32
If the frequency/rise time and distance is high enough to cause issues, then yes, you need termination.
Transmission-Line Model
At 97mm longest trace I think you will probably get away without them (given results of calculations below) If you have a PCB package that handles IBIS models and board level simulation (e.g. Altium and other expensive packages), ...
26
You need to be familiar with Transmission Line Theory to understand the deeper physics in play here. That said, here's the high-level overview:
How important termination is to your system is almost exclusively determined by how long the bus wires are. Here length is determined in terms of wavelengths. If your bus is shorter than one wavelength over 10, the ...
24
May be, a more mechanic explanation helps to understand:
Imagine you have a long rope, one end fixed to a wall, the other end held by you. By a short stroke upwards, you can create a wave travelling along the rope:
(from http://wwwex.physik.uni-ulm.de/lehre/physing1/node52.html)
Now, why is it so? Imagine the rope consists of many small pieces, each ...
20
Here's what finally helped me understand termination and reflections: Suppose you have a really, really long coaxial cable, with the far end shorted together. If you put current through it, what will the voltage be?
Because the cable is shorted at the far end, you'd expect the voltage to remain near 0. But, the far end is a long way away - if the voltage ...
18
That type of CAN bus is intended to implemented by a twisted pair of wires. The transmission line impedance of unspecified twisted pair isn't exact, but 120 Ω is going to be close most of the time for the relatively large wires commonly used for CAN.
The resistors also have another function in CAN. You can think of CAN as a open collector bus ...
17
At 1 MHz and 50 mm (2 inches) you don't need terminations. You have ordinary on-board digital signals, and not even very fast ones.
You don't need pullups or pulldowns on SPI lines. When used, SPI lines are always explicitly driven both directions. However, it can be good to put a pulldown (or pullup) on the MISO line. That is because this line is only ...
16
Since you are going short distances, I don't think termination resistors are a good idea. As you found, they have to be quite low to do the job, and then the line draws a lot of current and the voltage is attenuated by 2 if you also drive the line with the same impedance.
Your clock rate isn't all that high, so the frequencies you need to support even 4 ...
15
While, in theory, high impedances would reduce power dissipation for the same voltage swing, there are several important issues in practice.
1) It's the power, not the voltage, of a signal that determines signal to noise ratio. If you must swing the full rail, then you'd win by increasing the impedance. However if you launch a specific power, then low ...
13
Do not confuse power transfer with voltage transfer.
If the source is indeed 50 ohms, and your scope is set to 50 ohms, it becomes a voltage divider and what you see is half the voltage that you will see when the scope is set to high impedance.
simulate this circuit – Schematic created using CircuitLab
Power transfer has to do with the relationship ...
12
Each beginning has its end. This can't be answered, because the question is not correct. What you have is the transmission line - a twisted pair that has two ends, and that's the place where the termination has to be done. You can't choose the end of the bus, since there are only two of them.
11
For BNC terminators there are usually the following factors that contribute to the price:
Accuracy of the specified values (i.e. impedance is matching for the whole assembly)
Wattage of the resistor
Bandwidth of the whole assembly
Linearity of attenuation of the whole assembly
Contact quality
Some of these have similar influences to all kind of other BNC ...
11
It doesn't matter which nodes are at the ends, but it does matter that the terminators are at each end.
The bus is a transmission line. To keep edges from reflecting at the ends of the cable, the cable has to be terminated with its characteristic impedance. The common standard for CAN is twisted pair with 120 Ω impedance. You therefore need 120 &...
10
Take 50 MHz clock SPI.
50 MHz has a fundamental wavelength of 6 metres but you could argue that due to the fast rising edges everything up to the 5 th harmonic is potentially capable of disrupting the shape too much if terminations are not applied. So that's a wavelength of 1.2 metres.
On the other hand, on PCBs, the speed of signals is about 60% the ...
