Transmission Line Matching Impedance From a digital point of view

Thread Starter

Castrol

Joined Feb 15, 2020
7
Hi everyone,
I'm here 'cos I am having a hard time finding over the internet documentation related to transmission line design when digital signals are considered, instead of classical sinusoidal waves.
I would like to know how to treat the topic of "transmission line impedance matching" using stubs and the like when I can't define a specific frequency, i.e, every time we don't work with a sinusoidal carrier wave.
Smith charts are useless since I can't define the "wave-length" of digital signal who carries some bytes of information, I don't have even a clear foundamental component of that signal so what to do?
Please, point me to the right source of information on this matter, thanks :)
 

Papabravo

Joined Feb 24, 2006
14,412
As you may or may not know, a square wave is a combination of sine waves of various amplitudes starting with a fundamental that matches the frequency of the square wave. The other sine waves are harmonics of that fundamental frequency. Now you also need to be aware of the wavelength of your fundamental. The product of frequency and wavelength will be the speed of light (≈ 3 x 10^8) if the dimensions of your transmission line are much less than 1 wavelength you don't have to worry about matching or reflections
 

crutschow

Joined Mar 14, 2008
25,423
Stubs only work for matching a single frequency (or a narrow range of frequencies).
They won't work for a digital signal because of the wide frequency spectrum of a square-wave.

For digital signals the line is typically matched using a resistive load and/or source value equal to the line characteristic impedance.
 

Thread Starter

Castrol

Joined Feb 15, 2020
7
Stubs only work for matching a single frequency (or a narrow range of frequencies).
They won't work for a digital signal because of the wide frequency spectrum of a square-wave.

For digital signals the line is typically matched using a resistive load and/or source value equal to the line characteristic impedance.
So what you are saying is that I can't in any way using the classic transmission-line-theory to avoid reflections unless using resistive loads which bring power consumption, lowering the level of my signal, or becoming mad trying to put multiple stubs each one for a particular frequency of the square wave spectrum (I mean the frequency components which carry the most part of the energy) ?????
My problem is that I have a CAN bus where the signal has a really bad shape and I think this is related to reflections along the bus which is made of dozens of connectors, many un-terminated stubs, one for each node, and a bad choice of different cables with different characteristic impedance.

I'd want to try to match the impedance on some spots using non-resistive components because that would make the levels of my differential signals too low...
 
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crutschow

Joined Mar 14, 2008
25,423
I can't in any way using the classic transmission-line-theory to avoid reflections
Basically, no you can't for digital signals, and especially not for such a chaotic transmission line.

What is the impedance of the signal drivers?
Can you reduce that so the signal is not so adversely affected by output loading?

What is the maximum digital frequency of the bus?
What is the length of the bus?
 

nsaspook

Joined Aug 27, 2009
7,591
IMO it's a fools game to try frequency specific matching to a problem like yours.

http://www.ti.com/lit/an/snla027b/snla027b.pdf
The choice between using the single and fully matched system should be carefully considered because the fully matched system does sacrifice signal voltage magnitude to get a decreased dependence on absolute resistor values. If the load resistance for a matched driver circuit is made much greater than the line resistance, the initial wave arriving at the load at time t = τ will be almost double since ρL will be close to +1.0. Because source resistance is set equal to line resistance, ρ S becomes zero, the reflected voltage wave from the load is absorbed by the source at time t = 2τ, and steady state conditions prevail. Waveforms for this case are shown in Figure 17. This is called back matching or series termination.

The main advantage of series termination is a great reduction in steady state power consumption when compared with the parallel terminated case (RS ≪ R0 , RL = R0 ). At the same time, series termination provides the same signal fidelity to a receiver placed at the line end.
 

Papabravo

Joined Feb 24, 2006
14,412
You should be able to achieve a trunk line with the following length baudrate combinations
125 kbits/sec (8 usec bit time) 500 meters
250 kbits/sec (4 usec bit time) 250 Meters
500 kbits/sec (2 usec bit time) 100 meters

you should limit the number of nodes with transceivers similar to the 82C251 to 32

You can have unterminated stubs of up to 20 feet

Terminate both ends of the trunkline with 121 Ohms 1/4 watt resistors.

