{528} How To Repair SMPS || SMPS Repair Step By Step || Switch Mode Power Supply

This Video is Translated Online, it may have some Translation Errors. welcome to my channel. in this session  we will make a complete troubleshooting   for a power supply it is UC3842 based power supply   we will use the basic tips to find any fault in  this power supply or any switch mode power supply   i have a complete list of faults and how we can  troubleshoot it. sometime the input short circuit   the fuse will burn out, the switching of this  ic, startup circuit, vcc circuit, output circuit   feedback, fluctuation, low output  voltage and load handling problems let's start. i will request to watch this video  completely i guarantee after watching this video   you will be able to fix any power supply. in this diagram we see there is a fuse   then it have a first stage filter  class x capacitors class y capacitors   and some dual line filter then  these voltage are reaching to   bridge rectifier this portion it is called EMI electromagnetic interference circuit or RFI radio   frequency interference circuit protection circuit  over-Current protection, Over-frequency protection   over voltage protection if there are some spikes  so class x capacitor will remove that spikes   then we have dc rail. in dc rail there  is bridge rectifier and capacitors then input surge protection circuit  that is here NTC. so let's start   when we take a power supply first of all   okay i will use this clamp meter  or you can use any multimeter first of all set to ohms i have  this multimeter we can also use   this multimeter first of all we have to check  the input resistance it must be high 0.672 mega  Ohms 673 kilo ohms so that means  the input resistance is high if it is zero are near to 100 homes 200 ohms , that will  indicate that there is a input short circuit   that input short circuit  may be due to any failure of   parallel line in both lines what are that components class x capacitor the yellow one   if this capacitor will make a short circuit it  will allow our easiest path to electrical current   and it will cause to damage this fuse  if the fuse is damaged how we can verify set multimeter to continuity and connect across fuse both terminals it must give continuity   because the fuse is in series line from this point  to this point and this line is moving directly   here. these both lines are reaching here  across this capacitor this yellow capacitor   if it will make a short circuit what will happen we will take a tone on the multimeter we will  take a beep so that means there is a input short   circuit what is going on the current will flow  from this point to fuse and there from this path then we have dual line filter when we apply  voltage okay let's supply voltage as well always use series lamp for testing when  the current is applied the circuit is live   never touch any component we can see when i  will connect the lamp will glow for a while and then it will stable if the lamp eliminates  permanently that means there is something short   if this capacitor is short it will give a easiest  path so that means we have a short circuit here   in this portion or any component if we have  second stage filter that will also make problem   we can check ac voltage set  multimeter to ac voltage 231 VAC if you want to check fuse live check voltage  across this fuse it must be zero volt if the fuse   is open all ac voltage will drop here and we will find this supply voltage 231V across this   open fuse. if the fuse is open the fuse is blown at that time we will take full voltage here then these voltage reach at this  dual line filter that is here 233 volt 231 volt whatsoever and the same voltage must pass out from this inductor. in some cases this inductor basically this inductor is a piece  of two wires which are wound on this ferret core   the electrical current will enter from one  side and it will go to the second side the   same is other side so both wires will conduct the electrical parts to the second end in some cases this filter becomes open sometimes bad joints here sometime it is disconnected from the pcb   so if this inductor is fail. it will not allow to pass voltage to the second side   it is in two portions. one portion & the second portion one portion is here the second portion is here we must find continuity if we are checking voltage in that case   we will find ac voltage in the  output side 234 volt that's good then this voltage must reach at the bridge  rectifier here we are finding the same ac   voltage here in the bridge rectifier if you  cannot find voltage here that means there is some   open circuit what is open circuit fuse  sometimes there is ntc or ptc for protection   when any line will not give a electrical path  to the to this point or this line that means   we have something open circuit in  series there is a current protection   in parallel we will find voltage protection  might be any capacitor any component is causing   short circuit and it is causing to open this fuse  sometimes this ptc or ntc it is open circuit