10
Using a terminal block made with ordinary materials is quite sufficient for a relatively modest accuracy system as you're aiming for.
The cold junction compensation depends on the cold junction sensor (in this case, the chip itself) being at the same temperature as the two junctions where the thermocouple wire transitions to copper. In other words, all ...
10
From a professional point of view, this is a critical problem. Missing termination will cause energy bouncing back at the end which isn't terminated. This could lead to strange random noise on the line, such as for example transients or random pulses that seem like ok binary pulses, but with wrong voltage levels etc. CAN Hi and Lo don't necessarily behave ...
9
In general, for short cables (< 20-30m) and low baudrates (< 115200) you can leave them out without much trouble. But:
It is useful to put some kind of load on the signal lines to improve noise immunity (the RS485 driver will supply enough current to switch the voltage on the differential line, many noise sources will not). But you do not need this ...
9
Loop back? NO. If the line is long enough to need termination (longer than risetime*c/10 or so), then drive it strongly enough to end terminate it properly, and match the line and termination reasonably well. If the line is short enough not to need termination, then it won't need the 'extra' conductivity that you seem to be looking for with the looped back ...
8
The "terminators" of a CAN bus serve two purposes:
They terminate the transmission line. The type of CAN you are referring to is intended to be implemented as a twisted pair. 120 Ω is roughly the impedance of such twisted pair. By terminating the ends with the characteristic impedance of the cable, reflections from the ends are minimized.
They ...
7
CAN Bus is a differential bus. Each differential pair of wire is a transmission line.Basically, the terminating resistor should match with the Characteristic Impedance of the transmission line to avoid reflection. CAN bus have a nominal characteristic line impedance of 120Ω. Due to that we are using typical terminating resistor value of 120Ω at each end of ...
7
Note that if soldering wire ends and then clamping with a screw in a terminal block or connector post as seen in your photo then the majority of the wire end MUST NOT be soldered - even though this is common amateur practice and may seem to make sense. If the copper wire is solder filled so it is "solid" then the screw will compress the soldered bundle and ...
7
A presentation by Royce Bohnert clearly shows that termination has no effect.
Conditions of the experiment:
How much of an impact does this termination actually have on radiated emissions?
Marine VHF band (156 – 165 MHz) of particular interest
Evaluated by measuring CM current on cable
RF current probe place on ~20m long cable
Cable ...
7
It's not the frequency of the signal that's sent down a PCB trace, but the risetime that's important.
A 21cm PCB trace is about 30cm 'air equivalent', and as the speed of light is 1nS/ft (in units convenient for the engineer), that's about 2nS round trip. If the output gate risetime is longer than this, and most modern 'low speed' MCUs and SoCs will have ...
7
This termination resistance closes the current loop between two differential signal lines and creates a decent amount of power loss so that low voltage amplitudes are required to reduce power consumption.
I think there's a misconception here about how transmission lines work. The purpose of receiver-end termination is to dissipate all the power, in order to ...
6
Altera recommends their use with some types of SDRAM in this document, but says that they can be avoided by using internal termination for the FPGA and SDRAM, if it is offered. Neither of the FPGA boards I have with SDRAM have any external termination on the connections and the devices don't have internal termination. It looks like they should be used, ...
6
2 MHz gives a 500 ns bit period. 1 foot foot of ribbon cable would be about 1.5 ns (or less) propagation delay or 3 ns round trip.
Adding termination resistors will increase power consumption, and the CMOS-style chip you're using isn't really designed to work with a terminated line.
A cleaner solution is to slow down your rise and fall times. You could ...
6
I found the following examples of POE termination from around the internet:
Example 1 Figure 5
Example 2 Figure 4
Example 3 Figure 5
After posting the same question on TI's E2E forum, I got the following answer:
The majority of termination schemes will follow the examples you've cited and not the SLVU126 (TPS2384) and SLUU269 (TPS23841) TI EVM designs. ...
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