Help With LM338 Voltage Regulator

[Nusku]

Hi,

I would be grateful for any help in calculating how much heat an LM338 (TO-220 Package) voltage regulator will generate and what heatsink is required in the following application:

I want to build a variable voltage power supply for an electronic cigarette atomiser using items that I already have.

The input voltage for the LM338 circuit will be from an old laptop PSU rated at 16V, 4.5A. The load on the LM338 output will typically be a 2 Ohm Kanthal wire coil powered for, say, 5 second bursts with a minimum interval of 30 seconds.

This is the circuit:

Also, will there be any problem replacing R1 with a 2K resistor to limit the max voltage to 4.188V if I deem it necessary?

Thanks.

 

[(*steve*)]

Check the specs for the LM338. I think the use of a 1k resistor for R1 is already higher than recommended. You would be better off replacing R2 with a lower valued pot.

However the main problem is that if your heater is dissipating 8W at 4V, then the LM338 is dissipating 24W. That's a fairly large amount and you need a significant heatsink on the LM338.

If the ambient temperature is 30C and you want the heatsink temperature to be 80C or below, then you need a heatsink rated at about (80-30)/24 degrees C/W.

So that's a heatsink rated at about 2 degC/W.

This is a very approximate calculation assuming some reasonably conservative figures. However you'll need to insulate the 338 from the heatsink (most likely) and use thermal grease to ensure it has a good thermal contact with it.

If you do it well, you will be able to keep the die temperature below about 100C which leaves some leeway before the chip shuts itself down.

If you can run it from 12V or even 8V then the heatsinking requirements will be far less.

 

[Nusku]

Check the specs for the LM338. I think the use of a 1k resistor for R1 is already higher than recommended. You would be better off replacing R2 with a lower valued pot.

However the main problem is that if your heater is dissipating 8W at 4V, then the LM338 is dissipating 24W. That's a fairly large amount and you need a significant heatsink on the LM338.

If the ambient temperature is 30C and you want the heatsink temperature to be 80C or below, then you need a heatsink rated at about (80-30)/24 degrees C/W.

So that's a heatsink rated at about 2 degC/W.

This is a very approximate calculation assuming some reasonably conservative figures. However you'll need to insulate the 338 from the heatsink (most likely) and use thermal grease to ensure it has a good thermal contact with it.

If you do it well, you will be able to keep the die temperature below about 100C which leaves some leeway before the chip shuts itself down.

If you can run it from 12V or even 8V then the heatsinking requirements will be far less.

Thanks for the info, Steve. Could you tell me what effect a 1K resistor for R1 would have compared to using the typical 120-240 Ohm resistor?

Yes, I think it would be best to find another old PSU that has a lower voltage, and decent current capability, for the input.

 

[(*steve*)]

The 1k resistor will only provide a current of 1.25/1000 A (1.25mA). The datasheet tells you that the minimum output current to maintain regulation is between 3.5mA and 5mA. To be conservative, we tend to go for the high end, so the resistance required is 1.25/0.005 = 250 ohms. Normally a value of 220 ohms is specified.

In addition to this, the current through the ground terminal plays a part in the output voltage, and the variation in this current with load will affect the load regulation of the device. A smaller resistor will make both of these effects smaller.

In your case, the output load might always be present, so the loss of regulation without a load would not be a major problem. For heating a wire, the load is also constant, so the load regulation would likely not matter either.

A PC power supply that may be able to supply 15A to 50A and beyond on the 12V rail is pretty cheap and easily available.

 

[Nusku]

The 1k resistor will only provide a current of 1.25/1000 A (1.25mA). The datasheet tells you that the minimum output current to maintain regulation is between 3.5mA and 5mA. To be conservative, we tend to go for the high end, so the resistance required is 1.25/0.005 = 250 ohms. Normally a value of 220 ohms is specified.

In addition to this, the current through the ground terminal plays a part in the output voltage, and the variation in this current with load will affect the load regulation of the device. A smaller resistor will make both of these effects smaller.

In your case, the output load might always be present, so the loss of regulation without a load would not be a major problem. For heating a wire, the load is also constant, so the load regulation would likely not matter either.

A PC power supply that may be able to supply 15A to 50A and beyond on the 12V rail is pretty cheap and easily available.

Ah, I understand the resistor issue now.

And good call on the PC PSU, I've got a load of those knocking about...

Many thanks for your help.

 

[(*steve*)]

Be aware that some (generally older) PC power supplies need a load on the 5V rail to maintain regulation on the 12V rail.

If the power supply won't turn on, or the 12V rail is more than about 5% out, try placing a load (a 10 ohm 5W resistor should do) on the 5V rail.