- Linear and Non-linear devices
A linear electrical component is an electrical element that has a linear relationship between current and voltage in a circuit. Examples include resistors, capacitors and inductors.
Non-linear electrical elements do not have a linear relationship between the current and voltage. An example is a diode where the current is a non-linear function of the voltage.
The load line analysis is a graphical analysis of the characteristics of non-linear electronic circuits to show the constraints that is placed on the non-linear devices by the rest of the circuit.
Using a simple diode (a non-linear electronic device) to explain the load line as shown in the figure below, the circuit is analysed using Kirchoff’s Voltage Law (KVL) analysis.
VDC = IR + VD 1
Also from the characteristic equation of a diode we have that
Id= IseVdnVt- 1 2
If it is assumed that n, Vt and Is are known we substitute equation 2 into equation 1 we have
VD = VDC - ISReVDnVT- 1 3
A way to solve this equation is to use load line analysis by plotting the characteristic equation of the diode and the load line on the same graph as shown below.
The intersection of the two lines is the solution to equation 3 and it is the operating point of the non-linear diode.
For a non-linear device like the diode, the operating point or operating region of this device is the region between the cut-off point and saturation point of the diode. Also for a transistor, this is obtained when the Emitter-Base junction is forward biased and the Base-Collector junction is reverse biased.
The active operating region of the diode as shown in the figure above is the first quadrant showing the forward current with an exponential curve. In this region, the diode operates as an amplifier owing to the non-linear relationship between the current and voltage.
The current and voltage relationship in the diode in this region is
Id= IseVdVt- 1
Where Id is the diode current, Is is the saturation current, Vd is the diode voltage and Vt is the thermal voltage.
2. Capacitors and Inductors
A capacitor is made up of two conducting surfaces that are separated by an insulating material known as a dielectric. Capacitors store electric charges between these conducting materials in the form of an electric field between the conducting materials. When a voltage is applied across these conducting materials, an electric field is set up between then which allows a difference in free electrons between the materials to develop. This difference is equivalent to the energy storage in the capacitor and the potential charge between the conducting materials of the capacitor.
An inductor on the other hand stores electrical energy in the form of magnetic fields.
The time constant is the amount of time it will take the current (in inductors) or voltage (in capacitors) values to experience an approximate 63% change from their initial values to their final values. The time constant is normally expressed in seconds and represented with τ.
For an R-C circuit, the time constant is the time during which the voltage across the capacitor would have reached its maximum value if it maintained its initial rate of rise.
When the charging of a capacitor begins, the potential difference across it is zero therefore
The initial rate of rise of voltage across the capacitor is dvcdt = VCR = Vλ V/s
As defined above, if this rate is maintained, then the time to reach the voltage V would be
V + V/CR = CR. CR is thus the time constant of the circuit.
For a resistive – capacitive circuit, the time constant τ = RC.
For a resistive – inductive circuit, the time constant τ = L/R.
In order to derive how many time constants it would take for a capacitor to reach a voltage equal to 99% of the source voltage we apply the universal time constant formula as written below
% Change in value = 1- 1etτ × 100%
Where e is Euler’s number, t is the time in seconds and τ is the time constant also in seconds.
Plugging the figure 99% for the change in the voltage value of the capacitor, we have
99 = 1- 1etτ
98e-tτ = -1
Solving this equation for t we have t ≈ 5τ.
Thus it takes approximately 5 time constants (5τ) for the capacitor to charge to 99% of its initial value.
An RLC circuit consists of a combination of a Resistance, Inductance and Capacitance in a circuit connected either in series or parallel to one another. This circuit is able to resonate at a frequency known as the resonance frequency f0 due to the dual way energy is stored in the circuit - in the capacitance as electric field and in the inductance as magnetic field. When the source of excitation is removed, the circuit can still continue to oscillate naturally, which is referred to as the natural response of the circuit. Damping is offered by the resistance in the circuit. If the circuit will continue to resonate at its natural frequency even in the absence of a driving force, it is referred to as underdamped. If on the other hand, it will not resonate, it is referred to as overdamped. The damping factor represented as 'zeta' ζ is 1 when the circuit is critically damped.
The second order differential equation of the series RLC circuit is given generally as
d2i(t)dt2+ 2∝ di(t)dt+ ω02it=0
where α is the attenuation and ω0 is the angular resonance frequency and are both are angular frequency units.
The differential equation for the circuit has the characteristic equation
s2 + 2αs + ω02 = 0
And the roots of this equation are
s1= -∝ + ∝2- w02
s2= -∝ - ∝2- w02
The general solution to the differential equation is thus
it= A1es1t+ A2es2t
where A1 and A2 are determined by the current and voltage in the circuit at the beginning of the transient.
