How Current Limitation is happening in the circuit???

• How Current Limitation is happening in the circuit???
  Posted by ommsiva on 08 Jan, 2024 16:08
• Hi all,
  
  I went through this circuit in net , two current sense resistor(0.47 ohm) which is equivalent to 0.235ohm and current adjustable resistor 500 ohm were present in parallel is connected between current sense and current limit pin. As per the circuit , current can be varied between 0-10 amps.
  
  Can anyone explain how current limiting is working with the specified range???
  
  When the circuit would fail???
  
  I have attached source website:https://electronics-diy.com/30v-10a-variable-bench-power-supply.php

  ---

  
• #1 Reply
  Posted by coromonadalix on 08 Jan, 2024 17:23
• current sensing is done thru r5 r6
  
  the lm 723 calculations   is done thru their values  to get an 0.6v  drop,  once 0.6v drop is there   thru the r11 trimmer / pot,   you will  get a min or max current value 
  
  and the maximum should be around 14 amps
  
  2x 0.47ohms divided by 2,  multiplied by 0.6 = should be the maximum current limitter has
  
  R1 R2 R3 R4  are emitter balancing resistors,   you can have them on emitter or a different value on their base, and have all emitters common
  
  this is a basic circuit, and the lm723  wont go down to zero volts  by internal design,  you'll get around 1.2 volts at the output
  
  what do you mean by fail ?       if the pass transistor fail ??
• #2 Reply
  Posted by MrAl on 08 Jan, 2024 20:13
• Quote from: ommsiva on 08 Jan, 2024 16:08

  Hi all,
  
  I went through this circuit in net , two current sense resistor(0.47 ohm) which is equivalent to 0.235ohm and current adjustable resistor 500 ohm were present in parallel is connected between current sense and current limit pin. As per the circuit , current can be varied between 0-10 amps.
  
  Can anyone explain how current limiting is working with the specified range???
  
  When the circuit would fail???
  
  I have attached source website:https://electronics-diy.com/30v-10a-variable-bench-power-supply.php
  

  
  Hi,
  
  The 723 uses the base emitter junction of a transistor to provide for some rough current limit setting.
  As the voltage across pins 2 and 3 gets close to around 0.65 volts the internal transistor starts to turn on and shunt current away from the output and thus reduces the drive to the parallel power transistors, and that reduces the output current.
  
  There are two things that cause this action.  One is the two 0.47 Ohm resistors in parallel, which means the total resistance is roughly 0.25 Ohms.  To get to the 0.65 volts would then require 2.6 amps which would limit the current to around 2.6 amps.
  However, there is also that 500 Ohm pot.  That reduces the voltage getting to the base emitter of the transistor, so if you have the pot set to about 1/4 of its travel, that will mean it takes 4 times as much current to reach that 0.65 volts base emitter voltage.  That's about 10.4 amps.
  These are approximate values the max output current with that adjustment should be a bit higher like around 11 amps, but you can change that by turning the 500 Ohm pot while measuring the output current.  The current limit point will change with temperature though so you have to take that into account.  If that is not acceptable then you have to modify the design a little.
  
  Using the base emitter junction of a transistor to limit current was an old school method of limiting current.  It can still be used though when you do not need a precise current limit setting and can tolerate some change when the temperature of the chip starts to change.  This method was often used to protect the equipment from damage not to limit the current to a perfect level.
  
  
  
• #3 Reply
  Posted by ommsiva on 09 Jan, 2024 00:43
•  one fourth of 500 ohm pot means 125 ohm and current limiting at 2.6 amps.
  
  How we arrived at 10.4 amps?
• #4 Reply
  Posted by Electro Fan on 09 Jan, 2024 08:25
• looks interesting:
  
  https://electronicprojectsforfun.wordpress.com/power-supplies/a-collection-of-proper-design-practices-using-the-lm723-ic-regulator/
  
  some comments indicate the LM723 is legendary but maybe not optimum for a bench power supply
• #5 Reply
  Posted by MrAl on 09 Jan, 2024 09:15
• Quote from: ommsiva on 09 Jan, 2024 00:43

  one fourth of 500 ohm pot means 125 ohm and current limiting at 2.6 amps.
  
  How we arrived at 10.4 amps?
  

  
  Using the voltage divider formula.
  The resistance at 1/4 of travel for a 500 Ohm pot is 125 Ohms, and that divides the voltage by 125/500 which equals 1/4.
  That means when the voltage across the two parallel 0.47 Ohm resistors reaches 2.6 volts, the voltage at the transistor base emitter reaches 0.65 volts which is the point where it starts to cut the output current.  It takes 10.4 amps to reach 2.6 volts across those two resistors, hence the output is limited to roughly 10 amps.
  
  This is a rather old trick also.
  
  There is a drawback to this though.  With 2.6 volts across the two 0.47 Ohm resistors the total power is about 26 watts, which means EACH resistor has to be rated for about 26 watts or better.  The actual power in each is about 13 watts, but if the resistors are rated for 13 watts they would get super-hot.  With each rated at 25 watts for example, they will stay cooler.  They still will get hot though so there has to be free air flow to keep them cooler.  Note the values shown on the schematic are only 5 watts each, which cannot be correct.  They would get red hot and burn up.
  This is yet another reason why higher power power supplies started using a switching control method.  That keeps these side issues of power waste down to lower levels, which is also now a modern goal.  The 723 regulator was used back as far as the 1970's and a lot has changed since then.  They were used in some very, very expensive equipment though back then.
  
  There should be a resistor in series with the pot however, to keep the user from adjusting the output current to extremely high levels like 20 amps or even more.  At 1/8 the travel the output can reach around 20 amps, at 1/16 the travel it can go as high as 40 amps, either of these would probably blow something out and/or burn up those two 0.47 Ohm resistors.
  
  BTW if you can order from Amazon, they sell 10 Amp switching power supplies for around $60 USD now.
  
  
  
• #6 Reply
  Posted by ommsiva on 09 Jan, 2024 15:57
• Thank You sir.
  
  
  Under a load condition of 10 Amps, There is a possiblitiy that series pass transistor may go to breakdown. But under what condition by LM723 can go to breakdown condition?
• #7 Reply
  Posted by coromonadalix on 09 Jan, 2024 17:03
• like this  not sur if it would fail,  you have a pre-driver transistor   and the voltage pot is not directly on the output as some other uses the lm723 can get or be used
  
  and  you would need some power diodes on the output and one  on emitter collector of the power transistors
  
  and   you get over 37vdc on pin 11 and 12 of the lm723, way pass its input limit ...
• #8 Reply
  Posted by MrAl on 09 Jan, 2024 19:09
• Quote from: ommsiva on 09 Jan, 2024 15:57

  Thank You sir.
  
  
  Under a load condition of 10 Amps, There is a possiblitiy that series pass transistor may go to breakdown. But under what condition by LM723 can go to breakdown condition?
  

  
  Well for one you have to make sure the series pass transistors are made to handle the full current each one will have to pass.  You also have to make damn sure that the heatsink can handle the extreme power they will be dissipating as heat. If they get too hot they burn out.
  With an input power of say 30 volts and output set to 10 volts at 10 amps, the transistors and heatsinks will have to handle 20 volts at 10 amps, which is 200 watts.  That's a hell of a lot of power to dissipate and it's a waste of energy too.  If you have 4 transistors then each one will have to handle 50 watts, and if you have 5 transistors then each one will have to handle 40 watts, still a lot of power.  If you have 10 transistors then each one has to handle just 20 watts, a lot better, but together they still dissipate 200 watts.
  In cases like this where the power is very high, it is usual to use a switching power supply, or a linear like this one with an added switching front end that follows the output.  That way the power dissipation is much lower for any output voltage.  The switching front end limits the voltage to the linear back end, so the linear does not have to dissipate too much power.
  
  You may want to rethink all this.  The example above was for a 10 volt output at 10 amps, but for a 5 volt output at 10 amps it's even worse.
  
  To find out the possible failure conditions for the 723, you really have to look at the data sheet.  For one, the input voltage will be limited.
• #9 Reply
  Posted by ommsiva on 10 Jan, 2024 01:08
• Thank you sir.
  
  
  
  1)Which switching regulator will be suitable for 10amps current?
  
  2) why we need to feed the output  of switching regulator to linear regulator? Previous reply says to decrease power dissipation.  What are the other advantages of linear regulator system?
• #10 Reply
  Posted by MrAl on 10 Jan, 2024 11:41
• Quote from: ommsiva on 10 Jan, 2024 01:08

  Thank you sir.
  
  
  
  1)Which switching regulator will be suitable for 10amps current?
  
  2a) why we need to feed the output  of switching regulator to linear regulator? Previous reply says to decrease power dissipation.
  2b)  What are the other advantages of linear regulator system?
  

  
  [1]
  Do you want a ready made product or an IC chip of some kind to build a power supply yourself?
  
  [2a]
  The linear regulator dissipates too much heat when the output voltage is set to a mid or low value with high current.  The cause is because the input voltage is always high so you can get the higher output voltage when you need it.  By using a switching regulator front end, the output voltage of the switcher can be set close to the output voltage of the linear, and that reduces the power dissipation in the linear by a huge factor.
  
  [2b]
  A linear regulator has a smoother output with less noise.  This is why they are still used today, but in the higher power power supplies there is usually a switcher ahead of the linear which automatically adjusts it's output close to the output of the linear.  That keeps power dissipation way down.
  
  Power dissipation is a big problem with linear regulators that have to put out a lot of current at various voltages.  However, when the input voltage is close to the output voltage, the power dissipation goes down a lot.  A switcher is more efficient in that there is less power wasted, but sometimes we still want a linear output.  By combining the switcher with the linear, we get the best of both technologies.
  
• #11 Reply
  Posted by ommsiva on 10 Jan, 2024 15:59
• Do you want a ready made product or an IC chip of some kind to build a power supply yourself?
                    Yes Sir , It shall be IC which could do both CC CV
  
  
• #12 Reply
  Posted by MrAl on 12 Jan, 2024 06:35
• Quote from: ommsiva on 10 Jan, 2024 15:59

  Do you want a ready made product or an IC chip of some kind to build a power supply yourself?
                    Yes Sir , It shall be IC which could do both CC CV
  

  
  Hi,
  
  You can look up the LM3524D switching regulator controller to get some idea what is involved.  This is an older chip and easier to understand.
  When you take a look at the data sheet you will immediately see what you are dealing with even if you do not have a particular design ready yet.  This chip was used in a lot of equipment even computer power supplies and has been around at least as far back as the 1980's.
  
  For more modern controller chips, look on the TI website they have a ton of switching controllers.  You may find something you like better.  They are categorized into groups like buck, boost, then subdivided for various uses such as external or internal switch.
  
  Output current for the chips with external an switch is usually not limited because they use an external transistor switch (one or more).  That means you could design a controller that puts out 100 amps if you like.
  
  The ones that have the internal switch are usually not as efficient and they are limited on how much current you can get out of them.  You probably want to go with a controller chip that uses an external transistor switch.  You can really make a nice switcher like that.
• #13 Reply
  Posted by ommsiva on 16 Jan, 2024 12:04
• I had a good introduction from below video
  
  
  
  Can any one suggest an circuit with SG3524 with CC CV ..
  
  Thank you.
• #14 Reply
  Posted by MrAl on 17 Jan, 2024 00:26
• Quote from: ommsiva on 16 Jan, 2024 12:04

  I had a good introduction from below video
  
  
  
  Can any one suggest an circuit with SG3524 with CC CV ..
  
  Thank you.
  

  
  Hi,
  
  Did you look for app notes or reference designs on the data sheet?
  They usually have these somewhere direct from the manufacturer.
  You have to read all the specs though too so you know what the limitations are.
  
  Also, from what I have read the "D" suffix version is better than the no suffix version of the chip.  So 3524D not just 3524.
  
• #15 Reply
  Posted by coromonadalix on 17 Jan, 2024 02:20
• Quote from: MrAl on 12 Jan, 2024 06:35

  Quote from: ommsiva on 10 Jan, 2024 15:59

  Do you want a ready made product or an IC chip of some kind to build a power supply yourself?
                    Yes Sir , It shall be IC which could do both CC CV
  

  
  Hi,
  
  You can look up the LM3524D switching regulator controller to get some idea what is involved.  This is an older chip and easier to understand.
  When you take a look at the data sheet you will immediately see what you are dealing with even if you do not have a particular design ready yet.  This chip was used in a lot of equipment even computer power supplies and has been around at least as far back as the 1980's.
  
