ELECTRICAL ENGINEERING DEPARTMENT

California Polytechnic State University

EE 251 DC Transient Response - RC Circuit & DC Steady-State - RLC Circuit Pre-Lab 5

IMPORTANT – SHOW ALL WORK!

1. Show that resistance multiplied by capacitance has units of time.

2. For the circuit shown, at t=0 the switch is closed. R = 40KΩ, C = 0.02µF, Vs = 4 Volts

Note: VO (0) = 0V (initially uncharged)

a. Sketch Vc(t) for 0 ≤ t ≤ 5τ. Label y axis of plot with Vc values at t = τ, 2τ, 3τ, 4τ

and 5τ.

b. How long (ms) is required for the capacitor to reach steady-state?

c. What is the capacitor current at t = τ, t = 3τ, t = steady-state.

o2E 6 40ED 8E 4

140 e

02Mt te

4 p yet y te

it

a
004
I

4I E

3. An initially charged 0.02µF capacitor, VC(0) = 4V, is connected across a 40KΩ resistor.

a. At what time will the voltage across the capacitor equal 2V?

b. At what time will the voltage across the capacitor equal 0.5V?

35

ELECTRICAL ENGINEERING DEPARTMENT

California Polytechnic State University

EE 251 DC Transient Response - RC Circuit & DC Steady-State - RLC Circuit Pre-Lab 5

4. Inductors and capacitors; circle correct answer:

a. At steady-state, an inductor behaves like a/an (short / open) circuit relative to DC

current.

Short

b. At steady-state, a capacitor behaves like a/an (short / open) circuit relative to DC

current.

Open

5. Calculate I and V for the following circuit at steady-state: both resistors in ohms.

I
I I fo IA 25A

I
+
Rea
+

- V
REQ Is t s 20

5 2S s 3 75 V

36

ELECTRICAL ENGINEERING DEPARTMENT

California Polytechnic State University

EE 251 DC Transient Response - RC Circuit & DC Steady-State - RLC Circuit Lab 5

Objective

To investigate DC transient response of a RC circuit and to analyze a RLC circuit at DC steady-

state.

Workbench Equipment

• Function Generator, Agilent 33120A or Agilent 33220A

• DC Power Supply, Agilent E3640A

• Digital Multimeter, Agilent 34401A

• Digital Oscilloscope, Keysight InfiniiVision MSO-X2022A

• Resistor Box II, 10Ω/25Ω/40Ω/130Ω/269Ω/562Ω

• Resistor Decade Boxes, 10 kΩ step

• Inductor Decade Box, 10mH step: 0 – 100mH

• Capacitor Decade Box, 0.01µF step: 0 – 1.1µF

• Impedance Bridge, Gen Rad 1659

Check-out Equipment, 20-111 window

• Scope Probe (10:1), 2

• BNC to banana lead

• Banana to banana, 3 pairs, red/black

• Short leads, quantity 6, 1 bag

Background

Transient Response of RC Circuits

Capacitor Discharging: When an initially charged capacitor is connected to a resistance as shown

in Fig. 5-1, as time progresses, the capacitor’s stored energy is dissipated as heat by the resistor.

As stored energy is released, the voltage across the capacitor plates, as well as charge on the

plates ( V = Cq ) diminishes to zero at steady state. When plate voltage equals zero, there is no

electric field within the dielectric between the plates and stored energy is completely expended.

Fig. 5-1 RC Circuit (Capacitor Discharging)

KCL at top node of Fig. 5-1 with switch closed (t > 0) iC = iR

dVC (t ) dVC (t ) VC (t )

Recall: iC = -C (passive sign convention) C + =0 (5-1)

dt dt R

Current exits the higher voltage plate.

