Hi JOE, can you explain the maths behind the 3.6V and 1.4 Volts. Also if you short circuit CAN H and CAN L, won't you cause a dead short. Without the benefit of a schematic, I built a working model of a CAN signal generator but I had 47ohm resistors where your diodes are.
I doped the diodes so they would have a forward bias drop of 1.4V to look more like CAN voltages in this very simplified schematic. I could have put two diodes in series in place of each diode. Therefore for CAN H, 5V - 1.4V = 3.6V. For CAN L, it is just the 1.4V dropped across the diode when FETs are ON. For a CAN H to CAN L short, the diodes would each drop 2.5V to force Vdiff to be 0V. This is just a simplification to show the basic theory. The manufacturer may not show all the details such as additional components to handle failure modes like your short.
@@joebell6283 Thanks Joe. I had to do the same with my virtual lab. I've got the real circuit working using resistors for demonstration purposes, Its safer that way when I do the shorting out bit. And you're right, its only a schematic. Great videos Joe.
We have a one module transmitting a differential voltage at 2.5 volts and approximately 15 other modules transmitting differential voltage at 2.1 volts. We are generating continuous errors on a using a CAN Bus Monitor. Do you believe this is the cause?
Do you have a scope trace? Have you tried disconnecting modules one by one to see if the errors cease when a particular module is disconnected? What is the application?
@@joebell6283 Thanks for the response Joe! And yes, I have saved several waveforms at this point with a picoscope 4425, using pico 7 software. I am seeing high error counts using Nexiq Device Tester, but I am still learning how to decode on the pico. The device in question is a Bendix TPMS ECU, on a 500 kBaud powertrain network.
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Hi JOE, can you explain the maths behind the 3.6V and 1.4 Volts. Also if you short circuit CAN H and CAN L, won't you cause a dead short. Without the benefit of a schematic, I built a working model of a CAN signal generator but I had 47ohm resistors where your diodes are.
I doped the diodes so they would have a forward bias drop of 1.4V to look more like CAN voltages in this very simplified schematic. I could have put two diodes in series in place of each diode. Therefore for CAN H, 5V - 1.4V = 3.6V. For CAN L, it is just the 1.4V dropped across the diode when FETs are ON. For a CAN H to CAN L short, the diodes would each drop 2.5V to force Vdiff to be 0V. This is just a simplification to show the basic theory. The manufacturer may not show all the details such as additional components to handle failure modes like your short.
@@joebell6283 Thanks Joe. I had to do the same with my virtual lab. I've got the real circuit working using resistors for demonstration purposes, Its safer that way when I do the shorting out bit. And you're right, its only a schematic. Great videos Joe.
We have a one module transmitting a differential voltage at 2.5 volts and approximately 15 other modules transmitting differential voltage at 2.1 volts. We are generating continuous errors on a using a CAN Bus Monitor. Do you believe this is the cause?
Do you have a scope trace? Have you tried disconnecting modules one by one to see if the errors cease when a particular module is disconnected? What is the application?
@@joebell6283 Thanks for the response Joe! And yes, I have saved several waveforms at this point with a picoscope 4425, using pico 7 software. I am seeing high error counts using Nexiq Device Tester, but I am still learning how to decode on the pico. The device in question is a Bendix TPMS ECU, on a 500 kBaud powertrain network.