SECTION 5: Capacitance & Inductance

ENGR 201 – Electrical Fundamentals I 2 Fluid Capacitor

K. Webb ENGR 201 Fluid Capacitor 3

 Consider the following device:  Two rigid hemispherical shells  Separated by an impermeable elastic membrane  Modulus of elasticity,  Area,  Incompressible fluid 𝜆𝜆 𝐴𝐴  External pumps set pressure or flow rate at each port  Total volume inside shell is constant  Volume on either side of the membrane may vary

K. Webb ENGR 201 Fluid Capacitor – Equilibrium 4

 Equal pressures = = 0

 No fluidΔ𝑃𝑃 flow𝑃𝑃1 − 𝑃𝑃2 = = 0

 Membrane𝑄𝑄1 𝑄𝑄2 does not deform

 Equal volume on each side

= = 2 𝑉𝑉 𝑉𝑉1 𝑉𝑉2 K. Webb ENGR 201 Fluid Capacitor – > 5  Pressure differential 𝑃𝑃1 𝑃𝑃2 = > 0

 MembraneΔ𝑃𝑃 𝑃𝑃 deforms1 − 𝑃𝑃2

 Volume differential = > 0

 TransientΔ𝑉𝑉 flow𝑉𝑉1 − as𝑉𝑉 2 membrane stretches, but...

 No steady-state flow  As

𝑡𝑡 →=∞ = 0 𝑄𝑄1 𝑄𝑄2 K. Webb ENGR 201 Fluid Capacitor – < 6  Pressure differential 𝑃𝑃1 𝑃𝑃2 = < 0

 VolumeΔ𝑃𝑃 differential𝑃𝑃1 − 𝑃𝑃2 = < 0

 proportionalΔ𝑉𝑉 𝑉𝑉1 − 𝑉𝑉 to:2  Pressure differential Δ𝑉𝑉Physical properties, ,

 Total volume remains𝜆𝜆 constant𝐴𝐴 + =

 Again,𝑉𝑉 1no steady𝑉𝑉2 𝑉𝑉-state flow

K. Webb ENGR 201 Fluid Capacitor – Constant Flow Rate 7

 Constant flow rate forced into port 1 0

 Incompressible,𝑄𝑄1 ≠ so flows are equal and opposite =

1 2  Volume𝑄𝑄 on 𝑄𝑄each side proportional to time

= + 2 𝑉𝑉 𝑉𝑉1 𝑄𝑄1 ⋅ 𝑡𝑡 = = 2 2 2 𝑉𝑉 2 𝑉𝑉 1  Volume𝑉𝑉 differential− 𝑄𝑄 ⋅ 𝑡𝑡 proportional− 𝑄𝑄 ⋅ 𝑡𝑡 to time = = 2

K. Webb Δ𝑉𝑉 𝑉𝑉1 − 𝑉𝑉2 𝑄𝑄1 ⋅ 𝑡𝑡 ENGR 201 Fluid Capacitor – Capacitance 8

 Define a relationship between differential volume and pressure  Capacitance

= Δ𝑉𝑉 𝐶𝐶  Intrinsic deviceΔ𝑃𝑃 property

 Determined by physical parameters:

 Membrane area,

 Modulus of elasticity,𝐴𝐴

K. Webb 𝜆𝜆 ENGR 201 Fluid Capacitor – DC vs. AC 9

 In steady-state (DC), no fluid flows = = 0

 Consider𝑄𝑄1 sinusoidal𝑄𝑄2 (AC): = sin Δ𝑃𝑃  ResultingΔ𝑃𝑃 flow𝑃𝑃 rate𝜔𝜔𝜔𝜔 is proportional to:  Rate of change of differential pressure  Capacitance

= = = cos 𝑑𝑑𝑑𝑑 𝑄𝑄1 𝑄𝑄2 𝐶𝐶 𝜔𝜔𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 K. Webb 𝑑𝑑𝑑𝑑 ENGR 201 Fluid Capacitor – Time-Varying 10  Equal and opposite flow at both ports Δ𝑃𝑃 =

 Not the𝑄𝑄1 same𝑄𝑄2 fluid flowing at both ports  Fluid cannot permeate the membrane

 Fluid appears to flow through the device  Due to the displacement of the membrane  A displacement flow

 The faster changes, the higher the flow rate

Δ𝑃𝑃  The larger the capacitance,𝑄𝑄 ∝ 𝜔𝜔 the higher the flow rate

K. Webb ENGR 201 𝑄𝑄 ∝ 𝐶𝐶 Fluid Capacitor – Changing 11

 A given corresponds to a particular membrane displacement Δ𝑃𝑃  ForcesΔ must𝑃𝑃 balance  Membrane cannot instantaneously jump from one displacement to another

 Step change in displacement/pressure is impossible  Would require an infinite flow rate

 Pressure across a fluid capacitor cannot change instantaneously

K. Webb ENGR 201 Fluid Capacitor – Energy Storage 12

 Stretched membrane stores energy  Potential energy

 Stored energy proportional to: 

 Δ𝑃𝑃 Δ𝑉𝑉  Energy released as membrane returns  and are supplied

𝑃𝑃 𝑄𝑄

K. Webb ENGR 201 13 Electrical Capacitors

K. Webb ENGR 201 Electrical Capacitor 14

 In the electrical domain, our “working fluid” is positive electrical charge  In either domain, we have a potential-driven flow

