3.2 LOW-PASS CIRCUITS
Low-pass circuits derive their name from the fact that the output of these circuits is larger for lower frequencies and vice-versa. Figures 3.1(a) and (b) represent a low-pass RC circuit and a low-pass RL circuit, respectively.
In the RC circuit, shown in Fig. 3.1(a), at low frequencies, the reactance of C is large and decreases with increasing frequency. Hence, the output is smaller for higher frequencies and vice-versa. Similarly, in the RL circuit shown in Fig. 3.1(b), the inductive reactance is small for low frequencies and hence, the output is large at low frequencies. As the frequency increases, the inductive reactance increases; hence, the output decreases. Therefore, these circuits are called low-pass circuits. Let us consider the response of these low-pass circuits to different types of inputs.
FIGURE 3.1(a) A low-pass RC circuit; and (b) a low-pass RL circuit
For the circuit given in Fig. 3.1(a), if a sinusoidal signal is applied as the input, the output vo is given by the relation:
where, ω2 = 1/CR = 1/τ. From Eq. (3.1), the phase shift θ the signal undergoes is given as:
θ = tan−1(ω/ω2) = tan−1(τ/T)
Figure 3.2(a) shows a typical frequency vs. gain characteristic. Hence, f2 is the upper half-power frequency. At ω = ω2,
Figure 3.2(b) shows the variation of gain with frequency for different values of τ. As is evident from the figure, the half-power frequency, f2, increases with the decreasing values of τ, the time constant. The sinusoidal signal undergoes a change only in the amplitude but its shape remains preserved.
Figure 3.2(c) shows the variation of θ as a function of frequency. As (τ/T) becomes large, θ approaches 90°. This characteristic can be appreciated when we talk about an integrator later.
3.2.2 The Response of a Low-pass RC Circuit to Step Input
FIGURE 3.2(a) Typical frequency-vs-gain characteristic of a low-pass circuit to sinusoidal input
FIGURE 3.2(b) Gain-vs-frequency curves for different values of τ
FIGURE 3.2(c) Phase-vs-frequency curves for different values of τ
vo(t) = vf + (vi − vf) e−t/τ
Here, vf = V and vi = 0. Therefore,
As t → ∞, vo(t) → V.
Initially, as the capacitor behaves as a short circuit, the output voltage is zero. As the capacitor charges, the output reaches the steady-state value of V in a time interval that is dependent on the time constant, τ. On the other hand, the output of Eq. (3.2) can also be obtained by solving the following differential equation. From Fig. 3.1(a), For vi = V:
FIGURE 3.3 The response of a low-pass circuit to step input
We know that (1/C) ∫ idt = vo
Taking Laplace transforms:
Resolving into partial fractions:
Putting s = 0:
Putting s = −1/τ
Taking the Laplace inverse:
Now, for the circuit in Fig. 3.1(b):
From Eq. (3.8), it may be seen that the output reaches the steady-state value faster for smaller values of τ. Similarly, when τ is large, it takes a longer time for the output to reach the steady-state value.
Rise time: The time taken for the output to reach 90 per cent of its final value from 10 per cent of its final value is called the rise time. Using Eq. (3.8) to calculate the rise time for this circuit:
From Fig. 3.3 at t = t1, vo = 0.1 V. Therefore,
Similarly at t = t2, vo = 0.9 V:
Also f2 = 1/2πRC
Let a step voltage Vσ be applied to a low-pass circuit. The output does not reach the steady-state value Vσ instantaneously as desired. Rather, it takes a finite time delay for the output to reach Vσ, depending on the value of the time constant of the low-pass circuit employed. If this output is to drive a transistor from the OFF to the ON state, this change of state does not occur immediately, because the output of the low-pass circuit takes some time to reach Vσ. The transistor is thus said to be switched from the OFF state into the ON state only when the voltage at the output of the low-pass circuit is 90 per cent of Vσ. If this time delay is to be small, τ should be small. On the contrary, if the output is to be ramp, τ should be large. This is elucidated by Example 3.1.
Example 3.1: A 10-Volt step input is applied to a low-pass RC circuit. Plot the response of the circuit when: (a) RC = 1 ms, (b) RC = 10 ms and (c) RC = 100 ms.
