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Appendix 2

AC equation derivations (chapter 6)

Deriving charge over time

Chapter 1 showed that current is the flow of charge at a specific point for a specific length of time – this can now be defined mathematically as the rate of change of charge over time:

C u r r e n t, I = \frac{d Q}{d t}

$C u r r e n t, I = \frac{d Q}{d t}$

(6.1)

	$I$ $I$ is the current in amperes (A)
	$d Q$ $d Q$ is the rate of change of charge in coulombs (C)
	$d t$ $d t$ is the rate of change of time in seconds (s)

This equation may initially seem a little unusual when compared to other DC theory terms (like Ohm’s Law) that use more simple scalar relationships between quantities. With sine waves for AC analysis, the use of differentiation and integration is part of the standard mathematical mechanisms for manipulating them. AC circuits are time-varying circuits, and so changes in voltage, current and impedance must be analysed relative to how they change over time. The concept of rate of change is not discussed in detail in this book, but the substitution of the rate of change of charge over time ( $\frac{d Q}{d t}$ $\frac{d Q}{d t}$ ) for current (I) can be used to help derive some of the formulae relating to capacitors.

Deriving an RC time constant

Knowing the equation for capacitance and also knowing how a capacitor’s voltage and current changes over time, a time constant (τ) can be defined that describes the rate of change in a resistor/capacitor (RC) circuit (Appendix 2 Figure 1).

Appendix 2 Figure 1 Time constant in an RC circuit – from chapter 6, Figure 6.22. The diagram shows the charge/discharge graph for the series combination of a resistor and capacitor. The values of both components dictate how long it takes for the capacitor to reach ~63% of its final charge level, which is known as the time constant (τ).

The time constant (τ) denotes how long it takes for the capacitor to reach ~63% of the supply voltage, which is derived from the fact that the charging curve is exponential in shape – where $\frac{1}{e}$ $\frac{1}{e}$ approximates to 0.37, which is the reduction from maximum percentage in each time constant (100 − 37). It will take a total of 5 time constants to reach full charge, and a further 5 for the capacitor to discharge again – 6τ defines when the discharge cycle reaches ~37% (100 − 63). The mathematical proof for this exponential rise and fall involves solving the sum of the voltages (V_R and V_C) in the circuit as a differential equation, which is a more advanced mathematical process than needed for an introductory text. Nevertheless, this relationship can be stated to show how the time constant τ can be derived by first returning to Kirchoff’s Voltage Law from chapter 3:

The current equation from 6.1 can be rearranged with the equation for capacitor charge (Q = CV) because a capacitor varies over time. The differential of charge is current, so current must also be the differential of our capacitor voltage term in the equation (capacitance C is constant):

C h a r g e, Q = C V_{C}, C u r r e n t, I = \frac{d Q}{d t} = C \frac{d V_{C}}{d t}

$C h a r g e, Q = C V_{C}, C u r r e n t, I = \frac{d Q}{d t} = C \frac{d V_{C}}{d t}$

Current is constant in a series circuit, so Ohm’s Law can be used to factor in the resistor. This can be expressed in terms of the capacitor voltage to build an equation relating R and C:

C u r r e n t, I = C \frac{d V_{C}}{d t}, R e s i s t o r V o l t a g e, V_{R} = I R = R C \frac{d V_{C}}{d t}

$C u r r e n t, I = C \frac{d V_{C}}{d t}, R e s i s t o r V o l t a g e, V_{R} = I R = R C \frac{d V_{C}}{d t}$

Now use Kirchoff’s Voltage Law (KVL) to define all the voltages in the circuit (VS, VR and VC):

S u p p l y V o l t a g e, V_{S} = V_{R} + V_{C} = R C \frac{d V_{C}}{d t} + V_{C}

$S u p p l y V o l t a g e, V_{S} = V_{R} + V_{C} = R C \frac{d V_{C}}{d t} + V_{C}$

VC has not been changed to allow a first-order differential equation to be built, which can be solved by rearranging the terms:

V_{S} - V_{C} = R C \frac{d V_{C}}{d t} \to (V_{S} - V_{C}) d t = R C d V_{C} \to d t = \frac{R C d V_{C}}{(V_{S} - V_{C})}

$V_{S} - V_{C} = R C \frac{d V_{C}}{d t} \to (V_{S} - V_{C}) d t = R C d V_{C} \to d t = \frac{R C d V_{C}}{(V_{S} - V_{C})}$

∴ \frac{d t}{R C} = \frac{d V_{C}}{V_{S} - V_{C}}

$∴ \frac{d t}{R C} = \frac{d V_{C}}{V_{S} - V_{C}}$

With differential terms on both sides, integrate the entire equation to remove them:

