ii am making power supply and using a lm338 voltage regulator 2 to 32 volts but how i can find a current for this supply
here is the formula but what is I ADJ and how to give a value to it????

http://www.eleccircuit.com/wp-content/uploads/2007/12/voltage-regulator.png

 

If you read the datasheet, you'll find that Iadj is specified as being a fairly small current. It is the current that has to be sunk from the ADJ pin in order for the regulator to work and since it is sunk through R2, it adds to the output voltage. For most purposes, you can ignore it and just figure that the output voltage will be

Vout = 1.25V(1 + R2/R1)

Basically, the regulator adjusts Vout in order to maintain a difference of 1.25V between Vout and ADJ. Thus results in a defined current through R1, namely I=1.25V/R1, which this flows through R2, giving a voltage drop across it of I*R2=(1.25V/R1)R2. The total output voltage is then the sum of the two

Vout = 1.25V + (1.25V/R1)R2
Vout = 1.25V(1 + R2/R1)

Look at the datasheet to see the constraints on sizing R1. Generally something in the neighborhood of 220Ω is used, IIRC.

 

If you read the datasheet, you'll find that Iadj is specified as being a fairly small current. It is the current that has to be sunk from the ADJ pin in order for the regulator to work and since it is sunk through R2, it adds to the output voltage. For most purposes, you can ignore it and just figure that the output voltage will be

Vout = 1.25V(1 + R2/R1)

Basically, the regulator adjusts Vout in order to maintain a difference of 1.25V between Vout and ADJ. Thus results in a defined current through R1, namely I=1.25V/R1, which this flows through R2, giving a voltage drop across it of I*R2=(1.25V/R1)R2. The total output voltage is then the sum of the two

Vout = 1.25V + (1.25V/R1)R2
Vout = 1.25V(1 + R2/R1)

Look at the datasheet to see the constraints on sizing R1. Generally something in the neighborhood of 220Ω is used, IIRC.

thanks but how to find a out put current

 

That depends on the load you drive with it and you've given no indication of what that is.

 

That depends on the load you drive with it and you've given no indication of what that is.

hmm i have a transformer which gives 5 amp so in the output i will get 5 amp??

 

Please show a schematic (a schetch in Paint is fine) as what you are saying is making no sense.

I'm guessing that you have a transformer that is being used to provide power for the circuit and that the transformer is rated for a max current of 5 amps. Hopefully you are rectifying it and filtering it before trying to regulate it.

How much current is being output by a battery that is sitting on a table, unconnected to anything? Nothing. How much current is flowing if you power an MP3 player from it? Some amount. How much current is flowing if you also power an alarm clock from it at the same time? Some amount more. The amount of current flowing is dependent on the load that you connect to the output.

Again, please show some kind of a diagram.

 

The current that will be delivered by this circuit obeys Ohm's Law. For whatever voltage you set the regulator, that voltage divided by the resistance of the load will be the current. If you want to know the limits of how much current the regulator CAN deliver, that is related to the maximum on the datasheet and the situation where your rectified voltage is a lot more than the output voltage and so the regulator gets too hot to continue.

The LM338 comes in several sizes. Please say which one you are using.

 

Please show a schematic (a schetch in Paint is fine) as what you are saying is making no sense.

I'm guessing that you have a transformer that is being used to provide power for the circuit and that the transformer is rated for a max current of 5 amps. Hopefully you are rectifying it and filtering it before trying to regulate it.

How much current is being output by a battery that is sitting on a table, unconnected to anything? Nothing. How much current is flowing if you power an MP3 player from it? Some amount. How much current is flowing if you also power an alarm clock from it at the same time? Some amount more. The amount of current flowing is dependent on the load that you connect to the output.

Again, please show some kind of a diagram.

here is the image but i am making this circuit differently now i dont know much amperes on out put and alsoi want to know the formula for filtering

http://www.circuitstoday.com/wp-content/uploads/2009/08/13v5a-adjustable-powersupply-using-lm338.jpg

 

Nobody here knows how much amperes because you do not say what model of LM338 you are using, the voltage setting you will use, the resistance of the load, or how much heat sink you are providing.

 

Nobody here knows how much amperes because you do not say what model of LM338 you are using, the voltage setting you will use, the resistance of the load, or how much heat sink you are providing.

i am using this one regulator

 

You still have not said which package you will use, or any of the other things that are required to determine how much amperes.