  For more modern controller chips, look on the TI website they have a ton of switching controllers.  You may find something you like better.  They are categorized into groups like buck, boost, then subdivided for various uses such as external or internal switch.
  
  Output current for the chips with external an switch is usually not limited because they use an external transistor switch (one or more).  That means you could design a controller that puts out 100 amps if you like.
  
  The ones that have the internal switch are usually not as efficient and they are limited on how much current you can get out of them.  You probably want to go with a controller chip that uses an external transistor switch.  You can really make a nice switcher like that.
  

  
  the bests psu  have  both linear and smps, at least the Kikusui i have does
  
  they use the smps / phase controlled at the primary stage, to get a clean power source to drive the linear section, in some psu  the smps section will give  aroud 5 volts headroom for the linear section, that way you get a very low heat dissipation
  
  the disavantage is for sure, they are heavier than pure smps
  
  example :  i have a full linear Kepco model "80vdc at 8 amp", weight 51 pounds, and the kikusui who as almost the identical specs wight 26 pounds  and under a full load the full linear will dissipate 860 watts, and the other not even 120-150w
• #16 Reply
  Posted by MrAl on 18 Jan, 2024 13:17
• Quote from: coromonadalix on 17 Jan, 2024 02:20

  Quote from: MrAl on 12 Jan, 2024 06:35

  Quote from: ommsiva on 10 Jan, 2024 15:59

  Do you want a ready made product or an IC chip of some kind to build a power supply yourself?
                    Yes Sir , It shall be IC which could do both CC CV
  

  
  Hi,
  
  You can look up the LM3524D switching regulator controller to get some idea what is involved.  This is an older chip and easier to understand.
  When you take a look at the data sheet you will immediately see what you are dealing with even if you do not have a particular design ready yet.  This chip was used in a lot of equipment even computer power supplies and has been around at least as far back as the 1980's.
  
  For more modern controller chips, look on the TI website they have a ton of switching controllers.  You may find something you like better.  They are categorized into groups like buck, boost, then subdivided for various uses such as external or internal switch.
  
  Output current for the chips with external an switch is usually not limited because they use an external transistor switch (one or more).  That means you could design a controller that puts out 100 amps if you like.
  
  The ones that have the internal switch are usually not as efficient and they are limited on how much current you can get out of them.  You probably want to go with a controller chip that uses an external transistor switch.  You can really make a nice switcher like that.
  

  
  the bests psu  have  both linear and smps, at least the Kikusui i have does
  
  they use the smps / phase controlled at the primary stage, to get a clean power source to drive the linear section, in some psu  the smps section will give  aroud 5 volts headroom for the linear section, that way you get a very low heat dissipation
  
  the disavantage is for sure, they are heavier than pure smps
  
  example :  i have a full linear Kepco model "80vdc at 8 amp", weight 51 pounds, and the kikusui who as almost the identical specs wight 26 pounds  and under a full load the full linear will dissipate 860 watts, and the other not even 120-150w
  

  
  Hi,
  
  Well, if anyone knows about that stuff it's me, as I've worked in the power supply industry for many years designing and testing high power synthesized sine converters and other types.  From maybe 1 watt to 30000 watts.
  
  Until recently I also had two combination switcher/linear supplies 300 watts each.
  I've actually used the chip I mentioned too along with a couple other types that are similar.
  You'll find several posts where I talked about switcher front end and linear back end and the benefits.
• #17 Reply
  Posted by MathWizard on 19 Jan, 2024 14:48
• Quote

  example :  i have a full linear Kepco model "80vdc at 8 amp", weight 51 pounds, and the kikusui who as almost the identical specs wight 26 pounds  and under a full load the full linear will dissipate 860 watts, and the other not even 120-150w

  A few things like that 51lb PSU in a room and I'd want a warehouse or something with a good foundation for mounting it all. I don;t think my desk or end-tables would do.
• #18 Reply
  Posted by coromonadalix on 19 Jan, 2024 15:44
•  
• #19 Reply
  Posted by ommsiva on 19 Jan, 2024 16:25
• Hi,
  Correct me if i am wrong
  
  1)Inthis circuit R6  and P3 fixes voltage at the output and this pontential divider output is input of Non inverting input of opamp.
  
  2) R5 senses the current.
  
  3) My question is, How P1,P2 and R3 convert the current through R5 in to voltage and feed to opamp at inverting input of opamp?
  
  or
  
  How P1,P2 and R3 sense the current, convert in to voltage and give to inverting input of opamp for comparison?
  
  
  4)what is the function of R4?
  
  5) if inverting input goes high, that is current taken by the load is more than set current. Then output of opamp goes low that is -5V and R2 limits the current and LED glows.  My question is why we want D1 in the circuit?
  
  6) the output of 7905 is -5V. it is divided by R9 and P4. How the voltage is -1.2V? But it should be 5(100/200)= 2.5V
  
  Thank you all

  ---

  
• #20 Reply
  Posted by MrAl on 20 Jan, 2024 07:01
• Quote from: ommsiva on 19 Jan, 2024 16:25

  Hi,
  Correct me if i am wrong
  
  1)Inthis circuit R6  and P3 fixes voltage at the output and this pontential divider output is input of Non inverting input of opamp.
  
  2) R5 senses the current.
  
  3) My question is, How P1,P2 and R3 convert the current through R5 in to voltage and feed to opamp at inverting input of opamp?
  
  or
  
  How P1,P2 and R3 sense the current, convert in to voltage and give to inverting input of opamp for comparison?
  
  
  4)what is the function of R4?
  
  5) if inverting input goes high, that is current taken by the load is more than set current. Then output of opamp goes low that is -5V and R2 limits the current and LED glows.  My question is why we want D1 in the circuit?
  
  6) the output of 7905 is -5V. it is divided by R9 and P4. How the voltage is -1.2V? But it should be 5(100/200)= 2.5V
  
  Thank you all
  

  
  A few  things do not look right.
  
  First, there are too many parts for what this does.  It also looks like someone hung the schematic on the wall and covered it with glue, then threw a handful of capacitors at it, and whatever stuck they kept in the circuit.
  
  Then, the two pots P1 and  P2  are wired  in parallel.  That means they do the same function.  They are labeled in a way that says they do different functions, which does not make sense.
  
  Diode D1 does have a function, it prevents that 620 Ohm resistor current from affecting the LM317 feedback terminal.
  Also, the -1.2v line is typical for using an LM317 as that allows adjusting the voltage down to 0v rather than just +1.2 volts.
  P3 may be right, it allows adjustment of the output voltage.
  Would have to check if pin 7 of the 301 op amp is wired to the right place.
  
  I would look for a simpler, better design.
  
  
  
• #21 Reply
  Posted by ommsiva on 21 Jan, 2024 10:09
• Hi
  
  Now i am attached the circuit from Manufacturer data sheet
  
  1) In this circuit, how current through R3 (suppose 5 amps flow then voltage across R3 will be 1V) is related to R2?
  
  2) Resistance R6 and R8 set the output voltage, is it right?
  
  3) non Inverting input is connected to output voltage and inverting input is going to sense what?
  
  4)what is the function of R7?
  
  5) if inverting input goes high, that is current taken by the load is more than set current. what will happen to output voltage of regulator?
  
  6) What is the function of D1 and why 680 ohm is used as current limit resistor for LED?
  
  7) is opamp is comparator or Integrator(because of C5)?
  
  overall i want to understand feedback loop which contain opamp as comparator.
  
  thank you all
  

  ---

  
• #22 Reply
  Posted by MrAl on 21 Jan, 2024 13:49
• Quote from: ommsiva on 21 Jan, 2024 10:09

  Hi
  
  Now i am attached the circuit from Manufacturer data sheet
  
  1) In this circuit, how current through R3 (suppose 5 amps flow then voltage across R3 will be 1V) is related to R2?
  
  2) Resistance R6 and R8 set the output voltage, is it right?
  
  3) non Inverting input is connected to output voltage and inverting input is going to sense what?
  
  4)what is the function of R7?
  
  5) if inverting input goes high, that is current taken by the load is more than set current. what will happen to output voltage of regulator?
  
  6) What is the function of D1 and why 680 ohm is used as current limit resistor for LED?
  
  7) is opamp is comparator or Integrator(because of C5)?
  
  overall i want to understand feedback loop which contain opamp as comparator.
  
  thank you all
  

  
  [1]  Current flows through R3 and creates a voltage drop with the top more positive than the bottom.  When the voltage drop exceeds a certain amount the voltage at the inverting terminal of the op amp becomes greater than that at pin 3, and so the output of the LM301 begins to ramp down lower.  As it ramps down, it begins to pull current from the ADJ pin node which lowers the output voltage.  When this control reaches equilibrium, the output current stays at that one level.  That means the current has been adjusted to a certain level.
  
  [2] Yes.
  
  [3] Inverting input senses voltage across R3 and makes the op amp output go lower when the voltage across R3 gets to a certain level.  The level is adjustable with R2 as R2 and R5 create a voltage divider.
  
  [4] R7 is probably there to offset the effect of R6 allowing the sensing of the voltage across R3 but this should be simulated to be sure.
  
  [5] The output will go down until the equilibrium point.  That's the point of the current regulation.  To understand this better you should study a voltage regulator using an op amp in greater detail to see how the feedback works in keeping the system at one set point of operation.  It's not too difficult to understand either.
  
  [6] 680 Ohms is probably the right value to limit current through the LED which may be just 20ma.  D1 is there because when the output of the LM301 is higher, we don't want that output to interfere with the voltage regulation.  Thus it provides an "OR" function.  Either the voltage is regulated OR the current is regulated, but not both at the same time.  When current regulation takes over, D1 will conduct, but not until then.  Using a diode like this is typical in these circuits which have voltage regulation and a current limit setting.
  
  [7]  The op amp will act as a rather fast integrator, so it's like a slightly slowed down comparator, but C5 probably acts as compensation also that keeps the system stable.  A capacitor there is very typical in voltage regulators that use op amps as the control element.  Without it the system is usually unstable.
  
  Do you know how to do a simulation with any type of simulation software?  That can help answer a lot of these questions too.
• #23 Reply
  Posted by ommsiva on 22 Jan, 2024 00:35
• Sir,
  1) suppose I fix the output to 20V and that will be the non inverting input of opamp and on inverting input we have R2 and R5. In what way current sense resistance R3 is related to R2 and R5???
  I will carry out simulation.
• #24 Reply
  Posted by MrAl on 22 Jan, 2024 15:04
• Quote from: ommsiva on 22 Jan, 2024 00:35

  Sir,
  1) suppose I fix the output to 20V and that will be the non inverting input of opamp and on inverting input we have R2 and R5. In what way current sense resistance R3 is related to R2 and R5???
  I will carry out simulation.
  

  
  You can set the voltage to anything you want.  When the output current becomes great enough, there is a voltage across R3.  The bottom of R3 is connected to op amp pin 3, and the top of R3 is connected through the voltage divider to pin 2.  When the drop across R3 is great enough, the voltage at pin 2 becomes slightly greater than the voltage at pin 3, and so the output of the op amp ramps down more and more until the output falls.  When the output falls, it will fall to a level that allows just enough current through R3 to maintain the condition of the voltages at pin 2 and pin 3.  That means the circuit is not in current regulation and so the voltage may fall a little, down to 19 volts or even less if the output current is higher.
  If you were to set the pot to put out just 4 amps, then the output will put out 4 amps, and the voltage will be less than the voltage set point of 20v.
  
  Resistor R7 also sets the gain of the op amp part of the circuit.  It looks like the designer did not want to the gain too high, just high enough to get it to work well.
  