For more detailed information get a copy of the DeviceNet specification or
https://literature.rockwellautomation.com/idc/groups/literature/documents/um/dnet-um072_-en-p.pdf
https://en.wikipedia.org/wiki/DeviceNet
https://web.archive.org/web/20070128083016/http://www.odva.org/Portals/0/Library/Publications_Numbered/PUB00026R1.pdf
 
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Thread Starter

Castrol

Joined Feb 15, 2020
7
Basically, no you can't for digital signals, and especially not for such a chaotic transmission line.

What is the impedance of the signal drivers?
Can you reduce that so the signal is not so adversely affected by output loading?

What is the maximum digital frequency of the bus?
What is the length of the bus?
My CAN net is made of 22 nodes (but this should grow up to 27) 18-20 meters long bus.
I am trying to use a speed of 1Mbps.
Stubs are placed on PCB boards so the CAN bus connect to these stubs using 2 RJ45 connectors, i.e. the bus use a ethernet cable with 100 ohm of characteristic impedence but to branch a stub on a PCB I need to put a little piece of the bus on the PCB too, so I use one RJ45 to pass from cable to PCB, where I connect my stub, then I use another connector to come back to cable bus line line.
This happen for every node so I have a lot of RJ45 connectors, none of them is properly matched.
I use the Texas Instruments ISO1050DWG transeiver so the CAN controller should see a perfect square wave even when the CAN bus is noisy (the transceiver uses shmit triggers)
 

Papabravo

Joined Feb 24, 2006
14,412
My CAN net is made of 22 nodes (but this should grow up to 27) 18-20 meters long bus.
I am trying to use a speed of 1Mbps.
Stubs are placed on PCB boards so the CAN bus connect to these stubs using 2 RJ45 connectors, i.e. the bus use a ethernet cable with 100 ohm of characteristic impedence but to branch a stub on a PCB I need to put a little piece of the bus on the PCB too, so I use one RJ45 to pass from cable to PCB, where I connect my stub, then I use another connector to come back to cable bus line line.
This happen for every node so I have a lot of RJ45 connectors, none of them is properly matched.
I use the Texas Instruments ISO1050DWG transeiver so the CAN controller should see a perfect square wave even when the CAN bus is noisy (the transceiver uses shmit triggers)
You will never, I repeat never, see perfect square waves on a CAN bus, The reason is that recessive to dominant transitions are made by active devices pulling CAN_H high and CAN_L low. The dominant to recessive transitions are made by the termination resistors at each end of the bus. The short stubs you have shouls be no problem, but the Ethernet cable is a problem. The conductor size is too small and the conductors are too close together. This affects the distributed capacitance of the cable, and the transmit power versus the receiver threshold for nodes at the opposite ends of the cable. The critical parameter for using a cable in a CAN application is propagation delay. It should be at least 0.66c. Any less and you will start being crowded on the location of the sample point. If you are having difficulties you want to put the sample point as close to the end of the bit cell as possible and set the sjw to its minimum value. This also implies that all nodes must have an accurate crystal controlled clock.
 

Thread Starter

Castrol

Joined Feb 15, 2020
7
As you may or may not know, a square wave is a combination of sine waves of various amplitudes starting with a fundamental that matches the frequency of the square wave. The other sine waves are harmonics of that fundamental frequency. Now you also need to be aware of the wavelength of your fundamental. The product of frequency and wavelength will be the speed of light (≈ 3 x 10^8) if the dimensions of your transmission line are much less than 1 wavelength you don't have to worry about matching or reflections
I undertand all what you are saying here because I went to university and studied electronics.
I do not fully agree when you say I should worry only about the fundamental frequency, which is 1MHz in my application.
I found some application notes where they suggest to neglect propagation-related issues only if the rising/falling edge of your digital signal is slow enough to allow the propagated signal to run across all the bus line and back to the source three times before the rising/falling edge has completed.
This, always according to these documents I found, suggests to put the attention on the propagation delay of the signal, and this depends on the choice of the cable and the shape of PCB lines.
So we could have reflection problems even at 1MHz, or even lower.
 

