• State TI Thinks Resolved
  • Locked Locked
  • Replies 4 replies
  • Answers 2 answers
  • Subscribers 34 subscribers
  • Views 310 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

DS90LV011A: Application issues of chips

Tonyx.wu
Genius 3915 points
Part Number: DS90LV011A

Do I need impedance for the differential output OUT of this IC? How much impedance should be placed?

Why does this input end need to pull down 10K？

  • 0
    Guru 316690 points

    In general, LVDS uses an impedance of 100 Ω. (For very short traces, this does not matter.)

    The pull-down ensures a valid voltage level if there is no valid signal at the connector.

  • 0
    TI__Guru 72711 points

    Hi Tony,

    Clemens is correct.

    Tonyx.wu said:
    Why does this input end need to pull down 10K？

    TTLIN is an input pin which should be biased to something (don't leave it floating). The value of the resistor isn't too important but you wouldn't make it too strong (super low value) because it would force anything connected to it to be able to drive the current (a 3.3V output to a 100 ohm resistor to GND would drive 33mA for example). At the same time, a weak resistor (larger value like 100k) would be more susceptible to noise and be a weaker bias to GND. I'm sure you could get away with a 4.7k resistor or a 20k as well.

    Tonyx.wu said:
    Do I need impedance for the differential output OUT of this IC? How much impedance should be placed?

    Yes, impedance matching is important for differential signals (helps with reflections). For LVDS the most common impedance is 100 ohms.  

    -Bobby

  • 0 in reply to BOBBY
    Genius 3915 points

    What are the power consumption and static current of this chip, respectively?

    How did you design this chip for low power consumption?

    Is it directly controlling the power supply? Please provide me with a detailed explanation, as it is already in the design stage and is very important. Thank you!

  • 0 in reply to Tonyx.wu
    TI__Guru 72711 points

    You can estimate the power consumption when the device is not driving a load on the differential side (think of it as like a standby power) by looking at the electrical parameter: Idd (power supply current no load). The max current draw is 8mA, Power is calculated as P=I x V. If 'I' is 8mA and V=Vcc=3.3V then the power during this standby consumption is then 8mA x 3.3V = 26.4mW

    When the device is transmitting data across the differential outputs, the Idd goes up to 10mA max. So using the same equation 10mA x 3.3V would provide 33mW across the device. 

    If we calculate the power when the differential signal is driving (static power) then we can simplify this to be 450mV (VODmax) and this voltage is spec'd around LVDS standards which use 3.5mA going across a 100 ohm load. So 3.5mA x 450mV = ~1.6mW of power across the termination resistor (assuming you have one) which you could include in your power calculation since the current is being sourced/sunk by our device. 

    If we assume the device outputs the 3.3V and the load is 250mV (VOD min), then the power dissipation from our device's drivers would be (3.3V-0.25) x 3.5mA = ~10.7mW. 

    If we add the power from the idle/standby state with the power from the device's output drivers then we get 26.4mW + 10.7mW = 37.1mW (worst case for total device power dissipation). You can add the resistor's power load (1.6mW) if you want to add it to the total system's max power dissipation. 

    The device doesn't have a disable pin so the drivers are always active. I think the easiest way to conserve power is to powerdown the power rail that the device is connected to when it's not transmitting data. 

    -Bobby