Hero999 Posted June 12, 2010 Report Share Posted June 12, 2010 Audioguru answered your question.If I don't respond it's normally for any of the following reasons.I'm busy You haven't made it clear you're asking a question, respond clarifying it and add more information, if you can. I don't think you've given the question enough thought and feel that you could answer it yourself if you just gave it more thought.The LM338 has a drop out of 3V which means it requires a difference of 3 V between the input and output to regulate properly. When it's configured as a current regulator, the sense resistor between the output and the adjust pin will drop another 1.25V at full load. 3+1.25 = 4.25V. The current regulator is connected in series with the other LM338 which is configured as a voltage regulator. Quote Link to comment Share on other sites More sharing options...
Fight Posted June 13, 2010 Author Report Share Posted June 13, 2010 Audioguru answered your question.If I don't respond it's normally for any of the following reasons.I'm busy You haven't made it clear you're asking a question, respond clarifying it and add more information, if you can. I don't think you've given the question enough thought and feel that you could answer it yourself if you just gave it more thought.The LM338 has a drop out of 3V which means it requires a difference of 3 V between the input and output to regulate properly. When it's configured as a current regulator, the sense resistor between the output and the adjust pin will drop another 1.25V at full load. 3+1.25 = 4.25V. The current regulator is connected in series with the other LM338 which is configured as a voltage regulator. Ok1. your site silicontronics not open why Questioning about circuit diagram send at post 231. This circuit provide good current limiting and 2. 5A current provide at all voltage range 1.2 to 30V and3. good voltage regulation from 1.2 to 30V4. if 0-30V instead 1.2-30V than modification on circuit Quote Link to comment Share on other sites More sharing options...
Hero999 Posted June 13, 2010 Report Share Posted June 13, 2010 1. your site silicontronics not open whyIt isn't my site but Dazza's (another member of Electronics Lab), I just help to run it for him.It works for me, there must be something wrong with your Internet connection, try again later. Questioning about circuit diagram send at post 23There's no circuit diagram attached to post 23.1. This circuit provide good current limiting and 2. 5A current provide at all voltage range 1.2 to 30V and3. good voltage regulation from 1.2 to 30VThose aren't questions, they're statements.4. if 0-30V instead 1.2-30V than modification on circuitYou need to add an op-amp and a few other components.Here's a link to a PSU designed for 0 to 13.8V. The same method could be used with the LM338 and current booster to get 0 to 30V. If the link doesn't work I'll post an attachement.http://www.silicontronics.com/index.php?action=ezportal;sa=page;p=19 Quote Link to comment Share on other sites More sharing options...
Fight Posted June 13, 2010 Author Report Share Posted June 13, 2010 Quote1. This circuit provide good current limiting and 2. 5A current provide at all voltage range 1.2 to 30V and3. good voltage regulation from 1.2 to 30VThose aren't questions, they're statements.These statements ok or notHere's a link to a PSU designed for 0 to 13.8V. The same method could be used with the LM338 and current booster to get 0 to 30V. If the link doesn't work I'll post an attachement.http://www.silicontronics.com/index.php?action=ezportal;sa=page;p=19this link is not open please attatch Quote Link to comment Share on other sites More sharing options...
Fight Posted June 13, 2010 Author Report Share Posted June 13, 2010 What browser are you using?Internet explorerWhere do you live?Pakistanabout 2 or 3 days this site doenot open Quote Link to comment Share on other sites More sharing options...
Hero999 Posted June 13, 2010 Report Share Posted June 13, 2010 What version of Internet Explorer do you use?It works for me in IE 8.There is some Internet censorship in Pakistan but it's not as bad as some countries.http://en.wikipedia.org/wiki/Internet_censorship_in_PakistanThe statements you made a couple of posts ago are true.I've attached the circuit I was talking about. Quote Link to comment Share on other sites More sharing options...
Fight Posted June 14, 2010 Author Report Share Posted June 14, 2010 Very Very thanks a lot HeroOnly using Lm338 except lm317please explain this circuit how circuit work Quote Link to comment Share on other sites More sharing options...