Fluid Domain Electrical Domain

Pressure – P Voltage – V

Volumetric flow rate – Q Current – I

Volume – V Charge – Q

K. Webb ENGR 201 Electrical Capacitor 15

 Parallel-plate capacitor  Parallel metal plates  Separated by an insulator  Applied voltage creates charge differential  Equal and opposite charge =  Zero net charge 𝑄𝑄1 −𝑄𝑄2  Equal current =  What flows in one side 1 2  Schematic symbol: flows out𝐼𝐼 the𝐼𝐼 other  Units: Farads (F)

K. Webb ENGR 201 Electrical Capacitor – Electric Field 16

 Charge differential results in an electric field, , in the dielectric  Units: / 𝑬𝑬

 | | is inversely𝑉𝑉 𝑚𝑚 proportional to dielectric thickness, 𝑬𝑬  Above some maximum electric field strength, 𝑑𝑑 dielectric will break down  Conducts electrical current  Maximum capacitor voltage rating

K. Webb ENGR 201 Electrical Capacitor - Capacitance 17

 Capacitance  Ratio of charge to voltage

= 𝑄𝑄  𝐶𝐶 Intrinsic device𝑉𝑉 property  Proportional to physical parameters:  Dielectric thickness,  Dielectric constant, 𝑑𝑑  Area of electrodes, 𝜀𝜀 K. Webb ENGR 201 𝐴𝐴 Parallel-Plate Capacitor 18

 Capacitance

= 𝜀𝜀𝜀𝜀  : dielectric𝐶𝐶 permittivity 𝑑𝑑  : area of the plates 𝜀𝜀  : dielectric thickness 𝐴𝐴  Capacitance𝑑𝑑 is maximized by using:  High-dielectric-constant materials  Thin dielectric  Large-surface-area plates

K. Webb ENGR 201 Capacitors – Voltage and Current 19

 Current through a capacitor is proportional to  Capacitance  Rate of change of the voltage

= 𝑑𝑑𝑑𝑑 𝑖𝑖 𝑡𝑡 𝐶𝐶  Voltage across capacitor𝑑𝑑𝑑𝑑 results from an accumulation of charge differential  Capacitor integrates current 1 =

K. Webb 𝑣𝑣 𝑡𝑡 ∫ 𝑖𝑖 𝑡𝑡 𝑑𝑑𝑑𝑑 ENGR 201 𝐶𝐶 Voltage Change Across a Capacitor 20

 For a step change in voltage,

= 𝑑𝑑𝑑𝑑  The corresponding current would∞ be infinite 𝑑𝑑𝑑𝑑  Voltage across a capacitor cannot change instantaneously

 Current can change instantaneously, but voltage is the integral of current 1 lim = lim = 0 𝑡𝑡0+Δ𝑡𝑡

Δt→0 ΔV Δt→0 �𝑡𝑡0 𝑖𝑖 𝑡𝑡 𝑑𝑑𝑑𝑑 K. Webb 𝐶𝐶 ENGR 201 Capacitors – Open Circuits at DC 21

 Current through a capacitor is proportional to the time rate of change of the voltage across the capacitor = 𝑑𝑑𝑑𝑑 𝑖𝑖 𝑡𝑡 𝐶𝐶  A DC voltage does not𝑑𝑑 change𝑑𝑑 with time, so = 0 and = 0 𝑑𝑑𝑑𝑑  A capacitor is𝑑𝑑 an𝑑𝑑 open circuit𝑖𝑖 𝑡𝑡 at DC

K. Webb ENGR 201 Capacitors in Parallel 22

 Total charge on two parallel capacitors is = + = + 𝑄𝑄 𝑄𝑄1 𝑄𝑄2 = + 𝑄𝑄 𝐶𝐶1𝑉𝑉 𝐶𝐶2𝑉𝑉 = 𝑄𝑄 𝐶𝐶1 𝐶𝐶2 𝑉𝑉  Capacitances𝑄𝑄 𝐶𝐶𝑒𝑒𝑒𝑒 in𝑉𝑉 parallel add

= +

𝐶𝐶𝑒𝑒𝑒𝑒 𝐶𝐶1 𝐶𝐶2 K. Webb ENGR 201 Capacitors in Series 23

 Total voltage across the series combination is = +

1 2 𝑉𝑉 = 𝑉𝑉 + 𝑉𝑉 𝑄𝑄 𝑄𝑄 𝑉𝑉 1 1 2 1 = 𝐶𝐶 𝐶𝐶+ = 𝑄𝑄 𝑉𝑉 𝑄𝑄  The inverses 𝐶𝐶of1 capacitors𝐶𝐶2 𝐶𝐶𝑒𝑒𝑒𝑒 in series add

1 1 = + = −1 + 𝐶𝐶1𝐶𝐶2 𝐶𝐶𝑒𝑒𝑒𝑒 𝐶𝐶1 𝐶𝐶2 𝐶𝐶1 𝐶𝐶2

K. Webb ENGR 201 Constant Current Onto a Capacitor 24

 Capacitor voltage increases linearly for = constant current 𝐼𝐼 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝐶𝐶 = , 𝐼𝐼 𝑡𝑡−𝑡𝑡0 𝑣𝑣 𝑡𝑡 𝐶𝐶 𝑡𝑡 ≥ 𝑡𝑡0 K. Webb ENGR 201 Capacitor – Energy Storage 25

 Capacitors store electrical energy  Energy stored in the electric field  Stored energy is proportional to:  Voltage  Charge differential 1 1 1 = = = 2 2 2 2 2 𝑄𝑄 𝐸𝐸 𝑄𝑄𝑄𝑄 𝐶𝐶𝑉𝑉  Energy released as E-field𝐶𝐶 collapses  and supplied