  If you really want to understand this really well, look into a linear voltage regulator using an op amp as the control element.  Check out the voltages that come on the inputs and outputs.  In fact, you may gain insight by looking at just a voltage follower using an op amp.  That works in a similar manner.  As the output ramps up, the two voltages at the two inputs become nearly equal, and that stops the output from ramping up, and that's when it can be said to be regulating the output voltage.  A voltage follower is the simplest example of this kind of op amp action.
  For example, when the circuit starts up the output could be assumed to be at 0 volts, and then when say 2v is applied to the non inverting input, the output begins to ramp up.  Eventually the inverting input becomes nearly equal to the non inverting input, and that means the output stops ramping up.  It will ramp up to nearly 2 volts when this happens, then stop, holding the output at 2 volts.
  The simplest way to look at this is as the following expression shows:
  Vout=A*(vp-vn)
  where
  A is the internal gain (typically 100000),
  vp is the non inverting input which is the input voltage (say 2v),
  vn is the inverting input, which is connected directly to the output.
  This means the above comes out to be:
  Vout=A*(Vin-Vout)
  Solving this explicitly for Vout we end up with:
  Vout=(Vin*A)/(A+1)
  Now in pure theory, we take the limit as A goes toward infinity and get:
  Vout=Vin
  
  and so this shows how we get the same output as the input.  However, in real life we don't really have an infinite gain, so we stick with:
  Vout=Vin*A/(A+1)
  
  Now going back to that farther above:
  Vout=A*(Vin-Vout)
  if we start with Vout=1.999 (not quite 2.000 volts yet) we then have this:
  Vout=A*(2-1.999)
  or just:
  Vout=A*(0.001)
  and now with a gain of 1999 (just as an example) we have:
  Vout=1999*0.001
  or
  Vout=1.999
  and so that means the circuit had reached the equilibrium point because:
  1.999=1999*0.001
  
  It's good to think of this process as though the output could not ramp up quickly though as it takes some time.  As time progresses, eventually it would get to the point where the output was 1.999 volts and that would mean we had reached the regulation point and it will stay at that level.
  
  This might seem a little strange to you at first, but it makes sense when you go over it and try to understand what happens when the circuit is first turned on.  One of the main properties that comes into play here is the slew rate which limits how fast the output can ramp up to get to the right voltage so the feedback and internal gain end up maintaining that output at the same level thereafter.
  
• #25 Reply
  Posted by ommsiva on 22 Jan, 2024 17:43
• ""[6] 680 Ohms is probably the right value to limit current through the LED which may be just 20ma. ""
  
  Sir,
  
  if current taken by load is more, inverting input will be high. Making the output of opamp -6V.
  This -6v forward biases the LED because it is been connected to cathode and Anode is at ZERO. the adj pin and output pin has a loop with opamp which is open because D1 is short and I adj current  flows in D1 , D2 and to opamp.
  
  I=(6-3)/680--------------(3v is considered cut in potential of LED)
  I=4.41mA
  
  How it is 20mA?
  
  Correct me if i am wrong..
• #26 Reply
  Posted by MrAl on 23 Jan, 2024 11:17
• Quote from: ommsiva on 22 Jan, 2024 17:43

  ""[6] 680 Ohms is probably the right value to limit current through the LED which may be just 20ma. ""
  
  Sir,
  
  if current taken by load is more, inverting input will be high. Making the output of opamp -6V.
  This -6v forward biases the LED because it is been connected to cathode and Anode is at ZERO. the adj pin and output pin has a loop with opamp which is open because D1 is short and I adj current  flows in D1 , D2 and to opamp.
  
  I=(6-3)/680--------------(3v is considered cut in potential of LED)
  I=4.41mA
  
  How it is 20mA?
  
  Correct me if i am wrong..
  

  
  Hi,
  
  No that was just a quick guess about what they wanted to limit it to not that it would always be that level.
  With a 20ma LED you may even want to limit it to 10ma, but that does not mean it cannot be just 2ma, which it would probably be sometimes or maybe most of the time.  So that limit is just to protect the LED just in case it tries to go higher.  It may never go higher though, and the limit could even be much lower.
  I sometimes have limited my 20ma LEDs to just 5ma as that provides a lot longer life for the LED.  At 20ma the life is shorter, but even at 10ma the life is much longer, and at 5ma the half-life will be much, much longer.  If you need a lot of brightness (as in a flashlight) you may not be able to do this, but if not, it makes sense to limit the current to a lower level.
  
  BTW I should mention that this circuit is a very old circuit probably appearing in application books as far back as 1980 or even before that.  A more modern circuit would most likely limit current in a different way, conserving energy.  The 0.2 Ohm resistor at the full load of 5 amps will consume 5 watts, which is a lot by today's standards, and would really require a 10 watt rated resistor which is fairly large.  For a sarcastic view, imagine trying to install a 10 watt ceramic resistor inside a cell phone.
  You could look at other ways to do a current limit circuit which would use less power.  There are IC chips made today that are dedicated just for sensing current.  If you wanted to, you could look into that.  If you only use the power supply once in a while though or for short time periods, it may not be worth it to change anything though.  It's always up to you.
  We also have to keep in mind this is a LINEAR power supply, so at low to mid output voltages and higher current output the dissipation in the transistor will be quite high, which requires a larger heat sink.  With 30v input and 10v output at 5 amps, the power in the series pass transistor will be 20*5=100 watts.  That's really high.  At 5v output and 5amps, 125 watts, which is also very high. The heat sink has to be huge.  If you look up some heat sinks you can do a few calculations to find out how big it has to be.
  Back in the 1980's I worked on a program for Sandia Labs that had really bit heatsinks, about 12 inches wide and 18 inches long and about 4 inches high.  They are not that cheap, and they also take up a lot of room and need free air flow, so it's quite a job to get something like that to be practical.
  
  
  
  
• #27 Reply
  Posted by ommsiva on 04 Feb, 2024 05:35
• Sir,
  
  This circuit contain voltage regulator where the output voltage can be fixed by P2 and R4.
  current control is done by opamp 741 ,R1,R2 and P1  and it forms a closed loop with L200.
  
  My doubt is
  
  1) non inverting input (Pin3) is connected to output which is the second end of current sense resistor and inverting input(pin2) is connected to first end of current sense resistor?
  
          so always my inverting input will be greater than non inverting input. so results in pin 6 always moves towards ground potential
  
  2)Why we want R1 in the circuit?
  
  
  Thank you

  ---

  
• #28 Reply
  Posted by MrAl on 04 Feb, 2024 09:26
• Quote from: ommsiva on 04 Feb, 2024 05:35

  Sir,
  
  This circuit contain voltage regulator where the output voltage can be fixed by P2 and R4.
  current control is done by opamp 741 ,R1,R2 and P1  and it forms a closed loop with L200.
  
  My doubt is
  
  1) non inverting input (Pin3) is connected to output which is the second end of current sense resistor and inverting input(pin2) is connected to first end of current sense resistor?
  
          so always my inverting input will be greater than non inverting input. so results in pin 6 always moves towards ground potential
  
  2)Why we want R1 in the circuit?
  
  
  Thank you
  

  
  Hi,
  
  It is unclear why they still included R1 in the circuit when they already decided to go with the op amp.  Perhaps there is some transient or startup condition requirements.
  
  As to the output of the 741 going negative, that is easy to explain.
  As the voltage across pin 5 and pin 2 of the L200 becomes greater, the output current will be more limited.  This would mean that one way or another we would want to increase the voltage difference when we want the circuit to go into current limit.  Since pin 5 is somewhat constant when compared to pin 2, driving pin 2 lower would accomplish this, and thus the 741 would take over the current limit function.
  
  This must be a very old circuit though.  There are better choices for the replacement of the 741 op amp, most notably the LM358 assuming the response using the 741 was fast enough.  If the 741 was not really fast enough, then the LM358 will not be fast enough either, so an op amp with a faster slew rate might be used instead.  Of course stability is always an issue and has to be checked when using a faster component.
  
• #29 Reply
  Posted by ommsiva on 10 Feb, 2024 15:59
• How to analyse this circuit?
  
  We have to start from output or from input end. where to start?

  ---

  
• #30 Reply
  Posted by MrAl on 11 Feb, 2024 11:41
• Quote from: ommsiva on 10 Feb, 2024 15:59

  How to analyse this circuit?
  
  We have to start from output or from input end. where to start?
  

  
  Hello,
  
  You would start by learning to analyze much simpler power supply circuits first.
  You would also have to learn how to analyze circuits with bipolar transistors, such as in a differential amplifier.
  Also good to know would be how a basic ua723 voltage regulator works.
  Another point is to understand how power supply rectifier circuits work.
  
  You probably want to start with rectifier circuits.
  
• #31 Reply
  Posted by ommsiva on 20 Feb, 2024 17:06
• Sir,
  
  If my transformer is 24v-0-24v @10 ampere, then after bridge rectification and filtering I will get around 24 square root 2 which is 34V.
  
  My question is, I will get plus and minus 34V , How much current can I take from each line?
  
  It's a bipolar regulated power supply. Suggest some methods to get high current negative voltage regulation.
   
  
  Thank you all.
• #32 Reply
  Posted by Jwillis on 20 Feb, 2024 23:55
• When you calculate the total out put power of a linear power supply, calculate it from the VRMS of the transformer and not the peak voltage after rectification. Although the peak voltage is 34V, upon load that voltage will drop significantly down to the VRMS of the transformer. And depending on the transformers own regulation it will drop more at full load. So you will not get the expected 34V output but maybe 23V at full load.
  So if you want  to have say a 30V output then you would need at least a transformer above 30V taking in account its regulation. So for example to get 30V output at full load with a transformer with a 4% regulation you would need at least a 32V transformer. Assuming you can find a transformer with 4% regulation below 300 - 400VA. Most fall around 8 - 15% regulation some as much as 30%. Lower VA transformers will have poorer regulation compared to High VA transformers. 800 - 1000 VA transformers will have around 2 to 5% regulation.
  So when you look at common linear power supplies on the market you will find that the transformer may be a high as 10V or higher above the power supplies maximum output.
  I find that for every amp of output you need a transformer that has that many volts higher then the required output voltage. So for a 30V 10A power supply you would need a 40V transformer to hold regulation.
  
• #33 Reply
  Posted by ommsiva on 03 Mar, 2024 11:01
• Hi to all,
  
  In internet, they are two ways of increasing current handling capability of LM317 IC.
  
  first circuit: 10 ohm current sensing resistor is present to develop base emitter potential of the transistor which is going to supply extra current.
  
  Second circuit: 22 ohm current sensing resistor is present to develop base emitter potential of 2N2905 ,when collector current flows through 470 ohm resistor and voltage drop across it exceed the barrier potential of TIP73 , it will turn on.
  
  Which circuit is better for practical working condition?

  ---

  lm317.pdf

  
• #34 Reply
  Posted by MrAl on 03 Mar, 2024 18:41
• Quote from: ommsiva on 03 Mar, 2024 11:01

  Hi to all,
  
  In internet, they are two ways of increasing current handling capability of LM317 IC.
  
  first circuit: 10 ohm current sensing resistor is present to develop base emitter potential of the transistor which is going to supply extra current.
  
  Second circuit: 22 ohm current sensing resistor is present to develop base emitter potential of 2N2905 ,when collector current flows through 470 ohm resistor and voltage drop across it exceed the barrier potential of TIP73 , it will turn on.
  
  Which circuit is better for practical working condition?
  

  
  Hi,
  
  Probably neither unless you like really big heatsinks and pre-switcher era and pre-green era voltage regulator circuits 
  
  Either of those could waste a lot of power, something that is frowned upon in this day and age, and also gets hot which is undesirable as well.
  For short term use though it may make more sense.
  
  
• #35 Reply
  Posted by ommsiva on 06 Mar, 2024 04:53
• Dear sir,
  
  1)I found this circuit on web, TIP36C and TIP35C is going to do current boosting. Whether the 0.1 ohm /5W resistance should be in collector or emitter terminal?(to share equal current in all transistor)
  
  2) D2 and D3 are 1 amp diode but RPS supplies up to 10 amps of current. Is it correct to use 1AMP diode?
  
  3) Whether current limiting function of LM317 will happen when current drawn by base of TIP35C's exceed  at 2.2Amps?
  
  
  Thank you

  ---

  
• #36 Reply
  Posted by MrAl on 06 Mar, 2024 17:47
• Quote from: ommsiva on 06 Mar, 2024 04:53

  Dear sir,
  
  1)I found this circuit on web, TIP36C and TIP35C is going to do current boosting. Whether the 0.1 ohm /5W resistance should be in collector or emitter terminal?(to share equal current in all transistor)
  
  2) D2 and D3 are 1 amp diode but RPS supplies up to 10 amps of current. Is it correct to use 1AMP diode?
  
  3) Whether current limiting function of LM317 will happen when current drawn by base of TIP35C's exceed  at 2.2Amps?
  
  
  Thank you
  

  
  To understand how this circuit works and answer those questions it would help a lot to do a simulation that's what simulators are made for.  They help understand what happens and what could happen when things vary during the actual operation of the circuit.
  Without doing a simulation here are some ideas.
  