- l n (V_{S} - V_{C}) = \frac{t}{R C} + K

$- l n (V_{S} - V_{C}) = \frac{t}{R C} + K$

Where the term K is the constant of integration. Now remove the natural logarithm by changing the equation to exponential form (express the right-hand side as a power of e ). Also, split the integration constant term out (eK) to make it easier to solve:

V_{S} - V_{C} = e^{- \frac{t}{R C} + K} = e^{- \frac{t}{R C}} \times e^{K}

$V_{S} - V_{C} = e^{- \frac{t}{R C} + K} = e^{- \frac{t}{R C}} \times e^{K}$

All the terms of the equation have been derived, apart from the value of the integration constant (K). To do this, take a known time (time t = 0) and use the fact that the capacitor will not have begun to charge to define the value for capacitor voltage (VC = 0) at that point:

V_{S} - 0 = e^{- \frac{0}{R C}} \times e^{K} = 1 \times e^{K} ∴ e^{K} = V_{S}

$V_{S} - 0 = e^{- \frac{0}{R C}} \times e^{K} = 1 \times e^{K} ∴ e^{K} = V_{S}$

Substitute this into the full equation to state the full relationship between VC and VS:

V_{S} - V_{C} = e^{- \frac{t}{R C}} \times e^{K} = V_{S} e^{- \frac{t}{R C}} \to V_{C} = V_{S} - V_{S} e^{- \frac{t}{R C}}

$V_{S} - V_{C} = e^{- \frac{t}{R C}} \times e^{K} = V_{S} e^{- \frac{t}{R C}} \to V_{C} = V_{S} - V_{S} e^{- \frac{t}{R C}}$

∴ V_{C} = V_{S} (1 - e^{- \frac{t}{R C}})

$∴ V_{C} = V_{S} (1 - e^{- \frac{t}{R C}})$

The RC term dictates how much the exponent increases over time, and thus it is known as the time constant (τ):

V_{C} = V_{S} (1 - e^{- \frac{t}{τ}})

$V_{C} = V_{S} (1 - e^{- \frac{t}{τ}})$

Deriving capacitor voltage

Capacitor voltage varies over time, based on the time constant. Equation 6.1 can be used to show that current is equivalent to the derivative (rate of change) of charge over time, and then the terms of this equation can be rearranged to define charge (Q) as the integral of current over time:

From equation 6.7 for current as the derivative of charge over time, integrate current to define charge:

i f C u r r e n t, I = \frac{d Q}{d t}, t h e n d Q = I d t, ∴ Q = \int I d t

$i f C u r r e n t, I = \frac{d Q}{d t}, t h e n d Q = I d t, ∴ Q = \int I d t$

Then rearranging equation 6.7 for capacitance:

C a p a c i t a n c e, C = \frac{Q}{V} ∴ V = \frac{Q}{C}

$C a p a c i t a n c e, C = \frac{Q}{V} ∴ V = \frac{Q}{C}$

Substituting with the previous term for charge:

V o l t a g e, V = \frac{Q}{C} = \frac{\int I d t}{C} = \frac{1}{C} \int I d t

$V o l t a g e, V = \frac{Q}{C} = \frac{\int I d t}{C} = \frac{1}{C} \int I d t$

Using the value of AC current from equation 6.3:

I n s t a n t a n e o u s C u r r e n t, I = I_{p} s i n ω t

$I n s t a n t a n e o u s C u r r e n t, I = I_{p} s i n ω t$

Substituting with the previous term for voltage:

V o l t a g e, V = \frac{1}{C} \int I_{p} s i n (ω t) d t

$V o l t a g e, V = \frac{1}{C} \int I_{p} s i n (ω t) d t$

Now use integral substitution to solve

\int I_{p} s i n (ω t) d t

$\int I_{p} s i n (ω t) d t$

l e t u = ω t, d u = ω d t, ∴ d t = \frac{d u}{ω} = \frac{1}{ω} d u

$l e t u = ω t, d u = ω d t, ∴ d t = \frac{d u}{ω} = \frac{1}{ω} d u$

\int I_{p} s i n (ω t) d t = \int I_{p} s i n (u) \times \frac{1}{ω} d u

$\int I_{p} s i n (ω t) d t = \int I_{p} s i n (u) \times \frac{1}{ω} d u$

Split the terms (

\frac{1}{ω}, I_{p}

$\frac{1}{ω}, I_{p}$

) out from the integration:

\int I_{p} s i n (ω t) d t = I_{p} \times \frac{1}{ω} \times \int s i n (u) d u = \frac{I_{p}}{ω} \int s i n (u) d u

$\int I_{p} s i n (ω t) d t = I_{p} \times \frac{1}{ω} \times \int s i n (u) d u = \frac{I_{p}}{ω} \int s i n (u) d u$

