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.

 

Feedback and Suggestions Forum for providing feedback and suggestions about All About Circuits, including corrections to the e-book. This forum is not for getting help with technical questions.

This was neither feedback nor a suggestion, so I have moved it to a more appropriate forum.

 

This was neither feedback nor a suggestion, so I have moved it to a more appropriate forum.

Hi, my bad, I'm afraid I've double posted, please delete one of them. Thanks.

 

Hello,

The regulator will have to dissipate quite a deal of the power.
The voltage difference is 16 Volts - 4 Volts = 12 Volts.
The current used is 2 Amp.
The power dissipated in the regulator will be 12 X 2 = 24 Watts.
The power where the regulator is cutting back will be 25 watts (according to the datasheet).
If you really want to use this regulator a heatsink MUST be used.
With the heatsink given in the link below, the temperature rise will be 36 degrees Celsius above the ambient temperature:
http://www2.mouser.com/ProductDetai...OrhyHdtueg3JHY7cW3HDouTSq1CSHu33oPcQP%2bMfw==

Also lower the value of R1 to 120 Ohms as given in the datasheet for better stability.

A better way to reduce the voltage will be the use of a switching regulator.

Bertus

 

Hello,

The regulator will have to dissipate quite a deal of the power.
The voltage difference is 16 Volts - 4 Volts = 12 Volts.
The current used is 2 Amp.
The power dissipated in the regulator will be 12 X 2 = 24 Watts.
The power where the regulator is cutting back will be 25 watts (according to the datasheet).
If you really want to use this regulator a heatsink MUST be used.
With the heatsink given in the link below, the temperature rise will be 36 degrees Celsius above the ambient temperature:
http://www2.mouser.com/ProductDetail/Wakefield/680-125A/?qs=%2fha2pyFadugxlOrhyHdtueg3JHY7cW3HDouTSq1CSHu33oPcQP%252bMfw%3d%3d

Also lower the value of R1 to 120 Ohms as given in the datasheet for better stability.

A better way to reduce the voltage will be the use of a switching regulator.

Bertus

Excellent, thanks very much for the info, Bertus, that's exactly what I was after.

Yes, I'm planning to make a battery powered unit using a switching regulator in the future.

Also, when you say to use a 120 Ohm resistor for better "stability", what would that mean in practice as opposed to using a 1K or 2K resistor?

Thanks.

 

Hello,

For good regulation, the regulator needs a minimum amount of current.
A 120 Ohms resistor will take the current needed.
You can lower the value of R2 to have the 4 Volts output.

Bertus

 

Hello,

For good regulation, the regulator needs a minimum amount of current.
A 120 Ohms resistor will take the current needed.
You can lower the value of R2 to have the 4 Volts output.

Bertus

Hi Bertus,

Thanks for that.

With the components I have to hand I can make R1 135 Ohms (two 270R in parallel).

And for R2 I can put a 470R resistor in parallel with a 4.7K potentiometer, which (by my calculations) would give R2 a range of 0-427R and an output voltage of 1.25-5.2V. Which is fine for my needs.

Do you see any issues with this arrangement?

 

Hello,

It will work, but the potentiometer will be far from linear.
Just try to calculate some values on the different positions of the potentiometer.

Bertus

 

With this configuration it will be debatable as to if the Voltage regulator or the heater is the real heater in this case

 

With this configuration it will be debatable as to if the Voltage regulator or the heater is the real heater in this case

.. that is funny

A lower input voltage would help considerably.

 

As would PWM instead of regulating the voltage.

Bob

 

Here's an update for future reference...

I made the following circuit from the components I had knocking about:

The four "R1" resistors in parallel give an R1 value of 117.5R.

The "R2" resistor in parallel with the pot gives R2 a range of 0-427R.

I used 100uF electrolytic caps for C1 & C2.

I built it on perfboard and housed it in an empty old ATX PSU enclosure.

For the LM338 heatsink I used a heatsink taken from the aforementioned ATX PSU. It's basically a flat slab of aluminium 55 x 55 x 3mm.

I'm powering it from another old ATX PSU; which reduces the input voltage from 16V to 12V and the ATX PSU can easily handle the 2-3 Amp current demands.

I mounted a molex connector to create a port to insert multimeter probes so I have a digital display of output voltage.

I've put quite a few hours on it now and the unit works great, just as intended, very stable. The LM338 gets hot, it's dissipating 20W on the setup I'm using, but not hot enough that I can't hold it indefinitely. So, I'm guessing it gets into the 40-50°C range.

I've ordered a 470R linear pot to replace the R2 pot/resistor network so I can have easier fine control of the output voltage.

In the future I'm going to look into using a more efficient regulator.

Thanks for all the help.

 

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.

Power diss is (Vin - Vout) X Iload. If output min is actually 1.25 @ 4.5A with 16V input, power dissipation is about 70W which is impossible for a TO-220 device. Possible using a TO-3 device and huge heatsink.

 

You really need something like one of these:
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=380755888262

They can exceed 90% efficiency, and should be able to handle your 2A requirement. They are pretty compact at 1.715"L x 0.825"W x 0.530"H. It DOES take a while to get them to the USA from China. However, it would not be possible for you to "roll your own" regulator at these prices.

 

So, a longer term update for anyone searching on how to make an adjustable power supply for an e-cig, personal vapouriser, vape-tank, clearomiser, atomiser, whatever you want to call it.

I replaced the 4.7k log pot and 470R resistor with a 470R linear pot so the voltage is easy to adjust in the usual operating range.

I've been using the unit daily, sometimes all day long, and have had no issues at all. It functions excellently. No heat, or overheating, issues and no cut-outs.

I have two 6 foot lengths of 19 SWG wire (18 AWG, 1.02mm) running from the unit to a hand holder I made that houses the atomiser and momentary push switch. The switch is rated for 6 amps at 230V.

I make the atomiser coils from 0.2mm Kanthal wire.

The coils have a resistance of 1.6-2.0 Ohms. I run them at between 4-5 Volts (measured under load). So they are drawing 2-3 amps and dissipating 8-15 Watts depending on mood and what blend of juice I'm vaping.

It's so much better than vaping from an Ego style lithium-ion battery when you're not out and about.