if it   is damaged it will not allow the voltage to this  point now sometime the bridge rectifier is damaged first of all check dc voltage at  this capacitor if it has some voltage   discharge it properly how we can check  bridge rectifier set multimeter to diode in diode mod we have diode from  diode symbol here if we are using this type of multimeter we will find a  diode symbol here now we are in diode mode to test this bridge rectifier from this point to  this point we will verify all this circuit now   the black lead must be in the  common the red lead in the   diode voltage hertz pin locate the negative  terminal of the capacitor at negative terminal   connect this pin here red pin at negative  terminal a red pin of the multimeter   at negative terminal and red pin must be  here and black pin must be in the common port now red pin here black pin  here it is giving 0.57 volt   at input terminal where we are applying ac  voltage this time the ac voltage are disconnected   this point and this point so that means  the bridge rectifier two diodes are good   point five eight point five five okay now  move this red lead to positive terminal   and swap these leads here it must open that means   two diodes are forward biased two diodes are  reverse biased now swap these leads first red   lead was at negative terminal now black lead at  negative terminal and connect red lead one by one   here it is open it is open move blackly  to the positive terminal of the capacitor   0.55 0.55 that means the four  diodes are good if any pair   out of these four will short circuit at that  time we will take continuity and we will take   continuity in two conditions here it is giving  one diode here it is giving the second diode   but if we find the same time continuity  here that means the bridge rectifier is bad and then there is another  condition for this indication   that is this mosfet i will discuss later so  if this circuit is good until now we will find   dc voltage here if the ac voltage are available  here then we will find dc voltage here   in this circuit if we see this bridge  rectifier this end and this end   one end is at 90 degree and the second end it  have a slope it have a cut so this cut mark it is   positive if we see the bridge rectifier itself it  have a positive sign here under exactly this cut   so this is positive terminal and this end is  negative terminal negative terminal we have a ntc it is 5d11 sometime it is 5d 9 10 d11   so ntc it is a temperature coefficient  resistance so it will compensate the   temperature effect and it will protect this  capacitor from search from search ground in this diagram the ntc is shown here  and physically we have ntc here in the   output of this bridge rectifier sometimes  this capacitor is short circuit internally   if it is short circuit internally what happens it  will damage these diodes or it will damage this   fuse if the capacitor is short circuit it will  damage these diodes or might be the fuse will open   we replace the fuse fuse will open at  that time we need to check this capacitor   it is the short circuit fault then i will  discuss another fault for this capacitor   if we have two diodes short circuit in this bridge  at that time when one diode have to block and   second to conduct and the same time the second  diode will also conduct it will damage the fuse   so while cold testing we can verify the health of  this bridge rectifier now this negative voltage   are here and the same negative voltage are  applied here at this point here we have a   current sensor resistor and we have a mosfet  here most of the time this mosfet is damaged   positive voltage are reaching here and if  this mosfet is short circuit what will happen   it will apply easiest path to  this negative line from positive negative in normal conditions this  mosfet must be open here in continuity it is open if it gives continuity that  means the mosfet is short circuit and   this will also cause to damage this  fuse now let's check the first problem if it is short circuit fuse open number  one if second stage filter this capacitor   number two bridge circuit number  three capacitor number four   mosfet number five so these five  components will cause to burn this fuse   and then we have two capacitors here that are  class y capacitors so these are six components these two capacitors here this capacitor from one line second capacitor from  second line if these both capacitors in some cases   if these both capacitors get short circuit at  that time this ac line will come here and it   will reach back here sometime one capacitor is  damaged if one capacitor is damaged we will take   the live component ac voltage in the cases that  is one reason for that fault but if these both   capacitors become short circuits because the bad  behavior for a capacitor when it is open circuit   that means it have no effect but when it will  make short circuit it will make a big disaster   so these two capacitors are these two  capacitors will make the same situation now we will check voltage until here friends  