TCAN1051HG: Maximum power consumption on a isolated 5VDC supply

Hi Eric,

Eric Valin said:

Even though the spec indicates that this module can be used for isolated CAN transceiver application, the output power seems to be inadequate (the following paragraphs will show the calculation). As per the spec sheet, at a temperature of 105C, the VISO current is limited 85mA (VINP = 5V, 105C).  The VINP needs to be increased to 5.5VDC to get more current. Is that right?

Yes this is correct. To drive more power (current) VINP needs to be increased. For this device there is a 675mW maximum power limit. If a load exceeding maximum current is applied, VISO drops accordingly to meet the maximum power limit. 

Eric Valin said:

Which current is good? The power budget is very limited (500mW), and if the current is too high, the short circuit protection of the DC/DC converter will be triggered.  The following are snapshots of the spec where I took the values listed above.

Eric Valin said:

- The CAN transceiver will protect the output in 2 ways; using the DTO or limit the current on the CAN driver.  Do I let the CAN Driver limit the output current or do I let the DC/DC converter fall back to over-current or hiccup mode? 

- Should I increase the rating of the DC/DC converter to 1W (200mA)?

It depends on if your system has bus short fault and you still need the device to operate normally after the fault is cleared. If not then ISO6742DWR would be enough you don't need to concern about the maximum output current. A short current exceed that max output current can turn off your DC/DC converter, but you can do a power-cycle as well. Otherwise I would suggest to use increase the rating to 1W as you mentioned above.

Eric Valin said:

 Since the CAN bus is isolated, how are the other nodes able to read the differential voltages across CANH and CANL?  These are floating voltages that do not share the same ground.  Do I need to provide and connect the isolated ground to the ground of the other nodes through a cable shield or something else?

Yes you need connect all other grounds together. See our isolated CAN transceiver below:

Regards,

Sean

Hello Sean,

Thanks for answering my questions.  I am not sure I understand your answer to question 1. I am aware that the transceiver will not be operational during the fault on the bus and that there will be no communication possible during this time.  The goal here is that the short circuit does not damage the transceiver and the power supply.  I do expect the devices (power supply and transceiver) to operate normally once the fault is cleared.  The isolated power supply, the transceiver or both can limit the bus short circuit current, as long has the hardware is protected indefinitely during this time. Once the short circuit is removed, the communication will resume without damages to the system.  Is that what you mean? When you say “If not then ISO6742DWR would be enough”, you mean the UCC12040DVER power supply module?

You haven’t commented the current values in your answer but looking for price/availability/temperature of a 1W converter made me review these previous current values in an objective manner.  From the previous communication we had;

I(transceiver) = 180mA (Normal mode (dominant – bus fault)

I(transceiver) = 100mA (IOS(SS_DOM))

(And if I used an average short circuit current I(transceiver) = 28.8mA, using formula in the spec sheet)

I(transceiver) = 80mA (Communication Normal Mode max)

I(Isolator)= 6.4mA/2 +6.3mA/2 = 6.4mA max      

ITotal = I(transceiver) + I(Isolator) = 180mA +6.4mA = 186.4mA

This is the way I see it now.  The UCC12040DVER DC/DC converter is borderline for power because of operating temperature derating.  My first goal is to not overload the power supply during a “normal communication”.  Therefore, the estimated load current value for the design of the power supply will be the maximum current during normal mode communication + current for the digital isolator + some margin.

ITotal = Imax(transceiver) + I(Isolator) = 80mA +6.4mA = 86.4mA  --> 100ma

I have looked at many TI application notes using the transformer driver SN6501 to implement this CAN design.  All of them use a 100-150mA LDO regulator (>86mA) at the transformer output to protect the CAN circuitry.  Everything above this current will be limited by the LDO regulator by thermal shutdown.  Would it be ok to take this approach using a transformer driver and a 100-150mA LDO regulator and stay in the 500mW power range? Would the hardware be protected during the fault with this circuit?

Eric

Hi Eric,

Eric Valin said:

If not then ISO6742DWR would be enough

Sorry it was my mistake, I mean UCC12040DVER here.

The short current will not damage the transceiver and power supply. What I said above is at the worst case, the power supply can shut down and thus you need to do a power-cycle to make the device operate normally again. If it is an acceptable way for your system then UCC12040DVER can work well, no issue.

The datasheet specified we need to use average current for power supply (see section 8.3.6). In this case the total current will not exceed 100mA . However I would still tend to consider the maximum short current when choosing the power supply, if your system is very sensitive to short circuit bus fault. This will occur when the minimum fault voltage (-12V) is applied to the CANH pin while it is trying to drive dominant (to VCC). 

Eric Valin said:

Would it be ok to take this approach using a transformer driver and a 100-150mA LDO regulator and stay in the 500mW power range? Would the hardware be protected during the fault with this circuit?

The LDO approach looks good to me. As long as the working condition not exceeding abs max ratings we can guarantee no damage to the hardware.

Regards,

Sean

Hello Sean,

Thanks for answering my questions.  You said “What I said above is at the worst case, the power supply can shut down and thus you need to do a power-cycle to make the device operate normally again.”

From my understanding of the UCC12040DVER DC/DC converter spec, no power-cycle is necessary. Once the short circuit is removed, the IC will resume function and provide VISO unless the thermal shutdown is activated.  Do I understand the spec wrong?

Regards,

Eric