In the overdamped response, ζ > 1and the transient current decays without oscillating.
it= A1e-ω0(ζ+ ζ2- 1)t+ A2e-ω0(ζ- ζ2- 1)t
In underdamped response, ζ < 1the response is an oscillation at ωd that decays at the rate proportional to the attenuation α.
it= B1e-αtcos(ωdt)+ B1e-αtsin(ωdt)
Critically damped response
In critically damped response, ζ = 1the circuit decays fast without oscillation.
it= D1te-αt+ D2e-αt
A transfer function is a ratio of the output signal in a system to the input signal. It shows a mathematical relationship in frequency domain between the input and output of a linear time-invariant system. This is written mathematically as the ratio of the ratio of the signal output Y(s) in complex frequency to the signal input X(s).
Hs= Y(s)X(s) where s = σ + jω
In a linear circuit, for a sinusoidal signal, the input voltage can be written as
where Vi is the input phasor voltage with an amplitude and phase and the output voltage Vo is written as
The ratio of Vo to Vi is the voltage gain transfer function denoted as
T(jω) can be written in the form
Where A(ω) is the gain function and φ(ω) is the phase function.
In signal processing, a filter is an electronic device that is used to selectively remove unwanted frequency components from a signal. This removal of unwanted frequency components or noise is aimed at removing interference with the wanted frequency components and enhances the signal. Electronic filters are constructed using passive inductive and capacitive components either separately or combined alongside resistive components or active components using operational amplifiers.
A low-pass filter is an electronic circuit that selectively allows low frequency signals passage while blocking out high frequency signals. They are implemented by connecting an inductor in series with the load or a capacitor in parallel with it. At high frequencies, the resistance of the inductor increases and blocks out the signal while that of the capacitor reduces and shorts the signal out. Low-pass filters are rated at a cut-off frequency which is the frequency above which the output signal voltage falls below 70.7% of the input voltage.
Vout= 1jwCR+ 1jwC Vin = 1jwRC+1 Vin
Gain A = |VoutVin| = 11+ w2R2C2 and phase ϕ = tan-1(-ω/RC)
A high-pass filter selectively allows high frequency signals passage while blocking out low frequency signals. In a capacitive high pass filter, a capacitor is connected in series with the load or an inductor is connected in parallel with it. At frequencies above the cut-off frequency, the output voltage is greater than 70.7% of the input voltage.
Vout= RR+ 1jwC Vin = jwRCjwRC+1 Vin
Gain A = |VoutVin| = w2R2C21+ w2R2C2 and phase ϕ = tan-1(1/ωRC)
A band pass filter combines the properties of a high-pass and a low-pass filter for frequency selection. In this electronic circuit, the filter allows a range of frequencies between the cut-off frequencies of the high-pass and low-pass filters i.e. frequencies above the cut-off frequency of the high-pass filter and below the cut-off frequency of the low-pass filter. Band-pass filters are implemented by connecting a low-pass to a high-pass filter in series with each other or the other way round.
The transfer function for a band-pass filter can be written as
Hjω= K1+jQωω0- ω0ω
Where ωl is the lower cut-off frequency and ωu is the upper cut-off frequency.
ω0 is the center frequency ≈ √ωlωu
The magnitude is thus |H(jω)| = |K|1+jQ2ωω0- ω0ω2
The phase is -|K|K tan-1Qωω0- ω0ω
Bode plot for band pass filter is as shown below.
A band-stop filter also combines the properties of a high-pass and a low-pass filter but in order to prevent the passage of frequencies within a certain range. Band-stop filters block the passage of signals with frequencies above the cut-off frequency of the high-pass filter and below the cut-off frequency of the low-pass filter. It is also implemented by connecting a high-pass and a low-pass filter but in parallel with each other.
For a band stop filter, the frequency response is given by
Hjω= R2C2jω1+ R2C2jω (1+ R1C1jω+R2R1R1C1jω)
Hjω= K(ω02- ω2)ω02- ω2+ jω0Qω
|H(jω)| = |Kω|1+jQ2ωω0- ω0ω2
Bode plot for the band stop filter is shown below.
Operational Amplifiers (Op Amp)
The operational amplifier is an analog electronic device is a high-gain negative-feedback amplifier that can amplify signals and perform mathematical operations such as summation, addition, subtraction etc in analog computers. The general symbol is as shown in the figure below where the +V and –V represent the DC power supply to the Op Amp and the Input (inverting and non-inverting) and Output leads connect the input and output voltages to the Op Amp.
