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Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2002 The McGraw-Hill Companies Grob Schultz.

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Presentation on theme: "Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2002 The McGraw-Hill Companies Grob Schultz."— Presentation transcript:

1 Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2002 The McGraw-Hill Companies Grob Schultz

2 Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2003 The McGraw-Hill Companies 32 CHAPTER Integrated Circuits

3 Topics Covered in Chapter 32  Operational Amplifiers and Their Characteristics  Op Amp Circuits

4 +V CC -V CC Output Inverting input Non-inverting input IC Operational Amplifier 3 1 5 4 6 7 8 2 2 3 7 4 6

5 Op-Amp Gain The open-loop voltage gain (A VOL ) is what occurs with no negative feedback.  A typical value is 200,000 The closed-loop voltage gain (A VCL ) is what occurs with negative feedback.  Values of A VCL depend on the application.

6 +V CC -V CC Output = 10 V 2 3 7 4 6 Open-Loop Gain 50  V V out = 200,000 x 50  V = 10 V

7 Other Op-Amp Specifications Input bias current, I B+ and I B-  Typically on the order of 80 nA Slew rate, S R  Typically 0.5 V/  s Open-loop frequency response, f OL  The open-loop gain typically drops to 70.7% at 10 Hz, then decreases by a factor of 10 for each decade.

8 +V CC -V CC 2 3 7 4 6 Input Bias Currents Input offset current = |I B+ | – |I B- |  20 nA I B+  80 nA I B-  80 nA

9 Slew Rate +V CC -V CC 2 3 7 4 6 1  s 0.5 V Input signal > f max idthPowerBandw V S f p R   2 max VpVp

10 Open-Loop Frequency Response 1 10 100 1k 10k 100k 1M 100 dB 80 dB 60 dB 40 dB 20 dB 0 dB -20 dB Gain Freq. in Hz The voltage gain decreases by a factor of 10 (20 dB) for each decade increase beyond 10 Hz.

11 Op-Amp Circuits Inverting Amplifier  Output is 180° out of phase with the input.  Signal is applied to the inverting input.  A VCL = –R F /R i Noninverting Amplifier  Output is in phase with the input.  Signal is applied to the noninverting input.  A VCL = (R F /R i ) + 1

12 RFRF 50 mV 10 k  V1V1 R i = 1 k  V out = –500 mV Inverting Amplifier A VCL = –R F /R i = –10k/1k = –10

13 RFRF 10 k  R i = 1 k  Inverting Amplifier (ac input signal) A VCL = –R F /R i = –10k/1k = –10 (phase inversion)

14 Small-Signal Bandwidth with Negative Feedback RFRF 10 k  R i = 1 k  20 dB 100 kHz closed-loop 10 Hz open-loop

15 Non-Inverting Amplifier RFRF 10 k  R i = 1 k  A VCL = 11 A VCL = (R F /R i ) + 1

16 Op-Amp Circuits Voltage follower  No phase inversion  A VCL approximately 1  Very high input impedance  Very low output impedance Summing Amplifier  Produces the inverted algebraic sum of the input voltage levels.

17 Voltage Follower V out = 4 V V in = 4 V

18 RFRF R 3 = 3 k  Summing Amplifier 150 mV 20 mV 50 mV 10 k  mV k k k k k k V R R V R R V R R V FFF out 90015.0 3 10 02.0 2 10 05.0 1 10 )()( 3 3 2 2 1 1  V3V3 V2V2 V1V1 R 1 = 1 k  R 2 = 2 k  V out = –900 mV

19 Op-Amp Differential Amplifier One signal is applied to the inverting input and the second signal is applied to the non-inverting input. Where the inputs are V X and V Y, the output voltage is given by: Differential amplifiers amplify the difference in voltage between two input signals. )( 1 YX F out VV R R V 

20 Differential Amplifier RFRF 10 k  R 1 = 1 k  V out = –(R F /R 1 )(V X –V Y ) 150 mV 120 mV VXVX VYVY R 1 = 1 k  10 k  V out = –300 mV


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