Solution: For a step input, the output voltage is vo = V(1 − e−t/τ). Using this expression, the output voltage for different values of the time constant is shown in the Table 3.1. The waveforms are shown in Fig. 3.4.
TABLE 3.1 The output of a low-pass circuit for different values of τ
FIGURE 3.4 The response of a low-pass circuit for different values of τ for a step input
3.2.3 The Response of a Low-pass RC Circuit to Pulse Input
Let the input to the low-pass circuit be a positive pulse of duration tp and amplitude V as shown in Fig. 3.5(a). If this positive pulse is applied to drive an n–p–n transistor from the OFF state into the ON state, the transistor will be switched ON only after a time delay. Similarly, at the end of the pulse, the transistor will not be switched immediately into the OFF state, but will take a finite time delay. To know how quickly it is possible to switch a transistor from one state to the other, we have to consider the response of a low-pass circuit to the pulse input. During the period 0 to tp, the input is a step and the output is given by Eq. (3.8). At t = tp the input falls and the output decays exponentially as given in Eq. (3.13).
FIGURE 3.5 Response of a low-pass circuit for the pulse input for varying τ
For vi = V, the output for different values of τ is plotted in Fig. 3.5. It is seen here that the shape of the pulse at the output is preserved if the time constant of the circuit is much smaller than tp, i.e., τ tp. However, if a ramp is to be generated during the period of the pulse, τ is chosen such that τ tp. The method to compute the output is illustrated in Example 3.2.
Example 3.2: An ideal pulse of amplitude 10 V is fed to an RC low-pass integrator circuit. The width of the pulse is 3 μs. Draw the output waveforms for the following upper 3-dB frequencies: (a) 30 MHz, (b) 3 MHz and (c) 0.3 MHz.
Solution: Consider the low-pass circuit in Fig. 3.1(a).
- At f2 = 30 MHz
We know that f2 = 1/2πRC
tr = 2.2τ = 2.2 × 5.3 × 10−9 = 11.67 ns
At t = tp,
Vp = V(1 − e−tp/τ) = 10(1 − e−3×10−6/5.3×10−9) = 10 V
The output is plotted in Fig. 3.6(a).
FIGURE 3.6(a) Output waveform at f2 = 30 MHz
- At f2 = 3 MHz
tr = 2.2τ = 2.2 × 53 × 10−9 = 116.6 ns
At t = tp,
Vp = V(1 − e−tp/τ) = 10(1 − e−3×10−6/53×10−9) = 10 V
The output is plotted in Fig. 3.6(b).
FIGURE 3.6(b) Output waveform at f2 = 3 MHz
- At f2 = 0.3 MHz
tr = 2.2τ = 2.2 × 530 × 10−9 = 1.166 μs
At t = tp,
Vp = V(1 − e−tp/τ)
Vp = 10(1 − e−3×10−6/530×10−9) = 9.96 V
The output is plotted in Fig. 3.6(c).
FIGURE 3.6(c) The output waveform at f2= 0.3 MHz
3.2.4 The Response of a Low-pass RC Circuit to a Square-wave Input
Let the input to the low-pass circuit be a square wave as shown in Fig. 3.7 (a).
We have from Eq. (2.9):
vo1(t) = vf + (vi − vf)e−t/τ
From Fig. 3.7(c), at t = T1, vo1 = V2 and vi = V1 and vf = V′. Therefore:
FIGURE 3.7 The response of the low-pass circuit to a square-wave input for different values of τ
Again, at t = T2, Vo2 = V1 and we have vi = V2, Vf = V″
If the input is a symmetric square wave:
If τ T, then the wave shape is maintained. And if τ T, the wave shape is highly distorted, but the output of the low-pass circuit is now a triangular wave. So it is possible to derive a triangular wave from a square wave by choosing τ to be very large when compared to T/2 of the symmetric square wave. Let us consider an example.
Example 3.3: A symmetric square wave, whose peak-to-peak amplitude is 4 V and whose average value is zero is applied to a low-pass RC circuit shown in Fig. 3.1(a). The time constant equals the half-period of the square wave. Find the peak-to-peak output voltage of waveform.
FIGURE 3.8 The input and output waveforms at RC = T/2
Solution: As the input is a symmetric square wave, we have:
T1 = T2 = , V1 = −V2 and V′ = V″ = vo1 = V′ + (V1 − V′)e−t/τ
At t = T1:
where τ = T/2.
