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Fight
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thanks
very very thanks a lot ::) ??? -
Hello Hero and Audioguru
how are u
i sure that both are well
ok next
Hello Hero
in other threat named "Switch mode power supply" you tell me that if you buy a 12V 200W smps then you add a circuit for making power supply 0-30 and 0-5A variale voltage and variable conatant current power supply.
If i buy PC P-4 PSU then this psu use as a primary 12V 200W smps first tell me that
How idea
and how can add a circuit please if you have a spare time then send a diagram using IC (SG3524) beacause this IC is avail in our country easily. -
Hello Hero and Audioguru
how are touThe LM338 will only overheat after the current limit has kicked in, before then the voltage across it will be between 1.5V and 3V, depending on the current and ambient temperature.
if i add a circuit that prevent short circuit then this problem solved or not -
Hello Hero and audioguru
how r u
where are you in these days since about 2 weeks
i am waiting you
please come on internet and solve my problem -
When the current is below the current limit , the voltage across the first LM338 (used as a current limiter) will be very low: between 1.5V and 3V, depending on the current drawn and IC temperature so the power dissipation will be 15W maximum. When the current exceeds the limit set, the voltage across the first LM338 will increase but the current will remain the same causing a higher power dissipation. If the power dissipated is too high, causing the first LM338 to overheat, it will shut down in order to protect itself and the current will fall.
OkThis sort of current limiting is no good if you want to use your PSU as a constant current power supply to run high powered LEDs of but it's fine for a general purpose constant voltage power supply, when all you want is adjustable current limiting to protect sensitive components.
Then please tell me that how change in circuit that solve my problem -
Where are you going to buy a 12 Ohm pot. rated to 5A or 6.25W when it's set to 0.25Ω?
The maximum power rating for a pot. is always for the full track so when the wiper is in the middle it's halved, in this case 0.25Ω is just under 2% of the maximum resistance so the power rating will only be 2% of that specified by the datasheet.
i am using different fixed resistors and these resistors control with realy'sThe maximum output voltage with 40V in will be about 32V so why have you used a 5k pot which will give lots of dead band and no voltage regulation when the wiper is set too high? In other words with 120R and 5k the set voltage is 50V but the maximum output voltage is only 32V so when the pot is set for above 32V, there will not be any voltage regulation when 5A is drawn.
i am using about 3K resistor whose cove 32.5VThe CD4066 has a maximum voltage rating of only 18V and the maximum voltage it can switch is equal to its supply voltage so can't be used for this application.
Ok if i use this circuit for 0-18V then this is right or notThe voltage regulator uses many transistors for high power but the current regulator uses only a single LM338 that will get so hot and will shut-down when the current is higher than its setting then the circuit will not work.Quote
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
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!
as above quote this situation lm338 handle 15W if i use heat sink then this circuit ok or notThe LM338 will only overheat after the current limit has kicked in, before then the voltage across it will be between 1.5V and 3V, depending on the current and ambient temperature.
please explainWhen the current limit, the current regulator might shut down, depending on the load voltage and current setting, if it's a problem a diferent solution is required.
if i max current limit 5A then how increase wattage above 15W -
Hello hero and audioguru
how are you
my final power supply circuit diagram that attatch
Fig 1
the 40V dc supply make using transformer and diode and capacitor circuit
the first sec current limiter circuit
value of R2 vary from 0.24ohm to 12ohm (current varry 100mA to 5A) and wattage 10W
the wattage of R3 is V=0.8V and I = 400mA about 0.5W
the wattage of emmitter resistance is = ?
Please tell me that this circuit is right or any error in this circuit please under line those error
In fig 2 R5 replaced fixed resistors that control using CD4066 ic for open and close these resistors that right or not
very very thanks a lot
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These simple basic questions are about translating English into your language.
Maybe you should ask in your language on a website in your country.
in my mind that junction where transistor or IC leg sold
Case transistor or IC outer body
ambient mean air
am i right or not -
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 -
i am reading this pages http://sound.westhost.com/heatsinks.htm
and then questioning about heat sink
very very thanks audioguru and hero -
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
Please expalin these two factor
1. Qjc Thermal Resistance Junction to Case K Package 1 -
No, the power dissipation is equal to the voltage across the regulator multiplied by the current through it.
yesYour 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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This situation the heat sink size
Ok
This 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 6.1W
V = 5V and I = 197mA then P = 0.985W = ?
and how the transistors each dissipate 37.2W
V = ? and I = 5A then p = ?
please explain
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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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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 -
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 -
Very Very thanks a lot Hero
Only using Lm338 except lm317
please explain this circuit how circuit work -
What browser are you using?
Internet explorerWhere do you live?
Pakistan
about 2 or 3 days this site doenot open -
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 30VThose 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 -
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
-
hello hero where are you
your site silicontronics does not open why
and you are angry with me why
please answer my question -
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 circuitThe 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 = ? -
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 -
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
LM338 Power Suply Current Regulator
in Datasheet/Parts requests
Posted
Hello KevinIV
who are you
Hero not a moderator but
is a brilliant man
and my friend