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

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).

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[Q1]

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.)

[Q2]

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.)

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Regards,

Kanemaru

• 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,

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 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.

https://e2e.ti.com/support/logic/f/151/t/764984?tisearch=e2e-sitesearch&keymatch=faq%3Atrue

• Hi Dylan-san,