Part of Made in India,Design Innovation

Workshop,2015

A  Fresh  New  Experience  in  Project-­‐Making  &  Learning!

PHYSICS INVESTIGATORY PROJECT ON

JOULE THIEF

If  you  have  dead  or  discharged  baCeries  in  your  home,  don’t  throw  them  out.  Instead  use  them  to  make  a  Joule  Thief  circuit,  which  brings  back  the  dead  baCery  by  boosKng  its  energy  to  light  up  an  LED.  The  name  ‘joule  thief’  suggests  the  noKon  that  the  circuit  is  stealing  every  drop  of  energy  or  ‘joules’  from  the  source.

In  this  project,  you  will  need  a  breadboard,  a  resistor,  an  inductor,  a  1.2  V  cell  or  baCery,  a  capacitor,  two  transistors,    an  LED  and  a  few  connecKng  wires.  The  connecKons  are  made  such  that  the  inductor  (L),  the  capacitor  (C)  and  the  resistor  (R)  are  connected  in  series.  The  intersecKon  point  of  the  capacitor  and  the  resistor  is  further  connected  to  the  base  of  a  PNP  transistor  that  drives  another  NPN  transistor.  The  supply  is  given  aUer  the  compleKon  of  the  circuit.  On  giving  the  supply,  the  capacitor  starts  charging  and  the  LED  glows.  

We  know  that  one  cannot  glow  an  LED  using  a  1.2  V  cell  but  sKll  the  LED  glows.  The  concept  of  the  decreasing  current  with  Kme  through  an  inductor  will  help  you  understand  this  phenomenon.  This  decrease  in  current  leads  to  a  change  in  flux  causing  an  induced  voltage  (L  di/dt),  which  adds  up  to  1.2  V  (cell  voltage),  resulKng  in  a  sum  of  more  than  1.8  V  due  to  which  the  LED  glows.  The  net  voltage  across  the  LED  =  voltage  of  Cell  (1.2  V)  –  L  di/dt.  Since  the  current  is  decreasing  with  Kme,  so  di/dt  is  negaKve  and  hence,  -­‐  Ldi/dt  is  posiKve.  Note  that  during  the  charging  process,  the  transistors  remain  OFF.

When  the  capacitor  becomes  almost  charged,  the  LCR  circuit  becomes  open  and  the  PNP  transistor  becomes  ON  since  the  base  voltage  is  low.  This  transistor  reaches  its  saturaKon  state  due  to  which  its  emiCer  and  collector  get  shorted  and  its  collector  receives  Vcc.  Since  the  collector  of  the  PNP  transistor  is  connected  to  the  base  of  the  NPN  transistor,  therefore,  its  base  also  receives  Vcc.  This  makes  the  NPN  transistor  ON.  Like  wise,  the  transistor  is  in  the  saturaKon  region  and  its  emiCer  and  collector  also  get  shorted.  The  potenKal  of  the  collector  is  nearly  zero  volt  as  the  emiCer  of  the  NPN  transistor  is  grounded.  Since  the  posiKve  terminal  of  the  LED  is  connected  to  the  collector  of  this  transistor,  it  stops  glowing  due  to  insufficient  voltage.  Hence,  the  LED  is  OFF.

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