1. Chapter 4 The Operational Amplifier

2. 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

3. 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.

4. Basic concept of amplifier
Input resistance Ri
Output resistance Ro
Open loop gain: k

5. Ideal Amplifier: Ri = inf, Ro=0

6. Actual Gain is < k Voltage divider: Output voltage:
Gain = Vout/Vin :

7. 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

8. The Operational Amplifier
v+ and v- are node voltages relative to ground sometimes we use vp and vn
vo = A(v+ - v-) , ie. the voltage across the input
Vcc, -Vcc are power supply inputs, usually +/-15V

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

10. 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

11. Ideal Op Amp Model in Linear Range
Ri = infinity
Ro = 0
A = infinity
i+ = i- = 0
v+ = v-

12. 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

13. 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

14. “Closed Loop” Example: Unity Gain Buffer
Controlling Variable =
Solve For Buffer Gain
Thus The Amplification

15. 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)

16. 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]

18. Inverting Amplifier
Vo = -RfVs Rs
When in linear
region

20. Do Handout Problems 1,2

21. Summing Amplifier
Vo = - Rf (Va + Vb + Vc) (in linear region) Rs

23. Problem 2 Find the output voltage vo [900 V]

24. Non-Inverting Amplifier
Vo=(Rs+Rf) Vg Rs
In linear region

26. Problem 2 Calculate the output voltage vo [9V]

27. Difference Amplifier Vo= Rb (Vb – Va) in linear region AND
Ra IFF Ra/Rb = Rc/Rd

29. Problem 1 Find the output voltage vo [-80]

30. Handouts

31. 1) a) INVERTING OP_AMP Calculate Vo for Vs=0. 4, 2, 3. 5, -0. 6, -1
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]

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

33. 3) SUMMING OP-AMP Find Vo in the circuit shown if Va=0. 1V and Vb= 0
3) SUMMING OP-AMP Find Vo in the circuit shown if Va=0.1V and Vb= 0.25V [-7.5]

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

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