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When sending a dominant bit (0bit) on the CAN bus, I understand that CAN transceivers output CANH=3.5V and CANL=1.5V.

But when looking at the schematics of a CAN transceiver, I do not understand how the CANH=3.5V is achieved (figure as follows borrowed fro TI blog post).

enter image description here

Does the Schottky diode (in the picture) along with the transistor make the 1.5V drop? Based on the fact that Schottky diodes drop very low voltage (~0.5V), I can't imagine the MOSFET dropping 1.0V. Especially when the drain-to-source resistance is a few mOhms for CAN transceivers.

Anyone can explain how the 1.5V drop from Vcc=5V to CANH=3.5V is achieved?

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  • \$\begingroup\$ "I can't imagine the MOSFET dropping 1.0V" >.> They can drop whatever you tell them to drop, via the gate voltage. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 8 '16 at 23:37
  • \$\begingroup\$ I'm not really a circuit guy... but, then regardless of the drain-source resistance, it can drop to a preset value? \$\endgroup\$ – Kyong Tak Cho Jun 8 '16 at 23:52
  • \$\begingroup\$ In controlled situations, sure. \$\endgroup\$ – Ignacio Vazquez-Abrams Jun 8 '16 at 23:52
  • \$\begingroup\$ I have 3 CAN nodes all with identical hardware, but the actual peak output voltages are different: 3.45, 3.42, and 3.39 on CANH. Is this because all the resistance inside their circuits are slightly different? \$\endgroup\$ – Kyong Tak Cho Jun 8 '16 at 23:54
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A properly sized MOSFET can provide extremely low voltage drops, at the expense of increased die size. For instance, power MOSFETs routinely provide on resistances in the milliohm range. Of course, this is at currents of 10s of amps. For CANBus, the currents involved are far less, so the FETs don't need to be as large. But let's say the FET is sized for 0.1 volt drop at the desired current. Then the total drop within the driver will be about 1.2 volts (2 x 0.1 plus 2 x 0.5), leaving 3.8 with a 5 volt supply. This is comfortably greater than the minimum requirement of 3.5.

And yes, there is process variation between chips, so the actual voltage across the termination resistor(s) will vary a bit. But, of course, that's the beauty of digital - as long as the voltages are within tolerance nobody cares about the details.

EDIT - And the 1.5 limit represents the sum total of all of the leakage currents through the MOSFETs which have been turned off. Since there can theoretically be on the order of 127 such units, even small leakage currents could add up.

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  • \$\begingroup\$ What is the "2" and the "0.5" in your example equation? Is the 2 the number of mosfets and 0.5 the voltage drop from the diodes? \$\endgroup\$ – Kyong Tak Cho Jun 9 '16 at 0:12
  • \$\begingroup\$ @KyongTakCho - Right. 2 diodes at 0.5 each, and 2 MOSFETs at 0.2 each. \$\endgroup\$ – WhatRoughBeast Jun 9 '16 at 0:17
  • \$\begingroup\$ Between the Vcc and CANH, as there is only one diode and MOSFET, maybe your example should be... "saying FET is sized for 0.7 volt drop at the desired current. Then the total drop from Vcc to CANH would be 0.7+0.5=1.2 leaving 3.8 at CANH" right? \$\endgroup\$ – Kyong Tak Cho Jun 9 '16 at 0:20
  • \$\begingroup\$ On your new edit, what do you mean by the "1.5 limit"? 1.5V voltage drop? If so, is that a limit? \$\endgroup\$ – Kyong Tak Cho Jun 9 '16 at 0:24
  • \$\begingroup\$ @KyongTakCho - Yes, sorry about that. However, as far as I know your understanding of the standard voltage levels is off. See Table 1 in ww1.microchip.com/downloads/en/AppNotes/00228a.pdf for instance. \$\endgroup\$ – WhatRoughBeast Jun 9 '16 at 0:29

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