K. Webb 𝑉𝑉 𝐼𝐼 ENGR 201 26 Fluid Inductor

K. Webb ENGR 201 Fluid Inductor 27

 Consider the following device:  Lossless pipe  Heavy paddle-wheel/turbine  Frictionless bearing  Moment of inertia,  Incompressible 𝐼𝐼 fluid =

 Paddle𝑄𝑄1 wheel𝑄𝑄2 rotates at same rate as the flow

K. Webb ENGR 201 Fluid Inductor – Constant Flow Rate 28

 Constant flow rate = =

 Constant paddle-wheel𝑄𝑄1 angular𝑄𝑄2 velocity𝑄𝑄  Zero acceleration  Zero net applied force

 Frictionless bearing  No force required to maintain rotation  Zero pressure differential = = 0

1 2 K. WebbΔ𝑃𝑃 𝑃𝑃 − 𝑃𝑃 ENGR 201 Fluid Inductor – Constant 29  Constant pressure differential Δ𝑃𝑃 = 0

 Constant applied torqueΔ𝑃𝑃 𝑃𝑃1 − 𝑃𝑃2 ≠  Constant angular acceleration  Flow rate increases linearly with time

= = + Δ𝑃𝑃 𝑄𝑄1 𝑡𝑡 𝑄𝑄2 𝑡𝑡 ⋅ 𝑡𝑡 𝑄𝑄0  is inductance 𝐿𝐿  Intrinsic device 𝐿𝐿 property

K. Webb 𝐿𝐿 ∝ 𝐼𝐼 ENGR 201 Fluid Inductor – Changing Flow Rate 30

 Paddle wheel has inertia (inductance)  Does not want to change angular velocity  Changes in flow rate require:  Paddle-wheel acceleration  Torque  Pressure differential

 Pressure differential associated with changing flow rate:

= 𝑑𝑑𝑑𝑑 Δ𝑃𝑃 𝐿𝐿 𝑑𝑑𝑑𝑑

K. Webb ENGR 201 Fluid Inductor – AC vs. DC 31

 Steady state (DC) flow  No pressure differential = 0  Consider a sinusoidal (AC) flow rate:Δ𝑃𝑃 = = sin

 Sinusoidal𝑄𝑄1 𝑄𝑄2 acceleration𝑄𝑄 𝜔𝜔𝜔𝜔  Sinusoidal torque  Sinusoidal pressure differential:

= = cos 𝑑𝑑𝑑𝑑  proportional to:Δ𝑃𝑃 𝐿𝐿 𝜔𝜔𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 𝑑𝑑𝑑𝑑  Inductance Δ𝑃𝑃Frequency

K. Webb ENGR 201 Fluid Inductor – Changing Flow Rate 32

 Each flow rate has a corresponding angular velocity  Changes in flow rate require changes in angular velocity

 Angular velocity cannot change instantaneously from one value to another:  Must accelerate continuously through all intermediate values

 Flow rate through a fluid inductor cannot change instantaneously  Would require:  / =  = 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 ∞

K. Webb Δ𝑃𝑃 ∞ ENGR 201 33 Electrical Inductors

K. Webb ENGR 201 Electrical Inductors 34

 Inductance  Electrical property impeding changes in electrical current

 Ampere’s law  Electrical current induces a magnetic field surrounding the conductor in which it flows

 Electromagnetic induction  As current changes, the changing magnetic field induces a voltage that opposes the changing current

K. Webb ENGR 201 Inductors 35

 Inductors  Electrical components that store energy in a magnetic field  Coils of wire  Often wrapped around a magnetic core  Magnetic fields from current in adjacent turns sum  Inductance is proportional to the number of turns  Schematic symbol:

 Units: henries (H)

K. Webb ENGR 201 Inductors – Voltage and Current 36

 Voltage across an inductor is proportional to:  Inductance  Rate of change of the current

= 𝑑𝑑𝑑𝑑 𝑣𝑣 𝑡𝑡 𝐿𝐿  Current through inductor𝑑𝑑𝑑𝑑 builds gradually with applied voltage  Inductor integrates voltage 1 =

K. Webb 𝑖𝑖 𝑡𝑡 ∫ 𝑣𝑣 𝑡𝑡 𝑑𝑑𝑑𝑑 ENGR 201 𝐿𝐿 Current Change Through an Inductor 37

 For a step change in current,

= 𝑑𝑑𝑑𝑑  The corresponding voltage would∞ be infinite 𝑑𝑑𝑑𝑑  Current through an inductor cannot change instantaneously

 Voltage can change instantaneously, but current is the integral of voltage 1 lim = lim = 0 𝑡𝑡0+Δ𝑡𝑡

Δt→0 Δi Δt→0 �𝑡𝑡0 𝑣𝑣 𝑡𝑡 𝑑𝑑𝑑𝑑 K. Webb 𝐿𝐿 ENGR 201 Inductors – Short Circuits at DC 38

 Voltage across an inductor is proportional to the time rate of change of the current through the inductor = 𝑑𝑑𝑑𝑑 𝑣𝑣 𝑡𝑡 𝐿𝐿  A DC current does not𝑑𝑑 change𝑑𝑑 with time, so = 0 and = 0 𝑑𝑑𝑑𝑑  An inductor is𝑑𝑑 𝑑𝑑a short circuit𝑣𝑣 𝑡𝑡at DC

