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Help with dynamic test load

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by imix500, Feb 22, 2009.

  1. Roff

    Roff Well-Known Member

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    To get 10 amps, you use one of the same circuits you already have. Just limit the range of the control pot to 1V. (1V/0.1Ω)=10A). You can run the wiper of a single pot to all op amp inputs.
     
    Last edited: Mar 16, 2009
  2. imix500

    imix500 Member

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    Measured the heatsink tonight. I get .08 C/W with the fan I have on there now moving around 135 CFM.
    I think I found a really good deal on dual darlington devices.
    Part # is A50L-0001-0118/A
    Here's a question- if the device has twin sections, is it reasonable to think they are matched? If not it's no problem to add a couple resistors.
     
  3. Roff

    Roff Well-Known Member

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    That explains why your device died at 24V and 27A.:(
    How did you measure the thermal resistance?
    Can you post a link to the datasheet?
     
    Last edited: Mar 17, 2009
  4. dave

    Dave New Member

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  5. imix500

    imix500 Member

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    Really? I usually plan on a 30C rise and judge the relative safety of the devices on the heatsink by how long I can leave my hand on it. 55C is about the the most we can take comfortably, so if I can leave my hand on it I'm usually ok with it. I see what you are saying though for sure. The math doesn't lie.

    I measured the heatsink by attaching two 50w aluminum resistors and dissipating 100w through them and measuring temp rise relative to ambient.
    Rth = Tr / Ph

    Rth - Thermal resistance
    Tr - Temperature rise
    Ph - Power applied to heatsink
     
  6. imix500

    imix500 Member

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    Unfortunately, the dual darlington module I found seems to be a legacy model Faunc doesn't support any more. I have a couple emails out to companies that sell these to industry. I might have to experiment with it to see how much drive current it needs, but it is a regular npn darlington with a blocking diode.
     
  7. imix500

    imix500 Member

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    It was obvious that thermal coupling was very important with these high power devices. Since there is so much surface area, I was wondering if silpad would be better than grease because of the difficulty in achieving a very flat layer. Also, I think the layer of grease was too thick. It was difficult to get it thin enough so that the mounting points on the corners of the module didn't bow down and possibly create a tiny void under the module.

    I felt both the heatsink around the module and the connection tabs which are directly connected to the junctions on the bottom plate. (I took a bad one apart) They seemed to get nearly as warm as the heatsink. Since the bottom plate is only about 1/8" thick I couldn't touch or get my temp probe on it.
     
  8. Roff

    Roff Well-Known Member

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    Do you have a value for Θcs, the thermal resistance of whatever goop you are putting between the device and the heat sink? You could measure it. I think I would try to do it with the actual power device, run at low enough power that it is not in danger of dying.
    It would seem that Θcs is important in measuring Θsa (heat sink to ambient). If Θcs is infinite, Θsa will appear to be zero.
    For heatsinks where Θsa is much higher than Θcs, the error will be insignificant. I doubt this is the case for your setup. To accurately measure a low value of Θsa, the power would need to be dissipated within the heatsink.
    You can measure the series combination of Θcs+Θsa by putting one probe on the power device, but it would need to be the actual current source device, because it is a function of contact area, etc.
    Caveat: I may be FOS, because I don't really have experience with this. I'm just an old retired EE.:eek:

    Here is an informative (I think) and entertaining article about Θcs of various thermal compounds.
     
    Last edited: Mar 18, 2009
  9. imix500

    imix500 Member

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    I use Thermalcote from Avid-Thermalloy. Thermal conductivity of 0.765Wm-1°C-1

    Using an online calculator, I get .189 for thermal resistance with a thickness of .1mm

    Am I close to an actual measurement there?

    I'm not sure I can measure the junction while it's on the heatsink as my thermocouple is inside of a 1/4-20 bolt. I have a very small PTC thermistor I could measure though.
     
  10. Roff

    Roff Well-Known Member

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    I used their calculator and got .02. I used 108x62mm as the area of the KS621K30. What did I do wrong?
     
  11. imix500

    imix500 Member

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    Yeah, I did it wrong. Converted to square mm twice. Your number is way better.
     
  12. Roff

    Roff Well-Known Member

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    Did you understand the point I made above?
    In other words, if your heat source (the 100W heater) has a poor thermal connection to the heat sink, the heat sink will appear to be much better than it really is.
     