this video is good hit the like button if   you have not subscribed i can subscribe  it and press bell notification for all   so that you will get notified for my every  new video check ac voltage at input terminal   ac voltage at bridge rectifier input so that  means our this circuit is clear then dc voltage it is 319 volts ac voltage bridge rectifier inside 232 volt bridge  rectifier outer pins 319 approximately   we can say 100 volt difference how much are  the ac input voltage multiply it with 1.414 you will find the resultant voltage here if  we will find these voltage this dc voltage input ac voltage multiply it with 1.414 and  if you find that voltage here that means   the capacitor is good input circuit is good if  we cannot find the voltage approximately equal to   the resultant value that means there is something  wrong the first situation if this capacitor is bad   internally it is open at that  time it will not store the charge   and the multimeter will show the rms value of  the voltage it will not give a peak voltage so in that condition we must check  this capacitor this capacitor will open the second step here analyze this voltage 319.9 these are  stable voltage if these voltages are step   moving up down up down 315 300 to 325 like  that if it is making a jumping that means   there is something wrong in the power supply we  will discuss this point later here we discussed   input short circuit fusible burnout class x class  y capacitor bridge circuit capacitor and mosfet if this mosfet is bad what it will  do the basic duty of this mosfet   is to block this voltage when it is off it will  not allow any electrical current from ground to   the positive terminal ground is here mosfet and  winding if this mosfet is short circuit this   negative voltage will pass through this mosfet and  current will flow directly uncontrolled current it have to allow a control current the  pulses which are generated from this ic   but if it is allowing a full current uncontrolled  current that means it will not allow to build   voltage at this capacitor and it will draw  a maximum current because this transformer   it have a very low resistance in dc  it will work as a simple piece of wire   in that condition we will check this  mosfet if this mosfet is short circuit   it will not allow to build voltage here and it  will draw more current maximum current at in that   condition the fuse will burn out immediately  let's make short circuit for this mosfet   i connected a jumper between drain to source  what will happen the lamp will glow full so if this mosfet is short circuit most of the  time the fuse will burn out dc voltage at this   capacitor it is 317 volts stable then we will find  this voltage at this capacitor at this terminal   at this point it will reach to this transformer  primary winding and it will reach to this mosfet now we will measure voltage between mosfet drain  and source the source is connected to ground   through a through our resistor that is current  sensor resistor we will check voltage now at this   point and this point across this capacitor and  then we will check voltage across these two points   dc voltage i have 314 volt at capacitor why  we have more voltage here when the voltage are   applied at this point then we have winding  and we have a mosfet that is now switching   if this ic will receive voltage at  vcc pin that is pin number seven let's   discuss first of all pin number seven  we have smps controller ic here uc3842   and here we have a small capacitor  electrolyte chemistry 3842 uc four 38433844 uc three eight four  five these four ics are a family   the circuit configuration will remain  same just only the voltage level   and duty cycle will change from one to  another we have a 270 k resistor one watt   then we have a capacitor that is 47 microfarad  50 volts that is here and we have one resistor that is connected between positive line and this  capacitor that is here this resistor so it will   drop the current and it will charge this  capacitor for initial stage initial startup   because this ic needs some voltage to take a start first of all this capacitor will charge from this  point dc voltage it is 12.31 volt when this ic   will receive voltage at vcc pin at that time it  will start switching when it will start switching   now first of all we will check the frequency  between ground pin and the pin number six this   is pin number six so this ic is switching at 101  kilohertz or we can say 100 kilohertz that means   this ic is switching 100 kilohertz that it is  generating 100 000 pulses in one second to operate   this mosfet so this mosfet will allow a conduction  it will allow electrical current from ground to this winding when it will allow current  the current will change very speedy   the current will flow with a speed and it will  keep connecting disconnecting this connection   disconnection of this current will cause to  vary the magnetic field here in the winding   when this current will flow  the magnetic field will expand when this mosfet will turn off at that time