  [1]  It's a little difficult to say if the 0.1 Ohm resistors would work better in the emitters than in the collectors, but there does seem to be a strong possibility that idea would make the circuit work better.  For one, the collectors act more like constant currents which means the resistors will not do as much, and also if moved to the emitters a little more voltage drop in the emitter circuits may allow the smaller transistor to work better for limiting the total output current because that would loosen up the saturation voltage requirement of the smaller transistor.
  A more easily noticed problem though is the rating of the 0.06 Ohm resistors.  5 watts isnt enough because at some point they could be working too close to that power level.  10 watts (or even higher) would be better, and with free air flow all around.
  
  [2]
  Yes, I would say those two diodes should be at least of the 1N5400 series which are 3 amp diodes not 1 amp diodes.  The higher the rating the better as they are for reverse polarity protection against an external active load like a battery.  With the 1N5400 series, the protection relies partly on the behavior of the diode when it blows out.  They are banking on it shorting out.  If it happens to open up, the protection is lost.
  There are better ways to do this too but they are more complicated.  For example, the higher power diodes plus a series fuse.  If a reversed active load is connected to the output and the diode conducts, the fuse blows, and the load is disconnected.
  
  [3]
  The current to the base of the larger transistors is hard to predict exactly, but if we estimate using a minimum beta of 10 and with the LM317 drawing 1 amp, that means each base current is about 0.3 amps, ideally.  The problem then seems to be that R1 is not the right value if the goal is to get 10 amps  out of the output of this circuit.  With the emitters drawing 1 amp and the LM317 drawing 1 amp, we would want about 1.4 volts drop, which would mean we would need a resistor of about 0.7 Ohms.  This wouldn't be too much different if the LM317 was drawing 1.5 amps as then the emitters would be drawing less, but the total current might be around 2 amps again.  Obviously 2 amps through a 2 Ohm resistor is 4 volts, so the current limit would kick in too soon and limit the current to less than 10 amps, probably a lot less.
  This means R1 value would have to be decreased.  If it was 0.7 Ohms and with 2 amps through it, the power would be 2.8 watts, so a minimum of 5 watts would be a good starting point for the power rating of that resistor, with free air flow.
  
  These ideas should be verified with a simulation.  This circuit isn't too complicated so it's not hard to use with a simulator.  I might try this myself if I get a chance later today or tomorrow morning.  I'll update with any additional info I might find out.
  However, because of the more obvious problems lurking here and there in this circuit, I would study it out carefully before building a prototype.  A simulator would help a lot, followed by a bench test.  One of the things to pay attention to is the power dissipation in the transistors as well as the resistors.  I'd check every resistor power rating to make sure it can handle the required power.  I do not see any way around using somewhat larger heatsinks either for the transistors, and one for each LM3xx device.
  
  Linear power supplies are great for use for short term testing of other circuits and devices.  Not as good for long term use as they eat up too much energy in most cases.
  
  
  This idea should be studied in more depth with a simulation.
• #37 Reply
  Posted by ommsiva on 07 Mar, 2024 02:51
• Sir, I don't understand this calculation
  
  "With the emitters drawing 1 amp and the LM317 drawing 1 amp, we would want about 1.4 volts drop, which would mean we would need a resistor of about 0.7 Ohms.  This wouldn't be too much different if the LM317 was drawing 1.5 amps as then the emitters would be drawing less, but the total current might be around 2 amps again.  Obviously 2 amps through a 2 Ohm resistor is 4 volts, so the current limit would kick in too soon and limit the current to less than 10 amps, probably a lot less"
  
  if emitter takes 1 amp then base current is 0.1A, then three base current add together and makes 0.3A.
  Assume VEB of power transistor =1V @ 1A, VE=(Vin-V_0.06)=20-(0.06*3)=19.82V then Base voltage VB should be 1V less, VB=18.82V
• #38 Reply
  Posted by MrAl on 07 Mar, 2024 20:13
• Quote from: ommsiva on 07 Mar, 2024 02:51

  Sir, I don't understand this calculation
  
  "With the emitters drawing 1 amp and the LM317 drawing 1 amp, we would want about 1.4 volts drop, which would mean we would need a resistor of about 0.7 Ohms.  This wouldn't be too much different if the LM317 was drawing 1.5 amps as then the emitters would be drawing less, but the total current might be around 2 amps again.  Obviously 2 amps through a 2 Ohm resistor is 4 volts, so the current limit would kick in too soon and limit the current to less than 10 amps, probably a lot less"
  
  if emitter takes 1 amp then base current is 0.1A, then three base current add together and makes 0.3A.
  Assume VEB of power transistor =1V @ 1A, VE=(Vin-V_0.06)=20-(0.06*3)=19.82V then Base voltage VB should be 1V less, VB=18.82V
  

  
  Hi,
  
  Yes, I'm sorry the actual current would be 1/10 of the emitter current so if the emitter current was 1 amp the base current would be 0.1 amps and the total current 0.3 amps (not each base has 0.3 amps).  However, with 9 amps output each emitter would pass roughly 3 amps each, and 3/10 is 0.3 and the total would be close to 1 amp.
  This means if the current limit of the LMxxx device was 1 amp with 9 amps total through the emitters, that's 10 amps output, and that's 2 amps through resistor R1 approximately.
  
  If the current limit of the LMxxx device was 1.5 amps, then the emitter currents would total 8.5 amps, with total base currents 0.85 amps, which means current through R1 would be about 2.4 amps.
  
  If the current limit of the LMxxx device was 2 amps, then the emitter currents would total 8 amps, and that means about 0.8 amps into the bases, which is a total of 2.8 amps through the resistor R1. Thus R1 has to be sized based on something like that.
  
  Does this make more sense now?
  
  The main idea though is that the point of current limit is reached when the LMxxx device goes into current limit as that limits the current through the bases as well because the voltage drop across R1 will be limited.
  Because we are dealing with two base emitter drops though this could change significantly with temperature rise, and the temperature will rise.  This means it is a current limit but not a precision current limit at least until the whole circuit comes up to operating temperature and the ambient temperature is constant.
  
  I also recommended checking this circuit over very carefully because there were obvious errors.  Even with only 1 amp through that 2 Ohm resistor we see 2 watts, and the resistor is rated for 2 watts.  That means even with that unreal lower current the resistor would get way too hot.
• #39 Reply
  Posted by ommsiva on 13 Mar, 2024 06:39
• good morning to all,
  
  for past 2 months i was involved in studying power supply design and now I started building Lab power supply with LM317 and 337.
  Positive supply was adjustable from 1.2 to 28V, but negative was starting from -8 to -24v.
  
  Initially when i soldered and connected the circuit, my adjust pin of 337 was grounded and IC was heated, i have corrected that now and output was adjustable from -8 to -24V.
  
  My question,
  
  1) I have 7805 IC for powering peripheral circuit, after capacitive filtering the voltage was 35V.the output of 7805 is 10V instead of 5V, why?
  2) why my negative supply was from -8 to 24V?
  3) how to check LM337IC regulator?
  
  Thanks you all

  ---

  
• #40 Reply
  Posted by xavier60 on 13 Mar, 2024 07:27
• The outputs should be 27V and -27V with the values shown.
  Best way to check them is by function in a simple test circuit, https://uk.farnell.com/lm317-resistor-and-output-voltage-calculator
  If R2 is zero ohms, the output should regulate at close to 1.25V
  Note that the LM317 and LM337 pinouts are different.
  
  For the LM337, https://www.datasheets.com/de/reference-design/typical-application-circuit-for-lm337-1-5-a--adjustable-output--negative-voltage-regulator-125315
  
• #41 Reply
  Posted by ommsiva on 15 Mar, 2024 14:06
• Sir,
  
  Please suggest simple method of deriving High current negative adjustable regulator.
• #42 Reply
  Posted by xavier60 on 16 Mar, 2024 03:52
• The design in Reply #39 would be ok with current limiting added.
  Have you tested the LM337?
• #43 Reply
  Posted by ommsiva on 16 Mar, 2024 07:15
• Yes. Two times. LM337.
  
  
  Last time 22 ohm current sensing resistor had smoke.
  
  
  Any body can propose simple method of obtaining high current adjustable negative voltage
• #44 Reply
  Posted by MrAl on 17 Mar, 2024 15:57
• Quote from: ommsiva on 16 Mar, 2024 07:15

  Yes. Two times. LM337.
  
  
  Last time 22 ohm current sensing resistor had smoke.
  
  
  Any body can propose simple method of obtaining high current adjustable negative voltage
  

  
  Hi,
  
  You probably have to learn how to troubleshoot these circuits as well as build them.
  You use a meter and measure voltages and currents and try to figure out what is not right.
  Most of the time you have to just use a little power at first to test.  That would mean something like if you need 30 volts input you start with maybe 5 to 10 volts input and a light load, then try to figure out what currents are flowing and what voltages are present and see if anything looks wrong as you go.
  
  I say all this because you got the LM317 version going so you should be able to get the LM337 version going too.
  One of the big differences is the external transistors being used for the LM337 and the LM317.
  As you power up, compare the voltages and currents you measure in the LM317 circuit to the LM337 circuit and see what is different about the LM337 circuit, other than the polarities.
  
• #45 Reply
  Posted by ommsiva on 18 Mar, 2024 10:34
• Sir,
  
  please find the attached circuit, I connected this in bread board. The 120 ohm of 1Watt was fired two times. I cant find the reason for its breakdown. got frustrated after building this circuit.

  ---

  
• #46 Reply
  Posted by xavier60 on 18 Mar, 2024 10:49
• That circuit is correct. It should work properly if wired correctly and the LM337 and transistors aren't damaged. These parts can be unforgiving to wiring mistakes.
  Also, breadboard isnt good for this kind of circuitry. It's best to have the LM337 secured to a heatsink and the passive components soldered directly to it, including bypass capacitors to prevent possible oscillation.
   As suggested earlier, start simple. Wire up a test circuit with a new LM337 without the transistors and run it from a low voltage low power source.
  And show us some pictures.
• #47 Reply
  Posted by ommsiva on 18 Mar, 2024 16:19
• Sir,
  
  how much current can be obtained if I use 24v-0-24v , 10 ampere centre tapped transformer for positive and negative output each
  
  
  According to theory output would be maximum 23v@ 6ampere on positive and negative each regulated outputs.
• #48 Reply
  Posted by MrAl on 18 Mar, 2024 22:34
• Quote from: ommsiva on 18 Mar, 2024 16:19

  Sir,
  
  how much current can be obtained if I use 24v-0-24v , 10 ampere centre tapped transformer for positive and negative output each
  
  
  According to theory output would be maximum 23v@ 6ampere on positive and negative each regulated outputs.
  

  
  Hi,
  
  What theory is that?
  
  You have to know the Volt Ampere rating of the transformer.  It sounds like it could be a 240VA transformer but that's better obtained from the manufacturer.
  
  
• #49 Reply
  Posted by ommsiva on 19 Mar, 2024 00:55
• Sir,
  
  Rms power= 48*10=480va
  
  Average power=(.9*Vac)(.6*Iac)
  =(.9*24)(0.6*10)
  =21.6*6
  =129.6 watts
  On end of coil I could take=129.6 watts
  
  Overall I could take 260watt from 480va transformer.
• #50 Reply
  Posted by golden_labels on 19 Mar, 2024 01:23
• Even if LM337 or the transistors were wired wrong, there is a 2.5 kΩ resistor in the path. With 35 V directly across it, the current is limited to 14 mA. Power on this 120 Ω resistor is below 25 mW. Something is not adding up in that description.
  
  As for wiring: make sure LM337 is connected right. The component is very similar to LM317, but their pinout is different. TO-220 package for LM337 has input and output pins swapped compared to LM317.
• #51 Reply
  Posted by MrAl on 20 Mar, 2024 05:34
• Quote from: ommsiva on 19 Mar, 2024 00:55

  Sir,
  
  Rms power= 48*10=480va
  
  Average power=(.9*Vac)(.6*Iac)
  =(.9*24)(0.6*10)
  =21.6*6
  =129.6 watts
  On end of coil I could take=129.6 watts
  
  Overall I could take 260watt from 480va transformer.
  

  
  Hello again,
  
  Did you get that 480va from the calculation you did, or from the manufacturers data sheet?
  I mention this because you must get it from the data sheet you can't calculate this.  It's not just a matter of V*A because for some voltages it could have a different rating than for other voltages.  There's a chance this isn't the case for your transformer, but the data sheet is the tell-all of this specification.
  