The Op Amp is operated in either one of two modes – the open loop or the closed loop mode. In the open loop mode, the Op Amp has no resistor or capacitor connected from its output to any of its inputs. In this mode, it is said to be an ideal Op Amp and has the following characteristics
- the input resistance is if infinite (∞)
- the output resistance is zero (0)
- the open loop gain is infinite (∞)
- the bandwidth is infinite (∞)
The open loop characteristics thus serve as a good reference in characterising a real Op Amp with reference to this ideal one. In the closed loop operation of the Op Amp, there is a feedback resistor or capacitor connected between the output terminal of the Op Amp and any of its input terminals.
Operational amplifiers are connected in different configurations in order to achieve peculiar results. Some of these configurations are presented as follows
In an inverting application of the Op Amp, the non-inverting input terminal is connected to ground while the input signal is connected to the inverting input terminal and a feedback resistor connected from the output to the inverting input terminal of the Op Amp. In this arrangement, the closed loop gain of the amplifier is the ratio of the feedback resistor to the input resistor and the input is multiplied by a negative constant factor.
Vout= -RfRin Vin
In this arrangement, the input voltage is applied to the non-inverting terminal of the Op Amp and is such that negative feedback is still employed but an output which is a product of the input and a positive constant is produced.
Vout=( 1+ R2R1 )Vin
The differentiator application of the Op Amp is to provide an output voltage that is proportional to the rate of change of the input voltage.
Vout= -CR dVcdt
In the difference application of the Op Amp, an output voltage proportional to the difference of two input signals to the input terminals of the Op Amp is produced.
Vout=1+ R2R1V2- RfR1 V1
The comparator is used to compare the voltage levels of two input signals. The arrangement is the simplest Op Amp application and only connects the two voltages to be compared to the inverting and non-inverting terminals of the Op Amp. If the two voltages are the same, then the output voltage will be zero.
Vout = V1 – V2
The integrator amplifier is used to provide an output voltage that is proportional to the integral of the input voltage.
Superposition is a very useful concept in circuit analysis and is describes as this; Suppose a branch in an electrical circuit is connected to more than one current and voltage sources then the total current flowing through that branch in the circuit is the summation of all individual currents supplied by each current or voltage source. This is further explained with the illustration below
A current I flows through the resistance R as a result of the two independent voltage sources V1 and V2 acting simultaneously on the circuit. By replacing V2 with a short circuit and retaining V1 in its position in order to measure the current through the resistance R, we call this current I1. Doing similarly for V1 and measuring the current through R, we call it I2. The addition of the currents I1 and I2 gives us the actual current flowing through the resistance R.
I = I1 + I2
Thevenin's theorem simply states that any combination of batteries and resistances in a circuit can be replaced with a single voltage source Vth which is the open circuit voltage at the terminals and a series resistor which is the quotient of Vth and the short circuit current.
The figure above shows a network consisting of four resistances. The Thevenin voltage is the voltage of the circuit as seen from the terminals AB when the terminals AB is open circuited. In this case the Thevenin voltage will be
Vth = (R3 + R1// R2) Vs
R1 // R2 = R1 X R2R1+ R2
Rth = R3 + R1// R2
This equivalent circuit is depicted in the figure below
Maximum Power Output
For a battery or voltage source with internal resistance RTH and a load resistance RL is connected across this voltage source, the maximum power transfer theorem determines the value of the load resistance RL for which maximum power will be transferred to it from the voltage source.
For the Thevenin circuit above, the power in the load resistance is
PL = IL2RL = V2RLRTH+ RL 1
Thus IL = VL/RL 3
But the voltage across the load according to voltage division rule is
VL = RLRTH+ RL VTH 4
Substituting equation 4 into equation 1
PL = RL(RTH+ RL)2VTH2 5
Assuming PLVTH2 is P
Then equation 5 becomes
P = RL(RTH+ RL)2 6
In order to obtain the maximum power transferred, we differentiate equation 6 with respect to RL and then equate to 0.
dPdRL= ddRL(RL(RTH+ RL)2) 7
Solving equation 7 gives
RTH = RL
This shows that the maximum power is transferred when the load resistor is the same as the Thevenin resistance.
For the Thevenin equivalent circuit, the value of the maximum power absorbed by the load resistor RL is obtained by
PLmax= Vth24 X RL
Floyd, T.L. and Buchla, D. (1998). Basic Operational Amplifiers and Linear Integrated Circuits (2nd ed.). Prentice Hall, ISBN 978-0130829870.
Kuphaldt, T.R. (2005). Lessons In Electric Circuits, Volume I - DC. Available at http://www.ibiblio.org/obp
Theraja, B.L. and Theraja, A.K. (2007). A Textbook of Electrical Technology. S. Chand and Company Ltd., New Delhi, ISBN 81-219-2441-3.