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I'm looking to use the SN65HVD230 CAN transceiver but had a question about the current consumption.

Datasheet: http://www.ti.com/lit/ds/symlink/sn65hvd230.pdf

In table 8.3 Recommended Operating Conditions, it said the high level output current for the driver is -40mA and the low level output current is 48mA. However in Table 8.5, it is said that the supply current (Icc) has a max recommended current of 17mA. How can the driver current consumption be greater than the overall supply current ?

How do I calculate the overall current consumption of the bus ? The datasheet seems to only talk about the current consumption of the receiver and driver, but not the actual CANH and CANL lines?

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  • \$\begingroup\$ given the 120+120 ohm terminations on the standard CAN bus, or 60 ohms equivalent, that scaled by 40mA is 2.4 volts. Does 2.4 volts differential bus voltage agree with the CAN spec? \$\endgroup\$ – analogsystemsrf Sep 11 at 4:07
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The max supply current of 17mA is with "no load". That means there is nothing on the output to supply output current into. Any output current, up to the limits in Table 8.5, is in addition to this rating.

This approach is often used in datasheets because it gives you the values of current "consumed" by the device. Output current is consumed by external loads (in this case, the bus) so is useful to treat separately. The necessary output current is very dependant on the external environment, rather than the behaviour of the IC itself. Indeed, often these driver ICs have a separate supply pin just for the load. While this one does not, you can still see from the block diagram that the output could have its own rail.

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  • \$\begingroup\$ I see. In that case would it be more helpful to look at Figure 1 in Table 8.11 ? What I want to figure out is what the sum of all the transceiver supply currents (Icc) for 30-60 nodes on a single CAN bus is so I can spec my VCC power supply appropriately. \$\endgroup\$ – VanGo Sep 10 at 20:49
  • \$\begingroup\$ Upon further reading - the datasheet recommends the TPS76333 linear regulator which has a maximum continuous output current of 150 mA. Now that I know an upper bound on the current draw of the transceivers, how do I find the current draw of just CANH and CANL - since these differential signals will travel on a PCB, I need to know their current draw to make them the appropriate thickness. \$\endgroup\$ – VanGo Sep 10 at 21:17
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    \$\begingroup\$ No, Figure 1 must be in the recessive state and is a bit misleading. You need to add the ~17mA to the output current. You should allow the driver to drive the line as hard as it can for best signal integrity. From Figure 15 that's about 35mA, which just barely allows for a 2.1V dominant voltage across a 60Ω bus. So 53mA would be bare minimum I'd allow for and 100mA would be safer. The 150mA supply they recommend would do just fine. \$\endgroup\$ – Heath Raftery Sep 10 at 21:31
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    \$\begingroup\$ CANH and CANL will carry either 0A for recessive state, the current necessary to maintain the differential voltage in the dominant state (nominally 2.5V across 60Ω), and some transient currents to charge the bus capacitance. Feel free to ask a more specific question now that you're starting to formulate it. \$\endgroup\$ – Heath Raftery Sep 10 at 21:32
  • \$\begingroup\$ Posted the new question here: electronics.stackexchange.com/questions/458156/… \$\endgroup\$ – VanGo Sep 13 at 5:15

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