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AM26LS31: Estimate Power Consumption

Part Number: AM26LS31
Other Parts Discussed in Thread: AM26C31

Tool/software:

Hi,

Could you please tell me how to estimate the power consumption of AM26LS31?

There is an explanation at
https://e2e.ti.com/support/interface-group/interface/f/interface-forum/845314/am26ls31m-how-do-i-calculate-the-power-consumption-of-differential-line-drivers/3125917?tisearch=e2e-sitesearch&keymatch=AM26LS31%252520Cpd#3125917

but the link seems to be broken.

The customer calculated it as follows. Is it correct?

Pall = Pd_in + Pd + Pd_o
Where
Pall: Total current consumption
Pd_in: Input circuit current consumption
Pd_static: Static current consumption
Pd_o: Output circuit current consumption

Pd_in = = 3V x 0.36mA x 4ch = 4.32mW
Pd_static = 160pF ​​x (5V^2) x 1MHz x 4ch = 16mW
   Calculated with Vcc = 5V, Cpd = 160pF.

Pd_o = {(Vcc-Voh) x Io + Vol x Io} x 4ch
= {(5V-3.13V) x 29mA + 0.3V x 29mA} x 4
= 252mW

This formula is Vcc = (Vcc-Voh) + 100Ω x Io + Vol x Iol
From Fig. 5-6, Fig. 5-8, and 100Ω x Io in the data sheet rev. M,
if Io ≒ 29mA, the above formula becomes Vcc ≒ 5V, so use the Io value on the left.
Therefore,
Pall = 273mW (customer's calculation result)

On the other hand, in my understanding,
Pd_static = Vcc x Icc = 5V x 80mA = 400mW
Pd_o = Cpd x VCC^ 2 x f x 4ch = 160pF ​​x 5^2 x 1MHz x4ch = 16mW

I'm not sure if P_in can be ignored, but the total current consumption is
= 400mW + 16mW = 416mW (my calculation result)

I would appreciate any advice.
If you have any reference materials, please show them.

Best regard,
Hiroshi

  • Pd_o = {(Vcc-Voh) x Io + Vol x Io} x 4ch
    = {(5V-3.13V) x 29mA + 0.3V x 29mA} x 4
    = 252mW

    This basically gives the power of the RS422 resistors when the driver is idle (like a DC power).

    Pd_static = 160pF ​​x (5V^2) x 1MHz x 4ch = 16mW

    This not a static value. This is an AC power which tells the power loss due to the load capacitor's charging and discharging. I wouldn't add this power to the DC power (the one above). I would pick whichever one is the higher one. In this case it's the driver's DC power. But if the load capacitance or the switching frequency were higher, the AC power could become worse/larger.

    I would use do:

    total power = Device's supply current power + RS485 system power

    Device's supply current power (worst case) =  Vcc x Icc = 5V x 80mA = 400mW (this is probably an over estimate)

    Device's system power (includes driver and the resistor)= I_RS422 x Vcc x 4ch= 20mA x 5V x 4 = 400mW (I used 20mA as the current value but you can use larger to over estimate)

    The above is the DC idle power when it's driving low/high but not toggling. We will use this as the worst case but if we calculate the switching power, 

    AC switching power loss: Cpd x VCC^ 2 x f x 4ch , as a rule of thumb I usually say 1 meter of cable is 50pF. If you have 100 meters, you have 5000pF 

    If we use 1MHz (data rate is 2Mbps) and a 100 meter cable on all 4 channels then

    Cpd x VoD^ 2 x f x 4ch = 5000pF x 3^2 x 1Mhz x 4 = 180mW (VoD is the voltage across the resistor or Y-Z, I chose 3V this is the voltage the load capacitor will charge up to it won't charge up to Vcc)

    180mW > 400mW (DC we calculated earlier)

    So I would choose the DC power since it's larger (pick AC power loss or DC power loss, which ever is bigger for the final calculation)

    Power = 180mW (AC power) + 400mW (from device's supply current power) = 580mW

    -Bobby

  • Bobby-san,

    Thank you for your detail estimation.

    and Sorry for asking so many questions.

    >So I would choose the DC power since it's larger (pick AC power loss
    >or DC power loss, which ever is bigger for the final calculation)

    Since the DC power is larger, would the total power be
    400mW (DC power) + 5V * 80mA (Device's supply current power) = 800mW?

    Now the customer is comparing with the power consumption of AM26C31.

    In the case of C31, the system AC power and DC power are calculated the same, so
    Total power = 400mW + 5V * 3mA (maximum value of Data Sheet spec) = 415mW

    Is the calculation above correct?

    Best regard,
    Hiroshi

  • I have created a spreadsheet for customer convenience.

    Could you please check it?

    AM26LS31 C31 Power Estimation and junction Temp Cal.xlsx

  • This looks right from what I can see (I played with the AC section to see if it's bigger than the DC, the total changes which it does).

    You may want to include a note that Cpd is the load capacitance from the cables which we estimate to be 50pF/meter. 

    -Bobby

  • Hi Bobby-san,

    I appreciate your kind support. 
    I will add a comment about the load capacity from the cable and send it to my customer.

    Best regards,
    Hiroshi