The peak-to-peak output voltage = 2 × 0.924 = 1.848 V. V2 can also directly be calculated using Eq. 3.19.
The input and output are plotted as shown in Fig. 3.8.
Example 3.4: The periodic waveform applied to an RC low-pass circuit in Fig. 3.1(a), is a square wave with T1 = 0.1 s, T2 = 0.2 s and time constant = 0.1 s, [see Fig. 3.9(a)]. Calculate the output voltages and draw the output waveform.
FIGURE 3.9(a) The given input waveform
Solution: The capacitor charges and discharges to the same level for each cycle.
RC = τ = 0.1 s, T1 = 0.1 s, T2 = 0.2 s
At 0 < t < 0.1 s, vf = V′ = 10 V and vi = V2
vo1 = 10 − (10 − V2)e−t/τ
At t = 0.1 s:
For 0.1 < t < 0.3 s, vf = V″ = 0 V and vi = V1
vo2 = 0 − (0 − V1)e−(t−0.1)/τ
At t = 0.3 s:
V1 = 6.32 + 0.368 × 0.135V1 V1(1 − 0.05) = 6.32 0.95V1 = 6.32 V1 = 6.65 V
From Eq. (2),
V2 = 0.135 × 6.65 = 0.898 V
The input and output waveforms are plotted in Fig. 3.9(b).
FIGURE 3.9(b) The input and output waveforms of the low-pass circuit
where τ1 is the time constant of the circuit that generated this exponential signal. For the low-pass circuit, we have:
Using Eq. (3.20):
Case 1: When τ ≠ τ1
Applying Laplace transforms:
Resolving X into partial fractions:
Putting s = 0, A = V
Putting s = −1/τ, B = −V
FIGURE 3.10 The exponential input applied
Putting s = −1/τ
Taking the Laplace inverse, the solution of the differential Eq. (3.26) is:
FIGURE 3.11 The input and output of a low-pass RC circuit
Similarly, repeating for the case τ = τ1.
Case 2: When τ = τ1
- Here, if τ ≤ τ1, the output is similar to the input.
- If τ > τ1, then the output deviates from the input.
The choice of τ depends on the requirement. If τ is small, the distortion is negligible. However, if τ is decided by the circuit components at the input of an amplifier—over which there is no control—we can only predict as to how much the output deviates. Figure 3.11 shows the deviation of the output for different values of n (= τ/τ1).
Now let tr1 be the rise time of the exponential input tr1 = 2.2τ1 [using Eq. (3.11)], and let tr2 be the rise time of the low-pass circuit tr2 = 2.2τ. Thus, the rise time of the output tr is given approximately as:
where, η = tr2/tr1. If tr1 and tr2 are the rise times of the circuit that has generated the exponential input and of the low-pass circuit respectively, then Eq. (3.31) gives the rise time of the output of the low-pass circuit. Let us consider Example 3.5.
Example 3.5: An exponential input of Volts is applied to a low-pass RC circuit in Fig. 3.1(a). Plot the response for (i) RC = 10 ms and (ii) RC = 20 ms.
Case 1: τ = 10 ms and τ1 as specified is τ1 = τ = 10 ms
When the time constant of both input and the low-pass RC circuit are equal (n = 1), the output voltage is:
The output is calculated for different values of τ and is presented in Table 3.2.
Case 2: τ = 20 ms
When the time constant of input, τ1 and that of a low-pass circuit, τ are not equal (n = 2) the output voltage is:
|t (ms)||vi = 10(1 − e−t/τ1)(V)|
The output for n = 2 is presented in Table 3.3
TABLE 3.3 The output of a low-pass circuit when n = 2 (i.e. τ1 = 10 ms and τ = 20 ms)
|t (ms)||vi = 10(1 − e−t/τ1)(V)||vo(t) = 10[1 + e−t/τ1 − 2e−t/τ](V)|
FIGURE 3.12 The response of the low-pass circuit for n = 1 and for n = 2
3.2.6 The Response of a Low-pass RC Circuit to Ramp Input
The ramp input signal is described by the relation vi = αt. We have from Eq. (3.21) that vi = τ (dvo/dt) + vo.
