LM338 and power MOSFET overheating...

Discussion in 'The Projects Forum' started by jlf1200, May 20, 2011.

  1. jlf1200

    Thread Starter New Member

    May 20, 2011
    There are lots of posts out there dealing with power MOSFET switches and 12VDC lights (and/or motors). I'm experiencing some heating issues and I'm pretty stumped so I'll just describe the circuit and hopefully the mistake will be obvious to the more experienced readers:

    Table top power supply (from HP laptop) rated at 19.5VDC and 4.6A output is run through an LM338T regulator (TO-220 package with standard alu heatsink) set up as a current regulator to feed ~2.7A to the circuit (I=Vref/Rref=1.25V/0.47ohm). Three 12V/10W halogen lamps are connected in parallel between the LM338 adj pin and the drain of a IRLZ34N Hexfet power MOSFET, also fitted with a standard heatsink.

    With the FET gate set to high the lights come fully on and then fade quickly-- which I believe is caused by the LM338 overheating and turning itself off through protection circuitry. It gets very hot, as does the FET (as in, melt the breadboard hot).

    What confuses me is that the LM338 is rated at 5A (7A peak) and the FET is rated at some ungodly source current. I am way below the max for both. There is no official power dissipation rating for the 338 that I can find (the datasheet says "internally limited" in fine print). The power dissipation for the FET is 68W at 25C, which I'm sure is much lower at high temp, but the Rds drops to almost nothing at high current so my expectation was that a sub-3A circuit would be no problemo... What am I missing here?

  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    U need to post ur schematic
  3. Kingsparks


    May 17, 2011
    Hi jif1200
    From the information you have noted the unit should work. In other words, the specifications meet the power requirements but you still have problems. You say the pw. supp. is set up as a current limiter, post the schematic you have for this set up. Either you have a bad component, sort of unlikely as it would probably have gone up in smoke by now, still possible but...,or your circuit is the problem.

    BTW, the LM138/338 datasheet does have an internal power limit, not as such but on the first page it says; "5A over a 1.2 to 32V output range." Figure the power from that. ;)
  4. SgtWookie


    Jul 17, 2007
    19.5v-12v=7.5v dropped across the regulator; with 2.7A flowing through it, that will mean the regulator portion will have to dissipate 7.5v*2.7A = 20.25 Watts of power as heat; 1.25/7.5 * 20.25 = 3.375 Watts in the resistor (5.4W rating minimum required) and 16.875 Watts in the regulator itself. If your resistor is not rated for 5.4W or more, it's probably overheating itself, and might be increasing its' resistance.

    With no heat sink, thermal resistance of the junction to ambient air temp is 35°C/Watt. for the K (steel TO-3) package, and 50°C/Watt for the T (TO-220) package.
    Thermal resistance of the junction to case is 1°C/Watt for the K, and 4°C/Watt for the T packages.

    You're probably using the "T" package, so 50°C/Watt and 4°C/Watt are your numbers.
    An LM338's max temp is 125°C.

    If you start off at 25°C/77°F, then you have 125°C-25°C = 100°C before the regulator shuts down due to overheating.

    Worst case scenario, (no heat sink), that means 2 Watts of power dissipation before the regulator shuts down when operating at room temperature.

    Best case scenario (an ideal heat sink that always maintains room temperature, perfectly bonded to the regulator using heat sink compound) the regulator will handle 100°/4° 25 Watts before it overheats. As calculated above, your regulator currently has to dissipate 16.875 Watts of power, which means you will need about 68% of that ideal heat sink.

    You will need to use either a very large heat sink, a fan cooled heat sink, reduce your load current, or some combination of the above.

    BTW, three 10W, 12v bulbs in parallel is 30 Watts total; 30/12 = 2.5 Amperes, so you really need an 0.5 Ohm resistor rated for at least 5.4 Watts for reliability. You could use a pair of 1 Ohm 3 Watt (or higher rated) resistors in parallel instead.
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    Dang, I must have closed the window before I hit POST. Oh well, the Sarge had a much more thorough analysis of the LM338 then mine was anyway.

    As far as the FET goes, there is significant power being generated there in the slow turn off. Think of it this way: a constant current source generates the constant current (LDO!) up to the point the output voltage starts to starve it out. As the FET turns off slowly any increase in LM voltage drops across the FET, until the LM's maximum output voltage minus the FET voltage begins to work on the lamps (ie, is less then the 12 V the lamps want), only then do the lamps begin to dim. Thus the fet sees the full 2.7A current and an increasing voltage up to the point the lamp voltage goes to zero and they stop conducting.

    You don't say for how long or how often the lamps are turned off so I can't compute a power estimate, but it isn't negligible.
  6. jlf1200

    Thread Starter New Member

    May 20, 2011
    Thanks Wookie and everyone. That makes sense. It sounds like I'm just running into the limitations of an IC regulator that burns off energy as it regulates the output.

    Here are two schematics that I was considering-- one current regulation scheme and another with voltage regulation instead.