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dVC (t ) VC (t )

{1st-order homogeneous DEQ} + =0 (5-2)

{General solution} VC (t) = Aest (5-3)

Sub general solution into 1st-order homogeneous DEQ

sAest + Aest = 0 (5-4)

1 1 1
Solving for s: (s + ) Aest = 0 -> (s + ) = 0 -> s = - (5-5)

RC RC RC

VC(t) = Ae-t/RC (5-6)

Solve for A at t = 0: VC(0) = Ae-0/RC = A = Vo (5-7)

Natural Response (Due to Stored Energy Only): VC(t) = Voe-t/τ where τ = RC (5-8)

Capacitor voltage (5-8) decays exponentially as time advances. After five time constants (t =

5RC), the capacitor for all practical purposes has reached the steady-state voltage of zero (e-5 =

0.00674) and the capacitor is considered to be depleted of stored energy.

Capacitor Charging: When a source is applied to an RC circuit as shown in Fig. 5-2, VC < VS, the

electrical charge on the capacitor plates increases as time progresses. Therefore, voltage across

the capacitor plates ( V = Cq ) increases as well as charge. This increase in capacitor voltage

produces a stronger (more intense) electric field within the dielectric between the plates and the

capacitor stores increasing energy ( WC = CV 2 ) as time advances until steady-state is achieved.

Fig. 5-2 RC Circuit (Capacitor Charging)

Capacitor initially charged: VC(0) = Vo where Vo < VS

Complete Response = Natural Response + Forced Response

(due to stored energy) (due to independent source)

Natural Response: VC(t) = Voe -t/RC {Derived Previously (5-8)}

KCL at top node of Fig. 5-2 with switch closed (t > 0) iC = iR

dVC (t ) Vs − Vc dVc

Recall: iC = C (passive sign convention) -C =0 (5-9)

Current enters higher voltage plate.

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dVc Vc Vs

{VS is a constant, DC source} + = (5-10)

{Forced solution also a constant} VC (t) = XF (5-11)

XF Vs

Sub in XF: 0+ = -> XF = VS (5-12)

Complete Response: VC (t) = Ae -t/RC + VS (5-13)

Solve for A at t = 0: VC (0) = Vo = Ae -0/RC + VS -> A = Vo - VS (5-14)

VC (t) = VS + (Vo - VS)e -t/RC (5-15)

VC (t) = Voe -t/RC + VS(1 - e -t/RC) (5-16)

Complete Response = Natural Response + Forced Response

(due to stored energy) (due to independent source)

For initially uncharged Capacitor: Vo = 0 VC (t) = VS(1 - e -t/RC) (5-17)

As can be seen from (5-17), for all practical purposes, capacitor voltage increases from zero volts

to a steady-state value of Vs after five time constants (1 - e-5 = 0.99326) and the capacitor is

considered to be fully charged.

Capacitors & Inductors Under DC Steady-State Conditions

Inductors and capacitors have the following voltage – current relations:

dI (t )

VL = L (5-18)

dt

dV (t )

IC = C (5-19)
dt

For DC circuits, all time derivatives are zero since all currents and voltages are constant.

Therefore, under DC conditions:

V L = 0 for any current (same V-I relation as zero ohms = short) (5-20)

I C = 0 for any voltage (same V-I relation as infinite ohms = open) (5-21)

Hence, for DC conditions, inductors “act” as short-circuits and capacitors “act” as open-circuits.

Oscilloscope Basics

To view the transient response of an RC circuit, an oscilloscope (“scope”) is used. A scope

converts an electrical signal into a visible trace on a display screen. The display screen is a graph

of voltage (vertical axis) versus time (horizontal axis). Vertical grid lines (or divisions) on the

display screen have numerical values relative to the volts per division setting; horizontal screen

divisions have numerical values relative to the time per division setting. Both settings can be

manually adjusted or the Auto-Scale button can be pressed and the scope will automatically set

the volts per division setting and time per division setting for optimum signal visibility, see Fig.

5-3 for an example of a displayed sinusoidal voltage.

39

500us per horizontal division

1V per vertical division division

division

Fig. 5-3 Volts per Vertical Division & Time per Horizontal Division

A common scope measurement method is to use vertical cursors (dashed lines) to measure signal

amplitude such as peak or peak-to-peak or to use horizontal cursors (dashed lines) to measure the

time difference between two points on a signal such as measuring period, see Fig. 5-4.

