LM338 Power Suply Current Regulator

[audioguru2]

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. 

 

[Fight]

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 = ?

please explain

 

[Hero999]

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

 

[Fight]

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

 

[Hero999]

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.

 

[audioguru2]

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. 

 

[Fight]

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

 

[audioguru2]

Please expalin these two factor
1.   Qjc Thermal Resistance Junction to Case K Package 1

 

[Fight]

i am reading this pages http://sound.westhost.com/heatsinks.htm
and then questioning about heat sink

very very thanks audioguru and hero

 

[Fight]

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

 

[audioguru2]

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)

These simple basic questions are about translating English into your language.
Maybe you should ask in your language on a website in your country.

 

[Fight]

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

 

[Hero999]

The junction is the actual piece of silicon inside the case.

I'd recommend using an old Pentium 3 CPU cooler as the heat sink because it will have a thermal resistance of under 0.5

 

[Fight]

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

View attachment 41073

View attachment 41074

 

[Hero999]

The circuit is totally impractical.

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.

The 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.

The 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.

 

[audioguru2]

The 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.

 

[Hero999]

The 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.

The 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.

When 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.

 

[Fight]

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's

The 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 32.5V

The 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 not

The 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 not

The 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 explain

When 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

 

[Hero999]

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. This 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.

Why not use a real digital pot. rather than making a poor one with the CD4066?

 

[Fight]

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.

Ok

This 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