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Thread: Help with dynamic test load

  1. #91
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    Did you understand the point I made above?
    It would seem that Θcs is important in measuring Θsa (heat sink to ambient). If Θcs is infinite, Θsa will appear to be zero.
    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.
    Ron



  2. #92
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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.

  3. #93
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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.

  4. #94
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    Default 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.

  5. #95
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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.

  6. #96
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    Default 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 by imix500; 5th October 2009 at 01:42 AM.

  7. #97
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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 by Roff; 5th October 2009 at 02:26 AM.
    Ron


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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.

  9. #99
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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.

  10. #100
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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 by Roff; 6th October 2009 at 03:26 AM.
    Ron


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    I'm pretty sure I tried doing that in the beginning. Is it possible that the set load current will vary with applied voltage by tying the collector to IN+ ? That's what I seem to remember.

  12. #102
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    Quote Originally Posted by imix500 View Post
    I'm pretty sure I tried doing that in the beginning. Is it possible that the set load current will vary with applied voltage by tying the collector to IN+ ? That's what I seem to remember.
    No. The emitter currents from T1 and T2 flow through the current sense resistor (R19). You want the collector currents of T1 and T2 to flow through the load, so the only error is T1 base current. This assumes you are applying a known voltage on the control input (Vctrl), and want to be able to calculate the output current as Vctrl/R19. I realize you are not doing this.
    Nevertheless, it should still solve your high current problem caused by removing the load.
    Ron


  13. #103
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    Excellent. Thanks!

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