Download Analysis of Low Pass Filter and Voltage Dividers

Pre-­‐Experimental Analysis of a Low Pass Filter and Voltage Dividers for a Radon Detector Gogee Ghulam Introduction The purpose of a low pass filter is attenuate frequencies past some cutoff frequency. The ideal low pass filter allows for now passage of signals past the cutoff frequency, but this, of course, is unrealistic. The magnitude of the gain from the input to the output gradually decreases after the cutoff frequency. The goal of filter design is to determine the value of the cutoff frequency, the gain of the filter before the cutoff frequency, and how sharply the gain diminishes after the cutoff frequency. The purpose of this filter is to take out any possibility of high frequency fluctuations from a high voltage source (a source of 4 kV). The single source is to provide the proper DC voltage levels at the ends of the radon detector diode. The voltage levels can be controlled through the use of the voltage divider rule; selecting proper resistors can determine where the greatest voltage drops are in the circuit. The filter aspect of the circuit is an analog passive filter, which is a quite simple design. The passive aspect of the filter implies that there is no amplification of the input signal (i.e. an amplifier isn’t implemented). A basic analog low pass filter consists of resistors and capacitors. The central idea is change the circuit elements into the frequency domain. A capacitor’s relationship to an input frequency in can be described as follows: 𝑍=
1
𝐶𝑠
Where 𝑠 = 𝑗𝜔. Z is the impedance of the capacitor, which roughly translates to the resistance in what is called the phasor domain. One can note that as 𝑠 → ∞, 𝑍 → 0. One can also note as 𝑠 → 0, 𝑍 → ∞. Thus, a capacitor behaves as a short circuit for a high frequency signal and an open for a low frequency/DC signal. This is crucial for filter design; by placing a capacitor to ground high frequency signals will be attenuated because of the short, whereas low frequency signals will see an open circuit and not be grounded. Using the circuit elements in the frequency domain and applying techniques of circuit analysis, one can obtain a transfer function which shows the input output relationship of a circuit. The following circuit was to be analyzed: Figure 1: Each of these circuits have the same circuit layout, but the resistors vary to obtain different voltage differences across the Diode. In the following simulations the Diode was treated as an open circuit. The following transfer function was obtained (note Req =R2+R3): 𝐻 𝑠 =
𝑅! 𝑅!"
(𝑅! 𝑅!" − 𝑅! 𝑅! )(𝑅! + 𝑅!" + 𝐶𝑠𝑅! 𝑅!" )
A transfer function can be plotted on a logarithmic scale; with plots for both magnitude and phase. This is what is called a Bode plot. As there were little differences in the plots between graphs, there is only one Bode plot given (for the first circuit). An equivalent of the circuit can be obtained as well to determine the current through the circuit at any one time Figure 2: 𝑍!" =
𝑅! + 𝑅! + 𝑅! + 𝑅! 𝐶𝑗𝜔 𝑅! + 𝑅! + 𝑅! + 𝑅!
𝐶𝑗𝜔 𝑅! + 𝑅! + 𝑅! + 1
At low frequencies (i.e. the desired DC-­‐voltage): 𝑍!" = 𝑅! + 𝑅! + 𝑅! + 𝑅! Of course, by Ohm’s law: 𝑉 = 𝐼𝑍 The values for the components were determined via the transfer function as well as by the desired voltage difference between the diode. The voltage differences vary from 40 volt to 60 volt differences. Voltage divider equations were applied to the circuit and MATLAB code was implemented to help with selection of resistors. The code gave raw values, and then these values had to be matched to actual part values from the Allied Electronic Catalog. Bode plots were produced via MATLAB as well to ensure that the proper cutoff frequency was achieved. Circuit simulations in PSPICE and MultiSim were done in order to ensure that the correct DC voltage differences were obtained. The AC Analysis function was used to show the transfer curve at a desired input magnitude. Refer to the Appendix for a list of Codes that were implemented. Circuit A This circuit is designed to produce a voltage difference of about 50 Volts across the diode. Figure 3 The cutoff frequency obtained: 𝜔 = 53.4783 𝑟𝑎𝑑/𝑠 𝑓 = 8.5113 𝐻𝑧 𝑍!" = 2.615×10! 