FE EIEW OPEATIONAL AMPLIFIES (OPAMPS) 1 The Opamp An opamp has two nputs and one output. Note the opamp below. The termnal labeled wth the () sgn s the nvertng nput and the nput labeled wth the () sgn s the nonnvertng termnal. The standard way to show the devce s wth the nvertng termnal on top but ths s not always the case. BE CAEFUL!! deal The Ideal assumptons When we assume Ideal characterstcs or an opamp we are able to make several assumptons whch make analyss MUCH easer. The rst o these s that the nput resstance o each o the two termnals s Innte. n deal 1
Ideal assumptons contnued) The next major assumpton s that the output mpedance s equal to zero. deal out 0 Ths assumpton allows us to make the urther assumpton that ut s ndependent o the load. Ths means that t can not be loaded down. Ideal assumptons contnued) The next assumpton s that the openloop gan (A) s nnte (n real le s n the mllons so ths s a sae assumpton). Open Loop Gan deal A 5 Ideal assumptons contnued) The assumpton that A s nnte allows us to make a urther assumpton that the voltage on each termnal s equal to each other. Ths s concept s called a rtual Short. Ths assumpton turns out to be the KEY to opamp crcut analyss. ut A( ) Ad ut ut d so, lm A 0 deal A A t ollows that 6
Ideal assumptons contnued) 7 A rtual short s not lke a real short (or t wouldn t be vrtual). A vrtual short only aects voltage, not current. The Invertng Opamp = 0 volts due to the rtual 8 I Short to the ground I I = 0A Ideal Wth the addton o two resstors and a ground on the nonnvertng termnal we get an nvertng opamp. Note that snce the nonnvertng termnal s grounded the voltage on the nvertng termnal s also grounded due to the rtual short. The Invertng Opamp contnued) 9 Note that as long as there sn t a voltage source on the nonnvertng termnal there could be a dozen resstors on t and t stll would not aect the voltage on that termnal. Snce the current s assumed to be zero on the nvertng termnal, KCL proves that current I s equal to the eedback current I. 0 I I I I I I I 0
The Invertng Opamp contnued) Another aspect o the vrtual short s that the nput resstance o the nvertng opamp s equal to. 0 v I 10 = 0 volts I due to the rtual Short to the ground I I = 0A Ideal The Invertng Opamp contnued) The closedloop gan o the deal nvertng opamp s: I I I = 0A Ideal 11 A CL The FE exam uses a (A) or closedloop gan. The Invertng Opamp contnued) Fnd, I, I, and n or the crcut shown: k I o v 10k v 5v 10 k v v 0v I I 1mA k k n 1 10k Ideal I
The Nonnvertng opamp I you sht the nput voltage down to the nonnvertng termnal and ground you now have a Nonnvertng opamp. = I 1 10k 5k I 0A 1k deal The Nonnvertng opamp contnued) A The gan o the Nonnvertng opamp s: 1 cl 1 5k I Note that snce there s 0A on the nonnvertng termnal, there wll be 0 volts across the resstor so the voltage on the nvertng termnal wll be. = I 0A 1k 10k deal The Nonnvertng opamp contnued) The nput resstance o an deal nonnvertng opamp wll be nnte. 15 = I 10k n 5k I 0A 1k deal 5
The ltage Follower A ltage ollower s a specal case o the Nonnvertng opamp wth a voltage gan o 1. 16 A CL n 1 The Ideal Opamp Integrator I Ic C deal 17 1 d C 0 t t t o 0 v (t) 6 10 1 t, s wth: =10kΩ C=00 μ v o(t) v 6 10 1 t, s v The Ideal Opamp Derentator 18 d C t C I dt Ic deal wth 50k C.1n v (t) v o(t) 5 5 9 15 1 t, s t, s 6
19 0 1 Example What s the output voltage,, o the deal opamp crcut shown? The crcut s congured as An Invertng Opamp. 1k 10k deal o 10k 1k 0 Example What s the current, I, (nvertng termnal current) n I the crcut shown? 8 a) 0.88A v b) 0.5A 8 I c) 0A v deal d) 0.5A Snce the nput resstance o the nvertng and nonnvertng nputs s deally nnte, the current nto those nputs s deally 0A. So the answer s: c) 0A Example What s the output voltage o the crcut shown? 1 1 8 8 v v 8 v 1 7v o v v 8 I 8 I deal The crcut s an opamp summng crcut or a Lnear Combnaton Crcut. 7