Thread Starter

Castrol

Joined Feb 15, 2020
7
You will never, I repeat never, see perfect square waves on a CAN bus, The reason is that recessive to dominant transitions are made by active devices pulling CAN_H high and CAN_L low. The dominant to recessive transitions are made by the termination resistors at each end of the bus. The short stubs you have shouls be no problem, but the Ethernet cable is a problem. The conductor size is too small and the conductors are too close together. This affects the distributed capacitance of the cable, and the transmit power versus the receiver threshold for nodes at the opposite ends of the cable. The critical parameter for using a cable in a CAN application is propagation delay. It should be at least 0.66c. Any less and you will start being crowded on the location of the sample point. If you are having difficulties you want to put the sample point as close to the end of the bit cell as possible and set the sjw to its minimum value. This also implies that all nodes must have an accurate crystal controlled clock.
I am using ISO1050DWG which, as you say, drive CAN_H and CAN_L actively so recessive to dominant may be no problem. The opposite transition is achieved by weak polarization to 2.5V of the CAN_H and CAN_L lines. This one is done internally the 1050DWG as I read on a Texas Instruments blog where a guy explained how their isolated transceivers work (at least this is what I've understood).
I have already forced the sampling point to be at the 83% of the bit time, setting the dedicated register on the CAN controllers I am using.
 

Papabravo

Joined Feb 24, 2006
14,412
The transceiver has to work in such a way that the bus goes recessive iff (if and only if) all transceivers are in the recessive state, and in this state it is the termination resistors at each end that represent stronger influence. You say the transceivers are isolated. Have you taken the delay involved in the isolation into account for your bit time calculation at 1 Mbits/sec. It has the potential to reduce your maximum cable length to near zero or negative. At least that was my experience when we tried to do it.
 

Papabravo

Joined Feb 24, 2006
14,412
I undertand all what you are saying here because I went to university and studied electronics.
I do not fully agree when you say I should worry only about the fundamental frequency, which is 1MHz in my application.
I found some application notes where they suggest to neglect propagation-related issues only if the rising/falling edge of your digital signal is slow enough to allow the propagated signal to run across all the bus line and back to the source three times before the rising/falling edge has completed.
This, always according to these documents I found, suggests to put the attention on the propagation delay of the signal, and this depends on the choice of the cable and the shape of PCB lines.
So we could have reflection problems even at 1MHz, or even lower.
It is difficult to imagine that a reflection from a dominant bit could look like a recessive bit at the sample point. It is more likely that a reflection from a recessive bit could exceed the receiver threshold and look like a dominant bit. Remember each time there is a reflection there is an attenuation given by the reflection coefficient. Do you know what the propagation velocity is for your cable and do you know what the reflection coefficient from the termination resistors is.
 

magnatec

Joined Feb 26, 2020
2
To be pragmatic and observe the problem from a waveform quality point of view, can someone highlight references about the topic? Or can someone explain how to understand when a CAN BUS waveform can be considered good or bad?
 

Papabravo

Joined Feb 24, 2006
14,412
Only one thing matters. What is the waveform doing at the sample point. In other words, is the bit at the sample point at the receiver what it is supposed to be. The rest of the waveform can be garbage and it won't matter.

Check out the original papers on the CAN Physical Layer written by Bosch (ca. 1992)
 

magnatec

Joined Feb 26, 2020
2
Only one thing matters. What is the waveform doing at the sample point. In other words, is the bit at the sample point at the receiver what it is supposed to be. The rest of the waveform can be garbage and it won't matter.

Check out the original papers on the CAN Physical Layer written by Bosch (ca. 1992)
So, if the garbage before the sample point doesn't affect the receiving signal how the mismatch, and related problem, affect the communication?
 

Papabravo

Joined Feb 24, 2006
14,412
A reflection that passes the receiver at the sample point can change a recessive bit to a dominant bit. That is why cable length, velocity of propagation, and baudrate are the critical considerations. Each time there is a reflection there is attenuation. On a short cable system the reflections die quickly. On a longer cable system with a longer bit cell the reflections still die out but they have more time to do it.
 
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