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# LM338 Power Suply Current Regulator

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If I don't respond it's normally for any of the following reasons.

• I'm busy
• 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.

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If I don't respond it's normally for any of the following reasons.

• I'm busy
• 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.

Ok

1. your site silicontronics not open why

Questioning about circuit diagram send at post 23

1.  This circuit provide good current limiting and
2.  5A current provide at all voltage range 1.2 to 30V  and
3.  good voltage regulation from 1.2 to 30V

4.  if 0-30V instead 1.2-30V than modification on circuit

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1. your site silicontronics not open why

It 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 23

There'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  and
3.  good voltage regulation from 1.2 to 30V

Those aren't questions, they're statements.

4.  if 0-30V instead 1.2-30V than modification on circuit

You 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
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Quote
1.  This circuit provide good current limiting and
2.  5A current provide at all voltage range 1.2 to 30V  and
3.  good voltage regulation from 1.2 to 30V

Those aren't questions, they're statements.

These statements ok or not

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

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What browser are you using?

Internet explorer

Where do you live?

Pakistan

about 2 or 3 days this site doenot open
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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_Pakistan

The statements you made a couple of posts ago are true.

I've attached the circuit I was talking about.

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Very Very thanks a lot Hero

Only using Lm338 except lm317

please explain this circuit how circuit work

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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.

Regulator
This 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

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Very Very Thanks a lot

Question about Diagram which send ago
IC1(Which use for current regulator)
V = 3V    &    I = 5A    &    P = 15W
IC1 handle 15W power

IC2(Which use for Voltage regulator)
V = 30V    &    I = Depending upon BE resistor let 50mA    &    P = 1.5W
IC2 handle 1.5W power

Transistor(Which use for Current booster)
V = 30V    &    I = 5A    &    P = 150W
Transistors handle 150W power

am i reight or not

How the design heat sink from these ICs and transistors (Size Shape Material and other)

Thanks

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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.

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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 3V
and at this condition lm338 can handle 5A
than P=3*5 = 15W
am i right or not

than next
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if lm338 use as current regulator and input voltage 40V
the voltage drop across ic at 5A equal to approxomately 3V
and at this condition lm338 can handle 5A
than P=3*5 = 15W
am i right or not

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!
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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#msg92354

We're going round in circles.

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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 audioguru
if max draw 5A current than
max Power dissipation of IC 3V*5A =15W

if lm338 use as voltage regulator with transister use as current booster attatch diagram than
max power dissipation

V = 30V    &    I = Depending upon BE resistor let 50mA    &    P = 1.5W

am i right or not

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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.

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This situation the heat sink size

Ok
This situation the heat sink size and how design size of heat sink

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.

how the regulator didssipates 6.1W
V = 5V  and I = 197mA    then P = 0.985W  = ?

and how the transistors each dissipate 37.2W
V = ?  and I = 5A    then  p = ?

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No, the power dissipation is equal to the voltage across the regulator multiplied by the current through it.

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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 audioguru
how are you

how the regulator didssipates 6.1W
V = 31V  and I = 197mA    then P = 6.107W

am i right

and how the transistors each dissipate 37.2W
V = 36-0.925 = 35.075V  and I = 5A    then  p = 175.375W
each dissipate 175.375/4 = 43.84W

am i right

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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 = 30.875V

The current through all the transistors is 5 - 0.2A = 4.8

P = 30.875*4.8 = 148.2W

148.2/4 = 37.05W.

Don't worry, just design for a power dissipation of 40W per transistor and all will be well.

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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.

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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 = 30.875V

The current through all the transistors is 5 - 0.2A = 4.8

P = 30.875*4.8 = 148.2W

148.2/4 = 37.05W.

Don't worry, just design for a power dissipation of 40W per transistor and all will be well.

All Ok

1.  Qjc Thermal Resistance Junction to Case K Package 1
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1.   Qjc Thermal Resistance Junction to Case K Package 1
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and then questioning about heat sink

very very thanks audioguru and hero

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hello audioguru and hero
how are you
i read this topic in this topic describe that heat flow from
junction to case
case to heat sink
and heat sink to ambient

please tell me that what did mean
junction to case
case to heat sink
and heat sink to ambient
a 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

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please tell me that what did mean
junction to case
case to heat sink
and heat sink to ambient
a 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)

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