Does the Terminal Resistance of the CAN Bus Have to be 120Ω?

The resistance of the CAN  bus terminal is usually 120 ohms. Two 60 ohm resistors are linked in series in the design, and there are usually two 120 ohm nodes on the bus. Basically, persons who are familiar with the CAN  bus system. This is common knowledge.

Figure. 1

But, as a scumbag author, I'm well aware that this is a regularly used resistance value in different standards, datasheets, and application notes, but what are the functions of these two terminal resistors? Impedance matching has been discussed previously, but what does matching entail?

The following knowledge points are summarized in this article. In everyday work, knowing the purpose of the terminal resistor might help you locate the source of a problem sooner, such as unstable waveforms.

Ⅰ.  The role of terminating resistors 

The CAN  bus terminal resistor serves three purposes:

Enhance anti-interference performance by allowing high-frequency and low-energy signals to fade away fast.

Ensure that the bus quickly reaches the recessive state, allowing the parasitic capacitance's energy to dissipate more quickly.

Place these at both ends of the bus to reduce reflected radiation and increase signal quality.

01 Improve anti-interference ability

The "dominant" and "recessive" states of the CAN  bus are decided by the CAN  transceiver. "Dominant" symbolizes "0," "recessive" represents "1." The internal structure of a CAN  transceiver is shown in the diagram below, with CAN  H and  CANL linked to the bus.

Figure. 2

When the bus is dominant, Q1 and Q2 inside the transceiver are turned on, and a pressure difference is generated between CANH and CANL; when the bus is recessive, Q1 and Q2 are turned off, CANH and CANL are in a passive state, and the pressure difference is 0; when the bus is recessive, Q1 and Q2 are turned off, CANH and CANL are in a passive state, and the pressure difference is 0.

When there is no load on the bus, the differential resistance value is very high, and the internal MOS tube is in a high-impedance state. External interference requires very little energy to get the bus to enter dominant mode (the minimum voltage of the common transceiver dominant threshold). (Only 500mV). If there is differential mode interference at this moment, there will be noticeable fluctuations on the bus, and these fluctuations will have no place to absorb them, resulting in the creation of a dominating bit on the bus. As a result, a differential load resistor can be added to increase the anti-interference capabilities of the bus while it is recessive, with the resistance value as low as feasible to avoid the influence of most of the noise energy. However, the resistance should not be too low to avoid requiring too much current from the bus to become dominant.

02 Ensuring fast entry into recessive state

The parasitic capacitances of the bus are charged during the dominant state, and they must be discharged when the bus returns to the recessive state. The capacitor can only be discharged through the differential resistance inside the transceiver if no resistive load is applied between CANH and CANL. This is a significant impedance. The discharge time will be much longer due to the features of the RC filter circuit. To do the simulation test, we connect the transceiver's CANH and CANL with a 220PF capacitor. The bit rate is 500 kilobits per second. The waveform is depicted in the diagram. This waveform's falling edge is a relatively protracted condition.

Figure. 3

A load resistor must be connected between CANH and CANL in order to fast discharge the bus parasitic capacitance and ensure that the bus enters the recessive condition. The waveform after adding a 60 ohm resistor is illustrated in the diagram. The duration from dominant recovery to recessive is reduced to 128nS, which is similar to the dominant settling period, as seen in the figure.

Figure. 4

03 Improve signal quality

Signal reflection occurs when the signal edge energy hits an impedance mismatch at a high slew rate; if the geometry of the transmission cable's cross-section varies, the characteristic impedance of the cable changes as well, resulting in reflections.

The reflected waveform is superimposed on the original waveform when energy is reflected, resulting in ringing.

The quick change in impedance at the bus cable's end generates energy reflection of the signal edge, resulting in ringing on the bus signal. The communication quality will be harmed if the ringing amplitude is too big. At the end of the cable, a terminal resistance equal to the cable's characteristic impedance can absorb this part of the energy and prevent ringing.

The data rate is 1Mbit/s, the transceivers  CANH and CANL are connected to a twisted pair of roughly 10m, and the transceiver is terminated with a 120 resistor to assure the recessive conversion time, and the end is not loaded. The signal's waveform near the end is displayed in the picture, and there is a ringing on the signal's rising edge.

Figure. 5

When a 120 resistance is added to the twisted pair's end, the signal waveform improves dramatically and the ringing disappears.

Figure. 6

Because both ends of the cable are the sending and receiving ends in a linear topology, a terminating resistor must be placed at each end of the cable.

The CAN  bus is typically a hybrid structure of the bus and the star in the real application process and is not a flawless bus design. It imitates the  CAN bus's standard structure.

Ⅱ. Why choose 120Ω?

What exactly is impedance? The resistance to the current in a circuit is known as impedance in electrical. The ohm is the unit of impedance, which is commonly written as Z, which is a complex number Z= R+i(L–1/(C)). Impedance is separated into two parts: resistance (the real part) and reactance (the hypothetical part) (imaginary part). Capacitive and inductive reactances are included in the reactance. Capacitive reactance is the current obstruction produced by capacitance, while inductive reactance is the current obstruction caused by inductance. The term "impedance" relates to the Z mode.

Any cable's characteristic impedance can be determined experimentally. The cable is linked to a square wave generator on one end and an adjustable resistor on the other, with the waveform on the resistor being examined using an oscilloscope. Adjust the resistor's resistance value until the signal on the resistor is a good square wave with no ringing, and the resistance value is compatible with the cable's characteristic impedance at this point.

The characteristic impedance may be determined using two normal automotive cables and twisting them into twisted pairs using the above approach, which is about 120, which is also the resistance value of the terminal resistance recommended by the CAN standard, therefore this 120 is measured. It is not calculated; rather, it is calculated depending on the features of the real wiring harness. Of course, the ISO 11898-2 standard defines it as well.

Figure. 7

Ⅲ.  Why choose 0.25W for power?

This must be calculated in conjunction with a number of fault conditions. The short-circuit to the power supply and the short-circuit to the ground must be considered in all interfaces of the automobile ECU, so we must also consider the short-circuit of the CAN bus node to the power supply, as well as the short-circuit to 18V, as required by the standard. The current will flow to CANL through the terminating resistor if CANH is short-circuited to 18V, and the maximum injection current inside CANL is 50mA (indicated in the TJA1145) due to current limiting, and the power of the 120 resistor is 50mA at this moment. *50mA*120Ω=0.3W. The terminal resistor has a power of 0.5W while derating at high temperatures.