Example For the Derence Ampler crcut shown, determne the output voltage. a) 8.v 0v b) 6.07 5v c) 15.5v d) 18.1v We haven t talked about a derence amp but t can be easly solved wth KCL, the rtual short, and Ohm s law. 15 5 5v 5v 5 I 0 I deal Frst, the 5v n the equaton s the voltage on the nonnvertng leg voltage dvded between the 5 and the ohm resstors to gve us. Contnued on (Example contnued) Next, due to the vrtual short, the nvertng termnal also has 5 volts on t. Now we can determne I, whch s the derence between the voltages on each sde =5v dvded by 15 ohms. I 15 0v 5 5v 5v I 0 I I deal 15 0v 5v 15 1.667A (Example contnued) Now we note that snce the current nto the nvertng termnal s zero, I = 1.667A by KCL. 0v 5v 0 I I I 0 1.667A 0A I I 1.667A 8
(Example contnued) 5 0v 5v Fnally we wrte an equaton or n terms o I. I 0 5v 1.667A 0 5v 1.667A 0 5v.v 8.v Example For the deal operatonal ampler below, what should the value o be n order to obtan a gan o 5? 6 Example contnued on. (Example contnued) 7 1k k I k deal F IF Frst, note that the opamp s upsde down rom the way we normally look at t. Ths s the way that the exam seems to show t a Lot, but t s not ndustry standard. So, n the next slde we wll lp t over. Example contnued on 9
(Example contnued) 8 1k k I k deal F IF Now that t s lpped over, note that the voltage on the nonnvertng termnal s a voltage dvded. k 1k k Example contnued on (Example contnued) The voltage dvder has been replaced by the result o the dvder. Next, note that due to the vrtual short the voltage on the nvertng termnal, s also /. 9 due to the vrtual short Example contnued on (Example contnued) We can now nd an equaton or I. The voltage on the butt o the current arrow s / and the voltage at the tp o the current arrow s 0 volts. Dvde that derence voltage by k ohms and you get I. 0 I k k k 0 due to the vrtual short Example contnued on 10
(Example contnued) 1 F By KCL, 0 I I I 0 I I 0 I I k I I 0A deal Now, remember that the INFINITE mpedance o the Invertng and nonnvertng termnals means that the current out o those termnals s assumed to be zero. Usng ths and KCL (above) we nd that the eed back current, I s equal to I. Problem contnued on. (Example contnued) F I I k I I 0A deal Now we can dene ths current n terms o. The butt o the arrow s whle voltage at the tp o the arrow s (/). Problem contnued on. (Example contnued) The problem statement states that the gan s 5. So, that means that =5 I I 5 I I So, the answer to the queston s that t an 5 value o 19.5k ohms wll result n k a gan o 5. 5 5. 5k 19. 5k k 11
Example Evaluate the ollowng ampler to determne the value o requred to obtan a voltage gan o 10. 100 1M deal Contnued on (Example contnued) 5 Note that the nvertng termnal voltage s 0 volts due to vrtual short whch exsts between t and the nonnvertng termnal. Contnued on (Example contnued) We can now nd an equaton or the current I. Note that the voltage at the butt o the arrow s and the voltage at the tp o the arrow s 0v. We use ths voltage derence and Ohms law to nd an equaton or I. 6 0v I 1M 1M Contnued on 1
(Example contnued) Now, usng the act that the current nto both the nvertng and nonnvertng termnals s assumed to be 0A due to the assumed nnte nput resstance, we can use KCL to show that I equals I. 0 I I I 0 I I 0A I I 7 Contnued on We can use the same technque as beore to dene I n terms o x. and then equate agan to I. (Example contnued) 8 0 1M 0v x I I x I so x I I = 0 due to rtual short x I 100 I I deal Contnued on (Example contnued) 9 Next dene the current I n terms o and x. I x 100 I I I x 1M I deal Contnued on 1
(Example contnued) 0 And now use KCL to dene the currents n terms o voltage. I x 100 I I 0 I I I I I I 100 By KCL x o x x 1M I deal Contnued on (Example contnued) 1 From the problem statement: 10 10 and I = I 1 so we can expand that to x 500 k x 1M 1M x x x 100 10 1M 1M 1M 100 Contnued on (Example contnued) 10 1M 1M 1M 100 10 1M 1M 1M 1 M 100 10M 100 119.5M 119.5M.895k k 1 5k 1 5k So, the value o requred to have a voltage gan o 10 s k ohms. 1