this  magnetic field will collapse this collapsing   magnetic field will transfer voltage in the  secondary because it is a flyback in on time   it will store the energy in this winding and in  off time it will transfer voltage in the secondary   when it is transferring at the same time  the collapsing magnetic field will cause to   increase the voltage at this drain and these  voltage the collapsing magnetic field will   cause to generate voltage that  voltage will reverse in polarity   it is called snubber circuit it will  discharge that that voltage but it will cause   a voltage increase in the drain now if  we check voltage here it is 317 volt this mosfets the rightmost pin that is source  pin the center pin that is drain pin drain pin   is connected to the transformer primary winding  so we will check this same dc voltage here   320 volt and between drain to source  it is 373 volt why these are 373 volt at this point 320 but at this point 370 why   in the primary side in the hot side  we have transformer four terminals where is the transformer transformer is here we have four terminals here one two three and four this is capacitor positive terminal  so positive terminal is here and   it is reaching to the winding that is here transformer second side this point it is reaching to the  mosfet mosfet drain that is here   and then we have a diode resistor and capacitor   whereas diode at the same point where is mosfet  here and that's that point we have a diode here   that is here this diode this resistor and  this capacitor we have three components   it is called r c d resistor capacitor diode  snubber circuit what is the duty of this circuit   when mosfet is conducting in that  condition the negative voltage   are reaching here so it will  cause to reverse voice this diode and current have to flow in the  primary when this mosfet will open   the magnetic field stored here will collapse  when it it will collapse it will generate   reverse voltage equal to this capacitor if here  we have 316 volt 320 volt the same reverse voltage   will be generated through this pointing so the  voltage will reach at this point in reverse format   positive negative positive so mosfet this end  drain end will cause to bear a voltage 640 volt   to protect this mosfet from damage the excessive  voltage reverse polarity voltage in that condition   it will be discharged through this rcd cylinder  circuit if this this diode is short circuit most   of the time it is it becomes short circuit when  it is short circuit it will make two conditions   the first condition when the current will start to  flow it will give a parallel path to this winding   actually it have to stop the current but in  conduction mode it will allow the flow of current   through the diode to the resistor and the second  flow here at that time the ic will switch but   it will not create a strong magnetic field here so  that will cause to dropped voltage in the output   if we need 12 volt it will cause to produce  we can say 5 volt 6 volt 3 volt whatsoever so this is the basic problem for the  dropped voltage in the secondary side   so that is the reason here were  319 320 volt but here was 350 volt   30 volt more than this capacitor value that means  this mosfet is switching this ic is switching   the current is flowing in the primary if we  cannot find the voltage increase here that means   this ic is not working the first condition the  second condition might be the ic is switching   as we checked here at pin number six  we are finding 100 kilohertz signal   then we have our resistor we have a diode that  is to discharge this gate to give a voltage   at gate gate by sync so this zener sometime it is  short circuit if it is short circuit this mosfet   will not switch if the mosfet is open inside that  means it will not switch if it is short circuit   it will turn it will cause to burn out this fuse  but it is open when it is open it will not allow   a current in the winding ics switching it is  doing its job but it will not allow a current   flow in this part then we have a current sensor  resistor that is sometime less than one home all   the time sometime it is one of them point two home  point three ohm here it is using point seven ohm   if this resistor is open in most cases when  the mosfet is damaged this resistor also damage   because it is to protect the circuit that is  here it is always connected from negative line   to the source the rightmost pin of  this mosfet if this is open that means   it will not allow a ground path  to the source how we can check it it is giving 0.3 ohms now when we receive  voltage here at this ic pin number seven   pin number five of this ic is ground at that  time this ic will produce five volt reference how we can say this ic is  good set multimeter to dc it is ground and here is pin number seven   that is taking twelve point three  volt twelve point two nine volts and it will provide five volt reference  this ic have internal five volt regulator   if it is taking voltage and it  is not giving output voltage   5 volt here in the