K. Webb ENGR 201 Inductors in Series 39

 Total voltage across the series combination is = +

1 2 𝑉𝑉 = 𝑉𝑉 𝑉𝑉+ 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑉𝑉 𝐿𝐿1 𝐿𝐿2 = 𝑑𝑑𝑑𝑑 + 𝑑𝑑𝑑𝑑= 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑉𝑉 𝐿𝐿1 𝐿𝐿2 𝐿𝐿𝑒𝑒𝑒𝑒  Inductances𝑑𝑑𝑑𝑑 in series add 𝑑𝑑𝑑𝑑

= +

𝑒𝑒𝑒𝑒 1 2 K. Webb 𝐿𝐿 𝐿𝐿 𝐿𝐿 ENGR 201 Inductors in Parallel 40

 Voltage across the two parallel inductors:

= = so 𝑑𝑑𝑑𝑑1 𝑑𝑑𝑖𝑖2 𝑣𝑣 𝐿𝐿1 𝐿𝐿2 = 𝑑𝑑𝑑𝑑, 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑖𝑖1 𝑣𝑣 𝑑𝑑𝑖𝑖2 𝑣𝑣  Voltage𝑑𝑑 across𝑑𝑑 𝐿𝐿1 the equivalent𝑑𝑑𝑑𝑑 𝐿𝐿2 inductor:

= = + 𝑑𝑑𝑑𝑑 𝑑𝑑𝑖𝑖1 𝑑𝑑𝑖𝑖2 𝑣𝑣 𝐿𝐿𝑒𝑒𝑒𝑒 𝐿𝐿𝑒𝑒𝑒𝑒 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = + 𝑣𝑣 𝑣𝑣 𝑣𝑣 𝐿𝐿𝑒𝑒𝑒𝑒  Inverses of inductors𝐿𝐿1 𝐿𝐿 2in parallel add

1 1 = + −1 = 1+ 2 𝑒𝑒𝑒𝑒 𝐿𝐿 𝐿𝐿 𝐿𝐿 1 2 1 2 K. Webb 𝐿𝐿 𝐿𝐿 𝐿𝐿 𝐿𝐿ENGR 201 Inductor – Energy Storage 41

 Inductors store magnetic energy  Energy stored in the magnetic field

 Stored energy is proportional to:  Current  Inductance  Magnetic flux: =

1 𝜆𝜆 𝐿𝐿𝐿𝐿 = 2 2 𝐸𝐸 𝐿𝐿𝐼𝐼  Energy released as magnetic field collapses  and supplied

K. Webb ENGR 201 𝑉𝑉 𝐼𝐼 42 Example Problems

K. Webb ENGR 201 A typical US home consumes about 1000 kWh/month. a) To what voltage would a 1 F capacitor need to be charged in order to store this amount of energy? b) If the fully-charged voltage is limited to 200 V, how much capacitance would be required to store this amount of energy?

43 K. Webb ENGR 201 44 K. Webb ENGR 201 A 10 V, 1 kHz, sinusoidal voltage is applied across a 1 F capacitor. How much current flows through the capacitor? 𝜇𝜇

45 K. Webb ENGR 201 46 K. Webb ENGR 201 The following voltage is applied across a 1 F capacitor. Sketch the current through the capacitor. 𝜇𝜇

47 K. Webb ENGR 201 48 K. Webb ENGR 201 A typical US home consumes about 1000 kWh/month. a) How much current would be required in order to store this amount of energy in a 1 H inductor ? b) If the maximum current is limited to 200 A, how much inductance would be required to store this amount of energy?

49 K. Webb ENGR 201 50 K. Webb ENGR 201 If 10 VDC is applied across a 500 mH inductor, how long will it take the current to reach 10 A?

51 K. Webb ENGR 201 52 RC Circuits

K. Webb ENGR 201 Step Response 53

 Step response  Response of a dynamic system (not necessarily electrical) to a step function input  Unit step function or Heaviside step function:

0, < 0 = 1, 0 𝑡𝑡 𝑢𝑢 𝑡𝑡 � 𝑡𝑡 ≥  To characterize an electrical network, a voltage step can be applied as an input

K. Webb ENGR 201 RC Circuit – Step Response 54

 Step response of this RC circuit is the output voltage in response to a step input: =

𝑣𝑣𝑠𝑠 𝑡𝑡 1𝑉𝑉 ⋅ 𝑢𝑢 𝑡𝑡

 For 0,  DC  𝑡𝑡 ≫open circuit How does get from 0 = 𝑣𝑣𝑠𝑠 𝑡𝑡 → to = 1 ?  1 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑣𝑣𝑜𝑜 0𝑉𝑉 𝐶𝐶 → 𝑣𝑣𝑜𝑜 ∞ 𝑉𝑉 𝑣𝑣𝑜𝑜 𝑡𝑡 → 𝑉𝑉  For < 0,  = 0 𝑡𝑡  = 0 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉 K. Webb ENGR 201 RC Circuit – Step Response 55

