Voltage Divider

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PreExperimental Analysis of a Low Pass Filter and Voltage Dividers for a Radon Detector Gogee Ghulam

description

Dividing the input voltage

Transcript of Voltage Divider

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Pre-­‐Experimental  Analysis  of  a  Low  Pass  Filter  and  Voltage  Dividers  for  a  Radon  Detector  Gogee  Ghulam  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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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 =1𝐶𝑠

 

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:  

 

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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.    

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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:  

 

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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 = 53.1124𝑟𝑎𝑑𝑠

 

𝑓 = 8.4531  𝐻𝑧  

𝑍!" = 2.616×10!𝜴  

𝐼 = 1.529×10!!  𝐴  

The  following  DC  operating  points  were  obtained:  

node   Voltage   Voltage  Difference  

V(3)   3.09961   0.05216  

V(5)   3.04745        

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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   Voltage   Voltage  Difference  

V(3)   3.15097   0.05451  V(5)   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  

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𝜔 = 50.9836𝑟𝑎𝑑𝑠

 

𝑓 = 8.1143  𝐻𝑧  

𝑍𝑒𝑞 ≈ 2.622  ×10!  

𝐼 = 1.5256×10!!  

The  following  DC  operating  points  were  obtained  :  

node   Voltage   Voltage  Difference  V(3)   3.10436   0.06108  

V(5)   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 = 54.6429𝑟𝑎𝑑𝑠

 

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𝑓 = 8.6967  𝐻𝑧  

𝑍!" = 2.612×10!𝜴  

𝐼 = 1.5314×10!!  𝐴  

 

The  following  DC  operating  points  were  obtained:  

node   Voltage   Voltage  Difference  V(3)   3.10096   0.04599  

V(5)   3.05497        

Circuit  F  This  circuit  reduces  the  value  of  R3  to  27.4  Mega-­‐Ohm:  

Figure  8  

 

𝜔 = 55.7038𝑟𝑎𝑑𝑠

 

𝑓 = 8.8655  𝐻𝑧  

𝑍!" = 2.6084×10!𝜴  

𝐼 = 1.5329×10!!  𝐴  

node   Voltage   Voltage  Difference  

V(3)   3.10007   0.04205  

V(5)   3.05802        

In  total  circuits  A  through  F  can  be  described  by  the  following  table:  

 

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    Circuit  A   Circuit  B   Circuit  C  

Frequency  (Hz)   8.5113   8.4531   12.2087  Equivalent  Impedance  (Giga-­‐Ohms)   2.615   2.616   2.575  

Current  (micro-­‐Amperes)   1.5296   1.529   1.5534  

Voltage  Difference  (kilo-­‐Volts)   0.05052   0.05216   0.05451  

    Circuit  D   Circuit  E   Circuit  F  

Frequency  (Hz)   8.1143   8.6967   8.8655  Equivalent  Impedance  (Giga-­‐Ohms)   2.622   2.612   2.6084  

Current  (micro-­‐Amperes)   1.5256   1.5314   1.5329  

Voltage  Difference  (kilo-­‐Volts)   0.06108   0.04599   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  

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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 =𝑉!𝑉!=

𝑍!"𝑅!"(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!

1 + 𝑅!𝐶!𝑠𝑉!  

Multiplying  the  two  equations,  one  obtains  the  following  transfer  function:  

𝐻 𝑠 =𝑉!𝑉!=

𝑅!𝐶!𝑍!"𝑅!"(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  

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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:  

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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.    

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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.    

 

 

 

   

 

 

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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 = 750 * (10^-12); 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]

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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