output side this circuit  will never work we must find 5 volt here   ground and pin number eight must be clear for  five volt pin number eight must have five volt   pin number five must connect it to ground so that  this ic will work properly then we have rtct the   timing capacitor and timing resistor if these  voltage are not available the timing circuit will   not work because this 5 volt will cause to charge  and discharge this capacitor at pin number four   so these voltage are very very  important to make oscillation   then this ic is switching we tested the switching  if we find the voltage here drain to source   dc voltage that means this ic is switching now  how we can say this circuit is switching we can   also check frequency ground and gate pin it  is giving 100 kilohertz signal so that means we are finding the voltage here we are finding the  signal here we can also check with oscilloscope at   gate to ground i tested now great to ground we  are taking this voltage then in some conditions we find the voltage fluctuation at vcc  capacitor if this diode is open circuit in the auxiliary winding we have a diode and  current setting resistor this diode is here   so this diode must provide voltage here so this  current setting resistor and diode if we see here   as we discussed positive line positive voltage  the switching line that is coming from this drain   then we have two pins of this winding here the one  pin that is connected to the ground terminal the   second pin of this auxiliary winding that is here  it is connected to current setting resistor that   is under this transformer in this design it is  shown under here so this voltage will reach here   then we have a diode here at this point so  this diode is connected to vcc capacitor here   in some conditions this diode is damaged in  some conditions we have some open circuit here dc voltage check dc voltage across this diode   so we will verify that there is a switching  and we are finding this voltage here winding this end is grounded  this is output resistor and here we are taking 12.58 volt this diode is  giving voltage if we cannot find this voltage   so that means it will not charge this capacitor  here in initial stage the voltage were reaching   at this capacitor it was charging it was causing  to start when this ic will start switching   before this it will charge at 1 milli ampere but  the requirement of this ic for example it is 17   milli ampere so this capacitor will not drive  for a long time because it will discharge when   it will start discharging the voltage level will  become low to compensate this voltage we must find   the voltage from this side because this resistor  cannot provide enough current to drive this ic for   a long time so we will find a feedback voltage  we will find a operating voltage for this ic   that are called auxiliary voltage vcc voltage  or bias voltage it is providing biasing voltage   this resistor is very critical resistor in this  circuit we will discuss finally when we will   discuss the voltage issue now received voltage  at this point received voltage at this capacitor   it start to produce reference voltage and this  ic start to oscillate and provided drive signal   at gate of this mosfet and mosfet started  to switch its drain pin to the transformer   transformer now will provide voltage in the output in the output side we have this winding two pins one pin is  connected here at output pin that is   ground pin the second pin that is here it  is reaching to this diode this rectifier that is here sometime it is using this  diode sometime it is using a single diode   to take more current it is  using full wave rectifier   we will check dc voltage across this  rectifier it is giving 12.24 volts   here when we want to check output voltage of  this winding we cannot check voltage across   this winding here why because it is a source  it is a generator we will not find any voltage   dc voltage 67 millivolt if we want to check ac  voltage it will give zero volt why because the   transformer winding does not have any resistance  it is in micro ohms so we cannot test voltage here   because when it is operating it will provide  the easiest path so voltage will pass here   and in microvolt drop it will appear at this  winding so we cannot say there is a switching   so this diode it will rectify and it will  pass the voltage to the output capacitor   so we can find voltage here in the reverse voltage  when it is reverse cycle it will drop the voltage   in forward condition it will not  give voltage in a reverse biasing   it will give a voltage drop across  its junction so we will just only find   the voltage here the same voltage  will reach here in the output terminal 12.12 volt and some voltage for filtration  so if we cannot find a good voltage here   that means this diode is short  circuit this rectifier is bad if we have this situation at that  time we need to replace this diode   then we have output capacitors electrolyte  capacitors are the biggest problem in any   power supply because these capacitors have