9. Capacitor and Resistor Circuits

ElectronicsLab9.nb 1

Introduction Thus far we have consider resistors in various combinations with a power supply or battery which provide a constant voltage source or direct current (voltage) DC. Now we start to consider various combinations of components and much of the interesting behavior depends upon time so we will also consider AC or alternating current (voltage) sources which are signal generators. The first combination we consider is a resistor in series with a capacitor and a battery.

The RC Circuit Consider the resistor-capacitor circuit indicated below:

When the switch is closed, Kirchoff's loop equation for this circuit is

Q V = + iR (1) C for t>0 where both Q[t] and i[t] are functions of time. There are two unknown quantities Q[t] and i[t] in equation (1) and we need an additional equation namely ElectronicsLab9.nb 2

d i t = Q t (2) dt You can eliminate one of the unknowns between equations (1) and (2) by taking the derivative of equation (1)@ withD respect to@ timeD obtaining

1 d d 0 = Q t + R i t (3) C dt dt and using equation (2) to eliminate the derivative of the charge @ D @ D 1 d 0 = i t + R i t (4) C dt It is easy enough to solve equation (4) since by rearrangement @ D @ D d −1 i t = i t (5) dt RC Further @ D @ D 1 −1 âi = ât (6) i RC Integration yields ‡ ‡ i t t Log − (7) i0 RC where i0 is a@ constantD of integration which we will determine shortly. Using a property of the exponential function, Bwe obtainF from equation (7)

t i t = i0 Exp − (8) RC Initially at t=0, when the switch is closed, the capacitor has zero charge and therefore there is zero potential@ D across it. TheB currentF in the circuit is determined entirely by the battery potential V and the resistance R through Ohm's law

V i 0 R = V or i 0 = (9) R as initially the capacitor C play no role. @SettingD t=0 in equation@ (8)D and using equation (9) yields

V i 0 = i0 = (10) R so we have determined the constant of integration. Utilization of equation (10) in equation (8) yields finally@ theD solution as ElectronicsLab9.nb 3

V t ‡ i t = Exp − (11) R RC The product RC has units of time and usually is called the time constant  @ D B F τ = R C (12)

Graph of the Solution for the current Suppose the numerical values V=10 volts, R=8,000 ohms, and C=2.5 microfarads as indicated then

V = 10; R = 8000.; Cap = 2.5 ∗ 10−6; τ = R ∗ Cap; Print "Time Constant =", τ, " sec"

Time Constant =0.02 sec @ D IMPORTANT: C is a protect variable assigned to something specific in Mathematica so instead we use Cap as the symbol for capacitance.

V t i t_ := Exp − R τ

Plot@ Di t , t,B0, 4F∗ τ , AxesLabel → "time", "i t "

@ @ D 8 < i t8 @ D

0.001@ D

0.0008

0.0006

0.0004

0.0002

time 0.02 0.04 0.06 0.08

… Graphics …

So initially, right after the switch is closed, the current i[t] is a maximum and thereafter is decreases exponentially. The initial current is ElectronicsLab9.nb 4

So initially, right after the switch is closed, the current i[t] is a maximum and thereafter is decreases exponentially. The initial current is

i 0

0.00125 @ D which agrees with the graph above. The time constant (=0.02 seconds in this case) determines the rate of decay. After a time  the current has decreased to

i τ

0.000459849 @ D and this is equal to equation (11) with t= namely

i 0 ∗ ã−1

0.000459849 @ D Note that

ã−1.

0.367879

so after one time t= the current has dropped 37% in its value. After t=2, the current is

i 2 ∗ τ

0.000169169 @ D which is 14% of the original value of the current since

ã−2.

0.135335

and so on.