The integral of sin(u) is −cos(u) and the du term is removed by integration – then substitute $ω t$ $ω t$ back in for $u$ $u$ :

\frac{I_{p}}{ω} \int s i n (u) d u = \frac{I_{p}}{ω} \times - c o s (u)

$\frac{I_{p}}{ω} \int s i n (u) d u = \frac{I_{p}}{ω} \times - c o s (u)$

\int I_{p} s i n (ω t) d t = - \frac{I_{p}}{ω} c o s (ω t)

$\int I_{p} s i n (ω t) d t = - \frac{I_{p}}{ω} c o s (ω t)$

Now substitute this expression for AC current into the voltage equation:

V = \frac{1}{C} \times (- \frac{I_{p}}{ω} c o s (ω t))

$V = \frac{1}{C} \times (- \frac{I_{p}}{ω} c o s (ω t))$

Now state the equation for capacitor voltage in terms of current and capacitance:

C a p a c i t o r V o l t a g e, V_{C} = - \frac{I_{p}}{ω C} c o s ω t

$C a p a c i t o r V o l t a g e, V_{C} = - \frac{I_{p}}{ω C} c o s ω t$

(6.2)

	$I_{p}$ $I_{p}$ is the instantaneous current in amperes (A)
	$C$ $C$ is the capacitance in farads (F)
	$ω$ $ω$ is the angle of the sine wave signal in radians (rad)
	$t$ $t$ is the time in seconds (s)

Deriving capacitive reactance

A capacitor will react at different frequencies, which is defined as its reactance. The capacitance value must be fixed (the capacitor cannot expand or contract in size!) and the frequency of the input signal does not change (so $ω = 2 π f$ $ω = 2 π f$ is fixed) – thus the only terms in the equation that can vary are voltage, current and time. Ignoring the effect of time for now, an increase in the peak current will also increase the peak voltage across the capacitor (the peak value is not changed by time). This is no different from Ohm’s Law for a fixed resistance, where the same linear relationship between voltage and current exists:

V o l t a g e, V = I R

$V o l t a g e, V = I R$

(6.3)

	$I$ $I$ is the current in amperes (A)
	$R$ $R$ is the resistance in ohms (Ω)

In chapter 3 (equation 3.2), this equation was rearranged to define resistance in terms of voltage and current ( $R = \frac{V}{I}$ $R = \frac{V}{I}$ ). Ignoring the impact of time, the same approach can be taken to determine the reactance of a capacitor as the ratio of peak voltage over current:

The peak current terms cancel out to state the general equation for capacitive reactance (X_C):

C a p a c i t i v e R e a c t a n c e, X_{C} = \frac{1}{ω C}

$C a p a c i t i v e R e a c t a n c e, X_{C} = \frac{1}{ω C}$

(6.4)

	$X_{C}$ $X_{C}$ is the reactance of the capacitor in ohms (Ω)
	$C$ $C$ is the capacitance in farads (F)
	$ω$ $ω$ is the angle of the sine wave signal in radians (rad)

Looking at this equation, the reactance of a capacitor is effectively defined by its capacitance and also frequency (recall that $ω = 2 π f$ $ω = 2 π f$ ). This means that capacitors vary their reactance with frequency, which is a crucial point for audio electronics.

Deriving series impedance magnitude

The previous section showed that a capacitor has a reactance in ohms (Ω) that varies with the frequency of the input signal voltage applied to it. If frequency is ignored then reactance is the ratio of voltage over current ( $X_{C} = \frac{V}{I}$ $X_{C} = \frac{V}{I}$ ), in much the same way as it is for resistance ( $R = \frac{V}{I}$ $R = \frac{V}{I}$ ). This relationship can be used to determine the total impedance (symbol Z) for a series circuit that contains both a resistor and capacitor (known as an RC circuit). To do this, define each component voltage and its associated phase angle using a phasor diagram (Appendix 2 Figure 2).

Appendix 2 Figure 2 RC series circuit phasor diagram – from chapter 6, Figure 6.28. The left-hand diagram shows a resistor and capacitor in series, connected to an AC voltage supply (V_S). The right-hand diagram shows the resulting phasor diagram for this circuit, where the voltages in the circuit are combined to include the phase angle of each component. As it does not vary with frequency the resistor has phase angle 0°, whilst the capacitor has a phase angle of −90°.

