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SN74LV4T125: About VOH vs IOH Graph at VCC=5.0V

Guru 20470 points
Part Number: SN74LV4T125

Hi Team,


We understand that VOH is 4.4V(min) at VCC=5.0V, IOH=-16mA.

However, we would like to know Minimum VOH at VCC=5.0V and IOH= less than -16mA (so from 0mA to 15mA).




Could you please let us know if you have “VOH vs IOH Graph at VCC=5.0V”?

If you don’t have that Graph, could you please let us know the trend of VOH?

(For example, If IOH decrease, VOH decrease so there is Linearity.)



Could you please let us know if you have “Calculation method of VOH”?

For example, if VCC=5V and IHO=0/4/8/12/16mV, VOH=xx/xx/xx/xx/xxV.)





  • For currents in the allowed range, the output transistors behave like resistors, i.e., the voltage drop is linear with the current:

    VOH = VCC − 0.6 V × (IOH / −16 mA)

    There is no graph for the worst-case voltage drop, only for typical values:

  • Hi Clemens-san,


    Thank you for your prompt reply.

    I have three questions.




    Trend of “VOH” is different depending on "VCC".

    Is this a correct trend?

    -When Vcc=3.0V and IOH increase(From 5mA to 8mA), VOH decrease(From 2.7V to 2.6V).

    -When Vcc=4.5V and IOH increase(From 8mA to 16mA), VOH increase(From 3.7V to 3.8V).




    We would like to know “VOH = VCC − 0.6 V × (IOH / −16 mA)”.

    If IOH=-8mA and VCC=5V, VOH=2.2V

    Is this result correct?


    I am sorry if my interpretation is mistaken.




    We understand that Voltage drop at IOH=-40mA is around 1V.

    Is my understanding correct?


    And, could you please let us know the different of Line A and Line B?





  • When VCC is larger, the gate-to-source voltage at the output transistors also is larger, so the transistors' RDS(on) and voltage drop become smaller.

    VOH = 5 V − 0.6 V × (−8 mA / −16 mA) = 5 V − 0.6 V × 0.5 = 5 V − 0.3 V = 4.7 V

    The graph uses the opposite sign for the current; Line A is VOL vs. IOL, Line B is VOH vs. IOH. (And 40 mA is outside the recommended operating conditions.)

  • Hi Kanemaru,

    To add to to the info Clemens has provided, please refer to the FAQ below that covers this topic.

  • Hi Dylan-san,

    Thank you for your response.

    I understand. Thank you very much.