DS90LV804 Unused input pin (INn+/INn-)

F.N

Other Parts Discussed in Thread: DS90LV804

Hi team,

This customer is going to use DS90LV804 as 2-channel LVDS buffer.
And the the customer wants to reduce consumption current of unused 2-channel.
So, the customer asks us which input connection can reduce consumption current,
both INn+/INn- to GND or both to VDD.

I think both INn+/INn- to GND connection can reduce it.
Is this correct? Please let me know.

Best regards,
Fumio Nakano

over 7 years ago

0 Lee Sledjeski over 7 years ago

TI__Mastermind 43795 points

Since the DS90LV804 does not include internal termination resistors, the best approach is to pull IN+ = VDD and IN- = GND. This will prevent any extra current consumption within the DS90LV804.

If the input voltage on IN- and IN+ is the same, the input is at the threshold and the output is likely to switch and oscillate with system noise. This will increase the current within the DS90LV804.

Regards,

Lee

0 F.N over 7 years ago in reply to Lee Sledjeski

TI__Expert 8760 points

Lee-san,

Thank you for your prompt reply.

I have a question.

DS90LV804 datasheet FEATURES say " ・On-Chip 100Ω Input and Output Termination" .
Is this a typo?

・Datasheet : www.ti.com/.../ds90lv804

Best regards,
Fumio Nakano

0 Lee Sledjeski over 7 years ago in reply to F.N

TI__Mastermind 43795 points

Nakano-san,

Sorry for the confusion, I was thinking about a different device. Since there is an internal 100 ohm termination I would connect IN- directly to GND and IN+ to VDD with a 4.7K resistor. This will provide the input with approximately 65-70mV of offset. This is enough voltage to keep the unused channel at a static level.

Regards,

Lee

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DS90LV804 Unused input pin (INn+/INn-)