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

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

V_OH = V_CC − 0.6 V × (I_OH / −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.

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[Q1]

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).

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[Q2]

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.

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[Q3]

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?

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Regards,

Kanemaru

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

V_OH = 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 V_OL vs. I_OL, Line B is V_OH vs. I_OH. (And 40 mA is outside the recommended operating conditions.)

Hi Dylan-san,

Thank you for your response.

I understand. Thank you very much.

Regards,

Kanemaru