Malik: Formal Element ReportJunaid Malik

Formal Element,

Signals and Systems 1,

DT008/2 Group A

Objective:

Our objective for this laboratory was to investigate a Sallen and Key second order active low pass filter and to obtain the transient analysis using our modelled circuit and pspice simulation. The circuit has two RC networks, R1 and C1, R2 and C2. The frequency response of the second order low pass filter is similar to the first order type except that the roll-off will be twice that of the first order filters, at 40dB/decade as the operating frequency increases above the cut-off frequency.

Circuit Design:

U 1

u A 7 4 1

+3

-2

V+

7V

-4

O U T6

O S 11

O S 25

R 1

1 0 0 k

R 2

4 . 7 k

R 3

1 0 0 k

R 41 0 k

C 18 2 0 p

C 28 2 0 p

V 21 5 V d c

V 31 5 V d c

0

0

n e g

o u t

p o s

n e g

Junaid Malik, DT008/2, Group A, 6/11/12

V 4

F R E Q = 1 . 7 kV A M P L = 1V O F F = 0

A C = 1

p o s

VDB

Figure 1 Simulation settings to obtain transient analysis

Reference2

R1= 100k R2=100k C1= 820p C2 820p R4=4.7k R3=10k

Vin(s)→ →Vout(s)

Transfer Function

G(s)=218619869.126

s2+18658.5365854s+148720999.405

Cut-off frequency

fc = 1940.91394015 (Hz)

Quality factor Q = 0.653594771242

Figure 2: This was the layout which we followed while constructing the Sallen and Key Second-order active. Reference Paul Tobin.

Figure 3: This is a good layout example reference Paul Tobin.

Deriving the equation:

An active low-pass filter is required to meet the following specification:

The maximum passband loss Amax = 0.5dB

The minimum stopband loss Amin = 12dB

The passband edge frequency ωp = 100 rs−1( f p = ωp/2π = 15.9 Hz)

The stopband edge frequency ωs = 400 rs−1

We had to obtain a Butterworth transfer function so that the Sallen and Key active

filter circuit could meet the required specification.

The filter order is calculated as follows =>

n=log10 [

10(0.1 xAmin)−110( 0.1xAmax )−1 ]

2 log10(ws℘ )

n=log10 [

10(0.1 x 12)−110(0.1 x 0.5)−1 ]

2 log10(400100

)

n=log10[

14.84890.12207

]

2 log10(4)

n= 2.085291.204119

n=1.731

n≈2

Sincewe know that Rb=10kΩ∧Ra=4.7kΩ

k=1+RaRb

=1.47 k

Assuming that avalue for Rb=10k∧Ra=4.7k , thereforethe gain∈dB is

k∨dB=20 log (1.47)=3.346dB

The low pass filter transfer function

is obtainedby applying the potential divider technique .

Secondorder :$2+1.414$+1

H ($ )= 1$2+1.414$+1

H (s )= 1

( sωc )

2

+1.414 ( sωc )+1

= ωc2

s2+1.414 ωCs+ω c2

Tf= Vo ( s )Vin (s )

=( kC2R2 )

s2+( 3−kCR )s+( 1

C2 R2 )ωc= 1

CR=2 πfc= 1

CR=fc= 1

2 πCR

ωcQ

=3−kCR

=Q= 13−k

Results:

Table of ResultsFrequency Vin Vout Vout/Vin 20log(vout/vin)

1Hz 1 1 0 0 dB

10Hz 1 1.390V 1.390V 2.86 dB

100Hz 1 1.405V 1.405V 2.95 dB

1000Hz 1 1.431V 1.431V 3.11 dB

1.2kHz 1 1.361V 1.361V 2.677 dB

1.6kHz 1 1.269V 1.269V 2.062 dB

1.8kHz 1 1.170V 1.170V 1.363 dB

2.0kHz 1 1.064V 1.064V 0.531 dB

2.2kHz 1 962.3mV 962.3mV -0.33 dB

2.4kHz 1 864mV 864mV -1.269 dB

2.6kHz 1 776.6mV 776.6mV -2.19 dB

2.8kHz 1 695.4mV 695.4mV -3.15 dB

3.0kHz 1 626.4mV 626.4mV -4.06 dB

From this table of results we can see that as the frequency is increased, the out output voltage is reduced in effect.

Graph

First graph of frequency against voltage

Frequency

1.0Hz 3.0Hz 10Hz 30Hz 100Hz 300Hz 1.0KHz 3.0KHz 10KHz 30KHz 100KHzDB(V(OUT))

-80

-60

-40

-20

-0

20

Junaid Malik, DT008/2, Group A

passband gain = 1 + ra/rb = 1.47

(23.806K,-40.2dB decade)

(3.1331K,-6.01dB Decade)

GRAPH 1: The frequency response above is the same as for a first order filter. The difference this time is the steepness of the roll-off which is at -40dB/decade.

Second graph of volatge against frequency

Frequency

1.0Hz 3.0Hz 10Hz 30Hz 100Hz 300Hz 1.0KHz 3.0KHz 10KHz 30KHz 100KHzV(OUT)

0V

0.5V

1.0V

1.5V

Junaid Malik, DT008/2, Group A, 6/11/12

k = 1 +Ra/Rb=1.47k

(1v,1.47k)

GRAPH 2:

Conclusion:

In conclusion there are many filter types and ways to implement them but an active low-pass filter that’s greatly simplified if R1=R2 and the op amp stage is a unity gain follower (RB=short and RA=open). The Sallen and Key filter attenuates any input signal in the frequency range above the cut-off frequency to a point, but then the response turns around and starts to increase in gain with frequency. The response in the stop band should keep decreasing as the frequency increases. Instead, the response actually will begin rising again at some high frequency. A properly functioning op amp is needed for the filters operation; op amps lose voltage gain at some frequency because of their finite bandwidth.

References

Reference 1 : Paul Tobin

Reference 2: http://sim.okawa-denshi.jp/en/OPstool.php

Reference 3 : PSpice for Filters and Transmission Lines by Paul Tobin