 To determine the step response, apply KVL around the circuit = 0

 Current,𝑣𝑣 𝑠𝑠 𝑡𝑡 ,− is𝑖𝑖 the𝑡𝑡 𝑅𝑅 capacitor− 𝑣𝑣𝑜𝑜 𝑡𝑡 current

𝑖𝑖 𝑡𝑡 = 𝑑𝑑𝑣𝑣𝑜𝑜 𝑖𝑖 𝑡𝑡 𝐶𝐶  Substituting in for 𝑑𝑑and𝑑𝑑 1𝑖𝑖 𝑡𝑡 𝑣𝑣𝑠𝑠 𝑡𝑡 = 0 𝑑𝑑𝑣𝑣𝑜𝑜 𝑉𝑉 ⋅ 𝑢𝑢 𝑡𝑡 − 𝑅𝑅𝑅𝑅 − 𝑣𝑣𝑜𝑜 𝑡𝑡  Rearranging 𝑑𝑑𝑑𝑑 1 1 + = 𝑑𝑑𝑣𝑣𝑜𝑜 𝑉𝑉 𝑣𝑣𝑜𝑜 𝑡𝑡 ⋅ 𝑢𝑢 𝑡𝑡 K. Webb 𝑑𝑑𝑑𝑑 𝑅𝑅𝑅𝑅 𝑅𝑅𝑅𝑅 ENGR 201 RC Circuit – Step Response 56 1 1 + = 𝑑𝑑𝑣𝑣𝑜𝑜 𝑉𝑉 𝑣𝑣𝑜𝑜 𝑡𝑡 ⋅ 𝑢𝑢 𝑡𝑡  A first-order𝑑𝑑𝑑𝑑 linear,𝑅𝑅𝑅𝑅 ordinary,𝑅𝑅𝑅𝑅 non- homogeneous differential equation  Solution for is the sum of two solutions:  Complementary𝑣𝑣𝑜𝑜 𝑡𝑡 solution  Particular solution

 The complementary solution is the solution to the homogeneous equation  Set the input (forcing function) to zero  The circuit’s natural response 1 + = 0 𝑑𝑑𝑣𝑣𝑜𝑜 𝑣𝑣𝑜𝑜 𝑡𝑡 K. Webb 𝑑𝑑𝑑𝑑 𝑅𝑅𝑅𝑅 ENGR 201 RC Step Response – Homogeneous Solution

57 1 + = 0 𝑑𝑑𝑣𝑣𝑜𝑜 𝑣𝑣𝑜𝑜 𝑡𝑡  For a first-order𝑑𝑑𝑑𝑑 ODE𝑅𝑅𝑅𝑅 of this form, we assume a solution of the form = then 𝜆𝜆𝜆𝜆 𝑣𝑣𝑜𝑜𝑐𝑐 𝑡𝑡 𝐾𝐾0𝑒𝑒 = 𝑑𝑑𝑣𝑣𝑜𝑜 𝜆𝜆𝜆𝜆 and 𝜆𝜆𝐾𝐾0𝑒𝑒 𝑑𝑑𝑑𝑑 1 + = 0 𝜆𝜆𝜆𝜆 𝜆𝜆𝜆𝜆 so 𝜆𝜆𝐾𝐾0𝑒𝑒 𝐾𝐾0𝑒𝑒 1 𝑅𝑅𝑅𝑅 1 1 + = 0 = = RC and 𝜆𝜆 → 𝜆𝜆 − − 𝑅𝑅𝑅𝑅 𝜏𝜏 = = 𝑡𝑡 𝑡𝑡 − − 𝑐𝑐 𝑅𝑅𝑅𝑅 𝜏𝜏 K. Webb 𝑣𝑣𝑜𝑜 𝑡𝑡 𝐾𝐾0𝑒𝑒 𝐾𝐾0𝑒𝑒 ENGR 201 RC Step Response – Homogeneous Solution

= 𝑡𝑡 −𝜏𝜏  The complementary𝑣𝑣𝑜𝑜𝑐𝑐 𝑡𝑡 𝐾𝐾0𝑒𝑒 solution  is the circuit time constant = 𝜏𝜏  is an unknown𝜏𝜏 𝑅𝑅𝑅𝑅 constant  To be determined through 𝐾𝐾0application of initial conditions

 Next, find the particular solution,

K. Webb ENGR 201 𝑣𝑣𝑜𝑜𝑝𝑝 𝑡𝑡 RC Step Response – Particular Solution

 For a step input, the particular solution is the circuit’s steady-state response  As  Long after the input step 𝑡𝑡 → ∞  In steady state:  = 1 (DC)  Capacitor open circuit 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉  = 0 →  = = 1 𝑖𝑖 𝑡𝑡  The𝑣𝑣0 particular𝑡𝑡 𝑣𝑣𝑠𝑠 𝑡𝑡 solution𝑉𝑉 : = 1

𝑣𝑣𝑜𝑜𝑝𝑝 𝑡𝑡 𝑉𝑉 K. Webb ENGR 201 RC Step Response 60

 Step response  Solution to the non-homogeneous equation  Sum of the complementary and particular solutions = +

𝑜𝑜 𝑜𝑜𝑐𝑐 𝑜𝑜𝑝𝑝 𝑣𝑣 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 + 𝑣𝑣1 𝑡𝑡 𝑡𝑡 − 𝜏𝜏  Next, determine 𝑣𝑣by𝑜𝑜 𝑡𝑡 applying𝐾𝐾0𝑒𝑒 an initial𝑉𝑉 condition  For < 0 0  < 0 = 0 𝐾𝐾  𝑡𝑡 < 0 = 0 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉  At 𝑣𝑣𝑜𝑜=𝑡𝑡0 𝑉𝑉  0 = 1  𝑡𝑡Capacitor voltage cannot change instantaneously 𝑣𝑣𝑠𝑠 𝑉𝑉  0 = 0 – this is the initial condition

𝑣𝑣𝑜𝑜 𝑉𝑉 K. Webb ENGR 201 RC Step Response 61

 Apply the initial condition

0 = + 1 = 0 0 − 𝜏𝜏 𝑣𝑣𝑜𝑜 0 = 𝐾𝐾0𝑒𝑒+ 1 =𝑉𝑉0 𝑉𝑉 Note that the time axis is normalized to the 𝑣𝑣𝑜𝑜 = 1𝐾𝐾0 𝑉𝑉 𝑉𝑉 time constant, τ.