their  limited operation life with the passage of time   their life expires they reduce their capacitance  they reduce their specifications so these cannot   hold charge properly so that time the load will  not driven properly so we will find fluctuation   we will find some issues here so it is showing  voltage 12 volt the voltage positive voltage   are reaching here negative positive then  this voltage will pass through this inductor this inductor will remove ripples that is here  now we are finding voltage output voltage here   in the output terminal we have some resistor  and led and it is showing some voltage here   then we have a feedback circuit that is the most critical component in the  regulation one line that will reach to the   opto coupler one pin and then we will take a  voltage from this output terminal and it we   will make a voltage divider there is a variable  resistor to adjust the output voltage level so we will find a voltage level at pin number  one so here we have the voltage regulator   adjustable voltage regulator ground  terminal and this pin it is giving 2.47 volt we have voltage divider network here the  one line it is connected to led led is here   here we have led and a current  limit resistor for this led the same voltage are divided through this  voltage divider network this one resistor   this second resistor and these voltage  are reaching at this potentiometer so that   we will find 2.4 volt 2.478 when we  will find these voltage at this pin   at that time this power supply will  start regulation it will provide drive signal to the opto  coupler so optocoupler is here and the one line from this positive voltage is  reaching directly to the optocoupler one pin   now in the coupler one side it is giving 1.1  volt it is giving voltage here when it will   receive 2.47 volt or 2.5 volt here at pin  number one it is a condition at that time   it will give a cathode current to this  optocoupler it will generate optical light   from this led to its output transistor  that is phototransistor what will happen when there is no load we will receive some  regulated voltage here it will give a constant   light in that condition this reference voltage  v reference will reach here to the collector   and when it will switch it will give some voltage  and if this voltage will reach at pin number two   of this ic when the voltage level is maintained at  that time it will conduct it will provide voltage to pin number two now it is dropping 0.96 volts across its this transistor emitter to collector 0.9 volt   so it is now saturated it is  providing voltage to pin number two when the voltage level is dropped  at that time it will cause to   reduce the light when the light  will reduce the conduction of this   transistor will decrease when decreased internal  resistance will increase so these voltage when the voltage are regulated at 12 volt it will  give a light so this light will start conduction   so these reference voltage will reach here and we  will take output from here in that case it will   work in conduction mode so that this resistor  network it will give a feedback pin feedback   voltage to pin number two so that uh this  ic will stabilize its duty cycle if any time   the voltage level at this capacitor will decrease  it will cause to drop in the voltage at this   anode it will cause to drop the current it will  cause to decrease the light when the light will   decrease its internal junction resistance will  increase so that the voltage level the 5 volt here   coming here to this pin will change when it  will change this ic will increase its duty cycle   when it will increase its duty cycle the magnetic  field will increase increased magnetic field   will cause to more voltage in the secondary so  that decreased voltage level will compensated   and when it is compensated voltage level level  is maintained at that time this light will return   back to its good working conditions at that time  this conduction will become to normal so that this   ic will take feedback and it will work normally  it will increase or decrease its duty cycle with   the change in this light if this tl431 or  any voltage regulator is failed to provide   a cathode drive to this optocoupler here that is dc voltage 1.1 volt in this dc voltage there  are some pulses that is charging discharging   of this one this conductor the voltage  level here at this capacitor will charging   discharging charging discharging so charging  discharging charging discharging these pulses   will appear here how much the speedy pulses are  here that will determine the value of this load   if we are drawing two ampere current so that this  capacitor will char charge and discharge very   speedy so it will generate pulses these pulses  these ramps will reach here so that this ic will   adjust its duty cycle it will compensate the  voltage to maintain 12 volt here if the load   is removed this capacitor will stable when it  is stable it will not discharge rapidly at that   time the light will stable and ic will work at  its specified duty cycle if