• #52 Reply
  Posted by ommsiva on 20 Mar, 2024 13:29
• Sir,
  
  I got rms power by calculation. It is 24v-0-24v centre tapped transformer.  So I have taken peak to peak voltage 48V. Current rating of transformer is 10amps.
  
  Rms power = 48*10=480VA
  
  Average power for positive output = (0.9×24)*(0.6×10)=129 watts
  
  Average power for negative output =(0.9×24)*(0.6×10)=129 watts
  
  So from both lines i could take 260watt from 480voltampere.
  
  Is it correct sir?
• #53 Reply
  Posted by MrAl on 21 Mar, 2024 03:52
• Quote from: ommsiva on 20 Mar, 2024 13:29

  Sir,
  
  I got rms power by calculation. It is 24v-0-24v centre tapped transformer.  So I have taken peak to peak voltage 48V. Current rating of transformer is 10amps.
  
  Rms power = 48*10=480VA
  
  Average power for positive output = (0.9×24)*(0.6×10)=129 watts
  
  Average power for negative output =(0.9×24)*(0.6×10)=129 watts
  
  So from both lines i could take 260watt from 480voltampere.
  
  Is it correct sir?
  

  
  Hello again,
  
  What I was talking about was just the 480VA figure.
  Does the spec sheet tell you that the transformer is rated for 480VA?
  That's the only way to know for sure.
  For example, what if it is only 240VA.
  The DATA SHEET will tell you that for sure.
  Sometimes the VA rating will change with input voltage too and the only way you would know that is to look at the data sheet on the transformer.
  
• #54 Reply
  Posted by ommsiva on 22 Mar, 2024 15:09
• True sir, we don't have datasheet for transformer..
• #55 Reply
  Posted by Zero999 on 22 Mar, 2024 15:36
• How heavy is the transformer?
  
  What type is it i.e. toroid, E-core?
  
  That'll give you a good idea of the power rating.
• #56 Reply
  Posted by ommsiva on 23 Mar, 2024 07:05
• Sir,
  E core transformer,  weighing about 3.5 kilogram.
• #57 Reply
  Posted by Zero999 on 24 Mar, 2024 14:19
• It's probably around 240VA, so the maximum continuos current rating will be 5A into a purely resistive load, or around 3A from the DC side of the bridge rectifier.
  https://static.rapidonline.com/pdf/525323.pdf
  https://www.vigortronix.com/wp-content/uploads/2021/09/VTX-126-xxx-xxx-Chassis-Mains-D0003.pdf
• #58 Reply
  Posted by ommsiva on 26 Mar, 2024 10:20
• Sir,
  
  When I connect LM337 with out current boosting transistors, it works fine.
  
  After connecting current boosting transistors and turning on the circuit, the output goes to 30V and ic went to breakdown.
  
  This is same for lm317 section also.
  
  Now problem is with current boost transistor section. Even I cross checked the base , emitter junction of current boosting transistor.
  
  Suggest some solution to built high current bipolar power supply.
  
  Thank you

  ---

  
• #59 Reply
  Posted by xavier60 on 26 Mar, 2024 10:33
• I cant follow the wiring to check for mistakes. Even with no wiring mistakes, the wires are way too long and even if there are bypass capacitors, they wouldn't help because of the long wiring. Try to follow my previous advice.
  The power transistors are still good?
• #60 Reply
  Posted by iMo on 26 Mar, 2024 11:11
• Do not use solderless breadboards when messing with X Amps currents.
  The tiny contacts in there are not designed for such currents..
• #61 Reply
  Posted by xavier60 on 27 Mar, 2024 00:20
• I have ordered some parts on Aliexpress to test the negative regulator for myself.
  I have only ever tested LM317 positive designs.
• #62 Reply
  Posted by ommsiva on 27 Mar, 2024 12:47
• Sir,
  
  When there is a short between Collector and emitter of Power transistor 2N3055, I can consider the transistor went to breakdown?
• #63 Reply
  Posted by xavier60 on 27 Mar, 2024 12:55
• Quote from: ommsiva on 27 Mar, 2024 12:47

  Sir,
  
  When there is a short between Collector and emitter of Power transistor 2N3055, I can consider the transistor went to breakdown?
  

  Yes, but why did the output transistors overload? What kind of load was on the output?
• #64 Reply
  Posted by ommsiva on 27 Mar, 2024 13:04
• Sir,
  
  Automobile light 12V@60 Watts.
• #65 Reply
  Posted by xavier60 on 27 Mar, 2024 13:09
• Light bulbs have high inrush current. Was the voltage regulation working before the load was connected?
  The long wiring is likely adding to the problems.
• #66 Reply
  Posted by ommsiva on 27 Mar, 2024 13:12
• Sir,
  
  Only one time it went to 10V from 12V, After some time out become unregulated. This is consequence of breakdown on 2N3055.
  
  What is the proper load to test the circuit with out causing breakdown to Power transistor?
• #67 Reply
  Posted by xavier60 on 27 Mar, 2024 13:21
• Power resistors. I also use large rheostats. Before any more testing, the wiring needs to be shortened and bypass capacitors added.
  Damage can be caused by the circuit oscillating cussed by the long wires and no bypass capacitors.
• #68 Reply
  Posted by ommsiva on 27 Mar, 2024 13:24
• Sir,
  
  
  Sure. What is the optimum value of power resistor and its specification at 20V output. Current can be loaded up to 3A.
  
  Thank you.
• #69 Reply
  Posted by xavier60 on 28 Mar, 2024 03:32
• Quote from: ommsiva on 27 Mar, 2024 13:24

  Sir,
  
  
  Sure. What is the optimum value of power resistor and its specification at 20V output. Current can be loaded up to 3A.
  
  Thank you.
  

  Theoretically, 6.67Ω is needed. Just using 10W ceramic resistors, there are many ways of getting close to this. For example, parallelling 7 47Ω resistors gives 6.72Ω. The resistors need not be the same value as each other and need not be 10W. so long as the dissipation rating isn't greatly exceeded.   I used to glue ceramic resistors to an aluminum plate with RTV Silicone to increase their power handling.
  Do you know how to do voltage, resistance, current and power calculations?
• #70 Reply
  Posted by ommsiva on 28 Mar, 2024 04:41
• yes sir.
  
  
• #71 Reply
  Posted by ommsiva on 01 Apr, 2024 08:46
• Sir,
  
  Instead of voltage regulator , can we use TDA2050 opamp used for audio amplifier application, since it can source 5A current and it has protection towards short circuit ?
  
  why designers wont use this audio opamp for designing regulated power supplies?
• #72 Reply
  Posted by xavier60 on 01 Apr, 2024 09:24
• Quote from: ommsiva on 01 Apr, 2024 08:46

  Sir,
  
  Instead of voltage regulator , can we use TDA2050 opamp used for audio amplifier application, since it can source 5A current and it has protection towards short circuit ?
  
  why designers wont use this audio opamp for designing regulated power supplies?
  

  "Power dissipation at TCASE = 75 °C 25 W"
  At least 2 would be needed to satisfy dissipation rating, then current sharing becomes the next issue.
• #73 Reply
  Posted by ommsiva on 03 Apr, 2024 01:54
• Sir,
  
  LM338 regulator can provide 5Amp output, but it does not provide short circuit protection.
  
  how to make short circuit protection with LM338?
  
  when difference in voltage between input and output is more than 30, and if there is short at output then LM338 will go to breakdown.
• #74 Reply
  Posted by ommsiva on 03 Apr, 2024 14:27
• Sir,
  
  I got this circuit in net, I donot Know the terms cut, load.
  
  how this circuit is going to work during short circuit condition?

  ---

  
• #75 Reply
  Posted by xavier60 on 03 Apr, 2024 14:44
• That design has latching over current protection. The Cut switch triggers the protection circuit, disabling the PSU's output.
  The Load switch resets the protection, enabling the output.
  
• #76 Reply
  Posted by ommsiva on 03 Apr, 2024 14:50
• Sir,
  
  how this circuit is going to work during short circuit condition?
• #77 Reply
  Posted by xavier60 on 03 Apr, 2024 15:15
• When the current gets too high, the voltage across the 0R22 triggers the SCR to turn on, applying voltage across the relay's coil.
  The relay's NC contacts open, disconnecting the unregulated supply from the input to the regulation circuit.
• #78 Reply
  Posted by ommsiva on 03 Apr, 2024 15:20
• Sir,
  
  Whether this circuit is feasible for implementing short circuit protection?
  
  if yes i have a transformer of 24-0-24v @ 10 amps, if i put rectifier and filter i would get 34V and nearly 10 Ampere, What should be the specification of the relay?
  
  what is the specification of cut and load switch?
  
  Thank you.
• #79 Reply
  Posted by xavier60 on 03 Apr, 2024 15:40
• Quote from: ommsiva on 03 Apr, 2024 15:20

  Sir,
  
  Whether this circuit is feasible for implementing short circuit protection?
  
  if yes i have a transformer of 24-0-24v @ 10 amps, if i put rectifier and filter i would get 34V and nearly 10 Ampere, What should be the specification of the relay?
  
  what is the specification of cut and load switch?
  
  Thank you.
  

  Because there is no instantaneous current limiting, short circuit current will be very high. How high? I dont know.
  If the relay is too small, the contacts might weld. I'd use a 40A automotive relay.
  Also because of the high instantaneous current, the output transistors might fail.
• #80 Reply
  Posted by ommsiva on 03 Apr, 2024 16:11
• Sir,
  
  What would be the best circuit to implement short circuit protection for LM317 or LM338?
• #81 Reply
  Posted by xavier60 on 03 Apr, 2024 22:42
• Quote from: ommsiva on 03 Apr, 2024 16:11

  Sir,
  
  What would be the best circuit to implement short circuit protection for LM317 or LM338?
  

  I think something as simple as possible so it's easy to build and make work.
  From a previous post with some corrections, https://www.eevblog.com/forum/beginners/how-current-limitation-is-happening-in-the-circuit/msg5375117/#msg5375117
  The parts should be here soon so I can test it for myself.
• #82 Reply
  Posted by xavier60 on 08 Apr, 2024 08:36
• It definitely works but not in an altogether nice way.
  Because the current limiting has control only of the power transistors but not the LM337, the LM337 goes to its full current when there is an overload. It can get hot enough to go into thermal limiting. Maybe a good thing in a way because it will reduce the dissipation at the power transistors.
  It's also difficult to precisely set the current limit.
   The design is ok if brief overloads are expected.
  

  ---

  

  
• #83 Reply
  Posted by ommsiva on 08 Apr, 2024 17:31
• Sir,
  
  What is the function of Q1 transistor?
  
  what should be the voltage across collector to emitter in normal operation and short circuit condition in Q1?
  
  why there is 47 ohm resistance connected to the base of Q1?
  
  Q2 and Q3 are current boost transistor
• #84 Reply
  Posted by xavier60 on 08 Apr, 2024 23:42
• Quote from: ommsiva on 08 Apr, 2024 17:31

  Sir,
  
  What is the function of Q1 transistor?
  
  what should be the voltage across collector to emitter in normal operation and short circuit condition in Q1?
  
  why there is 47 ohm resistance connected to the base of Q1?
  
  Q2 and Q3 are current boost transistor
  

  Q1 is for current limiting. R2 and R3 are for current sharing of the output transistors and also serve as Current Sensing. R6 and R7 combine the voltage from the CS resistors and apply it to the Base of Q1.
  At 3.2A, the Vce of Q1 is 1.5V. It stays the same when the output is shorted because it was already limiting the current through the output transisitors, Q2 ans Q3.
   You don't need to use exactly the sane parts to test the design. If you have 0.47Ω resistors, use them in place of the 0.56Ω.
• #85 Reply
  Posted by ommsiva on 09 Apr, 2024 09:31
• Sir,
  
  I found this circuit on web. Whether this will perform as specified in the circuit.
  
  please advice me.

  ---

  
• #86 Reply
  Posted by xavier60 on 10 Apr, 2024 11:27
• Quote from: ommsiva on 09 Apr, 2024 09:31

  Sir,
  
  I found this circuit on web. Whether this will perform as specified in the circuit.
  
  please advice me.
  

  I'm not certain. I'd need to build it but I don't have enough time.
• #87 Reply
  Posted by ommsiva on 10 Apr, 2024 14:26
• ok sir
• #88 Reply
  Posted by ommsiva on 11 Apr, 2024 08:30
• Sir,
  
  Check this circuit and comment on the design.
  
