0 ECE 222 Electric Circuit Analysis II Chapter 4 The Capacitor Herbert G. Mayer, PSU Status Status 5/3/2016 For use at CCUT Spring 2016

1 Syllabus Capacitor Never Forget Power in Capacitor Energy in Capacitor Capacitor Example 1 Capacitor Example 2 Bibliography

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 capacitor is surface area of plates A, distance d of plates from one another, and dielectric ξ separating them Symbols:

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 [F], increasing with size A, increasing inversely proportional to the distance d, and depends on material of dielectric ξ Farad F is a large unit; mostly use pF, nF, or μF

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

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: Interestingly: [F H] = [s 2 ]

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

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 voltage at time t = 0, often v 0 = 0, then: v(t) = 1/C i(t) dt

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 –clearly a displacement current dv / dt is the rate of voltage change across capacitor C’s terminals Then “Ohm’s” Law for capacitors is: i(t) = C dv / dt

9 Capacitor – Check SI Units 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

10 Capacitors in Series & in Parallel

11 Series, 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 /R 2... simply stating that conductances add up We see an identical 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

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, extending 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 the identical voltage change dv / dt, we get: C eq = C 1 + C 2 + C 3 C eq = Σ C n, for n = 1..3

13 Parallel Capacitors To compute the equivalent capacitance C eq, use the analogous rule known from resistor conductances: C eq = C 1 + C 2 + C 3

14 Series Capacitors 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

15 Series Capacitors If n capacitors C n are connected in series, they can be replaced by a single, equivalent capacitor C eq Currents through n series capacitors 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 2 = 1 / C 2 i/dt... Finally: 1 / C eq = 1 / C / C / C n

16 Power & Energy in Capacitor

17 Power in Capacitor We know, power p is: p = v i 1. Substituting i = C dv/dt yields the power p equation as a function of v: p = v C dv/dt 2. Or substituting v = 1/C i dt + v 0 yields power p from t 0 to time t as a function of i: p = i ( 1/C i dt + v 0 )

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

19 Capacitor Example 1

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

21 Capacitor Example 1, Voltage Pulse Using Example 6.4 from p. 184 [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 or plot the following electric units in Example 1: 1. Plot the voltage v(t) 2. Compute the current i(t) 3. Compute and plot the power p(t) 4. Energy w(t)

22 1. Plot voltage v(t) for Example 1 Circuit parameters given via problem specification: Linear voltage v(t) from : v(t) = 4 t SI unit of Volt [V] Then v(t) from 1.. ∞ : v(t) = 4 e 1-t

23 1. Plot voltage v(t) for Example 1

24 2. Current i(t) for Example 1 Function for voltage v(t) is given. Then compute i(t): 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 * = 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 * = -2 e (1-t) [μA] i(t) = -2 e 1-t [μA]

25 2. Current i(t) for Example 1

26 3. Power p(t) for Example 1 With C = 0.5 μF, function for power p(t) is: p(t) = v i Compute p(t) for range [s], when v(t) = 4 t p(t) = v i = v C dv/dt = 0.5 * 4 * 4 * t * [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]

27 3. Plot power p(t) for Example 1

28 4. Energy w(t) for Example 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] 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]

29 4. Plot energy w(t) for Example 1

30 Energy Distribution in Example 1 When is energy stored in the capacitor? Happens, when a power function’s tangent p(t)’ is positive! Or when power increases! That is the time interval 0..1 [s] During the rest of the time for t > 1 s, energy is delivered by the capacitor

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 = = -4 [μJ] To be expected: total energy stored equals total energy being released!

32 Capacitor Example 2

33 Capacitor Example 2, Current Pulse Circuit with Independent Current Source i(t), Pulsed Range μs: i(t) = 5000 t [A]. Range μs: i(t) = t [A]

34 Capacitor Example 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 Example 2 circuit: 1. Plot the current i(t) 2. Compute the voltage v(t) 3. Power p(t) 4. Energy w(t) Discussions about computations:

35 1. Current i(t) of Example 2 Current i(t) is given per problem definition: A current pulse starting at time t = 0, continuing 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 of Example 2 for i(t)

36 1. Current i(t) of Example 2: mA over μ s

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

38 2. Voltage v(t) of Example 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 * ) 2 v(t 1 )=12.5 * 4 * v(t 1 )=5[V]

39 2. Voltage v(t) of Example 2 For range μs: v(t)=1 / C * i * dt + v 0 v 0 =5 v(t)=5 * 10 6 ( t ) dt Note v 0 = 5 V from preceding interval μs v(t)=10 6 t * 10 9 t 2

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

41 3. Power p(t) of Example 2 For range μs: p=v i p=62.5 * t 3 [W] For range μs: p=62.5 * t * 10 9 t * 10 5 t – 2 Students check: at time t = 40 μs: p(t) = 0

42 4. Energy w(t) of Example 2 For range μs: w=½ C v 2 w= * t 4 [J] For range μs: w=½ C v 2 Left as an exercise in computing

43 Bibliography  Electric Circuits, 10 nd edition, Nilsson and Riedel, Pearsons Publishers, © 2015 ISBN-13:  Wikipedia: