op-Amp). Very high differential gain High input impedance Low output impedance Provide voltage changes (amplitude and polarity) Used in oscillator, filter and instrumentation Accumulate a very high gain by multiple stages.

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Op-Amps

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  1. Op-Amps Microprocessor Interface

  2. Operational Amplifier (Op-Amp) • Very high differential gain • High input impedance • Low output impedance • Provide voltage changes (amplitude and polarity) • Used in oscillator, filter and instrumentation • Accumulate a very high gain by multiple stages

  3. IC Product DIP-741 Dual op-amp 1458 device

  4. Distortion The output voltage never excess the DC voltage supply of the Op-Amp

  5. Op-Amp Properties • 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 op-amp • (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 

  6. Frequency-Gain Relation • Ideally, signals are amplified from DC to the highest AC frequency • Practically, bandwidth is limited • 741 family op-amp have an limit bandwidth of few KHz. 20log(0.707)=3dB • Unity Gain frequency f1: the gain at unity • Cutoff frequency fc: the gain drop by 3dB from dc gain Gd GB Product : f1 = Gdfc

  7. ? Hz 10MHz 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 • Sol: • Since f1 = 10 MHz • By using GB production equation • f1 = Gdfc • fc = f1 / Gd = 10 MHz / 20 V/mV • = 10  106 / 20  103 • = 500 Hz

  8. Ideal Op-Amp Applications • Analysis Method : • Two ideal Op-Amp Properties: • The voltage between V+ and V is zero V+ = V • The current into both V+ and V termainals is zero • For ideal Op-Amp circuit: • Write the kirchhoff node equation at the noninverting terminal V+ • Write the kirchhoff node eqaution at the inverting terminal V • Set V+ = V and solve for the desired closed-loop gain

  9. Noninverting Amplifier • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields • or

  10. Noninverting amplifier Noninverting input with voltage divider Less than unity gain Voltage follower

  11. Inverting Amplifier • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields Notice: The closed-loop gainVo/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.

  12. Multiple Inputs • Kirchhoff node equation at V+ yields, • Kirchhoff node equation at Vyields, • Setting V+ = V– yields

  13. Inverting Integrator • Now replace resistors Ra and Rf by complex components Za and Zf, respectively, therefore • Supposing • The feedback component is a capacitor C, i.e., • The input component is a resistor R, Za = R • Therefore, the closed-loop gain (Vo/Vin) become: • where • What happens if Za = 1/jC whereas, Zf = R • Inverting differentiator

  14. Op-Amp Integrator • Example: • Determine the rate of change • of the output voltage. • Draw the output waveform. Solution: (a) Rate of change of the output voltage (b) In 100 s, the voltage decrease

  15. Op-Amp Differentiator

  16. Slew Rate The maximum possible rate at which an amplifier’s 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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