  Thank you.

  ---

  
• #89 Reply
  Posted by xavier60 on 11 Apr, 2024 09:16
• It will current limit so long as the output doesn't need to go below 1.25V such as when the output is short circuited.
• #90 Reply
  Posted by xavier60 on 11 Apr, 2024 09:25
• Correction. With a short on the output, the current can be no less than 5.7A. Does this suit your needs?
• #91 Reply
  Posted by ommsiva on 11 Apr, 2024 09:40
• Sir,
  
  
  At short circuit,  power dissipation = 5.7amps* 1.25V. And this will not be larger than 11 watts.
  
  
  Which could be easily dissipated by MJ15004 with heatsink.
  
  Why it is 5.7 amps at short at the output?
• #92 Reply
  Posted by xavier60 on 11 Apr, 2024 09:48
• You need to multiply the 5.7A by whatever the voltage is going to be across the power transistors.
  Also, put  a resistor in series with the Base of the current limiting transistor to protect it, like 470Ω.
• #93 Reply
  Posted by xavier60 on 11 Apr, 2024 09:52
• The LM317 can regulate only down to 1.25V which will be applied across the 0.22Ω resistor when the output is short circuited.
• #94 Reply
  Posted by ommsiva on 14 Apr, 2024 12:36
• Sir,
  
  I found this circuit in web and calculated node voltage.
  
  My doubt is
  
  1)when i move the p1 pot up to top position my node voltage will be 0.79V , which is enough to turn on my internal current limiting transistor. Then why i need those parallel 0.22ohm (4 nos)?
  
  2)which will be the negative terminal of power supply? either Right end or Left end(where filter capacitor were connected) of current sensing resistors(4x 0.22 ohm resistors). What is the technique employed?
  
  3) what will happen if there is no 4x0.22 ohm resistor?
  
  
  Thank you

  ---

  
• #95 Reply
  Posted by MrAl on 15 Apr, 2024 13:32
• Quote from: ommsiva on 14 Apr, 2024 12:36

  Sir,
  
  I found this circuit in web and calculated node voltage.
  
  My doubt is
  
  1)when i move the p1 pot up to top position my node voltage will be 0.79V , which is enough to turn on my internal current limiting transistor. Then why i need those parallel 0.22ohm (4 nos)?
  
  2)which will be the negative terminal of power supply? either Right end or Left end(where filter capacitor were connected) of current sensing resistors(4x 0.22 ohm resistors). What is the technique employed?
  
  3) what will happen if there is no 4x0.22 ohm resistor?
  
  
  Thank you
  

  
  [1]  The four 0.22 Ohm resistors in parallel create a resistor of value 0.055 Ohms, which senses current.  You need that to sense current.
  [2]  The output of the power supply is on the right (-) terminal.  The input is on the left.  The two can not be made into a single common.
  [3]   With no 0.22 Ohm resistors there is no current sense, hence there will be no current limit at all.
  
  You can analyze this by looking at the voltage divider created by the pot and the associated resistors, along with R5 the current sense resistors.
  If you look at all the top side resistors in series and call them RA, and all the bottom side resistors and call them RB, then the voltage at the base of the lower NPN (voltage referenced to the output common terminal) would be:
  Vbase=Vref*RB/(RA+RB)
  
  but it is also biased by the voltage across R5 and the output current because the emitter is tied to the input common not the output common. This means the output current affects Vbe of the lower NPN as:
  Vbe=Vref*RB/(RA+RB)+Io*R5
  
  where Io is the output current.
  
  Replacing RA and RB with the actual resistors we end up with:
  Vbe=(R5*(Io*R6+Io*R2+Io*Rp)+Vref*(R6+Rd))/(R6+R2+Rp)
  
  and here Rp is the total value of the pot (500 Ohms in the schematic) and Rd is the lower resistance part of the pot after adjustment.
  With the values substituted in for the variables and setting Vbe=0.7 volts for an approximation, we get current limit as follows.
  With the pot set all the way down, the current limit is around 11 amps.
  With the pot set all the way up, the theoretical current limit is actually about -1 amps, which is probably not possible, so the output current would probably be zero at that point.  If we adjust it slightly down from top, we get a theoretical zero amps output.
  
  Note Vbe may be slightly different than 0.7 volts, and changes with temperature (heating up of the IC chip and ambient temperature).
  
  The main point though is that R5 is definitely needed to sense current or you can not get current limit.  If you short it out, no current limit, if you open it up, no output at all except maybe some leakage current.  Either of these extremes is not an option that will work.
  
• #96 Reply
  Posted by ommsiva on 16 Apr, 2024 03:53
• Sir,
  
  thank you , what i understand is
  
  V_BE=V_B-V_E
  
  V_B=V_ref * RB/(RA+RB)
  
  V_E =I_oR₅
  
  R_A =R2+Partial resistance from p1
  
  R_B=R1+R3+R6+Partial resistance from p1
  
  I cannot correleate and understand .these equations
  
  Vbe=Vref*RB/(RA+RB)+Io*R5
  
  Rectify me.
• #97 Reply
  Posted by Zero999 on 16 Apr, 2024 07:59
• Since this topic has drifted.
  
  Please start from the beginning. What are you trying to do?
• #98 Reply
  Posted by MrAl on 16 Apr, 2024 11:20
• Quote from: ommsiva on 16 Apr, 2024 03:53

  Sir,
  
  thank you , what i understand is
  
  V_BE=V_B-V_E
  
  V_B=V_ref * RB/(RA+RB)
  
  V_E =I_oR₅
  
  R_A =R2+Partial resistance from p1
  
  R_B=R1+R3+R6+Partial resistance from p1
  
  I cannot correleate and understand .these equations
  
  Vbe=Vref*RB/(RA+RB)+Io*R5
  
  Rectify me.
  

  
  The total resistance of the pot P1 is Rp, and the resistance of the upper part of the pot is Ru, the lower part is Rd.
  
  The voltage divider formula plus the voltage due to the output current times the sense resistor R5:
  Vbe=(Vref*RB)/(RB+RA)+Io*R5
  
  Since RA=R2+Ru we have:
  Vbe=(Vref*RB)/(RB+R2+Ru)+Io*R5
  
  and since RB=Rd+R6 we have:
  Vbe=(Vref*(R6+Rd))/(R6+R2+Ru+Rd)+Io*R5
  
  and since Ru=Rp-Rd we have:
  Vbe=(Vref*(R6+Rd))/(R6+R2+Rp)+Io*R5
  
  This last result is the same as that given previously just factored differently.
  The formula shown on the updated schematic is the same also, also factored differently.
  
  Note R1 and R3 are not included in the calculation because the additional current Ix through R5 is so small.
  Ix=Vref/(R1+R3)
  Vx=R5*Ix
  If you like you can add that voltage to the calculation for Vbe also, although since Ix is so small (less than 700ua) the additional voltage will be very, very small (less than 40uv).
  There is also a small contribution from P2 and R10 which will also be small.
  
  The updated schematic is attached.
  
  
  

  ---

  
• #99 Reply
  Posted by ommsiva on 16 Apr, 2024 13:05
• Sir,
  
  1)Normally Base emitter voltage V_BE=V_B-V_E , But you have taken V_BE=V_B+V_E.
  What is the reason for that?
  
  2)What is the contribution of these terms interms of current limiting from these equation, Vbe=(Vref*(R6+Rd))/(R6+R2+Rp)-Io*R5 ?
  
  for example
  when load current is 10 amps,V_R5=10*0.055=0.55V, then what is the contribution of Vref*(R6+Rd))/(R6+R2+Rp) term?
  
                             the first term should be 1.25V (0.7V+0.55V) and second term should be 0.55V. is that correct?
  
  Thank you
• #100 Reply
  Posted by MrAl on 16 Apr, 2024 18:39
• Quote from: ommsiva on 16 Apr, 2024 13:05

  Sir,
  
  1)Normally Base emitter voltage V_BE=V_B-V_E , But you have taken V_BE=V_B+V_E.
  What is the reason for that?
  
  2)What is the contribution of these terms interms of current limiting from these equation, Vbe=(Vref*(R6+Rd))/(R6+R2+Rp)-Io*R5 ?
  
  for example
  when load current is 10 amps,V_R5=10*0.055=0.55V, then what is the contribution of Vref*(R6+Rd))/(R6+R2+Rp) term?
  
                             the first term should be 1.25V (0.7V+0.55V) and second term should be 0.55V. is that correct?
  
  Thank you
  

  
  
  Hello again,
  
  [1]
  I do not see Vb+Ve anywhere in my replies so you will have to explain what gave you that idea.
  The Vbe voltage is obviously the Vb minus the Ve, but those are implied in my equations.  If you calculate Vbe from my equations you will see that Ve does not have to be mentioned because Ve is at absolute ground (0v) and so calculating Vbe would be the same as calculating Vb because Vb-0v is Vbe.
  If you still think there is a mistake, just explain how you got the idea that I was adding Vb and Ve and I'll look it over.
  
  [2]
  The contribution from the other resistors and Vref come in as an addition to the voltage across the current sense resistor, as can be seen in the final formula.  In other words, we are not just sensing Io (Iout) we are biasing that with a small voltage in order to get an adjustment range using a rather larger value potentiometer.  Had we just used a potentiometer, it would have to be on the order of 0.1 Ohms at some 10 watts or something.  By using a biasing voltage, we can use a larger value potentiometer which is more common and more easily obtained.  We might even be able to up that to 1k instead of 500 Ohms if we change R2 to around 12k or something like that, and R6 up in value as well.  I'll leave this idea for later though, and of course all this has to be tested on the bench.
  When we add that bias however, there is the chance that the current regulation will become less stable, so that has to be tested in real life on the bench.
  As you said, the output current Io at 10 amps creates a voltage of 0.55v across the sense resistors, and so the bias only has to be about 0.15 volts (roughly) in order for the current limit to start to kick in (with Vbe 0.7 volts).  Had we adjusted the bias voltage to 0.425 volts with P1, then Vbe would reach 0.7 volts with only 0.275 volts across the sense resistors, meaning that the current would now be limited to just 5 amps instead of 10 amps.  So you can see the benefit of using a bias voltage.
  To get that action with a straight potentiometer, we'd have to use a pot that had total resistance of maybe 1 Ohms and adjust it to get 5 amps.
  There are other ways to try too though, such as with a 1 Ohm power resistor in parallel with a 100 Ohm (or so) potentiometer, but we lose more power that way because the power resistor has to be larger in order to get to the lower end of the adjustment range.
• #101 Reply
  Posted by ommsiva on 17 Apr, 2024 00:18
• Dear sir,
  
            1)now i understood that voltage developed across current sense resistor is applied to the base and emitter is at input common.
  
            2)"To get that action with a straight potentiometer, we'd have to use a pot that had total resistance of maybe 1 Ohms and adjust it to get 5 amps ", similar to the first circuit of this thread.
           
            3) But this 10Amp Load current will develop 0.55V at the bottom end of R6 and already we have calculated 0.13V top of R6 , P1 top will be .7V respectively.   How this 0.55V will affect my initial bias arrangement done using Vref ? then i have to apply Superposition Theorem?
  
  thanks a lot.
• #102 Reply
  Posted by Andy Chee on 17 Apr, 2024 05:49
• Quote from: Zero999 on 16 Apr, 2024 07:59

  Since this topic has drifted.
  
  Please start from the beginning. What are you trying to do?
  

  
  From what I can tell, they are a 3rd year university student, attempting to build an adjustable current limited laboratory power supply (or possibly a CC-CV lithium battery charger).
  
  However as a university project, they need to document all the fundamental theory and formulas that underpin its design.
  
  In a sense, this has the feel of a "homework" type question, but applied to a final year graduation/keystone project!
  
  
• #103 Reply
  Posted by MrAl on 17 Apr, 2024 10:04
• Quote from: ommsiva on 17 Apr, 2024 00:18

  Dear sir,
  
            1)now i understood that voltage developed across current sense resistor is applied to the base and emitter is at input common.
  
            2)"To get that action with a straight potentiometer, we'd have to use a pot that had total resistance of maybe 1 Ohms and adjust it to get 5 amps ", similar to the first circuit of this thread.
           
            3) But this 10Amp Load current will develop 0.55V at the bottom end of R6 and already we have calculated 0.13V top of R6 , P1 top will be .7V respectively.   How this 0.55V will affect my initial bias arrangement done using Vref ? then i have to apply Superposition Theorem?
  
  thanks a lot.
  