 The𝐾𝐾 step0 −response𝑉𝑉 is

= 1 + 1 𝑡𝑡 − 𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 − 𝑉𝑉𝑒𝑒 𝑉𝑉

K. Webb ENGR 201 Step Response – General Solution 62

= 1 + 1 𝑡𝑡 − 𝜏𝜏  This solution assumes an input𝑣𝑣𝑜𝑜 𝑡𝑡 that− steps𝑉𝑉𝑒𝑒 from 0𝑉𝑉 to 1 at = 0  These are also the initial and final values of  Suppose the input steps between two arbitrary voltage𝑉𝑉 levels:𝑉𝑉 𝑡𝑡 𝑣𝑣𝑜𝑜 < 0 = 0 𝑖𝑖 𝑖𝑖 𝑉𝑉 𝑡𝑡 𝑣𝑣 𝑡𝑡 � 𝑓𝑓  Now, the initial condition is 𝑉𝑉 𝑡𝑡 ≥ 0 =

 The particular solution is the𝑣𝑣𝑜𝑜 steady𝑉𝑉-𝑖𝑖state value, which is now = =

𝑜𝑜𝑝𝑝 𝑜𝑜 𝑓𝑓  Solution to the non-homogeneous𝑣𝑣 𝑡𝑡 equation𝑣𝑣 𝑡𝑡 → ∞ is 𝑉𝑉

= + 𝑡𝑡 − 𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝐾𝐾𝑜𝑜𝑒𝑒 𝑉𝑉𝑓𝑓 K. Webb ENGR 201 Step Response – General Solution 63

 Apply the initial condition to determine

0 = + = 𝐾𝐾0 0 − 𝜏𝜏 𝑣𝑣𝑜𝑜 = 𝐾𝐾𝑜𝑜𝑒𝑒 𝑉𝑉𝑓𝑓 𝑉𝑉𝑖𝑖

 Substituting in𝐾𝐾 0for 𝑉𝑉𝑖𝑖 −gives𝑉𝑉𝑓𝑓 the general voltage step response: 𝐾𝐾0 = + 𝑡𝑡 −𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉𝑓𝑓 𝑉𝑉𝑖𝑖 − 𝑉𝑉𝑓𝑓 𝑒𝑒

K. Webb ENGR 201 Step Response – General Solution 64

 General RC circuit step response:

= + 𝑡𝑡 − 𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉𝑓𝑓 𝑉𝑉𝑖𝑖 − 𝑉𝑉𝑓𝑓 𝑒𝑒  General step response for any first-order linear system with a finite steady-state value:

= + 𝑡𝑡 − 𝑓𝑓 𝑖𝑖 𝑓𝑓 𝜏𝜏  Not necessarily an 𝑦𝑦electrical𝑡𝑡 𝑌𝑌 system𝑌𝑌 − 𝑌𝑌 𝑒𝑒  is any quantity of interest (voltage, current, temperature, pressure, displacement, etc.)  𝑦𝑦 𝑡𝑡= 0 is the initial condition  = is the steady-state value 𝑌𝑌𝑖𝑖 𝑦𝑦 K. Webb 𝑌𝑌𝑓𝑓 𝑦𝑦 𝑡𝑡 → ∞ ENGR 201 First-Order Step Response 65

 Initial slope is inversely proportional to time constant

 Response completes 63% of transition after one time constant

 Almost completely settled after

7𝜏𝜏

K. Webb ENGR 201 RC Circuit Response – Example 66

 RC circuit driven with negative-going step 1 < 0 = 0 0 𝑠𝑠 𝑉𝑉 𝑡𝑡  <𝑣𝑣 0𝑡𝑡 � For 𝑉𝑉 𝑡𝑡 ≥  = 1 𝑡𝑡  = 1 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉  At 𝑣𝑣=𝑜𝑜 𝑡𝑡0 𝑉𝑉  0 = 0 𝑡𝑡  cannot change instantaneously 𝑣𝑣𝑠𝑠 𝑉𝑉  = 0 = 1 𝑣𝑣𝑜𝑜 𝑡𝑡  As 𝑉𝑉𝑖𝑖 𝑣𝑣𝑜𝑜 𝑉𝑉  = 0 DC  𝑡𝑡Capacitor→ ∞ open circuit 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉 →  = 0 →  = 0 𝑖𝑖 𝑡𝑡 → ∞ 𝐴𝐴  = 0 𝑣𝑣𝑜𝑜 𝑡𝑡 → ∞ 𝑉𝑉 K. Webb 𝑉𝑉𝑓𝑓 𝑉𝑉 ENGR 201 RC Circuit Response – Example 67

 Time constant = = 1 1 = 1 𝜏𝜏 𝑅𝑅𝑅𝑅 𝑘𝑘Ω ⋅ 𝜇𝜇𝜇𝜇  Voltage𝜏𝜏 step𝑚𝑚 response𝑚𝑚𝑚𝑚𝑚𝑚 = + 𝑡𝑡 − 𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉𝑓𝑓 𝑉𝑉𝑖𝑖 − 𝑉𝑉𝑓𝑓 𝑒𝑒 = 0 + (1 0 ) 𝑡𝑡 − 𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉 𝑉𝑉 − 𝑉𝑉 𝑒𝑒 Vo(t) reaches 63% of its final value = 1 after one time constant 𝑡𝑡 − 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉 ⋅ 𝑒𝑒

