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APPLICATIONS AND DESIGN OF INTEGRATED CIRCUITS

Module by: Dinh Sy Hien

Summary: As an example of the application of op-amps in area of active filters, we will discuss the Butterwort filter. The discussion is only an introduction to the subject of the filter theory design. We will also discuss various types of oscillators, Schmitt trigger circuits, and nonsinusoidal oscillators.

ACTIVE FILTERS

An important application of op-amp is the active filter. The word filter refers to the process of removing undesired portion of the frequency spectrum. The word active implies the use of one or more active devices, usually an operational amplifier, in the filter circuit. As an example of the application of op-amps in area of active filters, we will discuss the Butterworth filter. The discussion is only an introduction to the subject of the filter theory design.
Two advantages of active filters over passive filters are:
  1. The maximum gain or the maximum value of the transfer function may be greater than unity.
  2. The loading effect is minimum, which means that the output response or the filter is essentially independent of the load driven by the filter.

Active Network Design

From our discussion of frequency response, we know that RC-networks form filters. Figure 1a is a simple example of a coupling capacitor circuit. The voltage transfer function for this circuit is
T ( s ) = V 0 ( s ) V i ( s ) = R R + 1 sC = sRC 1 + sRC T ( s ) = V 0 ( s ) V i ( s ) = R R + 1 sC = sRC 1 + sRC size 12{T \( s \) = { {V rSub { size 8{0} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { {R} over {R+ { {1} over { ital "sC"} } } } = { { ital "sRC"} over {1+ ital "sRC"} } } {} (1)
Figure 1: a) Simple high-pass filter and b) Bode plot of transfer function amplitude
The Bode plot of the voltage gain magnitude /T(jω)//T(jω)/ size 12{ lline T \( jω \) rline } {} is shown in Figure 1a. The circuit is called a high-pass filter.
Figure 2(a) is another example of a simple RC network. Here, the voltage transfer function is
T ( s ) = V 0 ( s ) V i ( s ) = 1 sC 1 sC + R = 1 1 + sRC T ( s ) = V 0 ( s ) V i ( s ) = 1 sC 1 sC + R = 1 1 + sRC size 12{T \( s \) = { {V rSub { size 8{0} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { { { {1} over { ital "sC"} } } over { { {1} over { ital "sC"} } +R} } = { {1} over {1+ ital "sRC"} } } {} (2)
Figure 2: a) Simple low pass filter and b) Bode plot of transfer function
The Bode plot of the voltage gain magnitude /T(jω)//T(jω)/ size 12{ lline T \( jω \) rline } {} for this circuit is shown in Figure 2(b). This circuit is called a low-pass filter.
Figure 3: a) High-pass filter with voltage follower and b) low pass filter with voltage follower.
Although these circuits both perform a basic filtering function, they may suffer from loading effects, substantially reducing the magnitude gain from the unity value shown in Figure 1(b) and Figure 2(b). Also, the cutoff frequency fLfL size 12{f rSub { size 8{L} } } {} and fHfH size 12{f rSub { size 8{H} } } {} may change when a load is connected to the output. The loading effect can essentially be eliminated by using a voltage follower as shown in Figure 3. In addition, a non-inverting amplifier configuration can be incorporated to increase the gain, as well as eliminate the loading effects.
These two filter circuits are called one-pole filters; the slope of the voltage gain magnitude curve outside the passband is 6 dB/octave or 20 dB/decade. This characteristic is called the rolloff. The rolloff becomes sharper or steeper with higher-order filters and is usually one of the specifications given for active filters.
Two other categories of filters are bandpass and band-reject. The desired ideal frequency characteristics are shown in Figure 4
Figure 4: Ideal frequency characteristics: a) band pass filter and b) band reject filter.