The voltage across the resistor is i*R so we may graph this voltage as ElectronicsLab9.nb 5

Vr = Plot i t ∗ R, t, 0, 4 ∗ τ ,

AxesLabel → "time", "VR t "

VR@ t@ D 8 < 10 8 @ D

@ D 8

time 0.02 0.04 0.06 0.08

… Graphics …

Graph of the Charge Q[t] Combining equations (2) and (11) yields

d V t Q t = Exp − (13) dt R RC and integration yields @ D B F V t t' Q t = Q 0 + Exp − ât' (14) R 0 τ where the initial charge on the capacitor is zero Q[0]=0. The integral is easily obtained @ D @ D ‡ B F Clear τ ; t t' Exp − ât' τ 0 @ D

t − ‡τ − ã τ Bτ F

Combining this with equation (14) and recalling equation (12) we obtain

t Q t = VC 1 − Exp − (15) τ which we can also graph. @ D B F ElectronicsLab9.nb 6

V = 10; R = 8000.; Cap = 2.5 ∗ 10−6; τ = R ∗ Cap; t Q t_ := V ∗ Cap 1 − Exp − τ

Plot@ DQ t , t, 0, 4 ∗ τ B, F AxesLabel → "time", "Q t "

@ @ D 8 < Q8t @ D

0.00002@ D

0.000015

0.00001

5´10-6

time 0.02 0.04 0.06 0.08

… Graphics …

Initially the charge on the capacitor is zero

Q 0.

0. @ D and this agrees with the above graph. After a time t= the charge on the capacitor is

Q τ

0.000015803 @ D and this can also be obtained approximately from the graph. After a very long time the charge has its maximum value

Q 5 ∗ τ

0.0000248316 @ D which is almost the same as ElectronicsLab9.nb 7

V ∗ Cap

0.000025

Q The voltage across the capacitor is and we may graph this using C

Q t VC = Plot , t, 0, 4 ∗ τ , Cap

AxesLabel@ D→ "time", "VC t " B 8 <

VC t 10 8 @ D

8@ D

time 0.02 0.04 0.06 0.08

… Graphics …

Replace the Battery and Switch by a Signal Generator having a Square Wave The circuit diagram now appears ElectronicsLab9.nb 8

Suppose the square wave generator has a frequency f given by the square of the signal generator can be graph using

T = τ; 1 f = ; T Print "frequency =", f, "Hz"

frequency =50.Hz @ D Notice that the period of the signal generator is chosen to be the same as the time constant in the RC circuit. We will discuss this more later. Suppose the voltage amplitude of the signal generator is 10 volts (the same as the battery voltage) The square wave of the signal generator is graphed using ElectronicsLab9.nb 9

V0 = 10.; T SqWave t_ := If t < , V0, 0 2 Plot SqWave t , t, 0, T @ D B F 10 @ @ D 8

0.005 0.01 0.015 0.02

… Graphics …

Effectively having the signal generator in the circuit is the same as having the battery in the circuit for time 0

The Voltage Across the Capacitor We may graph the voltage across the capacitor together with the signal generator voltage and obtain ElectronicsLab9.nb 10

T = τ; Q t T Plot , SqWave t , t, 0, Cap 2

10@ D B: @ D> : >F

0.002 0.004 0.006 0.008 0.01

If the period T of the signal generator is longer, for example T=2*, then the capacitor has more time to charge

T = 2 τ; Q t T Plot , SqWave t , t, 0, Cap 2

10@ D B: @ D> : >F

0.005 0.01 0.015 0.02

Further if the period T of the signal generator is longer still, for example T=3*, then the capacitor has more time to charge ElectronicsLab9.nb 11

T = 3 τ; Q t T Plot , SqWave t , t, 0, Cap 2

10@ D B: @ D> : >F

0.005 0.01 0.015 0.02 0.025 0.03

and the capacitor almost has time to fully charge and have all the 10 volts appear across it. About 8 volts now appears across the capacitor.

The Voltage Across the Resistor We may graph the voltage across the resistor together with the signal generator voltage and obtain

T = τ; T Plot i t R, SqWave t , t, 0, , PlotRange → 0, 10 2 10 B8 @ D @ D< : > 8

0.002 0.004 0.006 0.008 0.01

If the period T of the signal generator is longer, for example T=2*, then the current gets smaller still and the voltage across the resistor is reduced further ElectronicsLab9.nb 12

If the period T of the signal generator is longer, for example T=2*, then the current gets smaller still and the voltage across the resistor is reduced further

T = 2 τ; T Plot i t R, SqWave t , t, 0, , PlotRange → 0, 10 2 10 B8 @ D @ D< : > 8

0.005 0.01 0.015 0.02

If the period T of the signal generator is longer, for example T=3*, then the current gets smaller still and the voltage across the resistor is reduced further

T = 3 τ; T Plot i t R, SqWave t , t, 0, , PlotRange → 0, 10 2 10 B8 @ D @ D< : > 8

0.005 0.01 0.015 0.02 0.025 0.03

About 2 volts now appears across the resistor.