𝜴 𝐼 = 1.5296×10!! 𝐴 The following DC operating points were obtained (in kilo-­‐volts): node Voltage Voltage Difference V(4) 3.05146 0.05052 V(3) 3.10198 The Following Bode Plot Was obtained: Circuit B This circuit contains a Resistor in series with the 33 Mega-­‐Ohm resistor. The voltage drop across the diode will be approximately 52 Volts. The value of Req changes in the transfer function. Figure 4 𝜔 = 53.1124
𝑟𝑎𝑑
𝑠
𝑓 = 8.4531 𝐻𝑧 𝑍!" = 2.616×10! 𝜴 𝐼 = 1.529×10!! 𝐴 The following DC operating points were obtained: node V(3) V(5) Voltage Voltage Difference 3.09961 0.05216 3.04745 Circuit C This circuit contains another 976 kilo-­‐Ohm resistor in series with the 33 Mega-­‐Ohm Resistor. The value of R2 was also reduced to 40 Mega-­‐Ohm: Figure 5 The cutoff frequency obtained: 𝜔 = 76.7094 𝑟𝑎𝑑/𝑠 𝑓 = 12.2087 𝐻𝑧 𝑍!" = 2.575×10! 𝜴 𝐼 = 1.5534×10!! 𝐴 The following DC operating points were obtained: node V(3) V(5) Voltage Voltage Difference 3.15097 0.05451 3.09646 Circuit D This circuit ups the value of R3 to 40 Mega-­‐Ohms. The value of R2 is kept at 82 Mega-­‐Ohms. Figure 6 𝜔 = 50.9836
𝑟𝑎𝑑
𝑠
𝑓 = 8.1143 𝐻𝑧 𝑍𝑒𝑞 ≈ 2.622 ×10! 𝐼 = 1.5256×10!! The following DC operating points were obtained : node V(3) V(5) Voltage Voltage Difference 3.10436 0.06108 3.04328 Circuit E This circuit reduces the value of R3 to 30 Mega-­‐Ohm in order to lower the voltage drop. Figure 7 𝜔 = 54.6429
𝑟𝑎𝑑
𝑠
𝑓 = 8.6967 𝐻𝑧 𝑍!" = 2.612×10! 𝜴 𝐼 = 1.5314×10!! 𝐴 The following DC operating points were obtained: node V(3) V(5) Voltage Voltage Difference 3.10096 0.04599 3.05497 Circuit F This circuit reduces the value of R3 to 27.4 Mega-­‐Ohm: Figure 8 𝜔 = 55.7038
𝑟𝑎𝑑
𝑠
𝑓 = 8.8655 𝐻𝑧 𝑍!" = 2.6084×10! 𝜴 𝐼 = 1.5329×10!! 𝐴 node V(3) V(5) Voltage Voltage Difference 3.10007 0.04205 3.05802 In total circuits A through F can be described by the following table: Circuit A Frequency (Hz) Equivalent Impedance (Giga-­‐Ohms) Current (micro-­‐Amperes) Voltage Difference (kilo-­‐Volts) Circuit D Frequency (Hz) Equivalent Impedance (Giga-­‐Ohms) Current (micro-­‐Amperes) Voltage Difference (kilo-­‐Volts) Circuit B 8.5113 2.615 1.5296 0.05052 Circuit C 8.4531 2.616 1.529 0.05216 Circuit E 8.1143 2.622 1.5256 0.06108 12.2087 2.575 1.5534 0.05451 Circuit F 8.6967 2.612 1.5314 0.04599 8.8655 2.6084 1.5329 0.04205 Adding a DC Decoupling Capacitor A decoupling Capacitor is to be added between the output of the filter circuit and the input to the preamp. The purpose of this capacitor is to cut out the high voltage DC and send through the signal. As previously stated the capacitor behaves as short for high frequencies and thus will allow the signal to pass. The additional capacitance will not cause a change in the DC operating points, but will create a new transfer function. The following circuit demonstrates this. The 2 Giga-­‐Ohm resistor is added as a “drain” for the decoupling capacitor. Figure 9 For completeness a new transfer function with the inclusion the decoupling capacitor and the drain resistor. Again, the diode was treated as an open circuit (refer to figure 10). Figure 10 In this case the voltage at Node 5 divides between the impedance from C2 and the R5. If one considers the two elements in series and then combines them with R4 in parallel. The equivalent impedance looking through terminal 5 is: 𝑍!" =
𝑅! (𝑅! 𝐶𝑠 + 1)
𝑅! 𝐶! + 1 + 𝑅! 𝐶!
If the equivalent impedance is substituted into the original transfer function, a new transfer function is obtained for node 5. 𝐻 𝑠 =
𝑍!" 𝑅!"
𝑉!
=
𝑉! (𝑍!" 𝑅!" − 𝑅! 𝑍!" )(𝑅! + 𝑅!" + 𝐶! 𝑠𝑅! 𝑅!" )
To obtain the output voltage, a voltage divider must be applied between the capacitor and the resistor. This gives the following relationship between V5 and Vo: 𝑉! =
𝑅! 𝐶!
𝑉 1 + 𝑅! 𝐶! 𝑠 !
Multiplying the two equations, one obtains the following transfer function: 𝐻 𝑠 =
𝑅! 𝐶! 𝑍!" 𝑅!"
𝑉!
