LM311P Comparator

Discussion in 'General Electronics Chat' started by whatisgoingon, Feb 6, 2013.

  1. whatisgoingon

    Thread Starter New Member

    Jan 30, 2013
    9
    0
    Hey guys,

    I need some help determining how to find the maximum current that this comparator can sink. I am using a pull up on the output of the comparator to a 5 V supply. I need to determine the resistance I need to limit the current appropriately.

    I cannot find anywhere on the datasheet the maximum collector/emitter current (or maximum current it can sink when grounded).

    Does anyone know where I can find this?
     
  2. SPQR

    Member

    Nov 4, 2011
    379
    48
    Hi,

    It seems to me that the primary issue there would not be max current, but rather typical current.
    If you look at the charts you'll see current when output high and when output low, and those are about 5mA.

    And assuming you are pulling up to 5V, you could do I=E/R, R=E/I and you would have 5/.005 which would be a 1k resistor.

    I think the major issue is not max current, because the circuit will only draw the current that it "needs" -
    but rather max volts, above which you'd toast the chip.

    The experts will be by in a flash to make comments.
     
    Last edited: Feb 6, 2013
  3. KnRele

    New Member

    Jan 7, 2013
    20
    8
    From the datasheet, there are several places where a load current of 50 mA is mentioned, once in the introduction, and there is a spec of the output low voltage (VOL) when the output current is 50 mA.

    That spec indicates that when the output is sinking 50 mA, the voltage on the output is typically 0.75 V and could be as much as 1.5 V.

    However, if you are only needing to pull the output to a good High level for subsequent logic circuits, you do not need to pull anywhere near 50 mA -- note that the VOL for 8 mA is 0.4 V max -- so aiming for an output load of 8mA will be more than sufficient. That will correspond to a pull-up resistor of (5 V - 0.4 V) / (8 mA) = 575 Ohms but even that will likely be using more power than is really needed.

    Look at the application examples, they show resistors of 1 kOhm or 2 kOhm to +5 V -- these will still give good output voltages if you don't need that much current. 1 kOhm will give 5 V / 1 kOhm = 5 mA or thereabouts.

    For a maximum current output, look at the Figure 8 on page 10. This shows how the output voltage rises with increased output current, up to 200 mA -- and associated power dissipation. So 200 mA looks like the max. that the transistor is going to be able to deliver, and it will be under severe strain. There is also the caveat that any such condition should not last for more than 10 seconds.

    A lot of this will depend on what you intend to connect to the output and how sensitive this is to various voltage levels.
     
    SPQR likes this.
  4. whatisgoingon

    Thread Starter New Member

    Jan 30, 2013
    9
    0
    I see,

    I actually needed to drive 3 thyristors which needs 8-80 mA (gate trigger).

    I guess I cannot trigger them directly with the comparator. Do you guys think a mosfet will be appropriate? See diagram.

    I used 30 ohm so I can get 5/30 = 166 mA (that is going to get divided by 3 for each of the gate currents). So hopefully around 50 mA for each thyristor gate trigger current.
     
  5. SPQR

    Member

    Nov 4, 2011
    379
    48
    By googling, there are many circuits that use a simple 2n2222 as a driver for a thyristor.
    Are you going to be switching it quickly or is it just an on/off switch?
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    A shunt driver like you drew is very inefficient. An active pullup, like an NPN emitter follower, or a PMOS common source, or PNP common emitter, would be much more efficient. That way, you only have to source current when the thyristor gates need it.
     
  7. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Look at the data sheet, that's a guaranteed spec for comparators.
     
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