Hi – I recently built a simple voltage regulator circuit using LM350, straight off the datasheet. (I’m looking to power a piece of equipment requiring 12.6V at 2.6 Amps.) I first tested it with about a 1 Amp load and everything was fine. I then hooked it to an automotive bulb (printed rating 27W). It stayed lit for a few seconds and then the chip’s thermal and/or current protection kicked in. I had the 350 on what I thought was a reasonable heatsink for the job, but to make sure I replaced it with a large 3”x4” heatsink and added a fan too. The bulb stayed on a few more seconds until it dimmed out. The big heatsink wasn’t that hot. The LM350 says it’s “guaranteed” to put out 3 Amps. What am I doing wrong?
- In a perfect world, 27W / 12.6V would be 2.1A. I understand that AC Watts may be derated based on efficiency, but I thought DC Watts equaled V x A? My ammeter shows just over 2 Amps while the bulb is lit.
- Do you think I might just have a bad LM350, or that there is some reason why it’s acting like a regular 1.5A LM317? Or do I really need gi-mongous cooling for such a seemingly modest circuit?
- Sure, I could redo the circuit with a pass transistor to share the load, but there would seem no reason to – the 350 says it can handle the Amps.
I would appreciate any ideas you might have. Thanks!
- Steve
- In a perfect world, 27W / 12.6V would be 2.1A. I understand that AC Watts may be derated based on efficiency, but I thought DC Watts equaled V x A? My ammeter shows just over 2 Amps while the bulb is lit.
- Do you think I might just have a bad LM350, or that there is some reason why it’s acting like a regular 1.5A LM317? Or do I really need gi-mongous cooling for such a seemingly modest circuit?
- Sure, I could redo the circuit with a pass transistor to share the load, but there would seem no reason to – the 350 says it can handle the Amps.
I would appreciate any ideas you might have. Thanks!
- Steve
What is the final application?
Bulbs tend to make poor loads due to variable resistance> unless current limit is set high in relation to bulbs cold resistance.
Remeasure all voltages with a 1A load and then again with the bulb. You're either in current limit or not enough V headroom? You must have around 3 V or more for dropout under full load.
The LM350K package has better thermals for heatsinking so you may need to claculate that at full load and worse ambient.
Bulbs tend to make poor loads due to variable resistance> unless current limit is set high in relation to bulbs cold resistance.
Remeasure all voltages with a 1A load and then again with the bulb. You're either in current limit or not enough V headroom? You must have around 3 V or more for dropout under full load.
The LM350K package has better thermals for heatsinking so you may need to claculate that at full load and worse ambient.
You didn't mention input voltage. This is important, as it will affect the power dissipated by the LM350; the power will be = (Vin - 12.6)*(I).
I would assume that if you ask the poor little device to dissipate more than 20 W, then it will shutdown quite quickly. So with a current of 2.6 A, that means a Vin of no more than 7.7 V higher than Vout, so Vin must be less than about 20 V. And if it was me, I'd want it even lower. Once you account for the heatsink and insulator thermal resistances, and add that to the 3 to 4 degC/W of the device itself (junction-to-case), you probably will have less than 20 W to play with.
I would assume that if you ask the poor little device to dissipate more than 20 W, then it will shutdown quite quickly. So with a current of 2.6 A, that means a Vin of no more than 7.7 V higher than Vout, so Vin must be less than about 20 V. And if it was me, I'd want it even lower. Once you account for the heatsink and insulator thermal resistances, and add that to the 3 to 4 degC/W of the device itself (junction-to-case), you probably will have less than 20 W to play with.
According to the National Semi (TI I guess now) it should do 4A.
Just.
However, you could add current boost PNP (or a few) as in the National Semi data sheet on page 11.
http://www.ti.com/lit/ds/symlink/lm150.pdf
Look at what they have done with the MJ4502. (they use the exact same cct in the LM317 datasheet)
This one is set up so that the MJ4502 starts to take over the current load at just under 20mA. ( 0.6v / 33R )
A little low for my taste but at that level you won't need a heat sink for the 350, just the MJ4502.
The MJ4502 is a TO-3 package...MJW1302AG will be easier to work with. TIP36C as well
Make sure there is a minimum load (LM350: 10mA) and add a few more mA for transistor leakage.
Just.
However, you could add current boost PNP (or a few) as in the National Semi data sheet on page 11.
http://www.ti.com/lit/ds/symlink/lm150.pdf
Look at what they have done with the MJ4502. (they use the exact same cct in the LM317 datasheet)
This one is set up so that the MJ4502 starts to take over the current load at just under 20mA. ( 0.6v / 33R )
A little low for my taste but at that level you won't need a heat sink for the 350, just the MJ4502.
The MJ4502 is a TO-3 package...MJW1302AG will be easier to work with. TIP36C as well
Make sure there is a minimum load (LM350: 10mA) and add a few more mA for transistor leakage.
I had a similar problem with a 317 driving a UHER 4400 drawing 420mA and later out of frustration I used the LM338Ksteel The problem is that if the package is the "T" package the maximum that it can handle is 15 (FIFTEEN) WATTS ONLY I could still hold the heatsink with large fins and it was just warm 100mm long by 150 mm approx but the chip overheated BEFORE the heat could reach the heatsink is my diagnosis as I tried three chips with the same results. The input was 15 volts and the output 7.5 volts
The tungsten filament in a light bulb is a PTC (positive temperature coefficient) resistor.
It increases in resistance from cold to hot by a factor of at minimum ~ 3times and for short life halogen approaching 10times.
A bulb rated at 27W 12V has a hot resistance of 5r33.
The cold resistance could be anywhere from 0r5 to 1r8
Measure the cold resistance !!!!
It increases in resistance from cold to hot by a factor of at minimum ~ 3times and for short life halogen approaching 10times.
A bulb rated at 27W 12V has a hot resistance of 5r33.
The cold resistance could be anywhere from 0r5 to 1r8
Measure the cold resistance !!!!
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