The Second Part of the Square Wave of the Signal Generator. ElectronicsLab9.nb 13 The Second Part of the Square Wave of the Signal Generator.

During the second part of the period of the signal generator for times T  t  T, the voltage is 2 zero in the original circuit. It helps make the analysis simpler to change the wave form a little and have

the signal generator voltage zero during the first part of the cycle and a constant V0during the second part of the cycle

T = τ;

V0 = 10.; T SqWave t_ := If t < , 0, V0 2 SigGen = Plot SqWave t , t, 0, T @ D B F 10 @ @ D 8

0.005 0.01 0.015 0.02

… Graphics …

This corresponds to the times t in the range 0.01 sec < t < 0.02 sec in the previous diagram. Effectively for the first part of the cycle the battery is removed from the circuit and replaced by a shorting wire and the circuit looks like ElectronicsLab9.nb 14

Kirchoff circuit law after the switch is closed is

Q 0 = + iR (16) C which is the same as equation (1) without the battery. Taking the time derivative of equation (16) and using equation (2) yields

d i i = − (17) dt R C Equation (17) can be solved using the same techniques as before and we obtain again equation (11)

d t Q t = i t = i0 Exp − (18) dt τ

However, the initial condition i0is different this time as we shall see. Equation (18) can be integrated for the charge Q[t]@ D obtaining@ D B F

t Q t = Q 0 + i0 τ 1 − Exp − (19) τ The capacitor is assumed fully charged initially (which can happen if the time constant  is short compared@ D with the@ Dperiod T of the squareB wave)F so initially

Q 0 = C V (20) and when t=0 the part of equation (19) involving the exponential function vanishes. For long times there is no charge on the capacitor Q[]=0. Since Exp − ∞ =0 and equation (19) reduces to @ D τ

Q ∞ = C V + i0 τ = 0 (21) A E and it follows that @ D C V V i0 = − = − (22) τ R Combining equations (20) and (22) with equation (19) yields

t t Q t = C V + VC Exp − − 1 = V C Exp − (23) τ τ Equation (23) should make intuitive sence, since during the second half of the square wave cycle, the @ D B F B F Q capacitor is discharging. The voltage across the capacitor is V= initially so C ElectronicsLab9.nb 15

T = τ; V = 10.; T SqWave t_ := If t < , 0, V 2 t T Plot V ∗ Exp − , SqWave t , t, 0, @ D τB F 2 10 B: B F @ D> : >F

0.002 0.004 0.006 0.008 0.01

… Graphics …

Further if the signal generator is longer say three times the time constant, T=3 then the capacitor has even more time to discharge ElectronicsLab9.nb 16

T = 3 ∗ τ; V = 10.; T SqWave t_ := If t < , V0, 0 2 t T Plot V ∗ Exp − , SqWave t , t, 0, @ D τB F 2 10 B: B F @ D> : >F

0.005 0.01 0.015 0.02 0.025 0.03

… Graphics …

The Voltage Across the Resistor The current in the circuit is obtained by taking the derivative of the charge equation (23) obtaining

V C t V t i t = − Exp − = − Exp − (24) τ τ R τ and the voltage across the resistor is just R*i[t]. Graphing the voltage across the capacitor and the voltage@ acrossD the resistorB for theF second half theB cycleF yields ElectronicsLab9.nb 17

T = τ; V = 10.; t t T Plot V ∗ Exp − , − V ∗ Exp − , t, 0, τ τ 2

B: 10 B F B F> : >F

0.002 0.004 0.006 0.008 0.01

-5

-10

… Graphics …

Notice the sum of the voltage of the capacitor and the voltage of the resistor is just zero as required by Kirchoff's law. If the signal generator period T is twice the time constant  then we obtain