 

You still have not said which package you will use, or any of the other things that are required to determine how much amperes.

What has package got to do with it?
He either uses a TO220 or the TO3.
Both output 5A continuous, though the TO3 package would dissipate heat much better than the TO220 counterpart.
Provided he uses a fairly good heatsink, I don't see any problems.
I've made a multitude of PSU's my whole life and never had an issue due to heat.

here is the image but i am making this circuit differently now i dont know much amperes on out put and alsoi want to know the formula for filtering

http://www.circuitstoday.com/wp-cont...sing-lm338.jpg

The circuit you've provided above will work quite well. Though a 10,000uF main filter cap is a bit overkill. You can use a 4,700uF and a 1,000uF caps instead.
Make sure which ever package of regulator IC you use, that you mount it on a good sized heatsink. Not a 'U' type either.
If you intend on making it variable to over 25 volts, then make sure you replace C4 and C5 with a higher rated voltage that will be above the maximum output you want. i.e. 32V max out, then use a 35V types.
EDIT: One last thing, make sure you mount C2 and C3 as close to the regulator IC as practical. This will help maintain stability with the regulator.
Regards,
Relayer

 

The package has a theta Jc which determines a maximum heat flow rate.
Both will output 5 amps until they overheat.

With no voltage, no load, and no heat sink declared, I can not calculate how many amps this circuit will provide without overheating. Then again, that's why many people help at this site. Some of them can do these things.

 

here is the image but i am making this circuit differently now i dont know much amperes on out put and alsoi want to know the formula for filtering

http://www.circuitstoday.com/wp-content/uploads/2009/08/13v5a-adjustable-powersupply-using-lm338.jpg

Well, we can talk about "Bob's" circuit (the only you linked) or we can talk about YOUR circuit. But if we want to talk about YOUR circuit then we need to see YOUR circuit, and not Bob's circuit.

It appears that you are wanting to have an adjustable output of 2V to 32V with an output current capacity of 5A. Since you've got a dropout voltage of 3V at that current, you will need a minimum voltage at Vin of 35V. If you are delivering 2V and 5A, you are dropping 33V across the regulator and thus dissipating 165W of power in it.

Given that the maximum junction temperature is 125°C and that the thermal resistance from the junction to the case is 4°C/W for the TO-220 case and 1°C/W for the TO-3 case, even with an absolutely perfect heatsink (no such thing, by the way) you would have to operate at an ambient temperature of -40°C for the TO-3 case and -535°C for the TO-220 case (and note that absolute zero is only -273°C).

So, contrary to what Relayer claims, it would appear that which case you use does make a difference since one is at least conceivably possible and the other is physically impossible. But, again contrary to what Relayer claims, I don't see this working too well just by "using a fairly good heatsink".

If you want to make this think work, you are going to need to either shunt the majority of the current around the regulator or incorporate a way to drop the voltage before the regulator. If you can keep the Vin-Vout voltage to, say, 5V, then you are asking the regulator to dissipate up to 25W. With a TO-220 case, you would still have to operate at ambient temperatures well below room temperature, but with a TO-3 case you should be able to mount a suitable heatsink without too much difficulty.

 

Given that the maximum junction temperature is 125°C and that the thermal resistance from the junction to the case is 4°C/W for the TO-220 case and 1°C/W for the TO-3 case, even with an absolutely perfect heatsink (no such thing, by the way) you would have to operate at an ambient temperature of -40°C for the TO-3 case and -535°C for the TO-220 case (and note that absolute zero is only -273°C).

I thought 4°C/W would mean a faster heat transfer than 1°C/W. If so, why does the TO-220 require such a low temperature?
Where can I find more information about heat transfer?
BTW, the recommended temperature range is 0°C to 125°C for the LM338.

 

I thought 4°C/W would mean a faster heat transfer than 1°C/W. If so, why does the TO-220 require such a low temperature?
Where can I find more information about heat transfer?
BTW, the recommended temperature range is 0°C to 125°C for the LM338.

Nope. Look at the units. You need a temperature difference of 4°C for each 1W of heat transfer. Just like 4Ω = 4V/A means that you need 4V of voltage difference for each 1A of current. If you want to get 10A of current flow, then you need 40V across the electrical resistance. Similarly, if you want 10W of heat flow, then you need 40°C across the termal resistance.

