Chapter 4The 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" the output voltage

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.

Basic concept of amplifier• Input resistance Ri

• Output resistance Ro

• Open loop gain: k

Ideal Amplifier: Ri = inf, Ro=0

• Voltage divider:

• Output voltage:

• Gain = Vout/Vin :

Actual Gain is < k

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

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

Linear Operation –vd is from -Vcc/A to +Vcc/A

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

Ideal Op Amp Model in Linear Range

• Ri = infinity

• Ro = 0

• A = infinity

• i+ = i- = 0

• v+ = v-

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

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

““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

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)

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]

Inverting Amplifier

Vo = -RfVs Rs

When in linear

region

Do Handout Problems 1,2

Summing Amplifier

Vo = - Rf (Va + Vb + Vc) (in linear region)

Rs

Problem 2 Find the output voltage vo [900 V]

Non-Inverting Amplifier

Vo=(Rs+Rf) Vg

Rs

In linear region

Problem 2 Calculate the output voltage vo [9V]

Difference Amplifier

Vo= Rb (Vb – Va) in linear region AND

Ra IFF Ra/Rb = Rc/Rd

Problem 1 Find the output voltage vo [-80]

Handouts

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]

2) Problem 1 INVERTING OP AMP find the voltage vo across the 1kΩ resistor. [-5V]

3) SUMMING OP-AMPFind Vo in the circuit shown if Va=0.1V and Vb= 0.25V [-7.5]

4) NON INVERTING Find the output Voltage when Rx is set to 60kWhat Rx will cause saturation? [4.8V, 75k ]

5) DIFFERENCE If Vb=4.0V, what values of Va will keep linear operation? [2<=va <=6]