General Two-Pole Active Filter

Consider Figure 5 with admittances Y1Y1 size 12{Y rSub { size 8{1} } } {} through Y4Y4 size 12{Y rSub { size 8{4} } } {} and an ideal voltage follower. We will derive the transfer function for the general network and will then apply specific admittances to obtain particular filter characteristics.
A KCL equation at node VaVa size 12{V rSub { size 8{a} } } {} yields
( V i − V a ) Y 1 = ( V a − V b ) Y 2 + ( V a − V 0 ) Y 3 ( V i − V a ) Y 1 = ( V a − V b ) Y 2 + ( V a − V 0 ) Y 3 size 12{ \( V rSub { size 8{i} } - V rSub { size 8{a} } \) Y rSub { size 8{1} } = \( V rSub { size 8{a} } - V rSub { size 8{b} } \) Y rSub { size 8{2} } + \( V rSub { size 8{a} } - V rSub { size 8{0} } \) Y rSub { size 8{3} } } {} (3)
A KCL equation at node VbVb size 12{V rSub { size 8{b} } } {} produces
( V a − V b ) Y 2 = V b Y 4 ( V a − V b ) Y 2 = V b Y 4 size 12{ \( V rSub { size 8{a} } - V rSub { size 8{b} } \) Y rSub { size 8{2} } =V rSub { size 8{b} } Y rSub { size 8{4} } } {} (4)
From the voltage follower characteristics, we have Vb=V0Vb=V0 size 12{V rSub { size 8{b} } =V rSub { size 8{0} } } {}. Therefore, Equation 4 becomes
V a = V b ( Y 2 + Y 4 Y 2 ) = V 0 ( Y 2 + Y 4 Y 2 ) V a = V b ( Y 2 + Y 4 Y 2 ) = V 0 ( Y 2 + Y 4 Y 2 ) size 12{V rSub { size 8{a} } =V rSub { size 8{b} } \( { {Y rSub { size 8{2} } +Y rSub { size 8{4} } } over {Y rSub { size 8{2} } } } \) =V rSub { size 8{0} } \( { {Y rSub { size 8{2} } +Y rSub { size 8{4} } } over {Y rSub { size 8{2} } } } \) } {} (5)
Substituting Equation 5 into Equation 3 and again noting that Vb=V0Vb=V0 size 12{V rSub { size 8{b} } =V rSub { size 8{0} } } {}, we have
V i Y 1 + V 0 ( Y 2 + Y 3 ) = V a ( Y 1 + Y 2 + Y 3 ) = V 0 ( Y 2 + Y 4 Y 2 ) ( Y 1 + Y 2 + Y 3 ) V i Y 1 + V 0 ( Y 2 + Y 3 ) = V a ( Y 1 + Y 2 + Y 3 ) = V 0 ( Y 2 + Y 4 Y 2 ) ( Y 1 + Y 2 + Y 3 ) size 12{V rSub { size 8{i} } Y rSub { size 8{1} } +V rSub { size 8{0} } \( Y rSub { size 8{2} } +Y rSub { size 8{3} } \) =V rSub { size 8{a} } \( Y rSub { size 8{1} } +Y rSub { size 8{2} } +Y rSub { size 8{3} } \) =V rSub { size 8{0} } \( { {Y rSub { size 8{2} } +Y rSub { size 8{4} } } over {Y rSub { size 8{2} } } } \) \( Y rSub { size 8{1} } +Y rSub { size 8{2} } +Y rSub { size 8{3} } \) } {} (6)
Multiplying Equation 6 and rearranging terms, we get the following expression for the transfer function:
T ( s ) = V 0 ( s ) V i ( s ) = Y 1 Y 2 Y 1 Y 2 + Y 4 ( Y 1 + Y 2 + Y 3 ) T ( s ) = V 0 ( s ) V i ( s ) = Y 1 Y 2 Y 1 Y 2 + Y 4 ( Y 1 + Y 2 + Y 3 ) size 12{T \( s \) = { {V rSub { size 8{0} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { {Y rSub { size 8{1} } Y rSub { size 8{2} } } over {Y rSub { size 8{1} } Y rSub { size 8{2} } +Y rSub { size 8{4} } \( Y rSub { size 8{1} } +Y rSub { size 8{2} } +Y rSub { size 8{3} } \) } } } {} (7)
Figure 5: General two pole active filter.
To obtain a low-pass filter, both Y1Y1 size 12{Y rSub { size 8{1} } } {} and Y2Y2 size 12{Y rSub { size 8{2} } } {} must be conductances, allowing the signal to pass into the voltage follower at low frequencies. If element Y4Y4 size 12{Y rSub { size 8{4} } } {} is a capacitor, then the output rolloff at high frequencies.
To produce a two-pole function, element Y3Y3 size 12{Y rSub { size 8{3} } } {} must also be a capacitor. On the other hand, if elements Y1Y1 size 12{Y rSub { size 8{1} } } {} and Y2Y2 size 12{Y rSub { size 8{2} } } {} are capacitors, then the signal will be blocked at low frequencies but will be passed into the voltage follower at high frequencies resulting in a high-pass filter. Therefore, admittances Y3Y3 size 12{Y rSub { size 8{3} } } {} and Y4Y4 size 12{Y rSub { size 8{4} } } {} must both be conductances to produce a two-pole high-pass transfer function.

