TF EIM Chapt2

Alternating Current (AC)

Thus far we have discussed direct current circuits which were built from a battery (constant voltage supply).

Another type of circuit is one in which the current is driven in an alternating fashion. Typically the current is increased so some positive maximum value, then decreased until it passes through zero and reverses direction until it reaches another maximum value and then it is decreased again, passing though zero once more on its way to a positive maximum value.

Definitions

frequency ([math]f[/math], [math]\nu[/math] )

The frequency is number of complete cycles which occur in 1 sec (cycles/sec = Hz)

Angular frequency =[math] \omega = 2\pi f[/math] = radians/sec

Period

The period is the time to complete one cycle = 1/[math]f[/math]

Amplitude

The change in the current from zero to its most positive value

peak-to-peak

The amount the current changes from its largest positive value to its most negative.

phase

The time a current takes to reaches its maximum value compared to another alternating current

Note

The above definitions can be used for voltage as easily as current.

Mathematical description

Trigonometric

[math]I(t) = \sin(\omega t + \phi) = \cos(\omega t + \phi - \pi/2)[/math]

Note

A 180 degree ([math]\pi[/math] radian) phase shift will put the above cosine wave completely out of sync with the original one such that if we add the two signals together the net current would be zero.

I_{RMS}

With this functional form for the alternating current we can now calculate another property, the RMS or root mean square.

The RMS quantifies an average fluctuation of the alternating current such that

[math]I_{RMS} \equiv \sqrt{\frac{1}{T} \int_0^T I^2(t) dt} = \sqrt{\frac{1}{T} \int_0^T I_0^2 \sin^2(\omega t) dt}= \frac{I_0}{\sqrt{2}} [/math]

where T represent the time interval over which the average is calculated ([math]\infty[/math])

[math]\int_0^T \sin^2(x) dx = \int_0^T \frac{1}{2} \left (1 - \cos(2x) \right) dx = \frac{1}{2} + \frac{1}{T} \int_0^T \cos(t)dt = \frac{1}{2}[/math] if T is infinite or an integer number of cycles

Power

The instantaneous power dissipated in a resister by an alternating current is given as

[math]P_{inst}(t) =I^2(t) R = R\sin^2(\omega t)[/math]

Average power

[math]P_{ave} = \frac{1}{T} \int_0^T P(t) dt = \frac{1}{T} \int_0^T RI^2(t) dt = \frac{R}{T} \int_0^T \sin^2(\omega t) dt = \frac{I_0^2R}{\sqrt{2}} = I_{RMS}^2 R[/math]

Plane waves

Another mathematical expression for an alternating current uses complex variables

[math]I\left ( \omega t \right )=I_0 \exp^{i \left ( \omega t \right )} = I_0 \left ( \cos \left ( \omega t \right) + i \sin \left ( \omega t \right) \right )[/math]

Expressing the current in this form will be useful later

Voltage sources of AC current

The circuit symbol for an emf used to drive alternating currents is given below

A mathematical form is for an AC voltage source is:

[math]V = V_0 \sin(\omega t)[/math]

Capacitors

A capacitor is a device used to store charge in a circuit. The simplest capacitor is built from two conducting plates placed close to each other.

Notice

The potential is constant over the face of the capacitor

[math]V_{AB} = - \int_A^B \vec{E} \cdot d \vec{l}[/math]

The symbol for a capacitor in a circuit is usually.

or

electrolytic symbol

The electrolytic capacitor

Electrolytic capacitors can be shorted out. Put the positive probe from the voltmeter on the positive side of the electrolytic capacitor, the capacitor should be good if you measure a large resistance [math](k \Omega)[/math] otherwise a shorted electrolytic capacitor (small resistance) is a bad one.

Capacitor Properties

1. [math]C\gt 0[/math] Measures ability to store charge
2. [math]V=Q/V[/math] : Double Q Doubles V
3. SI units = Farrads = Coul/Volt
4. Magnitude depends on the elements geometry (conductor area, separation of conductors, ...)

Capacitors in Parallel and Series

Series

Capacitors in Series add like resistors in Parallel.(Capacitance decreases when hooked up in series)

Note

The same amount of charge is pushed onto each capacitor

For capacitors in series you can apply the loop theorem for the above case and the one where all the capacitors are replaced by a single capacitor.

[math]V_{tot} = V_1+V_2+ ...[/math]

[math]Q/C_{tot} = Q/C_1+Q/C_2+ ...[/math]

[math]1/C_{tot} = 1/C_1+1/C_2+ ...[/math]

Parallel

Capacitors in parallel add like resistors in Series (Capacitance increases when hooked up in parallel)

If you were to replace all capacitors with an equivalent capacitance (C_{eq}) then the charge driven onto the equivalent capacitance would be equal to the sum of the charge driven onto each capacitor.

[math]Q_{tot} = Q_{eq} = Q_1 + Q_2 +[/math] ...

[math]VC_{tot} = VQ_{eq} = VQ_1 + VQ_2 +[/math] ...

[math]C_{tot} = Q_{eq} = Q_1 + Q_2 +[/math] ...