K. Webb ENGR 201 Current Step Response 68

 Now, consider the current through an RC circuit driven by a positive-going 0 … 1 step  At = 0:  0 = 1 𝑉𝑉 𝑉𝑉  𝑡𝑡 0 = 0 𝑣𝑣𝑠𝑠 𝑉𝑉  Voltage across resistor: 𝑣𝑣𝑜𝑜 𝑉𝑉 0 0 = 1

𝑠𝑠 𝑜𝑜  Current𝑣𝑣 through− 𝑣𝑣 resistor:𝑉𝑉

0 0 = 0 = 𝑠𝑠 𝑜𝑜 𝑖𝑖 𝑣𝑣 − 𝑣𝑣 𝐼𝐼 1𝑖𝑖 = 𝑅𝑅

𝑖𝑖 𝑉𝑉 K. Webb 𝐼𝐼 ENGR 201 𝑅𝑅 Current Step Response 69

 As :  Capacitor open circuit  Current𝑡𝑡 → ∞ 0 = 0 → → 𝐼𝐼𝑓𝑓  The current step ( ) is zero for < 0. It jumps instantaneously to response: 𝑖𝑖 0𝑡𝑡 = / =𝑡𝑡 1 (for unit step input and = 1 ) 𝑖𝑖 1𝑉𝑉 𝑅𝑅 𝐴𝐴 = + ( ) decays𝑅𝑅 toΩ 63% of 𝑡𝑡 − its final value after 𝜏𝜏 one𝑖𝑖 𝑡𝑡 time constant 𝑖𝑖 𝑡𝑡 𝐼𝐼1𝑓𝑓 𝐼𝐼𝑖𝑖 − 𝐼𝐼𝑓𝑓 𝑒𝑒 = 𝑡𝑡 𝑉𝑉 −𝜏𝜏 𝑖𝑖 𝑡𝑡 𝑒𝑒 𝑅𝑅

K. Webb ENGR 201 70 Example Problems

K. Webb ENGR 201 Determine for 0. The input source is: 𝑜𝑜 =𝑣𝑣2 𝑡𝑡 1𝑡𝑡 ≥

𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉 ⋅ 𝑢𝑢 𝑡𝑡 − 𝑉𝑉

71 K. Webb ENGR 201 72 K. Webb ENGR 201 Determine for 0. The input source is: 𝑜𝑜 𝑣𝑣= 1𝑡𝑡 𝑡𝑡 ≥

𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉 ⋅ 𝑢𝑢 𝑡𝑡

73 K. Webb ENGR 201 74 K. Webb ENGR 201 Determine for 0. The input source is: 𝑜𝑜 𝑣𝑣= 1𝑡𝑡 𝑡𝑡 ≥

𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉 ⋅ 𝑢𝑢 𝑡𝑡

75 K. Webb ENGR 201 76 K. Webb ENGR 201 77 K. Webb ENGR 201 78 K. Webb ENGR 201 79 RL Circuits

K. Webb ENGR 201 RL Circuit – Step Response 80

 For the RL circuit, we’ll first look at the current step response

How does get from to ?  For 0,  = 1 DC 𝑖𝑖 𝑓𝑓 𝑖𝑖 𝑡𝑡 𝐼𝐼 𝐼𝐼  𝑡𝑡 ≫short circuit  At = 0, 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉 →  Current cannot  1 / = change 𝐿𝐿 → instantaneously𝑡𝑡 𝑖𝑖 𝑡𝑡 → 𝑉𝑉 𝑅𝑅 𝐼𝐼𝑓𝑓  0 = 0 =  For < 0, 𝑖𝑖 𝑉𝑉 𝐼𝐼𝑖𝑖  = 0 𝑡𝑡  = 0 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉 K. Webb𝑖𝑖 𝑡𝑡 𝐴𝐴 ENGR 201 RL Circuit – Step Response 81

 To determine the step response, apply KVL around the circuit

= 0 𝑑𝑑𝑑𝑑  Here, we 𝑣𝑣have𝑠𝑠 𝑡𝑡 −a differential𝑖𝑖 𝑡𝑡 𝑅𝑅 − 𝐿𝐿 equation for 𝑑𝑑𝑑𝑑 1 + = 𝑖𝑖 𝑡𝑡 𝑑𝑑𝑑𝑑 𝑅𝑅 𝑖𝑖 𝑡𝑡 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑑𝑑𝑑𝑑 𝐿𝐿 𝐿𝐿  This is in the exact same form as the voltage ODE for the RC circuit  Same general solution applies = + 𝑡𝑡 −  𝜏𝜏 Where, now, the𝑖𝑖 time𝑡𝑡 constant𝐼𝐼𝑓𝑓 𝐼𝐼𝑖𝑖 −is 𝐼𝐼𝑓𝑓 𝑒𝑒 =

K. Webb 𝐿𝐿 ENGR 201 𝜏𝜏 𝑅𝑅 RL Circuit – Step Response 82

 Current step response:

= + 𝑡𝑡 − Note that the time axis is 𝑓𝑓 𝑖𝑖 𝑓𝑓 𝜏𝜏 𝑖𝑖 𝑡𝑡 1𝐼𝐼 𝐼𝐼1− 𝐼𝐼 𝑒𝑒 normalized to the time constant, τ. = 𝑡𝑡 − 𝑉𝑉 𝑉𝑉 𝜏𝜏 𝑖𝑖 𝑡𝑡 − 𝑒𝑒 1𝑅𝑅 𝑅𝑅 = 1 𝑡𝑡 − 𝑉𝑉 𝜏𝜏 𝑖𝑖 𝑡𝑡 − 𝑒𝑒 𝑅𝑅 for R = 1Ω