this tl431 is bad   it will drive this ic this optocoupler wrongly  then it will give a wrong command to this   optocoupler at that time the voltage level  will be any value might be it is very high   might be it is very low if this optocoupler is  bad its inside diode is giving good continuity   drop forward and reverse voltage we can say this  is good but it is if it is not working properly   at that time we have to replace this optocoupler  how we can say it is working in conduction mod here we have 1.1 volt and here we have  approximately 0.9 volt if this internal   transistor is bad at that time this optocoupler  will not switch this voltage and the ic will not   find the feedback many times these capacitors  become short circuits internally at that time   the ics switches on and off the voltage  fluctuates we will find random voltage here   here we have 12 volts if we have  some overload condition at that time the voltage will fluctuate like this at that time   the ic drives at full condition but we cannot  find the voltage output why if this capacitor   is short circuit internally the ic will drive but  all the voltage will discharge here entirely here   in that condition we will find fluctuation the  varying voltage the second thing at that time   if we have some problem here in the output  side we will find fluctuation here ground   and pin number seven our vcc capacitor here we  have 12 volt let me connect some load here now our connected load this  time we are taking 12 points 12.3 volts at vcc capacitor now i applied some  load when the load is applied the voltage at   vcc pin increased about 14.34 without load  12.2 volt 14 volts if i change another load   13.97 if i put overload what is  going on the voltage are increasing we can see this time this ic is working overload 12 volt and we can verify the both multimeters   when it becomes in drive at that time  this voltage at pin number 7 will increase   when it will increase it will increase from  its threshold that is 16 volt when the voltage   will reach above 16 volt this ic will turn off  when it will turn off it will disconnect its   drive at that time the vcc capacitor will  drive ic for a short time and then it will   off why it will off because that time when the  drive is disconnected this winding will turn off   because drive removed from here when it is  removed the voltage here in the primary will stop when the current flow in primary will stop it  will stop to generate voltage here in exaltery   in voice winding when voice winding will stop  at that time the ic will work on this capacitor   when this capacitor will discharge  about less than 12 volt 10 volt   what is its threshold at that time this ic  will start switching again when it will start   switching we have overload in the output side  at that time this voltage will increase again the multimeter is showing 14.44 volt but  if we will see on the oscilloscope it will   approximately 16 volt at 16 volt it is called olp  overload protection when overload protection will   occur at that time this ic will stop switching it  will remain off until uvlo under voltage lockout   that is less than 10 volt here it is showing 11  volt it is in milliseconds and this will reach   less than 8 volt 9 volt at that time this ic will  start switching again and we will find voltage   so this is the theory of  operation and fault finding   now we can see we must find 12.2 volt 12  volt at no load operation if we will find volt here with no load and we say we put some load   at that time this power supply goes in off  condition what is the reason how we can fix that   at that time we have to increase this  resistance that is in series of this diode it is 10 ohm we can apply 20 ohm ohms until without load if this 3842 will  receive 12 volt that time this ic will take   full load the second problem we  have a current limit resistor here in this current based smps ics we have  this current limit resistor if the value   of this current limit resistor will  increase at that time it will give a   more voltage feedback to this pin i  sense pin that is pin number three   in that condition this ic will not drive  for full load we have to reconsider the   value of this resistor or this resistor  we can adjust it to take full current   friends i hope so this video is informative  if it is informative hit the like button if   you have any question please let me know in the  comment box thanks for watching salaam alaika

Channel: Haseeb Electronics

Views: 238,412

Keywords: how to repair smps, how to repair switch mode power supply, smps practical troubleshooting, smps repair step by step, repair switch mode power supplies, haseeb electronics, smps repair Practical Troubleshooting Tutorial, how to, troubleshooting smps power supply, smps troubleshooting step by step, switching mode power supply, repair switching mode power supply, smps output fluctuating, switching mode regulator, power supply repair, how to repair switching mode power supply

Id: OwmN-N8kgHw

Length: 55min 32sec (3332 seconds)

Published: Sun Jan 02 2022