Hero999 Posted June 14, 2010 Report Share Posted June 14, 2010 I explained how the regulator works on Silicon Tronics which you can't access so I've coppied and pasted the text.The circuit compriese of two parts: the rectifer/voltage doubler and the actual regulator iteslf.Rectifier This part of the circuit converts the 15VAC from the transformer to +20V and -9VDC.The -9V rail it derived from connecting a negative voltage doubler (formed by D2, D3, C2 and C3) to the AC side of the rectifier an its output is restricted to 9V (see R1 and D4) because the op-amp is only rated for +/22V. R1 and D4 can be substituted for a negative regulator (LM78L09 or LM79L12) if you like.RegulatorThis part of the circuit is actually responsible for converting 20VDC to the lower output voltage.U2 forms a differential amplifier which looks at the voltage across R7; this should be a constant 1.2V to 1.3V depending on the LM317. The negative input of the differential amplifier is at the positive side and the and the positive input is at the negative side of R7 which causes U2's output to sit at -1.2V to -1.3V.The 'usual' negative point is connected to U2's output, therefore subtracting it from the output voltage meaning the output can go all the way to 0V.Now we might have a problem: we should only expect the 741's output to be able to supply a maximum of 10mA. The trouble is the current flowing through R6 can be as high as 13mA. The chances are it'll be all right since the datasheet of the 741 says it might be able to take 20mA but the minimum ia 10mA.Just in case the op-amp might not be able to handle 13mA I've connected a 1k pull-down resistor (R8) from U2's output to -9V, which will take 7.75mA away from the output so it only has to supply 5.25mA.NOTES:If you don't have the Quote Link to comment Share on other sites More sharing options...
Fight Posted June 15, 2010 Author Report Share Posted June 15, 2010 Very Very Thanks a lotQuestion about Diagram which send agoIC1(Which use for current regulator)V = 3V & I = 5A & P = 15WIC1 handle 15W powerIC2(Which use for Voltage regulator)V = 30V & I = Depending upon BE resistor let 50mA & P = 1.5WIC2 handle 1.5W powerTransistor(Which use for Current booster)V = 30V & I = 5A & P = 150WTransistors handle 150W poweram i reight or notHow the design heat sink from these ICs and transistors (Size Shape Material and other)Thanks Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 15, 2010 Report Share Posted June 15, 2010 The voltage regulator and current regulator will have 30V or more across them sometimes. An LM338 limits its current when it has more than 10V from its input to its output. Then its max current is only 1A.With a voltage of 30V and a current of 1A then it dissipates 30W, not 3W.A "150W" power transistor can dissipate 150W only if its case is cooled to 25 degrees C with liquid nitrogen. If it has a huge heatsink and a fan then it can dissipate only 75W when it will be at its absolute max allowed temperature.EDIT: Fixed a typo. Quote Link to comment Share on other sites More sharing options...
Fight Posted June 16, 2010 Author Report Share Posted June 16, 2010 The voltage regulator and current regulator will have 30V or more across them sometimes. An LM338 limits its current when it has more than 10V from its input to its output. Then its max current is only 1A.if lm338 use as current regulator and input voltage 40V the voltage drop across ic at 5A equal to approxomately 3Vand at this condition lm338 can handle 5Athan P=3*5 = 15Wam i right or notthan next Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 16, 2010 Report Share Posted June 16, 2010 if lm338 use as current regulator and input voltage 40V the voltage drop across ic at 5A equal to approxomately 3Vand at this condition lm338 can handle 5Athan P=3*5 = 15Wam i right or notYes but then its output voltage must never be less than 37V.If you connect a load that tries to draw more than 5A then the current regulation will reduce the output voltage until the current is 1A. Then the LM338 current regulator will dissipate as much as 40W! Quote Link to comment Share on other sites More sharing options...
Hero999 Posted June 16, 2010 Report Share Posted June 16, 2010 Fight,The circuit I posted in my first reply to this thread uses several power transistors so the power dissipated by each power transistor is only around 35W.http://www.electronics-lab.com/forum/index.php?topic=20224.msg92354#msg92354We're going round in circles. Quote Link to comment Share on other sites More sharing options...
Fight Posted June 17, 2010 Author Report Share Posted June 17, 2010 Yes but then its output voltage must never be less than 37V.If you connect a load that tries to draw more than 5A then the current regulation will reduce the output voltage until the current is 1A. Then the LM338 current regulator will dissipate as much as 40W!Hello audioguruif max draw 5A current thanmax Power dissipation of IC 3V*5A =15Wif lm338 use as voltage regulator with transister use as current booster attatch diagram thanmax power dissipation V = 30V & I = Depending upon BE resistor let 50mA & P = 1.5Wam i right or not Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 17, 2010 Report Share Posted June 17, 2010 If you set the current regulator to 5A and load the power supply with less than 5A then the current regulator is saturated and does not get hot.But if you have a load that tries to draw more than 5A (maybe the output is accidently shorted) then the current regulator will have its max current (which might drop to only 1A) and it will have the max voltage across it (maybe 36V) and it will be extremely hot.Your voltage regulator with the four current-boosting transistors has the base-emitter voltage of the transistors (maybe 0.8V) plus the voltage across the 0.1 ohm resistors (0.125V) for a total of 0.925V. The 4.7 ohm resistor with a voltage across it of 0.925V has a current of 197mA which is also the current of the regulator. If the input voltage is 36V, the output voltage is 5V and the current is 5A then the regulator didssipates 6.1W and the transistors each dissipate 37.2W. Quote Link to comment Share on other sites More sharing options...