Two-Pole Low-Pass Butterworth Filter

To form a low-pass filter, we set Y1=G1=1/R1Y1=G1=1/R1 size 12{Y rSub { size 8{1} } =G rSub { size 8{1} } = {1} slash {R rSub { size 8{1} } } } {}, Y2=G2=1/R2Y2=G2=1/R2 size 12{Y rSub { size 8{2} } =G rSub { size 8{2} } = {1} slash {R rSub { size 8{2} } } } {}, Y3=sC3Y3=sC3 size 12{Y rSub { size 8{3} } = ital "sC" rSub { size 8{3} } } {} and Y4=sC4Y4=sC4 size 12{Y rSub { size 8{4} } = ital "sC" rSub { size 8{4} } } {}, as shown in Figure 6. The transfer function, from Equation 7, becomes
T ( s ) = V 0 ( s ) V i ( s ) = G 1 G 2 G 1 G 2 + sC 4 ( G 1 + G 2 + sC 3 ) T ( s ) = V 0 ( s ) V i ( s ) = G 1 G 2 G 1 G 2 + sC 4 ( G 1 + G 2 + sC 3 ) size 12{T \( s \) = { {V rSub { size 8{0} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { {G rSub { size 8{1} } G rSub { size 8{2} } } over {G rSub { size 8{1} } G rSub { size 8{2} } + ital "sC" rSub { size 8{4} } \( G rSub { size 8{1} } +G rSub { size 8{2} } + ital "sC" rSub { size 8{3} } \) } } } {} (8)
At zero frequency, s = j ωω size 12{ω} {} = 0 and the transfer function is
T ( s = 0 ) = G 1 G 2 G 1 G 2 = 1 T ( s = 0 ) = G 1 G 2 G 1 G 2 = 1 size 12{T \( s=0 \) = { {G rSub { size 8{1} } G rSub { size 8{2} } } over {G rSub { size 8{1} } G rSub { size 8{2} } } } =1} {} (9)
In the high frequency limit, s=jω→∞s=jω→∞ size 12{s=jω rightarrow infinity } {} and the transfer function approaches zero. This circuit therefore acts as a low-pass filter.
Figure 6: General two pass filter.
A butterworth filter is a maximally flat magnitude filter. The transfer function is designed such that the magnitude of the transfer function is as flat as possible within the passband of the filter. This objective is achieved by taking the derivatives of the transfer function with respect to frequency and setting as many as possible equal to zero at the center of the passband, which is at zero frequency for the low-pass filter.
Let G1=G2=G=1/RG1=G2=G=1/R size 12{G rSub { size 8{1} } =G rSub { size 8{2} } =G= {1} slash {R} } {}. the transfer function is then
T ( s ) = 1 R 2 1 R 2 + sC 4 ( 2 R + sC 3 ) = 1 1 + sRC 4 ( 2 + sRC 3 ) T ( s ) = 1 R 2 1 R 2 + sC 4 ( 2 R + sC 3 ) = 1 1 + sRC 4 ( 2 + sRC 3 ) size 12{T \( s \) = { { { {1} over {R rSup { size 8{2} } } } } over { { {1} over {R rSup { size 8{2} } } } + ital "sC" rSub { size 8{4} } \( { {2} over {R} } + ital "sC" rSub { size 8{3} } \) } } = { {1} over {1+ ital "sRC" rSub { size 8{4} } \( 2+ ital "sRC" rSub { size 8{3} } \) } } } {} (10)
