Operational amplifiers (Op amps)


 Cynthia Stephens
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1 Operational amplifiers (Op amps) v R o R i v i Av i v View it as an ideal amp. Take the properties to the extreme: R i, R o 0, A.?!?!?!?! v v i Av i v A Consequences: No voltage dividers at input or output. (That s good.) No current flows into the input. (That s good.) The gain is infinite. (Is that good?) EE 230 op amps 1
2 How do we handle this infinite gain business? Av i In order to keep finite, then as A, v i 0. In other words, we must force the difference signal at the input to go to zero. How do we do that? With feedback, of course. Recall that the difference signal in a feedback arrangement must become very small if the gain is very big. 0, as A One input is connected to the source voltage in some fashion. The other input is connected to the feedback network. If the feedback is working properly, then v i v v _ 0. The condition of v v _ is called a virtual short at the input. This should be the case, if negative feedback is working in the circuit. EE 230 op amps 2
3 v s v v i 0 i 0 v s v v i 0 i 0 v v i Av i v A v negative feedback v i Av i v A negative feedback v s v v A v v i 0 i 0 negative feedback Don t really need the internal detail. Just use the rules. EE 230 op amps 3
4 Ideal op amp v v v noninverting input v _ inverting input When using an op amp in a circuit: v v _ virtual short (assuming a proper negative feedback configuration.) i i _ 0 due to infinite input resistance Because R o 0, there are no voltage divider effects at output. This means that we can connect anything to the output without worrying about loading effects. (We will see later that there are potential output problems.) EE 230 op amps 4
5 Noninverting amplifier v S Write a node equation at the inverting terminal. v v v s (virtual short due to feedback) i 0 (infinite input resistance) (Worth memorizing.) EE 230 op amps 5
6 Inverting amplifier v S Write a node equation at the inverting terminal. v v 0 (virtual ground!) i 0 (infinite input resistance) Note the negative sign! (Also worth memorizing.) EE 230 op amps 6
7 Noninverting with gain of 10 ( 9 kω and 1kΩ) Inverting with gain of 10 ( 10 kω and 1kΩ) EE 230 op amps 7
8 Don t panic if the feedback loop gets crazier. 10 k! 10 k! v x R 4 v in i R1 1 k! i R2 R 3 1 k! i R3 i R4 ut Use the usual circuit analysis along with the opamp rules. At the inverting terminal (note that v 0): i R1 i R2. v in v x at node x: i R2 i R3 i R4. v x v x R 3 v x R 4 ut Use the two equations to eliminate v x and solve for the gain: G ut v in R 4 R 4 R 3 Inserting the values: G 120. Big gain without big ratios. EE 230 op amps 8
9 Summing amp (weighted summer) v S3 v S2 v S1 R 3 R F At the inverting terminal: v v 0 (virtual ground). Write a node equation there. (Or use superposition.) Follow with second inverter if you don t like the negative sign. The virtually grounded inverting terminal becomes a summing node. EE 230 op amps 9
10 Difference amp We would like to amplify only the difference between va and vb. Anything this applied in common to both, will not be amplified. R2 At the inverting input: R1 vb va R3 R4 At the noninverting input: if EE 230 v v then ( ) Difference only! (Check it.) op amps 10
11 With a difference amp, we really want to amplify only the differences between the two inputs. Any voltage that is in common should not come through (i.e. the gain for commonmode voltages should be zero.) v com v b v com v a v dif v com v dif v com R 3 R 4 If resistors are perfectly matched: G d v dif G d But if matching is imperfect: G d v dif G c v com The commonmode rejection ratio is a measure of how much of the common signal can leak through to the output. CMRR G d G c CMRR of 10,000 or better is not hard to achieve. EE 230 op amps 11