K. Webb ENGR 201 RL Circuit – Step Response 83

 Now consider the voltage step response  Determine initial and final values:  For < 0  = 0 𝑡𝑡  = 0 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉  = 0 𝑖𝑖 𝑡𝑡 𝐴𝐴  At 𝑣𝑣𝑜𝑜=𝑡𝑡0 𝑉𝑉  0 = 1  𝑡𝑡Current through the inductor cannot change instantaneously,𝑣𝑣𝑠𝑠 𝑉𝑉 so 0 = 0  No voltage drop across the resistor, so 0 = 1 𝑖𝑖 𝐴𝐴  = 1 𝑣𝑣𝑜𝑜 𝑉𝑉  As 𝑉𝑉𝑖𝑖 𝑉𝑉  DC  𝑡𝑡Inductor→ ∞ short circuit, so = 0 𝑣𝑣𝑠𝑠 𝑡𝑡 →  = 0 → 𝑣𝑣𝑜𝑜 𝑡𝑡 → ∞ 𝑉𝑉 𝑓𝑓 K. Webb 𝑉𝑉 𝑉𝑉 ENGR 201 RL Circuit – Step Response 84

 Voltage step response: 0 = 1 = + 𝑣𝑣𝑜𝑜 𝑉𝑉 𝑡𝑡 − 𝑜𝑜 𝑓𝑓 𝑖𝑖 𝑓𝑓 𝜏𝜏 𝑣𝑣 𝑡𝑡 𝑉𝑉 𝑉𝑉 − 𝑉𝑉 𝑒𝑒 ( ) decays to 63% of its final = 0 + 1 0 value after one time constant 𝑡𝑡 𝑜𝑜 − 𝑣𝑣 𝑡𝑡 𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉 𝑉𝑉 − 𝑉𝑉 𝑒𝑒 = 1 𝑡𝑡 −𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉𝑒𝑒

K. Webb ENGR 201 RL Circuit Response – Example 85

 RL circuit driven with negative-going step 1 < 0 = 0.2 0 𝑉𝑉 𝑡𝑡 𝑣𝑣𝑠𝑠 𝑡𝑡 �  For < 0 𝑉𝑉 𝑡𝑡 ≥  = 1 (DC)  Inductor𝑡𝑡 is a short circuit (DC) 𝑣𝑣𝑠𝑠 𝑡𝑡 𝑉𝑉  = 0  = 1 /1 = 1 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉  At 𝑖𝑖 =𝑡𝑡 0 𝑉𝑉 𝑘𝑘Ω 𝑚𝑚𝑚𝑚  0 = 0.2  As  𝑡𝑡 cannot change instantaneously  = 0.2 DC 𝑠𝑠 𝑣𝑣 𝑉𝑉  𝑡𝑡Inductor→ ∞ short circuit  0 = 1 𝑠𝑠 𝑖𝑖 𝑡𝑡 𝑣𝑣 𝑡𝑡 𝑉𝑉 →  0 = 0 1 1 = 0.8  = 0 𝑖𝑖 𝑚𝑚𝑚𝑚 →  = 0.8  = 0 𝑣𝑣𝑜𝑜 𝑣𝑣𝑠𝑠 − 𝑚𝑚𝑚𝑚 ⋅ 𝑘𝑘Ω − 𝑉𝑉 𝑣𝑣𝑜𝑜 𝑡𝑡 → ∞ 𝑉𝑉 𝑉𝑉𝑖𝑖 − 𝑉𝑉 𝑓𝑓 K. Webb 𝑉𝑉 𝑉𝑉 ENGR 201 RL Circuit Response – Example 86

 Time constant

= = 1 1 𝐿𝐿 𝑚𝑚𝑚𝑚 𝜏𝜏 = 1 𝑅𝑅 𝑘𝑘Ω  Voltage𝜏𝜏 step𝜇𝜇 response𝜇𝜇𝜇𝜇𝜇𝜇 = + 𝑡𝑡 − 𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉𝑓𝑓 𝑉𝑉𝑖𝑖 − 𝑉𝑉𝑓𝑓 𝑒𝑒 = 0 + ( 0.8 0 ) 𝑡𝑡 − 𝜏𝜏 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑉𝑉 − 𝑉𝑉 − 𝑉𝑉 𝑒𝑒 = 0.8 𝑡𝑡 − 1 𝜇𝜇𝜇𝜇𝜇𝜇𝜇𝜇 𝑣𝑣𝑜𝑜 𝑡𝑡 − 𝑉𝑉 ⋅ 𝑒𝑒 Vo(t) reaches 63% of its final value after one time constant

K. Webb ENGR 201 87 Example Problems

K. Webb ENGR 201 Determine and for 0. The input source is: 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑖𝑖 𝑡𝑡 𝑡𝑡 ≥ = 2 + 4

𝑣𝑣𝑠𝑠 𝑡𝑡 − 𝑉𝑉 ⋅ 𝑢𝑢 𝑡𝑡 𝑉𝑉

88 K. Webb ENGR 201 89 K. Webb ENGR 201 90 K. Webb ENGR 201 Determine and ( ) for 0. 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑖𝑖 𝑡𝑡 𝑡𝑡 ≥

91 K. Webb ENGR 201 92 K. Webb ENGR 201 Determine and ( ) for 0. 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑖𝑖 𝑡𝑡 𝑡𝑡 ≥

93 K. Webb ENGR 201 94 K. Webb ENGR 201