The recommended temperature range for the junction is 0°C to 125°C. The ambient can be whatever it takes to keep the junction within that range. I don't see an operating temperature for the case, but note that there is a note saying that the specs are for power dissipations of up to 50W and a Vin-Vout voltage of up to 15V. Thus, for where the OP is trying to operate, all bets are off.

Consider the following heat sink, which is the only one that Digi-Key presently stocks that has a thermal resistance to free air of less than 1°C/W (it has 0.95°C/W).

So this would make the junction-ambient thermal resistance 1.95°C/W, so to dissipate 165W you would have to keep the ambient temperature 322°C below the junction temperature, which to keep the junction at 125°C would mean that the ambient would have to be -197°C. Now consider that the boiling point of liquid nitrogen is -196°C.

Oh, and that bad boy costs $144 for one of them.

If you go to forced air flow past the heat sink, you can get parts from Digi-Key that go down to about 0.3°C/W with about 500 LFM of air flow. But that air has to be at -90°C.

Even if you go to extreme measures to cool the case, you are stuck with the fact that, to dissipate 165W, you have to keep the case at -40°C.

And, even if you succeed, you now have a seriously thermally stressed part in which there is a temperature difference of 165°C over a very short distance and involving several different materials -- that's a recipe for cracked silicon.

 

You still have not said which package you will use, or any of the other things that are required to determine how much amperes.

i want 5 amperes on output

 

Well, we can talk about "Bob's" circuit (the only you linked) or we can talk about YOUR circuit. But if we want to talk about YOUR circuit then we need to see YOUR circuit, and not Bob's circuit.

It appears that you are wanting to have an adjustable output of 2V to 32V with an output current capacity of 5A. Since you've got a dropout voltage of 3V at that current, you will need a minimum voltage at Vin of 35V. If you are delivering 2V and 5A, you are dropping 33V across the regulator and thus dissipating 165W of power in it.

Given that the maximum junction temperature is 125°C and that the thermal resistance from the junction to the case is 4°C/W for the TO-220 case and 1°C/W for the TO-3 case, even with an absolutely perfect heatsink (no such thing, by the way) you would have to operate at an ambient temperature of -40°C for the TO-3 case and -535°C for the TO-220 case (and note that absolute zero is only -273°C).

So, contrary to what Relayer claims, it would appear that which case you use does make a difference since one is at least conceivably possible and the other is physically impossible. But, again contrary to what Relayer claims, I don't see this working too well just by "using a fairly good heatsink".

If you want to make this think work, you are going to need to either shunt the majority of the current around the regulator or incorporate a way to drop the voltage before the regulator. If you can keep the Vin-Vout voltage to, say, 5V, then you are asking the regulator to dissipate up to 25W. With a TO-220 case, you would still have to operate at ambient temperatures well below room temperature, but with a TO-3 case you should be able to mount a suitable heatsink without too much difficulty.

i have no circuit until i calculate the output of the circuit what will be the output current

 

i have no circuit until i calculate the output of the circuit what will be the output current

This is like saying that you have a bunch of lumber in the garage and asking how high the structure will be that you are going to build.

YOU CAN"T!!!

The circuit shown will output ZERO current (other than the voltage setting current) because there IS NO LOAD! If you attach a load, then it will pull a current consistent with the size of the load and the output voltage of the power supply.

If you WANT to be able to deliver a maximum of 5A out of a linear supply that can be adjusted from 2V to 32V, you has better be prepared to design a circuit that can dissipate over 150W without turning some part of the circuit into smoke. I've walked through the types of numbers you are looking at for a circuit similar to the one you linked.

 

This is like saying that you have a bunch of lumber in the garage and asking how high the structure will be that you are going to build.

YOU CAN"T!!!

The circuit shown will output ZERO current (other than the voltage setting current) because there IS NO LOAD! If you attach a load, then it will pull a current consistent with the size of the load and the output voltage of the power supply.

If you WANT to be able to deliver a maximum of 5A out of a linear supply that can be adjusted from 2V to 32V, you has better be prepared to design a circuit that can dissipate over 150W without turning some part of the circuit into smoke. I've walked through the types of numbers you are looking at for a circuit similar to the one you linked.

how can i save from turn into smoke??