Charging/Discharging Capacitor circuit

If

[math]V=V_0 cos(\omega t)[/math]

then applying loop theorem lead to

[math]V-\frac{Q}{C} = 0[/math]

[math]Q = VC= V_0 C \cos{\omega t)}[/math]

[math]I = \frac{dQ}{dt} = -V_0C \omega \sin(\omega t)= V_0 C \omega \cos(\omega t + \pi/2)[/math]

Notice that the current is 90 degrees out of phase with the driving voltage.

Phase shift using 562nF capacitor and a 200 mV sine wave voltage source at 65 Hz

Note

The current increases faster than the voltage. Below are a few scope pictures measuring the voltage coming out of the voltage source [math](V_{in})[/math] and the voltage on the ground side of the capacitor [math](V_{out})[/math] (the capacitor is grounded through the scope). This is actually a high pass filter because of the internal resistance of the scope used to measure [math]V_{out}[/math].

1 Hz	16Hz	kHz
Capacitor response at low frequency	16Hz frequency	kHz frequency

Reactance: Resistive nature of capacitance

To illustrate the resistive nature of capacitors, let's write the driving voltage in using complex notation

[math]V=V_0 \cos(\omega t) = V_0 Re \left [ e^{i\omega t} \right ][/math]

The notation

[math] e^{\omega t} = \cos(\omega t) + i \sin(\omega t)[/math]: Euler's equation

[math] Re \left [ e^{i\omega t} \right ]=Re \left [ \cos(\omega t) + i \sin(\omega t) \right ]= \cos(\omega t) [/math] The real part of a complex variable.

The current is written as

[math]I=\omega CV_0 Re \left [ e^{i \left (\omega t + \pi/2 \right ) } \right ][/math]

[math] =\omega C V_0 Re \left [ e^{i \left ( \omega t \right )} e^{i \left ( \pi/2 \right )} \right ][/math]

[math] =\omega C V_0 Re \left [ e^{i \left ( \omega t \right )} \left ( \cos(\pi/2) + i sin(\pi/2) \right ) \right ][/math]

[math] =\omega C V_0 Re \left [ e^{i \left ( \omega t \right )} \left ( 0 + i \right )\right ][/math]

[math] =\omega C V_0 Re \left [ i e^{i \left ( \omega t \right )} \right ][/math]

[math] =Re \left [ V_0 \frac{e^{i \left ( \omega t \right )}}{\frac{1}{i\omega C}} \right ]=Re \left [ V_0 \frac{e^{i \left ( \omega t \right )}}{X_C} \right ][/math]

where

[math]X_C = \frac{1}{i \omega C}[/math] Effective impedance of a Capacitor

Observe

At high frequencies [math](\omega)[/math] the effective impedance (X_C) is small so the driving voltage should pass right through.

At low frequencies [math](\omega)[/math] the effective impedance (X_C) is large so the driving voltage will be severely attenuated.

Capacitors block Alternating Currents so you can reduce ripples on DC voltages.

You could measure the capacitance by measuring [math](I/V)^2[/math] for several frequencies

RC circuit

Now add a resistor in series with the capacitor in the above capacitor charging circuit.

Quick Solution

Applying the loop theorem using the reactance[math] X_C[/math] of the capacitor instead of Q/C

[math]V - IR -X_CI = =V -(X_R +X_C) I = 0[/math]

[math]\Rightarrow[/math] the capacitor can be added in series with the other resistor

[math]X_{tot} = X_R + X_C[/math]

It looks like the voltage divider from the resistance section

[math]V_{out}= Re\left [\frac{X_C}{R+X_C} V_{in} \right ]= Re \left [\frac{X_C}{X_R+X_C} V_{in} \right ][/math]

To evaluate [math]Re[Z] = \sqrt{Z Z^*}[/math] where[math] Z = x+iy[/math] and [math]Z^* = x-iy[/math]

[math]\frac{V_{out}}{V_{in}} = Re \left [ \frac{\frac{1}{i \omega C}}{R+\frac{1}{i \omega C}} \right ] = \sqrt{\frac{\frac{1}{i \omega C}}{R+\frac{1}{i \omega C}} \frac{\frac{1}{-i \omega C}}{R+\frac{1}{-i \omega C}} } = \frac{1}{\sqrt{1 + \omega^2 R^2 C^2}}[/math]

Let

[math]\omega_b = \frac{1}{RC} =[/math] break point (cut off ) frequency

then

[math]\frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1 + \left ( \frac{\omega}{\omega_b}\right )^2}}[/math]

Note

The RC circuit is another way to measure the capacitance by measureing [math]V_{out}/V_{in}[/math] -vs- [math]\omega[/math] and do a linear extrapolation to determine the break point frequency.

Phasor Diagram

Phasor diagrams are a technique used to illustrate the phase difference between the input and output alternating voltages/currents.

Returning to the loop theorem for the low pass RC filter

[math]V_{in} = V_R + V_{out}[/math]

[math]V_{in} = IX_R + IX_C[/math]

[math]V_{in} = IR + I\frac{-i}{\omega C}[/math]

We can write the voltage as a complex variable using a coordinate system in which the "x-axis" represents real numbers and the "y-axis" represents the complex components. This is analogous to expressing 2-D vectors.

