Simulated Inductance

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Transcript Simulated Inductance

Simulated Inductance
Experiment 25
Modification from Lab Manual
• Write a MatLAB program
– Determine the transfer function
– Calculate the output voltage in phasor notation at the
corner frequency.
– obtain a Bode plot
• Plot of AC Sweep in PSpice simulation
– y-axis should be in dB
– Frequency range from 1 kHz to 10 MHz
• Construct a gyrator that has an effective inductance
of 10mH.
• Compare the operation of the gyrator with the 10mH
inductor in an RL circuit where R = 2 kW.
Gyrator
• Inductors are far from ideal devices
– Parasitic resistance because of the length of wire
used to form the inductor.
– Parasitic capacitance because of coupling between
parallel loops of wire.
Integrated Circuits
• Inductors are three dimensional devices,
usually free standing
– Very difficult to fabricate using standard
semiconducting processes and to integrate in a
silicon wafer.
Gyrator
• Operational Amplifier circuit that has a
frequency response similar to an inductor.
Low Pass RL Filter
At 0Hz, Vout = Vin. As the frequency increases and the
inductor begins to act like an open, Vout approaches 0V.
Bode Plot of Resistor Voltage vs. Vin
High Pass RL Circuit
At 0Hz, Vout = 0V. As the frequency increases and the
inductor begins to act like an open, Vout approaches Vin.
High Pass RL Filter
Vout
ZL

Vin
R  ZL
Vout
j 2 f L
H( f ) 

Vin R  j 2 f L
Let f B 
R
2 L
j( f B f )
H( f ) 
1  j( f B f )
where H ( f ) 
f fB
1  ( f f B )2
 f 
H ( f )  90  arctan  
 fB 
o
Voltage Transfer Function Plot:
High Pass RL Filter
Bode Plot:
High Pass RL Filter
Phase Plot:
High Pass RL Filter
Low Frequency Response:
As Expected
Higher Frequency Response:
Limited by Op Amp
Information from 741 Datasheet
Operational Amplifiers are designed to have large open loop gain.
But, the frequency response of the circuit has been sacrificed to achieve this.
Temperature and Frequency
Dependence of Open Loop Gain
Applications for Gyrator
• Low frequency (f < 500 kHz) applications in:
– Audio Engineering
– Biomedical Engineering
– Certain areas of power electronics
MatLAB
Defining a Vector
Three elements
t=[0,.1,3] or t=[0 .1 3]
t = 0 0.1000 3.0000
Four Elements
t=[0:3]
t= 0 1 2
3
Elements that include 0, then 0.1, and then increments of 1 until it reaches the
largest number ≤ 3
t=[0,.1:3]
t = 0 0.1000 1.1000 2.1000
Thirty one elements between 0 and 3 in increments of 0.1
t=[0:.1:3]
t=
0
0.9000
1.8000
2.7000
0.1000
1.0000
1.9000
2.8000
0.2000
1.1000
2.0000
2.9000
0.3000 0.4000 0.5000 0.6000 0.7000 0.8000
1.2000 1.3000 1.4000 1.5000 1.6000 1.7000
2.1000 2.2000 2.3000 2.4000 2.5000 2.6000
3.0000
MatLAB
Suppose you have determined that the
transfer function for your filter is:
• You can put this transfer function into MatLAB
– Define two vectors
• A = [An, An-1, ….., A1, A0]
• B = [Bm, Bm-1, ….., B1, B0]
– H = tf(A,B)
where n, m ≥ 0 and
n does not have to equal m
Example
High pass filter with a single-pole/single-zero
• Let RC = 104 s/rad
– A = [0, 10e3]
– B = [1 10e3]
– H = tf(A,B)
10000s
H = --------10000s + 1
is returned when you run the program
MatLAB: Bode Plots
• Once you have entered the transfer function
into MatLAB, you can use a predefined
function ‘bode’ to automatically generate
plots of the magnitude and phase vs.
frequency.
Enter: bode(H)
Bode Plot Parameters
Complex Numbers
• The default symbol for √-1 is i. However,
MatLAB does recognize that j is equivalent to i.
– The coefficient of the imaginary number must be
placed before the ‘i’ or ‘j’.
• If you typed c = 2-3j, MatLAB interprets it as
c = 2.0000 - 3.0000i
– To find components of complex number
• real(c) returns 2
• imag(c) returns -3
Magnitude and Phase
• Magnitude of a complex number
– Is the square root of the square of the real
component plus the square of the imaginary
component
• Phase of a complex number
– Is the arc tangent of the imaginary component
divided by the real component
• Must change to degrees if the output of the arc tangent
is given in radians