In this page, two applications are presented to show the usefulness of time response. Since the circuits presented are practical circuits, they are called applications. We take up the circuit of a half-wave rectifier first.
This page describes the half-wave rectifier circuit containing a sinusoidal source, a diode and a load. A circuit with a diode is a non-linear circuit, but the purpose is here to show how we can apply the knowledge gained about linear circuits to other areas.
Most of the power electronic applications operate at a relative high voltage and in such cases, the voltage drop across the power diode tends to be small. It is quite often justifiable to use the ideal diode model. An ideal diode has zero conduction drop when it is forward-biased and has zero current when it is reverse-biased. The explanation and the analysis presented below is based on the ideal diode model.
CIRCUIT OPERATION
A circuit with a single diode and an RL load is shown above. The source vs is an alternating sinusoidal source. If vs = E × sin (ω t), vs is positive when 0 < ωt < π , and vs is negative when π < ωt < 2π . When vs starts becoming positive, the diode starts conducting and the source keeps the diode in conduction till the source voltage is greater than the voltage across the load resistor. Before ωt reaches π radians, the source voltage equals the voltage across the load resistor at some instant. At this instant, there is current through the load circuit and the energy stored in the inductor peaks. After this instant, the inductor acts as the source and supplies power. When ωt reaches π radians, there is still some energy stored in the inductor. Since the energy stored in a system cannot change instantaneously, the current through the inductor does not fall to zero immediately, even if the source voltage becomes negative after ωt = π.
Let δ be the instant at which the source voltage is equal to the voltage across the load resistor. Let β the instant when the load current becomes zero. During the period, defined by δ < ωt < β, the inductor voltage that is the sum of the voltage across the load resistor and the source voltage and the inductor acts as the source. During the period π < ωt < β, the source voltage is negative and hence the inductor feeds power back to the source. On the other hand, during the period δ < ωt < π, part of the power consumed by the resistor is supplied by the source, whereas the rest is supplied by the inductor.
By convention, the voltage across an inductor is positive when the current through it is increasing and it becomes negative when the current through it tends to fall. When the voltage across the inductor is negative, it is in such a direction as to forward-bias the diode. The polarity of voltage across the inductor is as shown in the sketches shown below.
Mathematical Analysis
An expression for the current through the diode can be obtained as shown below. It is assumed that the current flows for 0 < ωt < β , where β > π . When the diode conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. During the period defined by β < ωt < 2π, the diode blocks current and acts as an open switch. For this period, there is no equation defining the behaviour of the circuit. For 0 < ωt < β , equation (1) defined below applies.
Given a linear differential equation, the solution is found out in two parts. The homogeneous equation is defined by equation (2). It is preferable to express the equation in terms of the angle θ instead of t. Since θ = ω t, we get that dθ = ω∙dt. Then equation (2) then gets converted to equation (3). Equation (4) shown above is the solution to this homogeneous equation and is called the complementary integral.
The value of constant A in the complimentary solution in equation (4) is to be evaluated, as shown below.
The particular solution is the steady-state response and equation (5) expresses the particular solution. The steady-state response is the current that would flow in steady-state in a circuit that contains only the source, the resistor and the inductor shown in the circuit above, the only element missing being the diode. This response can be obtained using the differential equation or the Laplace transform or the ac sinusoidal circuit analysis. The total solution is the sum of both the complimentary and the particular solution and it is shown as equation (6). The value of A is obtained using the initial condition. Since the diode starts conducting at ω t = 0 and the current starts building up from zero, i(0) = 0. The value of A can be evaluated and the total response is expressed by equation (7).
After evaluating A, current can be evaluated at different values of (ωt), starting from (ωt)= π. As ωt increases, the current keeps decreasing. For some value of ωt , say β , the current is zero. If(ωt)> β , the current calculated is a negative value. Since the diode blocks current in the reverse direction, the diode stops conducting when (ωt) reaches β. Then an expression for the average output voltage can be obtained. Since the average voltage across the inductor has to be zero, the average voltage across the resistor and the average voltage at the cathode of the diode are the same. This average value can be obtained as shown in equation (8).
Interactive Simulation
The operation of the circuit can be simulated as shown below. In order to simulate, the solution for current is presented in the following form.
Again it is preferable to normalize. Here E is set to unity and E/R is also set to unity. Then
To solve the expression, all we need to know is then the ratio τ . The applet shown below simulates this circuit. You have to key-in the ratio τ and then click on the button next to it. Do not key-in a NaN.
