SN65HVD09 Power consumption calculation

Hi Rajesh,

TI's policy is to not obsolete a product unless its manufacturing becomes technically infeasible.  You can read the official obsolescence statement here.

Regarding power dissipation, the total dissipated power is the sum of two components: the quiescent power of the device (fixed) and the output power of any drivers on the bus (variable).  The fixed power is simply I_CC * V_CC with no load.  This information can be found on page 5 of the SN65HVD09 datasheet.

The output power will be dependent on your bus loading, require a measurement of V_OH and V_OL, and necessitate the use of I_OH and I_OL, found on page 11 and 12 of the datasheet.

You can find a step by step description of how to use this information to compute the total power dissipation, as well as examples for RS-422 and RS-485 devices, in this application report.

Hi,

Thanks for the reply, It seems that VOD vs Io is not given in the data sheet(SN65HVD09) then how to derive this value.  As per the application  note An-805 whether the calculation is only applicable for B side alone or it includes everything(Logic drivers in the A side also) ? In my application I am planning to use channels are receivers(B side) all channels are transmitters in A side in this case how to calculate the power consumed by SN65HVD09.

Regards,

Rajesh

The "A" side I/O pins and logic (enable) pins are all high impedance and are assumed to be driving a high impedance load, so there is no "variable" power dissipation to calculate for those functions.  It is captured in the quiescent power calculation (I_CC * V_CC).

Similarly, when a channel is set to be a receiver, it also has no variable power component, as it is a high impedance receiver on the bus, not driving a load on the bus.  Thus for any combination of drivers and receivers, I_CC * V_CC gives you the total power dissipation for the "A" side pins, enable pins, receiver channels, and the quiescent power of the drivers (with no load).  Then you must only compute the output power for 1 driver channel, and multiply this by the number of drivers to calculate your output power dissipation.

You can produce the V_OD vs I_O plot from Figures 13 and 14; it is simply the difference of the two.

Now you need only to measure V_OD for one driver channel under your maximum load condition, and compute your output power per Equation 6 in AN-805.

Hope this clarifies things a bit --

Hi,

Thanks for the reply. Now some what clear, still i need some clarification on following thinks:

1. How you derived the above graph from figure:13 & 14 it seems confusing since both the graph are VOH vs IOH & VOL & IOL. ?

2. Just for clarification VOD= Vp-Vn(voltage difference between positive & negative terminals of the driver) currect me if i am wrong ?

3. If above statement is correct then most of the drivers will have positive will be at +5V & negative will be at 0V vice versa in most of the cases then why the above graph is drawn for VOD of 3V.

Regards,

Rajesh

1) With differential termination, I_OH and I_OL are the same current.  So for a given I_O, V_OD = V_OH - V_OL.

2) Yes.

3) If the driver were constructed with ideal transistors, the differential output would be 5V - 0V = 5V, as you say.  However they are not ideal, and have some resistance R_ds-ON, which causes a voltage drop and pulls V_OH and V_OL away from the rails.  So you have something like:

5V <== R_ds-ON1 (~30ohm) ==> V_OH <== R_term (100ohm) ==> V_OL <== R_ds-ON2 (~30ohm) ==> GND.