Applying Laplace transforms:
Resolving into partial fractions and solving:
Put s = 0, B = α
Put s = −1/τ, C = ατ
To get the value of A, from Eq. (3.34):
Equating the coefficients of s2 on both sides:
A + C = 0
Therefore, A = −C. As C = ατ, A = −ατ.
Taking the Laplace inverse:
At t = T:
Case 1: If τ T, then the deviation of the output from the input is very small since e−T/τ ≈ 0.
Case 2: If τ T, then e−T/τ can be expanded as series. The response is plotted in Fig. 3.13.
The response is plotted in Fig.3.13. When a ramp is applied as input to a low-pass circuit, the output deviates from the input. The transmission error, et, is calculated as:
Transmission error defines deviation from linearity. Thus, the smaller the value of et, the more linear the output. Let us consider an Example 3.6.
FIGURE 3.13 The response of a low-pass circuit to a ramp input
Example 3.6: A limited ramp, shown Fig. 3.14(a) (the pulse rises linearly and reaches V at T and remains at V beyond T), is applied to the low-pass RC circuit in Fig. 3.1(a). Plot the output waveforms when (a) T = τ, (b) T = 0.3τ and (c) T = 6τ.
FIGURE 3.14(a) The given input waveform
Solution: For a low-pass circuit, vo(t) = α(t − τ) + α τ e−t/τ, where α = V/T.
At t = T, vo(T) = α(T − τ) + α τ e−T/τ
- T = τ
vo(T) = α(τ − τ) + α τ e−τ/τ = α τ e−1
Beyond T, the output varies exponentially and reaches V
The output is as shown in Fig. 3.14(b).
FIGURE 3.14(b) The output waveform for T = τ
- T = 0.3τ
The output is as shown in Fig. 3.14(c).
FIGURE 3.14(c) The output waveform for T = 0.3 τ
- T = 6τ
The output is as in Fig. 3.14(d).
FIGURE 3.14(d) The output waveform for T = 6 τ
For the low-pass RC circuit in Fig. 3.1(a) to behave as an integrator, τ T. For T to be small when compared to τ, the frequency has to be high. At high frequencies, XC is very small when compared to R. Therefore, the voltage drop across R is very large when compared to the drop across C. Hence, vi ≅ iR.
The output is proportional to the integral of the input signal. Hence, a low-pass circuit with a large time constant produces an output that is proportional to the integral of the input. If the input to the circuit is vi(t) = Vm sin ωt, From Eq. (3.42):
where ω2 = 1/RC = 1/τ. For the low-pass circuit, we know:
The phase shift θ is given by:
θ = tan−1(ω/ω2)
When θ = 90°, the sine function at the input becomes a cosine function at the output, as is required in an integrator. When ω/ω2 = 100, θ = 89.4° which is nearly equal to 90°. Hence, only when RC T, a low-pass RC circuit behaves as a good integrator and the output is a cosinusiodally varying signal if the input is a sine wave. For a low-pass circuit, to behave as a reasonably good integrator, ω/ω2 ≥ 20.
3.2.8 An Op-amp as an Integrator
where A is the gain of the amplifier. If the capacitance C appears between the input and output terminals of the op-amp as shown in Fig. 3.15(a), then using Miller’s theorem, C can be replaced by Z1 and Z2.
FIGURE 3.15(a) An op-amp as an integrator
FIGURE 3.15(b) The op-amp integrator circuit using Miller’s theorem
where C′ = C(1 − A)
where C″ = (1 − 1/A)C ≈ C since A is large.
Hence, the integrator circuit is redrawn as shown in Fig. 3.15(b).
At the input of the amplifier, there appears a very large value of C′, which is called the Miller effect capacitor, resulting in a large value of τ, even while using a normal value of C.
Whereas in the case of a simple RC integrator, the physical value of C has to be very large to make τ large. An op-amp integrator is, therefore, a better integrator when compared to a simple RC integrator.
3.2.9 Low-pass RL Circuits
Consider the low-pass RL circuit in Fig. 3.1(b). For this low-pass circuit, vo = R/L ∫ vi dt. This circuit can also be used as an integrator when the time constant L/R T. The major limitations of high-pass and low-pass RL circuits are:
- For a large value of inductance, an iron-cored inductor has to be used. As such it is bulky and occupies more space.
- Inductors are more lossy elements, when compared to capacitors. So, it is possible to get ideal capacitors, but not ideal inductors.