1. 0 ECE 222 Electric Circuit Analysis II Chapter 4 The Capacitor Herbert G. Mayer, PSU Status 2/24/2016 For use at CCUT Spring 2016

2. 1 Syllabus Capacitor Never Forget Power in Capacitor Energy in Capacitor Capacitor Sample 1 Capacitor Sample 2 Bibliography

3. 2 Capacitor Capacitor consists of 2 conducting plates close to one another, separated by some dielectric Regular DC current cannot flow across a capacitor Yet applying DC to a capacitor’s terminals creates a brief displacement current immediately Soon no more current flows and the complete DC voltage drops across the 2 plates Important parameter for manufacturing is surface area of the plates A, distance d of the plates from one another, and the dielectric ξ separating them Symbols:

4. 3 Capacitor – Displacement Current When DC voltage is applied to plates, charge on one side accumulate, displacing similarly charged particles on the opposite plate Creating impression of a brief current, known as displacement current Yet no electron (or proton) moves across dielectric For AC this process is repeated per AC frequency, posing little to no resistance for high, but great resistance for low frequencies 0 Hz is very low frequency Key unit is capacity, measured in Farad, increasing with size A, increasing inversely proportional to the distance d, and depends on material of dielectric Farad F is a large unit; mostly use pF, nF, or μF

5. 4 Capacitor At terminals, displacement current is indistinguishable from conduction current And is proportional to the voltage rate change Also proportional to the capacity C in Farad [F] Key equation: i(t) = C dv / dt Passive Sign Convention: view reference direction of voltage across the capacitor plates the same as the direction of the current

6. 5 Capacitor – SI Units Key equation: i(t) = C dv / dt Unit of capacity is Farad [F] = [A s V -1] Duality of F with H for induction: [H] = [A -1 s V] Other units for [F] from Wikipedia: https://en.wikipedia.org/wiki/Farad Interestingly: [F H] = [s 2]

7. 6 Capacitor Capacitor named cap for short From i(t) = C dv / dt we see that the voltage cannot change instantaneously in a capacitor If it could, the resulting current would be infinite The equation also shows, if voltage at terminals is constant, the resulting current will be 0 A Or, in the presence of DC, cap acts like an open connection! Also useful to express the voltage as a function of the current: dv C = i dt

8. 7 Capacitor Voltage as a function of the current: C dv = i dt Divide by C, compute integral: dv = 1/C i dt v(t) = 1/C i(t) dt + v 0 With v 0 being the voltage at time t = 0, often v 0 = 0, then: v(t) = 1/C i(t) dt

9. 8 Never Forget If i(t) is the instantaneous current through an inductor –implying that it be time-variant C is the capacitance of some capacitor with current i through it –implying that it be a displacement current dv / dt is the instantaneous rate of change of voltage across terminals of capacitor C Then “Ohm’s” Law for capacitors is: i(t) = C dv / dt

10. 9 Capacitor – Check SI Units Earlier Left side of equation below, we have i, unit is A On the right we have capacity F times volt V, by time s Right hand side units is [C]=F=A s V -1 [dv]=V [1 / dt]=s -1 [C dv / dt]=A s V -1 V s -1 = A i(t) = C dv / dt

11. 10 Capacitors in Series & in Parallel

12. 11 Serial, Parallel Capacitors Duality simplifies the universe, and life for EEs Recall from resistors that R i resistances in series, i = 1.. n, can be replaced by one single, equivalent resistor R eq, with R eq = Σ R i Also, multiple parallel resistors R i can be replaced by an equivalent resistor R eq, where the following holds: 1 / R eq = 1 / R 1 + 1 /R 2... simply stating that conductances add up We see a dual relation with capacitors: With n capacitors C n in parallel, what is their equivalent single capacity C eq ? Clearly, all n voltages v p at the terminals of the various parallel C n are identical

13. 12 Parallel Capacitors Parallel capacitors L n all have the same voltage v p Partial, parallel currents through all C n add up; for example, with n = 3 here it follows : i eq = i = i 1 + i 2 + i 3 Starting at time t 0 to time t: i eq = C eq dv / dt i 1 = C 1 dv / dt i 2 = C 2 dv / dt i 3 = C 3 dv / dt Dividing by identical voltage change, we get C eq = C 1 + C 2 + C 3 C eq = Σ C n, for n = 1..3