Fight Posted June 18, 2010 Author Report Share Posted June 18, 2010 This situation the heat sink sizeOkThis situation the heat sink size and how design size of heat sinkYour voltage regulator with the four current-boosting transistors has the base-emitter voltage of the transistors (maybe 0.8V) plus the voltage across the 0.1 ohm resistors (0.125V) for a total of 0.925V. The 4.7 ohm resistor with a voltage across it of 0.925V has a current of 197mA which is also the current of the regulator. If the input voltage is 36V, the output voltage is 5V and the current is 5A then the regulator didssipates 6.1W and the transistors each dissipate 37.2W. how the regulator didssipates 6.1WV = 5V and I = 197mA then P = 0.985W = ?and how the transistors each dissipate 37.2WV = ? and I = 5A then p = ?please explain Quote Link to comment Share on other sites More sharing options...
Hero999 Posted June 18, 2010 Report Share Posted June 18, 2010 No, the power dissipation is equal to the voltage across the regulator multiplied by the current through it. Quote Link to comment Share on other sites More sharing options...
Fight Posted June 19, 2010 Author Report Share Posted June 19, 2010 No, the power dissipation is equal to the voltage across the regulator multiplied by the current through it.yes Your voltage regulator with the four current-boosting transistors has the base-emitter voltage of the transistors (maybe 0.8V) plus the voltage across the 0.1 ohm resistors (0.125V) for a total of 0.925V. The 4.7 ohm resistor with a voltage across it of 0.925V has a current of 197mA which is also the current of the regulator. If the input voltage is 36V, the output voltage is 5V and the current is 5A then the regulator didssipates 6.1W and the transistors each dissipate 37.2W. Hello Hero and audioguruhow are youhow the regulator didssipates 6.1WV = 31V and I = 197mA then P = 6.107Wam i rightand how the transistors each dissipate 37.2WV = 36-0.925 = 35.075V and I = 5A then p = 175.375Weach dissipate 175.375/4 = 43.84Wam i right Quote Link to comment Share on other sites More sharing options...
Hero999 Posted June 19, 2010 Report Share Posted June 19, 2010 Maybe audioguru made a mistake, I make it 43W per transistor.It's not 5A shared between the transistors, remember 0.2A goes through the regulator.If the voltage across the emitter resistors is 0.125V.The voltage across the transistors is 31-0.125 = 30.875VThe current through all the transistors is 5 - 0.2A = 4.8P = 30.875*4.8 = 148.2W148.2/4 = 37.05W.Don't worry, just design for a power dissipation of 40W per transistor and all will be well. Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 19, 2010 Report Share Posted June 19, 2010 We forgot that the regulator has the 197mA current in the 4.7 ohm resistor plus the 160mA base current of the transistors. So the regulator dissipates 11W and each transistor dissipates 35.8W each. Quote Link to comment Share on other sites More sharing options...
Fight Posted June 19, 2010 Author Report Share Posted June 19, 2010 We forgot that the regulator has the 197mA current in the 4.7 ohm resistor plus the 160mA base current of the transistors. So the regulator dissipates 11W and each transistor dissipates 35.8W each. Maybe audioguru made a mistake, I make it 43W per transistor.It's not 5A shared between the transistors, remember 0.2A goes through the regulator.If the voltage across the emitter resistors is 0.125V.The voltage across the transistors is 31-0.125 = 30.875VThe current through all the transistors is 5 - 0.2A = 4.8P = 30.875*4.8 = 148.2W148.2/4 = 37.05W.Don't worry, just design for a power dissipation of 40W per transistor and all will be well.All OkPlease expalin these two factor1. Qjc Thermal Resistance Junction to Case K Package 1 Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 19, 2010 Report Share Posted June 19, 2010 Please expalin these two factor1. Qjc Thermal Resistance Junction to Case K Package 1 Quote Link to comment Share on other sites More sharing options...
Fight Posted June 20, 2010 Author Report Share Posted June 20, 2010 i am reading this pages http://sound.westhost.com/heatsinks.htm and then questioning about heat sinkvery very thanks audioguru and hero Quote Link to comment Share on other sites More sharing options...
Fight Posted June 20, 2010 Author Report Share Posted June 20, 2010 hello audioguru and herohow are youi read this topic in this topic describe that heat flow from junction to case case to heat sinkand heat sink to ambientplease tell me that what did meanjunction to case case to heat sinkand heat sink to ambienta transistor sold on pcb where transistor leg sold this is junction or other and what did mean case and waht did mean ambient (ROOM Temprature or other)thanks Quote Link to comment Share on other sites More sharing options...
audioguru Posted June 20, 2010 Report Share Posted June 20, 2010 please tell me that what did meanjunction to case case to heat sinkand heat sink to ambienta transistor sold on pcb where transistor leg sold this is junction or other and what did mean case and waht did mean ambient (ROOM Temprature or other)These simple basic questions are about translating English into your language.Maybe you should ask in your language on a website in your country. Quote Link to comment Share on other sites More sharing options...
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