[SOLVED] Power supply filtering for the CAN transceiver

[Tomas1707]

Hi,
I have a question about power supply filtering. We have a HF system working on 8 MHz. It consists of transmitter and receiver. A couple of these devices have together power supply of 24 V. The devices are also connected with CAN bus for the communication with each other and to set parameters from the PC. We had encountered problem with receiver signal interfered by the transmitting on the CAN bus and discovered, that the cause of the interference is the power supply of the CAN transceiver, that ''pulls" the together 24 supply and also 24 V power supply for HF power amplifier on the transmitter pcb. We have solved this problem by "filtering" the input of the voltage regulator for the CAN tranceiver (see fig.) with 100 ohm resistor in series and 470 uF capacitor. Because the input voltage is 24 V and output voltage is 5V, there is enough reserve for the voltage drop on the 100 ohm resistor. Resistor limits the current from the 24V input supply when CAN transceiver transmitts on the bus and capacitor supplies the "missing" current, so there aren't current pulses on together 24 V power supply and on the 24 V supply on the transmitter pcb.
CAN speed is 125 kbps.
My question is, if there can be some problem with this solution?
The only thing I can think of is the warming up of the capacitor due to the ripple current and so the shortening of its lifetime and capacitance, but the transmitting is not so often, so the temperature shouldn't rise much.

 

[Gozaki]

I would say : " how much current is flowing through de 100 ohm resistance " ==> RI²
i'll put a 100nf ceramic cap for HF filtering.
How much wire length between power supply and transmitter/receiver ?

 

[Tomas1707]

The current through the 100 ohm resistor is about 50 mA in "transmitt" state, and about 10 mA in "receive" state, so the max. power loss in resistor is 250 mW, but it is the max. instantaneous power loss, the average power loss is less than 100*((0,05+0,01)/2)^2 = 90 mW, that is nothing that wouldn't make 1206 resistor.
There is already the 100 nF capacitor in parallel with the big 470 uF one (see fig.). Because of the speed of the CAN of 125 kHz, 100 nF is not enough to "average" the supply current of the CAN transceiver.
The lengths of the wire is about 5 to 10 meters from the power supply to the receiver/transmitter and than about 1,5 m between receiver and transmitter.
The main problem is, that the receiver is sensitive to the changes of the amplitude of transmitted signal, so when there is something, that changes the power supply voltage, it affects the transmitter's signal, because the transmitter power amplifier shares the same supply. So the CAN current "modulates" the amplitude of the transmitter signal and so interferes in the receiver.

 

[BradtheRad]

Would it improve anything to run two separate power lines, one from each unit to the power supply?

 

[Tomas1707]

I tried the solution with separated power supply for the CAN transceivers in all devices on all pcbs and it worked, but that is not possible in some cases, because separated power supply needs another pair of wires. So I tried "filtering" and it works, but I need confirmation, that it would work in the longer term. CAN transceivers are separated from the microcontroller by optocouplers already but share input power supply on the pcb.

 

[BradtheRad]

It will work to use the 100 ohm resistor with smoothing capacitor.

Have you checked using an oscilloscope, as to whether there is noise or dips in the supply V to that part of the circuitry?

At the output pin of U41 voltage regulator?

When transmitting?

Etc.

 

[Tomas1707]

There is voltage drop only on the input of U41 regulator, the output is stable and quiet during transmitting. I know, that this solution is working properly right now, but what bothers me is, if there isn't some factor, that I've missed, that could cause problems. Also I'm wondering, why this filtering isn't used commonly. I don't remember any app note or eval board schematic using this filtering.

 

[BradtheRad]

Also I'm wondering, why this filtering isn't used commonly. I don't remember any app note or eval board schematic using this filtering.

It's one of those clever stratagems, that if we didn't know to use it, we might instead add heavier-duty filter capacitors on the general supply.

I have seen a few schematics using this method. It's cousin to the way we use a zener diode by putting a safety resistor between the supply and the zener.

Similarly your 100-ohm resistor provides some degree of isolation for the sensitive circuit. Hence the 470 uF capacitor is more effective at its job.

 

[Tomas1707]

Thank you all for the comments.