  13. imix500

    imix500 Member

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    I do. Lessons learned will help when it comes time to mount up new devices.

    On another note, I think I've answered my own question about IGBT's. It seems they do not like being operated in the linear region at all and would be a bad choice for this.
     
  14. imix500

    imix500 Member

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    I should clarify- I'm confident with the thermal coupling of the heater during measurement. (I did monitor the case temp of the resistors as well)
    I think there were improvements to be made with the Powerex modules though.
     
  15. power58

    power58 Member

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    load

    The best load Ive seen and used is a Maxim design showcase titled
    "Dummy Load Maintains Contant Current" If you can't find the MAX480 op amp an LM324 works fine for duplicating 4 load cells or an LM358 for 2 load cells. An ICL8069 provides the 1.3 V program voltage that goes to the current program pot. A 9 volt battery provides Op amp power and ref volt power. The battery eliminates ground loops. Each Load cell can sink 10 amps depending on the MOSFET and Sense Resistor wattage.
     
  16. imix500

    imix500 Member

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    I actually have that article (collected quite a bit of info over the past year) but was turned off by the unavailability of the MAX480. Unfortunate as it looks like it was a great little opamp. I did try using the circuit with the OPA703 chips I had but gave up because of instability problems caused by my less than stellar design. This was long before I found this forum and the excellent help of its members.

    It's very similar to the topology I have now and once I get some new parts I look forward to see how it does around a LM324. I had to increase the power handling of the output device because of my load specifications. The cost of these high power devices calls for very creative sourcing as I have a very small budget. High power mosfets are still hard to find inexpensively so darlingtons had to be substituted.
     
  17. imix500

    imix500 Member

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    Back from the dead

    Hey everyone, bringing this one back from the dead. I had to lay this project down for a while, but now I'm back at it.
    I have rebuilt the whole thing to what we were talking about back in the spring. I now have two LM324 driving 6 power darlingtons.
    So far I have tested it up to 10v @ 30A and it's behaving extremely well. The only issue so far is this:
    Normally the whole thing draws about 50mA at 30A. If the load is turned off or disconnected while R4 is up at all the current consumption jumps to about 1A and slowly increases. The outputs of the opamps jump from about 3V to over 11V and saturate the Tip transistors. The old versions didn't do this so I was looking for reasons why this new version does and also for possible solutions.

    EDIT: Or, maybe the old versions did behave like this and I never really noticed. Any way around it?
    Thanks!
     

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    Last edited: Oct 4, 2009
  18. Roff

    Roff Well-Known Member

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    When you disconnect the load, the darlington becomes two diodes (b-e junctions) between the emitter of T1 and the current sense resistor (R19). T1 tries to drive R19. through these two diodes. In this attempt, the op amp must go to the positive rail in an attempt to provide the base current that T1 needs.
    Determine the maximum base current that T2 needs under normal operation, and the base voltage of T1 under the same conditions. Add a resistor in series with the collector of T2 that is as large as possible while not saturating T2 (Vc>Vb by a couple of volts). Then, when the load is removed, the op amp output will still saturate, and so will T2, which should limit the current.
    This may or may not be sufficient. I would need to see the entire circuit before I could say definitively whether or not it would work.
     
    Last edited: Oct 4, 2009
  19. power58

    power58 Member

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    You can eleiminate T1 , R7 and direct drive T2 from the op amp. I would change R5,R9 to 1K. Take C4 out of the circuit also. It is possible to go down to 100 ohms on R5,R9 as some MOSFET circuits go this low. Thanks for your latest work.
     
  20. imix500

    imix500 Member

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    Thanks for the replies.

    Roff, I'll try adding those resistors and see how they affect things. I'm not so worried about the opamp going to a rail as its driving a pretty high impedance, but I don't have room for heatsinks for the TIP31's and under normal operation they dont need them. When they saturate obviously they dissipate a lot more heat. If they still get too warm with the collector resistors I'll figure out something for cooling them.

    Power58, I'd love to drive the power modules directly from the LM324, but this condition of the load being removed makes me think I'd be blowing a lot of output stages. When load is removed base drive current of T2 goes from about 15mA to about 150mA.
     
  21. Roff

    Roff Well-Known Member

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    Is there a reason why you connect the collector of T1 to +12V instead of to the output? That would solve your problem, I believe.
     
    Last edited: Oct 5, 2009

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