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Quick Question About a LM2901N

Discussion in 'General Electronics Discussion' started by Supercap2F, Jun 21, 2014.

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  1. Supercap2F

    Supercap2F

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    Mar 22, 2014
    Hey Everyone!

    First of all it appears that the LM2901N is not able to put out a high level voltage. Is that true? If so then I will need a pull up resistor to drive a CMOS input. What value should the pull up resistor be? Is there a formula to calculate it?

    Thanks Everyone. :)

    Dan
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    The LM2901 is a so called open-collector component. This means that it can pull to ground (low), but positive potential (high) has to be provided externally, typically by a resistor.

    Advantages of this type of circuit are:
    • can be used to level-shift by using a different voltage for the output circuit
    • can be used to vreate a so called wired-or (Google) to combine several outputs into one logic signal
    The value of the pull-up resistor is entirely up to you. Probably the two most important issues to consider are:
    • power requirements
    • speed
    • output current in high state
    Power requirements:
    In the low state of the output the resistor takes the full voltage from Vcc to Gnd and will consequently dissipate power. You want to have a large resistance to dissipate very little power.

    Speed:
    The resistor forms a low pass filter together with all the capacitances that are connected to the output (wire traces, input circuits etc.). You want the output to switch fast, therefore you make the resistance small to minimize the RC time constant.

    These requirements are contradictory: a large resistance will dissipate little power, but slow down the circuit. And vice versa. Therefore you have to find a suitable compromise, taking into account factors as e.g. the expected mean on-time of the output transistor (aka duty ciycle of the output signal). If the output is on for a comparatively short time and off for a long time, you may be able to tolerate higher dissipation during that short time much easier than if the output is low for a long time.

    Output current in high state:
    The load current (from other inputs connected to the output) will drop some voltage across the load resistor. This voltage drop has to be small enough to still ensure a logic high level. With CMOS gates the input current is (almost) negligible and therefore not an issue. If you drive a load that needs considerable current (e.g. a bipolar inpu circuit) you have to take into account the input current of this circuit as it will typically not be negligible.
     
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  3. Supercap2F

    Supercap2F

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    Mar 22, 2014
    It does not need to be very fast... So how does a 5K6 resistor sound?

    Thanks

    Dan
     
  4. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Hi Dan
    The device can sink 20mA and has a internal base current of 80uA so that's a gain of 250. The value of resistor will depend on the desired response time of the output. But there are min and max values, you will have to choose one in between that fit the requirements. Say 4K7 to start with.

    To work out the min and max values as follows.

    VinH = ( High Level of gate your driving)
    VinL = (Low Level of gate your driving)

    I pull up min = Leakage through comparator when off + Leakage current through logic gate your driving

    Rmax = V+ - VinH / I pull Up min
    Rmin = V+ / 20mA(max current of comparator) - Leakage through logic gate.

    Thanks
    Adam
     
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  5. Arouse1973

    Arouse1973 Adam

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    You beat me again Harald :) I must have been typing this as you posted.
     
  6. Supercap2F

    Supercap2F

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    Mar 22, 2014
    Hey Adam!

    Can you explain where to find the "leakage through comparator when off" and "leakage current through logic gate" on a data sheet? :D

    All the best

    Dan
     
  7. Harald Kapp

    Harald Kapp Moderator Moderator

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    The "leakage through comparator output when off" would be the high-level output current, 0.1nA for this comparator.
    The "leakage current through logic gate" would be the input current of the logic gate, which in the case of a CMOS gate is a leakage current because the input is voltage drivven and input current is ideally 0mA. Other logic families, e.g. the venerable 74xx series, may have much higher input currents in the mA range.
     
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  8. Supercap2F

    Supercap2F

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    Mar 22, 2014
    OK heres what I came up with.

    Gate #1 = 1uA leakage, 3.5V high, 1.5V low
    Gate #2 = 1uA leakage, 4V high, 1V low
    Supply = 5VDC

    Gate #1 max resistance = 999,900Ω
    Gate #2 max resistance = 1,499,850Ω
    Gate #1 min resistance = 250Ω
    Gate #2 min resistance = 250Ω

    So a 2KΩ resistor will work good. I picked that value because I think the bands on it are nice looking. LOL Remember that Harald? :p

    Thanks so much for the help! :D

    Dan
     
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    I'd prefer black - red - yellow, unfortunately this is an invalid color code for resistors ;)
     
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  10. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Thanks Harald
    Adam
     
  11. Supercap2F

    Supercap2F

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    Mar 22, 2014
    Not any more. :D

    Resistor.jpg

    Dan
     
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  12. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Are they made in Deutschland by any chance :)
     
    Harald Kapp likes this.
  13. Supercap2F

    Supercap2F

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    Mar 22, 2014
    No, they're made in photoshopville.:p

    Dan
     
  14. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    :rolleyes::rolleyes::rolleyes:
     
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