1. 0 ECE 222 Electric Circuit Analysis II Chapter 11 Op Amp Cascade Herbert G. Mayer, PSU Status 5/10/2016 For use at CCUT Spring 2016

2. 1 Syllabus Review Goal Op Amp Cascade Example 1 Cascaded Integrating Op Amps Bibliography

3. 2 Review Ideal Op Amp Focus and limitation of ideal Op Amp: In ideal Op Amp, no currents flowing into --or out of-- inputs pins: i p = i n = 0 [A] Voltages v n and v p at negative and positive input terminals are identical: v n = v p Hence if one input pin is connected to ground, both voltages are zero: v n = v p = 0 [V] Holds even if ground connection is via resistor R, as no current flows through R, hence no voltage drop! Op Amp easily saturates, unless feedback resistor is included from output v o to one of the input pins

4. 3 Review Ideal Op Amp i n and i p are 0 [A], and v n = v p in Ideal Op Amp

5. 4 Goal to Integrate Use cascade of Op Amps to generate v o as a function of signal input voltage v g v o = f( v g ) Doing so in cascade of 2 stages! Goal to have f( v g ) be the integral function! How can we do this? And if done twice, then we solve Second-Order DE Other circuit parameters: resistors R 1 and R 2 at Op Amp input pins, and C 1 and C 2 capacitors at output pin, to feed back signal Note that voltage function with capacitor includes first derivative, this is how integral function can be built! Hence the name: integrating Op Amp!

6. 5 Op Amp Cascade Circuit and Equations

7. 6 Op Amp Cascade – The Circuit Below cascade of 2 Op Amps: The output of the first Op Amp becomes input to second Op Amp In both cases, use feedback via capacitor: first derivative function

8. 7 Op Amp Cascade – Current Equation In left Op Amp of The Circuit, with i p = i n = 0 from KCL In left Op Amp of The Circuit, with i p = i n = 0 from KCL i R1 + i C1 = 0 v g / R 1 + C 1 dv o1 / dt = 0 dv o1 / dt = - v g / (R 1 C 1 ) Similarly in the right Op Amp, using KCL: Similarly in the right Op Amp, using KCL: v o1 / R 2 + C 2 dv o / dt = 0 dv o / dt = - v o1 / (R 2 C 2 )

9. 8 Op Amp Cascade – Current Equation Differentiating dv o /dt = -v o1 /(R 2 C 2 ) one time yields: Differentiating dv o /dt = -v o1 /(R 2 C 2 ) one time yields: d 2 v o / dt 2 = -dv o1 / dt * 1/(R 2 C 2 ) And substituting dv o1 /dt = -v g /(R 1 C 1 ) yields: And substituting dv o1 /dt = -v g /(R 1 C 1 ) yields: d 2 v o / dt 2 = v g / (R 1 C 1 ) * 1/(R 2 C 2 ) That is the key, second order differential equation, for the step response of circuit with two cascading Op Amps That is the key, second order differential equation, for the step response of circuit with two cascading Op Amps Each of these two is an integrating Op Amp Each of these two is an integrating Op Amp v o is a second order DE of v g, with other electric parameters, such as: R 1, C 1, R 2, C 2 v o is a second order DE of v g, with other electric parameters, such as: R 1, C 1, R 2, C 2

10. 9 Example 1 Cascaded Integrating Op Amps

11. 10 Example 1 Example 1, a cascade of 2 integrating Op Amps, with a step function for v g When t < 0, no energy is stored in capacitors v g = 0 V At t = 0, v g jumps from 0 V to v g = 25 mV R 1 = 250 kΩ, R 2 = 500 kΩ, C 1 = 0.1 μF, C 2 = 1 μF Now compute: 1. Intermediate unit: 1 / R 1 C 1 2. Intermediate unit: 1 / R 2 C 2 3. Formula for: d 2 v o /dt 2 4. Formula for v o (t) 5. Time t s to saturation 6. Which Op Amp saturates first

12. 11 Example 1 – Integrating Circuit Apply voltage step function v g = 25 mV at time t = 0 Express v o as function of v g

13. 12 Example 1 1.1 / R 1 C 1 =1 / ( 250 * 10 3 * 0.1 * 10 -6 )=40[s -1 ] 2.1 / R 2 C 2 =1 / ( 500 * 10 3 * 1 * 10 -6 )=2[s -1 ] 3.d 2 v o / dt 2 =v g / (R 1 C 1 ) * 1/(R 2 C 2 ) d 2 v o / dt 2 =v g / ( 40 * 2 ) d 2 v o / dt 2 =25 * 10 -3 * 80 d 2 v o / dt 2 =2

14. 13 Example 1 4. Now integrate twice, given that d 2 v o / dt 2 = 2 dv o / dt =2 t v o =2 * 1/2 t 2 + v o -- v o at t = 0, thus v o (0) = 0 V v o =t 2 -- max v o = 9 V 5. Compute time t S to saturation: t S =3[s]-- at 3 s Op Amp saturates

15. 14 Example 1 6.Analyze, if first Op Amp saturates earlier: dv o1 / dt=- v g / (R 1 C 1 ) dv o1 / dt=- 25 10 -3 * 40=-1 Now integrate: v o1 =-1 * t=-t But for t = t S = 3 s, we see that v o1 = -3 V This is legal range, hence the first Op Amp does NOT saturate, and the second Op Amp is the one saturating first, at time t S = 3 [s]

16. 15 Bibliography  Nilsson, James W., and Susan A. Riedel, Electric Circuits, © 2015 Pearson Education Inc., ISBN 13: 9780-13-376003-3  Differentiation rules: http://www.codeproject.com/KB/recipes/Differentiati on.aspx  Euler’s Identities: https://en.wikipedia.org/wiki/Euler%27s_identity  Table of integrals: http://integral- table.com/downloads/single-page-integral-table.pdf