The phasor diagram shown in this figure is a common way of analysing an AC circuit by combining component voltages (or currents) with their associated AC phase angles. Phasors can be used to represent the phase component of a capacitor to accurately model either the voltage levels or current flow within an RC circuit. Thus, a phasor diagram can show how to derive the overall impedance (Z) of the circuit. The resistor voltage (V_R) has a phase angle of 0°, as resistive components do not vary with frequency. The previous section derived the voltage across a capacitor as $V_{C} = - \frac{I_{p}}{ω C} c o s ω t$ $V_{C} = - \frac{I_{p}}{ω C} c o s ω t$ and the negative cosine component of this equation explains why the capacitor voltage in the diagram has a phase angle of −90°. The overall voltage (V_S) of the circuit forms the hypotenuse of this triangle, defining the magnitude of this voltage using trigonometry as follows:

Divide throughout by current (I) to state the equation for the magnitude of the total circuit impedance (Z) in an RC series circuit:

I m p e d a n c e M a g n i t u d e, Z = \sqrt{R^{2} + X_{C}^{2}}

$I m p e d a n c e M a g n i t u d e, Z = \sqrt{R^{2} + X_{C}^{2}}$

(6.5)

	$R$ $R$ is the resistance in ohms (Ω)
	$X_{C}$ $X_{C}$ is the reactance of the capacitor in ohms (Ω)

This equation is very useful, allowing a series AC circuit to be analysed in much the same way as DC series circuits were in chapter 3.

Deriving series capacitance

Returning to Kirchoff’s Voltage Law from chapter 3:

From KVL, the total voltage must equal the sum of the voltage drops across all the resistors. In chapter 3 (equation 3), this law proved that the total resistance ( $R_{t o t} = \frac{V t o t}{I}$ $R_{t o t} = \frac{V t o t}{I}$ ) is the sum of all the resistances in a series circuit:

In this case, the voltage across each component is added to equal the total voltage in the circuit – which is the same thing that happens when capacitors are connected together in series (Appendix 2 Figure 3).

Appendix 2 Figure 3 Series capacitor circuit. The diagram shows two capacitors connected in series, where the sum of the voltages across each capacitor (V_C1 and V_C2) is equivalent to the total voltage (V_tot) for the circuit. An AC supply is shown in this diagram, but time is not included in magnitude calculations.

This series circuit shows the same sum of voltages used to derive total series resistance in chapter 3, where time is omitted from the calculations to determine the magnitude only (effectively at the positive peak of the input signal). For capacitors in series, equation 6.7 can be used to define the total capacitance for the circuit in terms of the overall charge and voltage in the circuit ( $C_{t o t} = \frac{Q_{t o t}}{V_{t o t}}$ $C_{t o t} = \frac{Q_{t o t}}{V_{t o t}}$ ). Rearrange the terms of this equation to define each capacitor in terms of its voltage to perform the same sum of the voltage drops calculation for a capacitor series circuit:

Cancel out the charge (Q) terms, to derive the equation for total series circuit capacitance:

T o t a l S e r i e s C a p a c i t a n c e, \frac{1}{C_{t o t}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \dots \frac{1}{C_{n}}

$T o t a l S e r i e s C a p a c i t a n c e, \frac{1}{C_{t o t}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \dots \frac{1}{C_{n}}$

(6.6)

	$C_{t o t}$ $C_{t o t}$ is the total capacitance in ohms (Ω)
	$n$ $n$ is the number of capacitors in series

Deriving parallel capacitance

For parallel capacitors, use Kirchoff’s Current Law from chapter 3 to calculate the total parallel capacitance:

Kirchoff’s Current Law (KCL) states that the total charge (or current) entering any junction (or node) in a circuit will be the same as the current leaving that junction. Recall from chapter 3 that there will always be the same level of charge present in a circuit – free electrons cannot enter or leave other than through a voltage source. This principle can be used to derive the equation for total capacitance when capacitors are connected together in parallel (Appendix 2 Figure 4).

Appendix 2 Figure 4 Parallel capacitor circuit. The diagram shows two capacitors connected in parallel, where the total charge (Q_tot) is the sum of all branch charges (Q₁ and Q₂) in the circuit. Again, we show an AC supply but we do not include time in our magnitude calculations.

This parallel circuit has two capacitors that draw current from the AC supply (V_S), where the sum of all branch charges is equal to the total charge in the circuit (Q_tot). Although this is an AC supply, if time is again ignored to focus on the magnitude of the capacitance, the total charge in the circuit is equal to the sum of the charge across each capacitor:

Cancel out the voltage (V) terms to derive the equation for total parallel circuit capacitance:

T o t a l P a r a l l e l C a p a c i t a n c e, C_{t o t} = C_{1} + C_{2} + \dots C_{n}

$T o t a l P a r a l l e l C a p a c i t a n c e, C_{t o t} = C_{1} + C_{2} + \dots C_{n}$

(6.7)

	$C_{t o t}$ $C_{t o t}$ is the total capacitance in ohms (Ω)
	$n$ $n$ is the number of capacitors in series

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Table of Contents for Appendix 2 AC equation derivations (chapter 6)

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Table of Contents for
Appendix 2 AC equation derivations (chapter 6)