14. 13 Parallel Capacitors To compute the equivalent capacitance C eq, use the similar, dual rule known from resistor conductance: C eq = C 1 + C 2 + C 3

15. 14 Serial Capacitors The voltages of a series of capacitors add up; if we view n = 3 capacitors, total voltage v s is: v s = v eq = v 1 + v 2 + v 3

16. 15 Serial Capacitors If n capacitors C n are connected in series, they can be replaced by a single, equivalent capacitor C eq The currents through all in series are identical i eq = i 1 = i 2 =... = i n And voltages add up to v s : v s = v eq = v 1 + v 2 + v 3 v 1 = 1 / C 1 i/dt + v 1 (0) v 2 = 1 / C 2 i/dt + v 2 (0)... Finally: 1 / C eq = 1 / C 1 + 1 / C 2... 1 / C n

17. 16 Power & Energy in Capacitor

18. 17 Power in Capacitor We know, power p is: p = v i Substituting i yields: p = v C dv / dt Or substituting v yields: p = i ( 1/C i dt + v 0 )

19. 18 Energy in Capacitor As used before, the power p in a capacitor is: p = v i And also power is: p = dw / dt Hence: p dt = dw dw = C v dv Integrating yields energy: dw = w = C v dv w = ½ C v 2 With voltage at t 0 = 0 V, and energy at time t 0 also 0

20. 19 Capacitor Sample 1

21. 20 Capacitor Sample 1, Voltage Pulse Circuit with Independent Voltage Source, Pulsed Range: 0..1 [s] linear = 4 t, and 1.. ∞ e-function: 4 e (1-t)

22. 21 Capacitor Sample 1, Voltage Pulse Using Example 6.4 from [1]: Given a voltage pulse v(t) = 4 t, starting at time t = 0, and v(t) = 4 e 1-t for t > 1 second Compute and plot the following electric units in the Sample 1 circuit 1. Voltage v(t) 2. Current i(t) 3. Power p(t) 4. Energy w(t) Discussions about computations and graphs follow:

23. 22 1. Voltage v(t) for Sample 1 Circuit parameters given via problem specification: Linear voltage v(t) from 0.. 1 : v(t) = 4 t Note: unit of Volt [V] Then v(t) from 1.. ∞ : v(t) = 4 e 1-t

24. 23 1. Voltage v(t) for Sample 1

25. 24 2. Current i(t) for Sample 1 Function for voltage v(t) is given: With C = 0.5 μF, function of current i(t) is: i(t) = C dv / dt Compute for range 0.. 1, and v = 4 t i(t) = C d( 4t ) / dt = i(t) = 0.5 * 4 * 10 -6 = 2 μA Check units: [A s V -1 V s -1 ] = [A] i(t) = 2 μA Compute for range 1.. ∞, and v = 4 e 1-t i(t) = 4 C d( e 1-t ) / dt i(t) = -1 * 4 * 0.5 e (1-t) * 10 -6 = -2 e (1-t) μA i(t) = -2 e 1-t μA

26. 25 2. Current i(t) for Sample 1

27. 26 3. Power p(t) for Sample 1 With C = 0.5 μF, function for power p(t) is: p(t) = v i Compute p(t) for range 0.. 1 [s], when v(t) = 4 t p(t) = v i = v C dv/dt = 0.5 * 4 * 4 * t * 10 -6 [W] p = 8 t [μW] Compute p(t) for range 1.. ∞ [s], when v(t) = 4 e 1-t p = v C dv / dt p = v C d( e 1-t ) / dt p = -1 * 0.5 * 4 e 1-t 4 e 1-t p(t) = -8 e 2*(1-t) [μW]

28. 27 3. Power p(t) for Sample 1

29. 28 4. Energy w(t) for Sample 1 Function for energy w(t), with C = 0.5 μF, is: w = ½ C v 2 Compute for range 0.. 1, and v = 4 t w = ½ C v 2 = ½ * 0.5 * 16 t 2 [J] 10 -6 p = 4 t 2 [μJ] Compute for range 1.. ∞ seconds, and v = 4 e 1-t w = ½ C v 2 w = ½ * 0.5 *16 e 2-2t w = 4 e 2-2t w(t) = 4 e 2(1-t) [μJ]