Below is a figure representing the complex plane.

The angle [math]\phi[/math] is the angle between[math] V_{out} = \frac{-i}{\omega C}[/math] and [math]V_{in} = IR + I\frac{-i}{\omega C}[/math] Based on the above Phasor diagram

[math]\tan(\phi) =\tan \left (\frac{R}{\frac{-1}{\omega C}} \right )= \tan \left(\omega RC \right )[/math]

High Pass Filter

A high pass filter switches the order of the resistor and the capacitor.

Using the same reactance based analysis as above

[math]V_{in} - IX_C - IR = 0[/math]

[math]V_{out} = IR[/math]

[math]\frac{V_{out}}{V_{in}} = \frac{IR}{I \left ( X_C + R \right ) }= \frac{R}{\frac{1}{i \omega C} + R} = \frac{ i \omega RC}{1 + i \omega RC}[/math]

[math]\left |\frac{V_{out}}{V_{in}} \right | = \sqrt{\left ( \frac{ i \omega RC}{1 + i \omega RC}\right ) \left ( \frac{- i \omega RC}{1 - i \omega RC}\right )} = \frac{\omega RC}{\sqrt{(1 + \omega^2 R^2C^2}}[/math]

Phasor diagram

The phasor diagram is very similar except this time V_{out} is along the x-axis

Returning to the loop theorem for the low pass RC filter

[math]V_{in} = V_R + V_{out}[/math]

[math]V_{in} = IX_C + IX_R [/math]

[math]V_{in} = I\frac{-i}{\omega C}+IR[/math]

We can write the voltage as a complex variable using a coordinate system in which the "x-axis" represents real numbers and the "y-axis" represents the complex components. This is analogous to expressing 2-D vectors.

Below is a figure representing the complex plane.

The angle [math]\phi[/math] is the angle between [math]V_{out} =IR[/math] and [math]V_{in} = I\frac{-i}{\omega C} + IR [/math] Based on the above Phasor diagram

[math]\tan(\phi) = \tan \left (\frac{\frac{-1}{\omega C}}{R} \right ) = \tan(\frac{1}{\omega RC})[/math]

High Pass -vs- Low Pass graphs

if

[math]\omega_B=\frac{1}{RC}=10[/math]

Low pass	High Pass
Gain -vs- [math]\omega[/math]	Gain -vs- [math]\omega[/math]
Phase Shift -vs- [math]\omega[/math]	Phase Shift -vs- [math]\omega[/math]

exact solution

Again apply the loop theorem

[math]V -IR -\frac{Q}{C} =0[/math]

or

[math]R \frac{dQ}{dt} + \frac{1}{C} Q =V(t)[/math] First order non-homogeneous linear differential equation

First solve homogeneous part

[math]R \frac{dQ}{dt} + \frac{1}{C} Q =0[/math] First order homogeneous linear differential equation

The above is integratable

[math] \frac{dQ}{dt} = -\frac{1}{RC} Q [/math]

[math]\int \frac{dQ}{Q} = -\int \frac{1}{RC} dt [/math]

[math]\ln Q = -\frac{t}{RC} -Constant[/math]

[math] Q(t) = e^{-\frac{t}{RC}} e^{Constant} = Q_0e^{-\frac{t}{RC}}[/math]

[math] I(t)= \frac{d Q}{d t} = \frac{d}{dt} Q_0 e^{-\frac{t}{RC}} = \frac{-Q_0}{RC}e^{-\frac{t}{RC}}[/math]

This gives you the general solution for the homogeneous problem

In this case it represents the solution for a discharging capacitor.

Trial solution method for Non-Homogeneous Solution

A common and more general technique to solving such a differential equation is to use the solution from the discharging equation (the homogeneous diff. eq.) as a trial solution by inserting constants.

[math]Q_{trial} = A + Be^{-t/RC}[/math]

You then substitute this trial solution into the differential equation and apply boundary conditions to determine the coefficients.

[math]\frac{d}{dt} \left ( A + Be^{-t/RC} \right ) + \frac{1}{RC} \left ( A + Be^{-t/RC} \right ) = \frac{dV}{Rdt}[/math]

[math] \left ( B \frac{-1}{RC}e^{-t/RC} \right ) + \frac{1}{RC} \left ( A + Be^{-t/RC} \right ) = \frac{dV}{Rdt}[/math]

[math]\Rightarrow \frac{1}{RC} \left ( A \right ) = \frac{dV}{Rdt}[/math]

[math]A = C \frac{dV}{dt} = C \frac{d V_0 e^{i \omega t}}{dt}[/math]

To determine the other coefficient you apply the boundary condition that q = 0 when t=0

[math]0= A + Be^{-0/RC} = A+B[/math]

[math]A=-B[/math]

[math]Q=CV (1 - e^{-t/RC}) =Q_0 (1 - e^{-t/RC})[/math]

Forest_Electronic_Instrumentation_and_Measurement