We define time constant as τ3=RC3τ3=RC3 size 12{τ rSub { size 8{3} } = ital "RC" rSub { size 8{3} } } {} and τ4=RC4τ4=RC4 size 12{τ rSub { size 8{4} } = ital "RC" rSub { size 8{4} } } {}. If we then set s = j ωω size 12{ω} {}, we obtain
T ( jω ) = 1 1 + j ωτ 4 ( 2 + j ωτ 3 ) = 1 ( 1 − ω 2 τ 3 τ 4 ) + j ( 2 ωτ 4 ) T ( jω ) = 1 1 + j ωτ 4 ( 2 + j ωτ 3 ) = 1 ( 1 − ω 2 τ 3 τ 4 ) + j ( 2 ωτ 4 ) size 12{T \( jω \) = { {1} over {1+j ital "ωτ" rSub { size 8{4} } \( 2+j ital "ωτ" rSub { size 8{3} } \) } } = { {1} over { \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) +j \( 2 ital "ωτ" rSub { size 8{4} } \) } } } {} (11)
The magnitude of the transfer function is therefore
/ T ( jω ) / = [ ( 1 − ω 2 τ 3 τ 4 ) 2 ] − 1 / 2 / T ( jω ) / = [ ( 1 − ω 2 τ 3 τ 4 ) 2 ] − 1 / 2 size 12{ lline T \( jω \) rline = \[ \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) rSup { size 8{2} } \] rSup { size 8{ - 1/2} } } {} (12)
For the maximally flat filter (that is, a filter with a minimum rate of change), which defines a Butterworth filter, we set
d / T / dω / ω = 0 = 0 d / T / dω / ω = 0 = 0 size 12{ { {d lline T rline } over {dω} } \rline rSub { size 8{ω=0} } =0} {} (13)
Taking the derivative, we find
d / T / dω = 1 2 [ ( 1 − ω 2 τ 3 τ 4 ) 2 + ( 2 ωτ 4 ) 2 ] − 3 / 2 [ − 4 ωτ 3 τ 4 ( 1 − ω 2 τ 3 τ 4 ) + 8 ωτ 4 2 ] d / T / dω = 1 2 [ ( 1 − ω 2 τ 3 τ 4 ) 2 + ( 2 ωτ 4 ) 2 ] − 3 / 2 [ − 4 ωτ 3 τ 4 ( 1 − ω 2 τ 3 τ 4 ) + 8 ωτ 4 2 ] size 12{ { {d lline T rline } over {dω} } = { {1} over {2} } \[ \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) rSup { size 8{2} } + \( 2 ital "ωτ" rSub { size 8{4} } \) rSup { size 8{2} } \] rSup { size 8{ - 3/2} } \[ - 4 ital "ωτ" rSub { size 8{3} } τ rSub { size 8{4} } \( 1 - ω rSup { size 8{2} } τ rSub { size 8{3} } τ rSub { size 8{4} } \) +8 ital "ωτ" rSub { size 8{4} } rSup { size 8{2} } \] } {} (14)
Setting the derivative equal to zero at ω=0ω=0 size 12{ω=0} {} yields
d / T / dω / ω = 0 = [ − 4 ωτ 3 τ 4 ( 1 − ω 2 τ 3 τ 4 ) + 8 ωτ 4 2 ] = 4 ωτ 4 [ − τ 3 ( 1 − ω 2 τ 3 τ 4 ) + 2τ 4 ] d / T / dω / ω = 0 = [ − 4 ωτ 3 τ 4 ( 1 − ω 2 τ 3