LM338 Power Suply Current Regulator

[Fight]

than any idea for increasing current

and tell me that this circuit provide good current limiting and
5A current provide at all voltage range 1.2 to 30V  and
good voltage regulation from 1.2 to 30V

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

 

[Fight]

It's not a very good idea because the extra regulator will increase the drop out voltage by 4.2V.

where 4.2V drop
2V drop across IC 1 that for current limiting/regulating
0.7V drop across BE resistor
2V drop across IC 2 that for voltage regulating
i am right or not

if input voltage 38V than 4.7V drop acrosss these the output will vary easily 1.2 to 38-4.7 = 33.3V
yes or not

 

[Hero999]

The LM338 has a drop out voltage of 3V at full current and when configured as a current regulator, the voltage reference of 1.25V across the current sense resistor, adds to it making 4.25V.

 

[Fight]

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

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

The LM338 has a drop out voltage of 3V at full current and when configured as a current regulator

if lm338 configured as voltage reglator than voltage drop across it = ?

 

[audioguru2]

It looks like you did not read the datasheet for the LM338:
1) It shows a graph of the typical dropout voltage at various currents but does not list a max dropout voltage.
2) It lists a max output current of 5A only when the Vin - Vout is 10V or less (then your minimum output voltage is 23V. It lists its max output current to be only 1A when Vin - Vout is 30V. 

 

[Fight]

hello hero where are you
your site silicontronics does not open why
and you are angry with me why
please answer my question

 

[Hero999]

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.

 

[Fight]

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.

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

 

[Hero999]

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

 

[Fight]

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

this link is not open please attatch

 

[Fight]

What browser are you using?

Internet explorer

Where do you live?

Pakistan

about 2 or 3 days this site doenot open

 

[Hero999]

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.

View attachment 41056

 

[Fight]

Very Very thanks a lot Hero

Only using Lm338 except lm317

please explain this circuit how circuit work

 

[Hero999]

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

 

[Fight]

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

 

[audioguru2]

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.

 

[Fight]

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

 

[audioguru2]

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!

 

[Hero999]

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.

 

[Fight]

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 notView attachment 41064