  
  Hi,
  
  I guess you could say that, but I thought it was simpler to just realize that the voltage at the bottom of R6 depends on the output current because of the drop across R5, and that means that the top of R6 changes as a result of the output current to such that it will increase as the output current increases.
  But yes, you can look at it as a superposition too because if we consider the voltage across R6 separately from the voltage across R5, then we have to add them to get the final voltage at the top of R6 and add that to the voltage across the bottom section of the potentiometer to get the voltage at the arm of the potentiometer and thus the Vbe voltage which is a major part of figuring out what the current limit set points will be.
  
  We also have the small contributions from other parts of the circuit which we ignore because the currents are so small as compared to the output current, but we could add them into the equation as well if desired.
  
• #104 Reply
  Posted by ommsiva on 27 Apr, 2024 07:17
• Hello All,
  
  I started constructing prototype of the below circuit, But 68ohm,2Watts resistance went to breakdown.
  again i changed to 470ohm,2Watts , again it started smoking.
  
  
  I also changed the MJE2955, but 470 ohm started smoking, What is the problem in the circuit?
  
  How to find my transistor has went to breakdown?
  
  Thank you all.

  ---

  
• #105 Reply
  Posted by MrAl on 28 Apr, 2024 16:05
• Quote from: ommsiva on 27 Apr, 2024 07:17

  Hello All,
  
  I started constructing prototype of the below circuit, But 68ohm,2Watts resistance went to breakdown.
  again i changed to 470ohm,2Watts , again it started smoking.
  
  
  I also changed the MJE2955, but 470 ohm started smoking, What is the problem in the circuit?
  
  How to find my transistor has went to breakdown?
  
  Thank you all.
  

  
  
  Hello there,
  
  You are probably going to have to learn to troubleshoot these circuits yourself.  That means making measurements with different loads, and possibly running the input voltage up a little at a time to see when the power dissipation starts to get too high in the resistor(s).  You keep having this same problem with just about every circuit you work on like this.  A resistor smokes in every circuit you posted so far.
  To find out the problem you often have to take voltage and/or current measurements before anything smokes, and that means possibly with a light load and slowly increasing that load.  If it happens even with light load, then you might have to turn the input voltage up slowly until the problem shows itself but does not blow anything out.
  
  So to start see if it runs with no load.  You don't seem to say what the load is when this happens, or even if there is a load or not.
  
• #106 Reply
  Posted by Xena E on 28 Apr, 2024 22:40
• Quote from: ommsiva on 27 Apr, 2024 07:17

  Hello All,
  
  I started constructing prototype of the below circuit, But 68ohm,2Watts resistance went to breakdown.
  again i changed to 470ohm, 2Watts , again it started smoking.
  
  
  I also changed the MJE2955, but 470 ohm started smoking, What is the problem in the circuit?
  
  How to find my transistor has went to breakdown?
  
  Thank you all.
  

  
  Incredible!
  
  Taking 470 ohms as the value of the resistor, you have to have over 30 Volts across it to even max out its design dissipation of 2 Watts.
  
  Now, assuming that the base emitter junctions of the pass transistors are intact that means that the output would have to be shorted to ground and in that case almost all of that hypothetical voltage has to appear across the 0.1 Ohm emitter degeneration resistors, that will cause them to try to dissipate over 8.5kW each.
  
  So as this isn't possible, (the transformer isn't that big is it?), it's either ignorance or Bullshit.
  
  The circuit diagram as drawn won't work, in fact if the practical circuit is wired as drawn it would be the transformer that has smoke coming out of it, (you show the output of the rectifier on the 5.1Volt supply shorted to ground for starters).
  
  If you really want help you need to furnish facts and details.
  
  Regards,
  Xena.
  
  
  
  
  
  

  ---

  
• #107 Reply
  Posted by ommsiva on 29 Apr, 2024 00:49
• It is not an auxiliary coil that is used for deriving 5V for volt amp meter.
  
  
  From the main transformer after full wave rectification,  I have used it.
  
  I want to derive variable power supply up to 30V and current limiting adjustable up to 5 amp.
• #108 Reply
  Posted by Xena E on 29 Apr, 2024 07:01
• Quote from: ommsiva on 29 Apr, 2024 00:49

  It is not an auxiliary coil that is used for deriving 5V for volt amp meter.
  
  
  From the main transformer after full wave rectification,  I have used it.
  

  
   
  
  No, you haven't. Not connected like that, its a dead short.
  
  If you want help, please provide detailed photographs of your work so far. If you are a student, a tutor would throw out what you have provided so far as being inadequate.
  
  

  Quote from: ommsiva on 29 Apr, 2024 00:49

  I want to derive variable power supply up to 30V and current limiting adjustable up to 5 amp.
  

  
  The regulator chip you are using is ubiquitous, why not just copy an application note from one of the manufacturers?
  
  I repeat, if you want help post photos and details.
  
  Regards,
  Xena.
• #109 Reply
  Posted by ommsiva on 29 Apr, 2024 07:28
• MAM,
  
  PLEASE FIND THE ATTACHMENT

  ---

  
• #110 Reply
  Posted by Xena E on 29 Apr, 2024 10:08
• Is that stripboard?
  
   
  
  Where's the rest of it?
  
  (Transformer, load, volt amp meter, pass devices, as you have it connected).
  
  I don't want to be rude, however I or anyone else here haven't got a frigging crystal ball! If a circuit is burning up components, there is either something fundamentally wrong, with the design or you have mis-connected the circuit.
  
  Vague comments about things burning up and asking what's wrong as if we're at the bench with you is a waste of time.
  
  OK, here's a stab in the dark. If I am to understand from your previous post that the meter you are using is connected to that 5.1V stabiliser and that is being powered from the same transformer winding/rectifier as the main circuit....
  
  

  Quote

  It is not an auxiliary coil that is used for deriving 5V for volt amp meter...
  ...From the main transformer after full wave rectification,  I have used it.

  
  ...if there is no form of galvanic isolation between the two there is a possible source of your burning up. The circuit as drawn does not support that though...
  
  If you want something that just works buy one. If this is a project that demonstrates your skill level you have already achieved that.
  
  REGARDS,
  XENA.
  
• #111 Reply
  Posted by MrAl on 29 Apr, 2024 11:20
• Quote from: ommsiva on 29 Apr, 2024 00:49

  It is not an auxiliary coil that is used for deriving 5V for volt amp meter.
  
  
  From the main transformer after full wave rectification,  I have used it.
  
  I want to derive variable power supply up to 30V and current limiting adjustable up to 5 amp.
  

  
  
  Here is a cleaned up drawing.  You should try to post drawings that are very clear so people can help easier.
  
  As mentioned by Xena, there is a short in the diagram indicated here by a red "X".
  
  Also, you seem to be suggesting that there is only one transformer.  In the new diagram the way you have it connected, are the transformers T1 and T2 actually two different transformers or are they the same?  Alternately, is T2 just a separate winding on T1 perhaps?
  
  

  ---

  
• #112 Reply
  Posted by ommsiva on 29 Apr, 2024 11:41
• Hi to all,
  
  Mam,
  
  Its my mistake. now I have attached, and they contain
  
  1) 6 number of current sharing series pass transistor.
  
  2) two potentiometer, one for voltage and other for current
  
  3) two fuse, one for main 30V adjustable @ 5 amps and other for 5V.
  
  MY thanks to good hearts present here and  name a few to mention their kindness
  
  1) Mr Xavier who himself has bought LM337 and Power transistor(build circuits and given results) and shared his design ideas for building High current Negative regulator and given valuable feedback for the circuits.
  
  2) Mr AI ,my Guru who always inspired me by his ideas and  way of thinking.
  
  Thank you all.

  ---

  

  

  

  
• #113 Reply
  Posted by Xena E on 29 Apr, 2024 14:48
• I'm very sorry Ommsiva your pictures seem deliberately not to show any detail, at all.
  
  Pictures of the tops of T03 transistors may be useful in identifying the type numbers but don't convey anything about the connections made to them.
  
  Two front panel photos ... of similar value.
  
  A general internal shot of a disassembled rats nest of wiring.
  
  All of the above at a resolution such that even if it were tidy construction and complete it couldn't be of any less use in assisting with your query.
  
  As you seem to not like answering direct questions about your project, then the purpose of this thread is clearly intended as a piss-take of the kind people who have tried to help you.
  
  Have a nice life.
  
  X
  
  
• #114 Reply
  Posted by ommsiva on 29 Apr, 2024 15:29
• Mam,
  
  Sorry,   absolutely you are correct.  It's my mistake, each picture were nearly 6mb and our site allows maximum of 8mb totally for a reply.so i reduced the image size .Now I have disconnected power transistor from heat sink out of frustration.
  
• #115 Reply
  Posted by MrAl on 29 Apr, 2024 18:35
• Quote from: ommsiva on 29 Apr, 2024 11:41

  Hi to all,
  
  Mam,
  
  Its my mistake. now I have attached, and they contain
  
  1) 6 number of current sharing series pass transistor.
  
  2) two potentiometer, one for voltage and other for current
  
  3) two fuse, one for main 30V adjustable @ 5 amps and other for 5V.
  
  MY thanks to good hearts present here and  name a few to mention their kindness
  
  1) Mr Xavier who himself has bought LM337 and Power transistor(build circuits and given results) and shared his design ideas for building High current Negative regulator and given valuable feedback for the circuits.
  
  2) Mr AI ,my Guru who always inspired me by his ideas and  way of thinking.
  
  Thank you all.
  

  
  Hi,
  
  I meant to ask, was this circuit working before this or did you just build it for the first time?
  
• #116 Reply
  Posted by ommsiva on 30 Apr, 2024 01:09
• Sir,
  
  Second time.
  
  It was working well with 0.33 ohm 10watt.
  
  Now 470 ohm got smoke.
  
  
  Thank you
• #117 Reply
  Posted by xavier60 on 30 Apr, 2024 01:19
• Quote from: ommsiva on 30 Apr, 2024 01:09

  Sir,
  
  Second time.
  
  It was working well with 0.33 ohm 10watt.
  
  Now 470 ohm got smoke.
  
  
  Thank you
  

  Is the 470Ω the resistor in parallel with the CS resistor?
  The 0.33Ω resistor you mention is the CS resistor?
  What load were you testing with?
• #118 Reply
  Posted by ommsiva on 30 Apr, 2024 01:39
• Sir,
  
  I will check whether they are parallel.
  
  i had 10ohm 50watt. i had an idea of keeping 20V output and check.
  
  thank you.
• #119 Reply
  Posted by MrAl on 30 Apr, 2024 10:16
• Quote from: ommsiva on 30 Apr, 2024 01:09

  Sir,
  
  Second time.
  
  It was working well with 0.33 ohm 10watt.
  
  Now 470 ohm got smoke.
  
  
  Thank you
  

  
  Hi,
  
  If you are going to change values it would be best to update the schematic.  Please use the cleaner diagram I posted so we can read it better.
  
• #120 Reply
  Posted by MrAl on 30 Apr, 2024 21:37
• Hello again,
  
  Here is an updated schematic with some of the part numbers enumerated so it would be easier to discuss this circuit.
  
  There are also two test conditions, with the values shown in green.  The pots are shown to have a fixed value for the two tests.
  The two tests are with RL=4 and with RL=1.8, nothing else is changed.
  The results are show on the diagram.
  When RL=4 the circuit is not in current limit and puts out the set voltage level around 20 volts or so, and about 5 amps out.
  When RL=1.8 the circuit is just starting to current limit so the output voltage level goes down a little to 18.6 volts and the current goes up to the set limit point set by R9 and R10.  The limit is set to be around 10 amps.
  
  This means the circuit looks like it works OK so maybe it is a simpler problem that came up such as a blown out part or bad or wrong connection somewhere.  Note however that I did not check the LM317 part of the circuit.
  Oh there is one problem, the 1000uf cap has to be made a lot bigger.
  
  See the attached diagram for details.
  
  Added later:
  About that 1000uf cap, even 10000uf (10 times higher than 1000uf) might not be enough.  There would be a lot of ripple voltage with 10 amps output.  The problem is, if the ripple is too high then you need a higher input voltage in order to get a smooth output voltage of a fairly high value like 30 volts DC.  You can not go too high because of the limitations of the 723, so the burden becomes making the input filter capacitor value large enough.  20000uf would be better for example.
  
