Chapter 4 The Operational Amplifier. Recall Voltage divider with Load RL "no-load" vo = 75V attach...
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Transcript of Chapter 4 The Operational Amplifier. Recall Voltage divider with Load RL "no-load" vo = 75V attach...
![Page 1: Chapter 4 The Operational Amplifier. Recall Voltage divider with Load RL "no-load" vo = 75V attach RL = 150k, vo drops to 66.6 V The load "pulls down"](https://reader035.fdocuments.us/reader035/viewer/2022081512/56649de45503460f94adba01/html5/thumbnails/1.jpg)
Chapter 4The Operational Amplifier
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Recall Voltage divider with Load RL• "no-load" vo = 75V
• attach RL = 150k,
vo drops to 66.6 V
• The load "pulls down" the output voltage
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Amplifiers
• Amplifiers are devices that magnify signals, and also remain mostly unaffected by changing load resistance.
• Amplifiers are used in many instruments and electronic devices (iPod, cell phone, EEG) to boost signals (music, brainwaves) and buffer (isolate) them from loads.
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Basic concept of amplifier• Input resistance Ri
• Output resistance Ro
• Open loop gain: k
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Ideal Amplifier: Ri = inf, Ro=0
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• Voltage divider:
• Output voltage:
• Gain = Vout/Vin :
Actual Gain is < k
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Open Loop Gain (k) is fixed but Feedback lets us vary circuit gain
• we are given:
• KCL at node A:
(k inf)
• Resulting Circuit gain: = R2/R1
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The Operational Amplifier
• v+ and v- are node voltages relative to groundsometimes we use vp and vn
• vo = A(v+ - v-) , ie. the voltage across the input
• Vcc, -Vcc are power supply inputs, usually +/-15V
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Linear Operation –vd is from -Vcc/A to +Vcc/A
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Op-Amp model, dependent V source
• Typically:– Ri is very large 1M-ohm– Ro is small– A is 105 – 106
– model applies to
linear range only
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Ideal Op Amp Model in Linear Range
• Ri = infinity
• Ro = 0
• A = infinity
• i+ = i- = 0
• v+ = v-
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Op Amps can be used "open loop"outside linear range, v+ ≠ v-
• Ideal Comparator and Transfer Characteristic
“Zero-Cross” Detector → Heart of Solid State Relay Cnrtl
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2 Ways of Using Op-Amps
• “Open Loop”: very high gain amplifier– Useful for comparing 2 voltages– Fixed gain, always at MAX OUTPUT!!
• “Closed Loop” with negative feedback– Useful for amplifying, adding, subtracting,
differentiation and integration (using capacitors)– Variable gain, controlled by resistor selection
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““Closed Loop” Example: Unity Gain Buffer Closed Loop” Example: Unity Gain Buffer
Controlling Variable = IRV iin
Solve For Buffer Gain
O
iOO
is
out A
RARRV
Vrecall
1
1 Thus The Amplification
1S
outO V
VA
0 inOOis VAIRIRV :KVL
0 inOO VAIRoutV- :KVL
Op-Amp BUFFER GAINLM324 0.99999LMC6492 0.9998MAX4240 0.99995
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Consequences for Vp-Vn• Normally, A is 10,000 or more, so to avoid
saturation, abs(Vp-Vn) must be < Vcc/10000, or, if Vcc = 20V, about 2 mV which is negligible for most circuits
• With an Ideal Op-amp, A = infinity, so Vp = Vn to avoid saturation
• Negative Feedback resistors “force” Vp = Vn i.e. if Vp-Vn gets large, A(Vp-Vn) pulls back toward zero (more on this later)
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1 a) Calculate Vo if va = 1 and vb = 0, b) Repeat for va=1 and vb=2, c) for va=1.5, specify the range of vb to avoid saturation
•
[-4,6,-.8<=vb<=3.2]
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Inverting Amplifier
Vo = -RfVs Rs
When in linear
region
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Do Handout Problems 1,2
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Summing Amplifier
Vo = - Rf (Va + Vb + Vc) (in linear region)
Rs
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Problem 2 Find the output voltage vo [900 V]
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Non-Inverting Amplifier
Vo=(Rs+Rf) Vg
Rs
In linear region
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Problem 2 Calculate the output voltage vo [9V]
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Difference Amplifier
Vo= Rb (Vb – Va) in linear region AND
Ra IFF Ra/Rb = Rc/Rd
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Problem 1 Find the output voltage vo [-80]
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Handouts
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1) a) INVERTING OP_AMP Calculate Vo for Vs=0.4, 2, 3.5, -0.6, -1.6, -2.4 [-2,-10,15,3,8,10]
b) Specify the range of Vs required to avoid saturation [-2<= Vs <=3]
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2) Problem 1 INVERTING OP AMP find the voltage vo across the 1kΩ resistor. [-5V]
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3) SUMMING OP-AMPFind Vo in the circuit shown if Va=0.1V and Vb= 0.25V [-7.5]
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4) NON INVERTING Find the output Voltage when Rx is set to 60kWhat Rx will cause saturation? [4.8V, 75k ]
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5) DIFFERENCE If Vb=4.0V, what values of Va will keep linear operation? [2<=va <=6]