12 Unity gain buffer v s Noninverting amp with 0 and. So G 1, meaning v s. What good is that? v 1 R o1 R i2 R o1 v 1 R i2 Connecting two circuits. If R o1 is not much smaller R i2, then much of the voltage is lost in the voltage divider, v i2 << v 1. High input resistance of op amp makes v v 1. Zero output resistance of op amp makes v i2. Since v, then v i2 v 1. The op amp served as a buffer between the two circuits, eliminated the voltage divider problem. EE 230 op amps 12
13 Simple application of unitygain buffer You would like to use a potentiometer as a volume control or to provide a reference voltage. V S R P V S R P R P v adj R L v V S V R P adj S v adj R P R L V S R P v adj R L v RL v adj V S R P EE 230 op amps 13
14 Cascading amps The various types of circuits can serve as building blocks for more complicated circuits. R 4 v a v b R 3 1 inverting noninverting 2 R 5 R 6 R 7 summing 3 EE 230 op amps 14
15 20 k! R5 10 k! R2 R1 vs 1 k! vo1 vo2 8 k! summing 2 k!. want vo2 / vs... Feedback loop around feedback loops!! EE 230 R3 R4 noninverting. op amps 15
16 Integrating amplifier Use an inverting amp with a capacitor as the feedback element. vc C R vi(t) ir ic vo(t) () ( ) ( ) () () () ( ) ( ) () EE 230 () () () () ( ) ( ) The output signal, as a function of time, is proportional to the time integral of the input. op amps 16
17 v i (t) R C (t) ( ) ( ) ( ) If the input is a constant voltage, v i (t) V 1, then ( ) ( ) The output starts at whatever value it has t 0, and then ramps downward in time with slope V 1 /RC. If V 1 is negative, then the output ramps upward in time. If the input switches back and forth between two constant values (square wave), then the output ramps up and down correspondingly (sawtooth). V 1 v i (t) (t) t t V 2 t 0 EE 230 op amps 17 t 0
18 A practical concern with an integrating amp: If there is a small, but constant, DC voltage at the input (and we will see later that most opamps have DC error voltages built in), then that will be integrated forever and the output will go to infinity. (In reality, it will saturate at the power supply limit.) to infinity C (t) and beyond! v R error 1 mv v DC o (t) v error RC t t This is a problem because at DC, the capacitor is an open circuit and the amplifier has essentially infinite DC gain. To make it better, put a resistor in parallel with the cap; >>. EE 230 op amps 18 v i (t) C (t)
19 Differentiating amplifier Can also differentiate. Switch resistor and capacitor. R v i (t) C i R i C (t) i C i R C dv C dt v R v C v 0 (t) RC dv i (t) dt However, not used much. If there is noise at the input, the differentiator tends to make it worse. The integrator, on the other hand, tends to average out the noise. EE 230 op amps 19
20 Instrumentation amplifier Recall the difference amp: R v 1 b v v a o R 3 R 4 If the ratios are matched: R 4 /R 3 /, then the circuit becomes a perfect difference amp with G d (v a v b ), where G d / and commonmode voltages are completely rejected (G c 0). But it could be better. First, the input resistances depend on the resistor values. Secondly it is difficult to adjust the gain, because the resistance ratios much remain matched. EE 230 op amps 20
21 An improvement Buffer amps can be added to increase the input resistance. Since the buffer amps have unitygain, their outputs are equal to their input. The differencing amp on the right works exactly as before. v b v a v b v a v o (v a v b ) EE 230 op amps 21
22 Instrumentation amp A final modification turns this into a typical instrumentation amplifier. vy v vy ir3y vr4 vx vy ir4 vx R4 vy R3 v vx vx Use KVL: va va ir3x R1 R2 vo still true: R2 vo (va R1 vb ) ir3x ir3y ir4 vb ir4 (2R3 R4 ) vb R1 vb ir3xr3 ir4r4 ir3yr3 va since i 0 : EE 230 ir4 vb R3 R4 va R2 2R3 1 R4 vx vo R2 R1 2R3 1 R4 vx vy vy op amps 22
23 Adjustable instrumentation amp v y R 4 R R 3 1 R 3 v o v x 2R 3 R 4 1 v x v y EE 230 op amps 23
D is the voltage difference = (V +  V  ).
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