  

  ---

  
• #121 Reply
  Posted by xavier60 on 02 May, 2024 11:55
• Quote from: ommsiva on 30 Apr, 2024 01:39

  Sir,
  
  I will check whether they are parallel.
  
  i had 10ohm 50watt. i had an idea of keeping 20V output and check.
  
  thank you.
  

  Also include what test you might have in your profile or signature. It will save anyone having to ask.
• #122 Reply
  Posted by Zero999 on 03 May, 2024 17:41
• I admit. I don't have time to read the entire thread in detail.
  
  Some notes about the '723
  
  The op-amp only works with its inputs 2V above the negative rail. This means that, in the usual configurations, it will only work down to a minimum output voltage of 2V. There are two possible ways round this:
  
  1) Use a negative supply of at least 2V. This adds some complexity, but if it's run off a transformer, a simple voltage doubler circuit will do. Another downside is it effectively increases the total supply voltage to the '723 by 2V.
  
  2) Use a potential divider to scale the control voltage, so it starts above 2V and design the op-amp circuit so it subtracts 2V off the input. This is more complicated, but doesn't increase the supply voltage to the '723.
  
  The current limit is approximate because it depends on the base-emitter voltage of a transistor. It's only meant to protect the circuit, not as a constant current supply. The solution is to add an op-amp error amplifier and current sense resistor.
  
  In another thread I sketched out a design for accurate current limiting an operation with an output voltage down to 5V. It's not quite finished and requires some frequency compensation, to prevent oscillation.
  
  https://www.eevblog.com/forum/projects/lm-723-opamp-addon/msg5403488/#msg5403488
  
  
• #123 Reply
  Posted by MrAl on 05 May, 2024 20:44
• Here are some charts that illustrate the effect of the capacitor value and the total input impedance Rs.
  
  
  
  All of these are with an input sine wave peak of 30 volts and two diode drops 1.15 volts each,
  frequency was 50Hz, Rload=3 Ohms.
  
  Units: Volts, Ohms, Farads.
  
  
  C=0.001000
  Rs=0.00  Vmax=19.00  Vmin= 6.60  Vpp=12.398
  Rs=0.10  Vmax=18.38  Vmin= 6.60  Vpp=11.782
  Rs=0.20  Vmax=17.78  Vmin= 6.58  Vpp=11.200
  Rs=0.30  Vmax=17.21  Vmin= 6.56  Vpp=10.651
  Rs=0.40  Vmax=16.66  Vmin= 6.53  Vpp=10.134
  Rs=0.50  Vmax=16.14  Vmin= 6.49  Vpp= 9.648
  Rs=0.60  Vmax=15.64  Vmin= 6.45  Vpp= 9.191
  Rs=0.70  Vmax=15.17  Vmin= 6.40  Vpp= 8.762
  Rs=0.80  Vmax=14.71  Vmin= 6.35  Vpp= 8.358
  Rs=0.90  Vmax=14.28  Vmin= 6.30  Vpp= 7.979
  Rs=1.00  Vmax=13.87  Vmin= 6.25  Vpp= 7.623
  
  C=0.005000
  Rs=0.00  Vmax=27.10  Vmin=17.55  Vpp= 9.550
  Rs=0.10  Vmax=25.92  Vmin=17.37  Vpp= 8.560
  Rs=0.20  Vmax=24.37  Vmin=16.88  Vpp= 7.489
  Rs=0.30  Vmax=22.64  Vmin=16.19  Vpp= 6.454
  Rs=0.40  Vmax=20.91  Vmin=15.39  Vpp= 5.521
  Rs=0.50  Vmax=19.27  Vmin=14.55  Vpp= 4.714
  Rs=0.60  Vmax=17.76  Vmin=13.72  Vpp= 4.034
  Rs=0.70  Vmax=16.40  Vmin=12.93  Vpp= 3.467
  Rs=0.80  Vmax=15.19  Vmin=12.19  Vpp= 2.997
  Rs=0.90  Vmax=14.11  Vmin=11.50  Vpp= 2.607
  Rs=1.00  Vmax=13.15  Vmin=10.87  Vpp= 2.282
  
  C=0.010000
  Rs=0.00  Vmax=27.55  Vmin=21.46  Vpp= 6.082
  Rs=0.10  Vmax=25.50  Vmin=20.61  Vpp= 4.896
  Rs=0.20  Vmax=22.25  Vmin=18.64  Vpp= 3.614
  Rs=0.30  Vmax=19.02  Vmin=16.41  Vpp= 2.609
  Rs=0.40  Vmax=16.28  Vmin=14.37  Vpp= 1.906
  Rs=0.50  Vmax=14.08  Vmin=12.65  Vpp= 1.427
  Rs=0.60  Vmax=12.33  Vmin=11.23  Vpp= 1.097
  Rs=0.70  Vmax=10.92  Vmin=10.06  Vpp= 0.865
  Rs=0.80  Vmax= 9.79  Vmin= 9.09  Vpp= 0.696
  Rs=0.90  Vmax= 8.85  Vmin= 8.28  Vpp= 0.571
  Rs=1.00  Vmax= 8.07  Vmin= 7.59  Vpp= 0.476
  
  C=0.015000
  Rs=0.00  Vmax=27.63  Vmin=23.17  Vpp= 4.464
  Rs=0.10  Vmax=24.33  Vmin=21.20  Vpp= 3.126
  Rs=0.20  Vmax=19.41  Vmin=17.51  Vpp= 1.906
  Rs=0.30  Vmax=15.43  Vmin=14.24  Vpp= 1.186
  Rs=0.40  Vmax=12.56  Vmin=11.78  Vpp= 0.783
  Rs=0.50  Vmax=10.51  Vmin= 9.96  Vpp= 0.548
  Rs=0.60  Vmax= 9.00  Vmin= 8.59  Vpp= 0.402
  Rs=0.70  Vmax= 7.85  Vmin= 7.54  Vpp= 0.306
  Rs=0.80  Vmax= 6.95  Vmin= 6.71  Vpp= 0.241
  Rs=0.90  Vmax= 6.23  Vmin= 6.04  Vpp= 0.194
  Rs=1.00  Vmax= 5.64  Vmin= 5.48  Vpp= 0.159
  
  C=0.020000
  Rs=0.00  Vmax=27.66  Vmin=24.13  Vpp= 3.532
  Rs=0.10  Vmax=22.87  Vmin=20.77  Vpp= 2.104
  Rs=0.20  Vmax=16.78  Vmin=15.70  Vpp= 1.080
  Rs=0.30  Vmax=12.67  Vmin=12.07  Vpp= 0.608
  Rs=0.40  Vmax=10.03  Vmin= 9.65  Vpp= 0.379
  Rs=0.50  Vmax= 8.25  Vmin= 8.00  Vpp= 0.257
  Rs=0.60  Vmax= 6.99  Vmin= 6.81  Vpp= 0.184
  Rs=0.70  Vmax= 6.06  Vmin= 5.92  Vpp= 0.138
  Rs=0.80  Vmax= 5.34  Vmin= 5.23  Vpp= 0.108
  Rs=0.90  Vmax= 4.77  Vmin= 4.68  Vpp= 0.086
  Rs=1.00  Vmax= 4.31  Vmin= 4.24  Vpp= 0.070
  
  C=0.025000
  Rs=0.00  Vmax=27.68  Vmin=24.75  Vpp= 2.926
  Rs=0.10  Vmax=21.32  Vmin=19.86  Vpp= 1.465
  Rs=0.20  Vmax=14.58  Vmin=13.92  Vpp= 0.653
  Rs=0.30  Vmax=10.64  Vmin=10.30  Vpp= 0.345
  Rs=0.40  Vmax= 8.29  Vmin= 8.08  Vpp= 0.208
  Rs=0.50  Vmax= 6.76  Vmin= 6.62  Vpp= 0.138
  Rs=0.60  Vmax= 5.69  Vmin= 5.59  Vpp= 0.098
  Rs=0.70  Vmax= 4.91  Vmin= 4.84  Vpp= 0.073
  Rs=0.80  Vmax= 4.32  Vmin= 4.26  Vpp= 0.057
  Rs=0.90  Vmax= 3.85  Vmin= 3.81  Vpp= 0.045
  Rs=1.00  Vmax= 3.47  Vmin= 3.44  Vpp= 0.037
  
  C=0.030000
  Rs=0.00  Vmax=27.68  Vmin=25.18  Vpp= 2.500
  Rs=0.10  Vmax=19.79  Vmin=18.74  Vpp= 1.049
  Rs=0.20  Vmax=12.78  Vmin=12.36  Vpp= 0.419
  Rs=0.30  Vmax= 9.12  Vmin= 8.91  Vpp= 0.212
  Rs=0.40  Vmax= 7.03  Vmin= 6.91  Vpp= 0.125
  Rs=0.50  Vmax= 5.70  Vmin= 5.62  Vpp= 0.082
  Rs=0.60  Vmax= 4.79  Vmin= 4.73  Vpp= 0.058
  Rs=0.70  Vmax= 4.12  Vmin= 4.08  Vpp= 0.043
  Rs=0.80  Vmax= 3.62  Vmin= 3.59  Vpp= 0.033
  Rs=0.90  Vmax= 3.23  Vmin= 3.20  Vpp= 0.026
  Rs=1.00  Vmax= 2.91  Vmin= 2.89  Vpp= 0.021
  
  C=0.050000
  Rs=0.00  Vmax=27.69  Vmin=26.11  Vpp= 1.586
  Rs=0.10  Vmax=14.73  Vmin=14.39  Vpp= 0.342
  Rs=0.20  Vmax= 8.35  Vmin= 8.24  Vpp= 0.107
  Rs=0.30  Vmax= 5.72  Vmin= 5.67  Vpp= 0.050
  Rs=0.40  Vmax= 4.34  Vmin= 4.31  Vpp= 0.029
  Rs=0.50  Vmax= 3.49  Vmin= 3.47  Vpp= 0.019
  Rs=0.60  Vmax= 2.91  Vmin= 2.90  Vpp= 0.013
  Rs=0.70  Vmax= 2.50  Vmin= 2.49  Vpp= 0.010
  Rs=0.80  Vmax= 2.19  Vmin= 2.19  Vpp= 0.007
  Rs=0.90  Vmax= 1.95  Vmin= 1.94  Vpp= 0.006
  Rs=1.00  Vmax= 1.76  Vmin= 1.75  Vpp= 0.005
  
  C=0.075000
  Rs=0.00  Vmax=27.70  Vmin=26.60  Vpp= 1.093
  Rs=0.10  Vmax=10.77  Vmin=10.65  Vpp= 0.120
  Rs=0.20  Vmax= 5.73  Vmin= 5.70  Vpp= 0.034
  Rs=0.30  Vmax= 3.87  Vmin= 3.86  Vpp= 0.015
  Rs=0.40  Vmax= 2.92  Vmin= 2.91  Vpp= 0.009
  Rs=0.50  Vmax= 2.34  Vmin= 2.33  Vpp= 0.006
  Rs=0.60  Vmax= 1.95  Vmin= 1.95  Vpp= 0.004
  Rs=0.70  Vmax= 1.67  Vmin= 1.67  Vpp= 0.003
  Rs=0.80  Vmax= 1.47  Vmin= 1.46  Vpp= 0.002
  Rs=0.90  Vmax= 1.30  Vmin= 1.30  Vpp= 0.002
  Rs=1.00  Vmax= 1.17  Vmin= 1.17  Vpp= 0.001
  
  C=0.100000
  Rs=0.00  Vmax=27.70  Vmin=26.86  Vpp= 0.835
  Rs=0.10  Vmax= 8.38  Vmin= 8.32  Vpp= 0.054
  Rs=0.20  Vmax= 4.35  Vmin= 4.33  Vpp= 0.014
  Rs=0.30  Vmax= 2.92  Vmin= 2.91  Vpp= 0.007
  Rs=0.40  Vmax= 2.20  Vmin= 2.19  Vpp= 0.004
  Rs=0.50  Vmax= 1.76  Vmin= 1.76  Vpp= 0.002
  Rs=0.60  Vmax= 1.47  Vmin= 1.46  Vpp= 0.002
  Rs=0.70  Vmax= 1.26  Vmin= 1.26  Vpp= 0.001
  Rs=0.80  Vmax= 1.10  Vmin= 1.10  Vpp= 0.001
  Rs=0.90  Vmax= 0.98  Vmin= 0.98  Vpp= 0.001
  Rs=1.00  Vmax= 0.88  Vmin= 0.88  Vpp= 0.001