    Perhaps I need to ditch the linear regulator and reconfigure the circuit to use the energy from the power supply more effectively?... or just purchase a different supply that delivers what I actually need.

    Still-- this doesn't explain why the power FET is getting crazy-hot, right? I just can't imagine it sinking 20A or more when less than 3A makes it hot enough to melt the plastic bushing on the heat sink screw. Do I just need to go get a HUGE heat sink to work with this kind of power circuit?
  7. jlf1200

    Thread Starter New Member

    May 20, 2011
    ...and BTW, I was testing this circuit withOUT PWM, just giving the FET gate 5V to simulate "on." So whatever is happening, switching issues are not a factor.
  8. SgtWookie


    Jul 17, 2007
    In your 1st schematic, you have the 0.47 Ohm resistor in the wrong place. It needs to be between the OUT and ADJ terminals, and the regulated current output is taken from the ADJ terminal. The way you have it wired, the regulator would output maximum current continuously, as the load is on the wrong side of the 0.47 Ohm resistor.

    If you try to limit the current using either the current regulator or the voltage regulator, you'll still be dissipating power as heat in the regulator portion.

    What you might be able to get away with is straight PWM of the MOSFET, but never exceed 62% duty cycle, and use a base frequency of ~500Hz or higher. 19.5v * 0.62 = 12.09v - or if you prefer, 70.5% for effectively 13.75v.

    The IRLZ034 is rated for Vdss=60v, Rds(on)=0.05 Ohms, and Id=30A; the total gate charge is 35nC.

    With an average of 2.7A flowing across 0.05 Ohms, you should have ~0.135 Watts power dissipation in the MOSFET, plus the power dissipated while being switched on/off.

    Note that while you CAN use a 555 timer to drive a MOSFET gate, you will need to power it with ~6.5v or so, as the output of a BJT (transistorized) 555 timer won't go higher than about Vcc-1.3v even under light load; if you used 5v for Vcc, the output would only be 3.7v or slightly less, which would likely toast the MOSFET.

    MOSFETs are very sensitive to static electricity, and you will destroy them instantly if Vgs goes higher than +20v or below -20v.

    You need to get the Vgs (voltage on the gate relative to the source terminal) to 5v quickly in order to avoid the "linear region", and from 5v back to 0v quickly as well. Power dissipation is high when the MOSFET is changing states from ON to OFF or vice-versa; which is why you want it to change state quickly. Vgs being between ~1.5v and ~4v is the "danger zone".

    While the gate needs to be driven quickly, it also needs to be free of "ringing", which usually requires a resistor of around 22 Ohms between the MOSFET gate driver and the MOSFET gate, and the wiring from whatever is driving the gate to the gate itself should be as short as possible. Otherwise, the capacitance of the gate and the inductance of the wiring creates a series LC circuit, which will resonate like a struck bell when the MOSFET is switched on or off. This will cause a great deal of power dissipation in the MOSFET causing heat, burning up your MOSFET quickly.
  9. jlf1200

    Thread Starter New Member

    May 20, 2011
    Firstly, thanks so much for your help. A few more points, if you have time:

    Oops-- the drawing is incorrect. The circuit was built correctly, however, with the output taken from the Adj terminal.

    This is pretty clever-- essentially doing away with the need to adjust or clamp voltage across the lamps. I am probably going to switch to 20W lamps which draw ~1.5A each, meaning I could run the power supply (4.5A) without the need to regulate anything.

    I should have mentioned-- I'm going to be using a microcontroller to send the gate signal (either an Arduino or a PIC), so theoretically I should be able to get a clean 5V signal.

    So if I understand you correctly, you're saying that the FET will cook if it spends a lot of time in "linear region" as a result of Vgs being too low. I thought getting the FET into saturation had more to do with Vds. Where on the sheet do I determine the appropriate Vgs to stay out of linear operation? (I only see the threshold Vgs listed between 1 and 2 volts).

    Yes, I plan on wiring the controller out to the gate in very close proximity. I'll use the small resistor as you advise.
  10. Audioguru

    New Member

    Dec 20, 2007
    The Vgs determines if the Mosfet is turned on or is turned off.

    Why do so many people look at the threshold voltage rating without seeing that then the current through the Mosfet is only 0.00025A which is almost nothing and the Mosfet is almost turned off?

    Instead the on-resistance rating must be seen where the gate voltage is 4.0V and 5.0V which is much higher than the threshold voltage then the Mosfet is turned on hard with a very low resistance.
  11. DC_Kid

    Distinguished Member

    Feb 25, 2008
    what exactly does the PWM signal look like? any scope of it? those halogen bulbs are also inductors, coupled to the C of the FET there's likely some ringing going on there too. make sure there are no voltage spikes hitting the regulator, try also placing a diode across the load.
  12. SgtWookie


    Jul 17, 2007
    You need to look at the Rds(on) specification, as the Vgs and usually Id will be shown on the same line. In some cases, you may see multiple Rds(on) and Id specs for different Vgs.

    The threshold voltage is really not terribly useful.