Vpp Period

Frequency
Delta X = Period

Delta Y = Vpp

Fig. 5-4 Measuring Amplitude & Time Difference with Cursors

In addition, the Keysight scope has a feature, Meas (button below the horizontal position knob),

that automatically displays amplitude, period, and /or frequency, as well as many other signal

parameters without the use of cursors.

Most often, a scope probe is used to transfer the signal under test to the scope input. The most

popular scope probe (and used in this lab) is the 10:1 scope probe. The ratio 10:1 refers to the

factor of attenuation (loss) between the scope probe tip and the scope input. For example, if the

scope probe is connected to a 10 Vpeak signal, the signal amplitude at the scope input is 1 Vpeak. In

other words, the scope probe has attenuated (decreased amplitude) by a factor of ten.

The scope has a probe adjust setting which compensates for this loss in amplitude; the displayed

amplitude matches the signal amplitude at the scope tip.

A BNC to BNC lead or BNC to banana lead can also be used to connect a signal to a scope input.

These leads have minimal attenuation and are considered ideally to be shorts. The main

advantage a scope probe has over a lead is higher input impedance which means less loading.

Recall from experiment one; an ideal voltmeter has infinite internal resistance. Scopes can be

thought of as sophisticated voltmeters and likewise have ideal infinite internal resistance /

impedance.

40

Scopes commonly have multiple inputs (also called channels) allowing multiple simultaneous

signals to be displayed. Our lab scopes are dual channel (two inputs) and in this experiment one

input will be used as a “triggering” source or signal.

Scope triggering can be thought of as synchronized picture taking. A signal displayed (“picture”)

on a scope screen is the result of many consecutive digitized samples. This process of picture

taking (i.e. producing signal on scope screen) must be synchronized to a unique point on a

waveform that repeats, otherwise the displayed signal becomes unstable. The triggering source

provides the unique point, most often the triggering signal has a sharp transition, such as a square

wave does, and either the high to low transition (falling-edge or trailing-edge) or the low to high

transition (rising-edge or leading-edge) is used as the unique point for synchronization.

Procedure 1: RC Transient Response

• Use impedance bridge set at 100 Hz to measure capacitor value and use ohmmeter to

measure resistor value used to construct circuit of Fig. 5-5 and record in Table 5-1.

• Set function generator (FG) to high-Z output:

To set Agilent 33120A function generator to high-Z output:

o Press shift-menu, right arrow to the SYS menu, down arrow to the OUTPUT TERM

menu, down arrow to ‘50 ohms,’ right arrow to select Hi Z, enter.

To set Agilent 33220A function generator to high-Z output:

o Press Utility, then choose Output Setup, select High Z.

• Set function generator to a square wave output with a minimum value of 0 V and a maximum

value of 4 V (amplitude = 4Vpp with DC offset = 2V). Set frequency of square wave to

100Hz and use a duty cycle of 50%.

o Duty cycle refers to the amount of time a square wave is equal to its largest amplitude

(4V in this case) compared to its period.

• Connect channel 1 scope probe between node A and ground (black clip to ground).

• Connect channel 2 scope probe between node B and ground (black clip to ground).

• NOTE: Probe adjust needs to be set to 10:1!

• To check probe adjust: Press channel button, channel menu appears at bottom of scope

screen and then press Probe soft key. Probe ratio will appear at bottom of screen to the

right of Units.

• To adjust probe adjust: Turn knob located directly below time per division knob.

• Set both scope channels to DC coupling (on channel menu) and set both channel vertical

sensitivity to 1V/div.

• Set horizontal sensitivity to 1millisec/division.

• Set trigger settings as follows:

o Press Mode Coupling button and select auto.

o Press Trigger button and choose Source 1 and rising edge .

o Note: May need to adjust Trigger Level knob to stabilize displayed signal. Adjust

Trigger Level until horizontal line with T at far left is within the amplitude of

source square-wave signal. If signal is still unstable, consult instructor.