30. 29 4. Energy w(t) for Sample 1

31. 30 Energy Distribution in Sample 1 When is energy stored in the capacitor? Happens, when function p(t) is positive! That is the time interval 0..1 [s] During the rest of the time for t > 1 s, energy is delivered by the capacitor

32. 31 Integrate Power Curve p(t) Compute integral I 1 for the power p(t) dt for ranges 0..1, and > 1 s I 1 for range t = 0..1, with p(t) = 8 t I 1 = p(t) dt = 8 t dt = 4 t 2 --for range 0..1 I 1 = 4 [μJ] I 2 for range t > 1, with p(t) = -8 e 2(1-t) I 2 = p(t) dt = -8 e 2(1-t) dt I 2 = -8 (-1) ½ e 2(1-t) = 4 e 2(1-t) --for range 1..∞ I 2 = 0 - 4 = -4 [μJ] To be expected that total energy stored equals total energy being released!

33. 32 Capacitor Sample 2

34. 33 Capacitor Sample 2, Current Pulse Circuit with Independent Current Source i(t), Pulsed Range 0..20 μs: i(t) = 5000 t [A]. Range 20..40 μs: i(t) = 0.2 - 5000 t [A]

35. 34 Capacitor Sample 2, Current Pulse Given a linear current pulse i(t) = 5000 t, starting at time t = 0 to 20 μs And i(t) = 0.2 – 5000 t [A] from 20 to 40 μs Compute or plot the following electric units in the Sample 2 circuit: 1. Current i(t) 2. Voltage v(t) 3. Power p(t) 4. Energy w(t) Discussions about computations:

36. 35 1. Current i(t) of Sample 2 The current i(t) is per problem definition: A current pulse starting at time t = 0 to 20 μs is i(t) = 5000 t And i(t) = 0.2 – 5000 t [A] from 20 to 40 μs Before and after i(t) = 0 [A] See Plot 1. for i(t)

37. 36 1. Current i(t) of Sample 2

38. 37 2. Voltage v(t) of Sample 2 For range 0.. 20 μs: v(t)=1 / C * i * dt + v 0 v 0 =0 [V] v(t)=5 * 10 6 5000 t dt v(t)=25 * ½ * 10 6+3 t 2 v(t)=12.5 * 10 9 t 2 [V]

39. 38 2. Voltage v(t) of Sample 2 Voltage at time t 1 = 20 μs is: v(t 1 )=12.5 * 10 9 t 2 v(t 1 )=12.5 * 10 9 ( 20 * 10 -6 ) 2 v(t 1 )=12.5 * 4 * 10 9+2-12 v(t 1 )=5[V]

40. 39 2. Voltage v(t) of Sample 2 For range 20.. 40 μs: v(t)=1 / C * i * dt + v 0 v 0 =5 v(t)=5 * 10 6 ( 0.2 - 5000 t ) dt Note the 5 V from preceding interval 0.. 20 μs v(t)=10 6 t - 12.5 * 10 9 t 2

41. 40 2. Voltage v(t) of Sample 2, up to 20 μs

42. 41 3. Power p(t) of Sample 2 For range 0.. 20 μs: p=v i p=62.5 * 10 12 t 3 [W] For range 20.. 40 μs: p=62.5 * 10 12 t 3 -7.5 * 10 9 t 2 + 2.5 * 10 5 t – 2 Students check: at time t = 40 μs: p(t) = 0

43. 42 4. Energy w(t) of Sample 2 For range 0.. 20 μs: w=½ C v 2 w=15.625 * 10 12 t 4 [J] For range 20.. 40 μs: w=½ C v 2 Left as an exercise in computing

44. 43 Bibliography  Electric Circuits, 10 nd edition, Nilsson and Riedel, Pearsons Publishers, © 2015 ISBN-13: 978-0-13- 376003-3  Wikipedia: https://en.wikipedia.org/wiki/Farad