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SN65HVD3083E: Total current in operation

na na78
Guru 16770 points
Part Number: SN65HVD3083E
Other Parts Discussed in Thread: THVD1550,

Hi

We want to estimate total current consumption in two case.

1. Normal operation (VCC=5V, 1Mbps)

2. Output short

>1.Normal operation

The device has driver and receiver.  So I think the both should be considered.

According to the Figure 14 of datasheet, Icc can be obtained as ~16mA.

Pq = 5V * 16mA = 80mW.

Po_driver =(VCC- Vod) * Iout = (5V-2.7V) * 23mA = ~53mW  (Referring Figure 16 with RL=120ohm)

Po_receiver = (VCC-(VOH-VOL)) * Iout = (5V-(4.6V-0.15)) * 10mA = 5.5mW

Ptotal = Pq+Po_driver+Po_receiver = 138.5mW

So the required current capability would be 138.5mW/5V = 27.7mA 

>2. Output short

It is 250mA as maximum from datasheet.

Especially, I want you to check calculation for normal operation.

Is it reasonable? Should I take 60mA as driver output current?

I appreciate if you could reply.

BestRegards

  • 0
    TI__Mastermind 25705 points

    Hi,

    You can refer to this blog for the RS-485 power dissipation calculation. The quiescent current of the device can be found at the Icc table (page 3). With 120Ohm load, you can use 4V max Vod for current estimation. The Figure 14 shows the power consumption of AC load, which is proportional to data rate. Please let me know if you have more questions.

    Regards,

    Hao

  • 0 in reply to Hao L
    Guru 16770 points
    Hi

    Thank you for your reply.

    Could you please answer three additional questions?

    No1.
    SN65HVD3083E is dull-duplex but THVD1550 is half-duplex.
    In the blog, power dissipation of receiver does not seems to be considered.

    Regarding to receiver, Is it enough to take account of PDic and PDac?
    PDic and PDac for both of driver and receiver would be obtained from
    ICC table (page 3) and Figure 14.

    No2.
    How did you get 4V max Vod with 120ohm?
    I appreciate if you can show me the way.

    No3.
    >The Figure 14 shows the power consumption of AC load..
    Do you mean we can obtain PDac from Figure 4?
    e.g. PDac @( 1Mbps, 5V) = 5V * 16mA = 80mW

    BestRegards
  • 0 in reply to na na78
    TI__Mastermind 25705 points

    Hi,

    No1. In general the receiver takes much less power consumption. You can apply the same methodology from the blog for the receiver for more accurate evaluation. For example, use the Icc table for PDic and estimate PDac by a small capacitive load value, then compare your result with Figure 14.   

    No2. 4V is a rough estimate about max Vod, since it's not provided in the datasheet.

    No3. What I meant was "the power dissipation of AC load is proportional to the data rate" for Figure 14. Since it's lab data, it includes some non-ideal factors beyond the estimation.

    Please let me know if you have more questions.

    Regards,

    Hao

  • 0 in reply to Hao L
    Guru 16770 points
    Hi Hao

    Thank you for your reply. I understand.
    I will open another thread if I would have additional question.

    BestRegards