T = 2 ∗ τ; V = 10.; t t T Plot V ∗ Exp − , − V ∗ Exp − , t, 0, τ τ 2

B: 10 B F B F> : >F

0.005 0.01 0.015 0.02

-5

-10

… Graphics … ElectronicsLab9.nb 18

If the signal generator period T is three the time constant  then we obtain

T = 3 ∗ τ; V = 10.; t t T Plot V ∗ Exp − , − V ∗ Exp − , t, 0, τ τ 2

B: 10 B F B F> : >F

0.005 0.01 0.015 0.02 0.025 0.03

-5

-10

… Graphics …

Laboratory Exercises PART A: Place a signal generator in series with a resistor and capacitor. Pretty much any output level (the output voltage) of the signal generator will do OK but after you get the oscilloscope working prop- erly make a note of the maximum voltage in your lab notebook. Choose a square wave and make the frequency f of the signal generator such that f= 1 with T==RC at first. With channel 1 of the oscillo- T scope, measure the voltage across the signal generator and with channel 2 measure the voltage across the capacitor. Compare with the graphs of the first example above. Make the frequency f of the signal generator smaller (T larger) so the capacitor has more time to charge. Keep decreasing f. Sketch the oscilloscope figures you get and indicate the values of the voltage on the vertical scale and the time on the horizontal scales.

Example: Suppose C=0.1 F and R=6.8 k then the time constant =RC=0.00068 sec. as indicated below: ElectronicsLab9.nb 19

R = 6.8 × 103; c = 0.1 × 10−6; τ = R × c

0.00068

NOTE: The value of R and C you use need not be the values given above. Use your digital ohmmeter to measure the value of the resistor and make sure it is the same as given by the color code. Use your digital capacitor meter to measured the value of the capacitor and it should agree with the capacitor code (which is not that standardized so check with the maker of the capacitor and use your meter For example, a capacitor labeled 250 B means 1 is the first digit and 5 is the second digit for the capacitance. 2 is the multiplier in powers of 10 so this capacitor is C=25×100 F  25 F where F=106 F. Capacitors can be much smaller and pF  1012 F is often used

 is the time it take the capacitor to charge to 67% of the maximum voltage (which is the maximum voltage of the signal generator). The signal generator frequency should be set to have a period T= at first but what you actually control is the frequency f of the signal generator where f=1/T. For the example above,

T = τ; f = 1 T

1470.59 ê So the frequency f=1,470 Hz= 1.5 kHz corresponds to one time constant . The horizontal time scale of the oscilloscope had better be something like this frequency f. If the oscilloscope is set at too high a frequency, the time will be too short to see the voltage rise. On the other hand, if the oscilloscope is set at too low a frequency, there will not be enough time to see the voltage rise across the capacitor. You also must make sure to set the voltage scale at roughly the output voltage of the oscilloscope which you should have measured first before connecting the capacitor and resistor in the circuit.

PART B: With channel 1 of the oscilloscope, measure the voltage across the signal generator and with channel 2 measure the voltage across the resistor. Compare with the graphs of the second example above. Make the frequency f of the signal generator smaller (T larger) so the capacitor has more time to charge. Keep decreasing f the frequency of the signal generator. Sketch the oscilloscope figures you get.

PART C: Call the capacitor used above C1. Take a second capacitor and call it C2. Combine the two capacitors in SERIES without the signal generator and oscilloscope attached. The effective capacitance is given by

1 1 1 = + Ceff C1 C2 Compute the numerical value of the effective capacitance and check it with the digital capacitance meter.

Note the effective capacitance of two capacitors in SERIES is less than both C1 and C2. Use the SERIES

combination of C1 and C2 together with the signal generator and oscilloscope and repeat the measurements of PART A above. ElectronicsLab9.nb 20 Compute the numerical value of the effective capacitance and check it with the digital capacitance meter.

Note the effective capacitance of two capacitors in SERIES is less than both C1 and C2. Use the SERIES

combination of C1 and C2 together with the signal generator and oscilloscope and repeat the measurements of PART A above.

PART D: Call the capacitor used above C1. Take a second capacitor and call it C2. Combine the two capacitors in PARALLEL without the signal generator and oscilloscope attached. The effective capacitance is given by

Ceff = C1 + C2 Compute the numerical value of the effective capacitance and check it with the digital capacitance meter.

Note the effective capacitance of two capacitors in PARALLEL is less than both C1 and C2. Use the

PARALLEL combination of C1 and C2 together with the signal generator and oscilloscope and repeat the measurements of PART A above.