It is possible to build a square-wave oscillator using a single op-amp. The transfer characteristic of an op-amp makes it possible to realize the circuit of a square-wave oscillator.
When an op-amp is used in the open-loop, its transfer characteristic is as shown in Fig. 73. When the voltage at the input terminal input is negative, the output is at positive maximum and when the input is positive, the output is at its lowest value. The maximum and the minimum levels depend on the power supplies and the op-amp. It has been assumed that the maximum output voltage is E and that the minimum output voltage is - E. For a practical op-amp, the magnitude of its maximum output voltage may be different from that of its minimum output voltage. In this case, we can have equal clamping voltage in both directions, by the use of a couple of zeners and a current-limiting resistor.
The oscillator circuit
The circuit of the square-wave oscillator is presented in Fig. 74. Its operation is as follows. Equation (13 ) presents the assumptions made. This means that the output stays at E so long as vi remains negative. Since the output vo is at + E volts and vC is at - E/2, the capacitor starts getting charged. For 0 < t < T/2 , the capacitor gets charged.
For the output to remain positive, vi, calculated as shown by equation (13), should remain negative. At t = 0, vi = - E, as shown by equation (14). At t = 0, the output voltage is at E, and this voltage acts as the step input to R and C connected in series. Hence the capacitor voltage varies, as described by equation (15). When vi becomes zero and tends to become positive, the output will swing from + E to - E. Let vi become zero at t = T/2. Then T/2 = (RC) × lne 3 . That is, T = 2 (RC)× lne 3 . The frequency of operation is 1/T.
The behaviour in the other half-cycle is similar. The output is at - E and the capacitor is at E/2 at the start of this half cycle. Hence vi remains positive till the capacitor voltage changes from + E/2 to - E/2. The time taken for this process will also be T/2 and at the end of this period, vi becomes zero and tends to become negative. The output swings from - E to + E. Now we are back to where we started and the cycle repeats again.
A Matlab program has been used to obtain the responses shown above. The program is presented below.
% Square-wave oscillator % RC time constant = 0.1 % step size is calculated as 1% of cycle period % cycle period = 6 log (RC) % Output varies from + 10 V to - 10V and vice versa % Output positive for the first half cycle and negative for the % rest of the cycle. % initial capacitor voltage is - 5 V % time is denoted as zeit.
clear; step =0.002*log(3.0); zeit(1)=0.0; n2=0; halfper=0.0; n1=0; vinop(1)=-10; outputV(1)=10; capV(1)=-5.0; n3=1;
while (n1 == 0)
n3=n3+1;
zeit(n3)=zeit(n3-1)+step;
if (n2==0)
capV(n3)=10.0 -15.0*exp(-10.0*zeit(n3));
noninv(n3)=5;
outputV(n3)=10;
vinop(n3)=capV(n3)-noninv(n3);
halfper=halfper+step;
if (vinop(n3)>=0.0) n2=1; end;
else
capV(n3)=-10.0 +15.0*exp(-10.0*(zeit(n3)-halfper));
noninv(n3)=-5;
outputV(n3)=-10;
vinop(n3)=capV(n3)-noninv(n3);
if (vinop(n3)<=0.0) n1=1; end;
end;
end;
for n4=n3+1:(2*n3); zeit(n4)=zeit(n4-1)+step; capV(n4)=capV(n4-n3); outputV(n4)=outputV(n4-n3); vinop(n4)=vinop(n4-n3); end;
subplot(3,1,1)
plot(zeit,outputV)
ylabel('Output Volt')
axis([0 0.45 -12 12])
grid on;
subplot(3,1,2)
plot(zeit,capV)
ylabel('Capacitor Volt.')
grid on;
subplot(3,1,3)
plot(zeit,vinop)
ylabel('vi: Input Volt')
xlabel('Time')
grid on;
The plots obtained are presented below.
It can be seen from the plots that the output is at + 10 V, when vi remains negative, and that the output is at - 10 V, when vi remains positive. The change over in output occurs when vi tends to cross zero value. The change in output is reflected in vi as a step change. The capacitor voltage varies in an exponential manner.
This page has presented two applications, to illustrate how analysis of first-order circuits can be applied. The next topic describes how the state-variable approach can be used to solve differential equations.