=
𝑉! (𝑍!" 𝑅!" − 𝑅! 𝑍!" )(𝑅! + 𝑅!" + 𝐶! 𝑠𝑅! 𝑅!" )(1 + 𝑅! 𝐶! 𝑠)
Though the transfer function is more complex, the additional capacitor does not do much to affect the frequency response of the system. An input signal is filtered by the first capacitor (C1). So even though the decoupling capacitor behaves as a short in AC, there is no AC signal present from the initial voltage source to go through anyway. Essentially, the decoupling capacitor allows for just the signal from the diode to pass (i.e. cuts out the DC signal, hence the name decoupling capacitor). To demonstrate the importance of the decoupling capacitor, a switch attached to a voltage source was placed in position of the diode. The voltage source was given a voltage of 600 mV, which is generally the voltage associated with a diode. The following was created using iCircuit, an Apple circuit simulating software. Figure 11 When the voltage source is first closed, the circuit must reach the steady state and thus the circuit goes through a transient response. Both the filter capacitor and the decoupling capacitor are assumed to have no initial voltage. The voltage source turn on can be represented as a negative impulse step function. The voltage then decays from the capacitor exponentially. The following diagram represents the initial impulse and decay: When the switch is closed, the circuit has (assumedly) reached the steady state and thus the capacitor must undergo transience again. The circuit on the left represents the capacitor’s voltage right after switch has been closed; the circuit on the right represents the voltage after the switch is opened again. Discussions This pre-­‐experimental analysis shows that the voltage divides, low pass filter, and the decoupling capacitor ought to work effectively. Several different analytical tools were used in order to ensure that the circuit will be able to function properly. These tools include hand calculations, SPICE, MultiSim, MATLAB, and iCircuit. MATLAB was quite useful in checking hand calculations for arithmetic errors as well as simulating Bode plots. The analysis could not have been completed as effectively without the circuit simulation software. It will be interesting to see how the experimentally implemented results will match up with these simulations. The MultiSim analysis was conducted with a voltage source of 4 Volts, whereas the experiment will be conducted with a voltage source of 4 kV. The simulation could not be done with 4 kV, as some of the resistors would burn out as soon as the simulation began. This, of course, is because the design criteria assumed that the resistors were not high voltage. MultiSim is an offshoot of SPICE and has the same design parameters as SPICE, but for completeness, simulations were run in both. It was not too time consuming to do both, as the programs are generally easy to work with. The transient analysis was simulated using iCircuit. This circuit program is new software that was developed a few years ago. It must be noted that it is generally better to use for qualitative analysis. MultiSim and SPICE are much more precise tools. Thus, this analysis only used it for testing the response of the DC decoupling capacitor. iCircuit is easy to use, and contains a ‘switch’ element, which is generally how a diode is described in circuit analysis. There are two basic models for diode analysis; the constant drop model and one in which the diode can be a short or an open. In the constant drop model, the diode is assumed to have a drop of 600 mV when on and to be an open when off. In the other model, the diode is a short when on and an open when off. This simulation used the constant drop model, but it should be noted that it doesn’t make too big a difference in which model is used because the 4 kV source is so much bigger than the diode voltage. Thus, the impulse sent to the decoupling capacitor will not depend on the voltage drop of the diode, but rather the voltage across the resistor in parallel with the diode. The circuit simulation illustrates this clearly. It should be interesting to see how the circuit holds up to this pre-­‐experimental analysis. It might be worthwhile to use a pulse generator to test the capabilities of the decoupling capacitor. It also might be equally important to use an AC source to demonstrate the effectiveness of the filter. The cutoff frequency given in this report represents the moment at which the value of the output begins to decrease. The Bode Plot shows the magnitude of the output decreasing rather sharply. Appendix: Functions Used in Analysis MATLAB code used to test hand-­‐calculations and Bode Plot done by hand: R1 = input('value of R1: ');
R2 = input('value of R2: ');
R3 = input('value of R3: ');
R4 = input('value of R4: ');
C = 200 * (10^-12);
Rx = R2 + R3;
% H(s) = (R4*Rx)/((R4+Req-R1*R4)*(R1+Rx+C*s*R1*Rx))
a = R4*Rx;
b = R4+Rx - (R1*R4);
d = R1+Rx;
e = C*R1*Rx;
% H(s) = (a/be)/((bd/be)+ s);
x = (b*d)/(b*e);
y = a/(b*e);
X = [1 x]
Y = [0 y]
bode(Y,X)
grid on
MATLAB Code to determine resistor values for specific voltage drops V = input('voltage drop across diode (in Volts): ');
R4 = input('value of R4: ')
R3 = ((45/3000)*(R4))/(1-(45/3000))
Rxx = R3 + R4;
Req = ((1/4)*(Rxx))/(1-(1/4))
Rx = Req;
R1 = Req/2;
C
a
b
d
e
%
x
y
= 750 * (10^-12);
= R4*Rx;
= R4+Rx - (R1*R4);
= R1+Rx;
= C*R1*Rx;
H(s) = (a/be)/((bd/be)+ s);
= (b*d)/(b*e);
= a/(b*e);
X = [1 x]
Y = [0 y]
bode(Y,X)
grid on
MATLAB Code to determine cutoff frequencies, current, and equivalent impedance R1 = input('value of R1: ')
R2 = input('value of R2: ')
R3 = input('value of R3: ')
C = 200 * (10^-12);
Req = R2+R3;
s = (R1+Req)/(R1*Req*C)
f = s/(2*pi)
R4 = 2*10^9;
Rx = R1+R2+R3+R4
Vin = 4*10^3;
I = Vin/Rx