LM339 Help

Discussion in 'General Electronics Chat' started by crash563, Apr 12, 2013.

  1. crash563

    Thread Starter Member

    Feb 25, 2013
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  2. #12

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    The top row is 3 constant current generators controlled by R2, T12, R1.
    The left side is a Darlington differential pair like an op amp input.
    T5 and T6 split the current through the differential pair.
    Collector T4 drives T7 to T8 which is the output "open collector" for the chip.
     
  3. WBahn

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    Mar 31, 2012
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    Is it just me, or does it seem like the base of T7 is hard tied to ground?

    Also, and I'm a lot less sure on this one, but it seems like the collectors of T1 and T4 be tied to the collectors of T2 and T3, respectively instead of tied to ground?
     
  4. patricktoday

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    Feb 12, 2013
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  5. ScottWang

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    Aug 23, 2012
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    The wires connection of T6 and T7 could be wrong.

    The correct wiring should be like below:

    [​IMG]
     
  6. Ron H

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    Tying the collectors of T1 and T4 to ground allows the input common mode range to include the negative rail. If they were connected to the T2 and T3 collectors, T1 and T4 would be saturated when the inputs are at the negative rail.
     
  7. #12

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  8. patricktoday

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    Well it's definitely an error but I merely followed up on WBahn's and apparently OP's catch. Wouldn't do a whole lot of good as pictured, would it? ;)
     
  9. #12

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    I guess we'll have to wait and see if crash spotted the error or was just wanting to know the general functions inside the chip.
     
  10. crash563

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    Feb 25, 2013
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    After loking at the ti datasheet it looks likd a connection error but I just wanted a complete circuit operation
     
    #12 likes this.
  11. crash563

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    Feb 25, 2013
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    So how does that diff pair work
     
  12. WBahn

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    Makes sense. It probably also improves the matching of the two channels since one of the sources for mismatch is no longer in the channel.
     
  13. Ron H

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    It works the same as any diff pair. The two NPNs at the bottom are a current mirror.
     
  14. WBahn

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    The current source at the top makes it so that the total current going through both transistors in the pair is a constant. The two input signals determine how that constant current is split between the two paths. Imagine relacing the current mirror at the bottom with two resistors, one in each path. In the output branch (the one that the base of the next stage input is connected to) as the input signals change the current that goes down that branch, the voltage at the input to the next stage changes. The bigger the resistor, the more the voltage changes by for the same change in current. The current mirror is an "active load" and is basically a current source with a high output impedance. It's output impedance acts as the load resistor that the output voltage (the input voltage to the next stage) is developed across.
     
  15. crash563

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    Feb 25, 2013
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    The open collector (T8) needs a pull-up resistor, but I have a hard time understanding its operation. Is it when the comparator is following the +VSAT or HIGH, the transistor is off or vice versa?
     
  16. #12

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    When the output is "high" T8 is off and the pull up resistor does all the work. When the output is "low", T8 is on and it dumps both the current from the pull up resistor and the current from the load.
     
  17. crash563

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    Feb 25, 2013
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    I am confused about the open-collector still. When the comparator is following the +VSAT due to the noninverting input being higher than the inverting, you are saying that the transistor is off allowing the voltage to flow across the pull-up resistor to the load.
     
  18. #12

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    Yes.......
     
  19. crash563

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    Feb 25, 2013
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    Now is this due to the current generated by T10
     
  20. #12

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    Yes. When T7 is off, T10 current goes to T8.
     
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