• Use cursors to determine time constant of Vc during charging, may need to change horizontal

sensitivity for accuracy, and record in Table 5-1.

• Capture scope display of Vc using Keysight BenchVue desktop icon (ask instructor for steps

after double-clicking icon).

41

• Observe the effect on the charge curve as R is changed to 30KΩ and then changed to 50KΩ.

o Capture scope displays for both above R values to help answer a post-lab

question.

• Set R back to 40KΩ and change square wave to a minimum value of 0 V and a maximum

value of 5 V.

o Observe effect this has on charge curve and capture to help answer a post-lab

question.

• Change Source 1 trigger setting to falling edge and determine time constant of Vc during

discharging and record in Table 5-1.

• Capture scope display of Vc.

• Observe the effect on the discharge curve as R is changed to 30KΩ and then changed to

50KΩ.

o Capture scope displays for both R values to answer a post-lab question.

• Set R back to 40KΩ and change square wave to a minimum value of 0 V and a maximum

value of 5 V.

o Observe effect on discharge curve and capture to answer a post-lab question.

• Calculate percent error between theoretical (done in prelab #2 & #3) and experimental

charging and discharging time constants. Record in Table 5-1.

RE 4454.021nF
D

Set to Hi Z

Fig. 5-5 RC Transient Circuit

R(40KΩ) (Ω) C(0.02µF) (µF)

Measured

Charging (µs) Discharging (µs)

Measured Time

Constant

Calculated Time

Constant

Percent Error (%)

Table 5-1 R & C Measured Values / RC Transient Time Constant Measurements & Calculations

Procedure 2: Capacitors & Inductors at DC Steady-state

• Measure all resistors (Ω) used to construct the circuit of Fig. 5-6 and record in Table 5-2a.

• Calculate and record REQ at nodes a-b using ideal DC equivalents for inductors (short) and

capacitors (open).

• Construct the circuit of Fig. 5-6, except for the voltage source (power supply).

• Measure the open circuit resistance across terminals a-b. This yields the equivalent resistance

REQ.

42

+
40088 µ

- 99.008 0928,66

Fig. 5-6 Capacitors & Inductors at DC steady-state Circuit

Resistances R1 (25Ω) (Ω) R2 (10Ω) (Ω) R3 (130Ω) (Ω) REQ = Rab (Ω)

Measured

Calculated Rab = Percent Error Rab =

Table 5-2a DC Inductor and Capacitor Circuit: Resistance Measurements

• Calculate the voltage across each resistor using ideal DC equivalents for inductors (short)

and capacitors (open). Record calculated voltages in Table 5-2b

• Connect the 10V power supply (0.5A current limit) and measure the voltages across each

resistor. Record measured voltages in Table 5-2b.

• Calculate percent error between measured and calculated voltages and record in Table 5-2b.

R1 (25Ω) R2 (10Ω) R3 (130Ω)

Measured Voltage (V)

Calculated Voltage (V)

Percent Error (%)

Table 5-2b DC Inductor and Capacitor Circuit: Voltage Measurements and Calculations

Discussion

1. In Procedure 1, for all practical purposes, how long (in ms) does it take for the capacitor to

fully discharge? Use measured values.

2. Describe the effect changing R in procedure 1 had on charging and discharging curves.

3. For procedure 1, describe the effect changing square wave amplitude from 0V to 5V had on

charging and discharging curves.

4. In Procedure 1 when R = 40KΩ and the square wave has minimum value 0V and maximum

value 5V, if the frequency of the square wave is increased to 200 Hz, will the capacitor

charge fully to 5V? Why or why not?

5. In Procedure 2, what is the primary reason for differences between measured and calculated

values of resistor voltages? Hint: non-ideal storage element.

43

1) 5 ms

2) In procedure 1, increasing the Resistance made the waves more and more flat while decreasing the resistance made the
waves more and more square shaped.

3)

4) No it won’t, because the capacitor would not have enough time to reach steady state. In other words, if the voltage begins
decreasing while the capacitor is still transient, then the ca