Selecting a terminating resistor for AM IRIG-B is slightly different to that of a DCLS signal. It’s
better to think of the terminating resistor as a voltage divider, which is used to match the
line voltage to the input requirements of the slave devices.
To explain this, let’s work through an example calculation! Starting by looking at the image below,
you can see that the terminating resistor is attached across the line, at the end of the IRIG-B
bus. By attaching it here it’s effectively creating a divider for which the ratio is defined by the
total resistance of the line, as well as the clocks’ internal resistance.
Before starting this calculation you will need to know the following information:
- The internal impedance of the clock’s output. In the case of Tekron’s range this is 120 Ω
- The input impedance of each IED that is connected to the IRIG-B bus. For most relays
this will be in the range of kΩ’s. For this example we are assuming all the relays have
a 6 kΩ input impedance. This can be found on most relay manufacturers datasheets. - The input voltage requirements of the IED’s. This is where you need to determine the
maximum voltage input that the relay allows. This can be anywhere from 5 Vdc to 10
Vdc. This can be found on most relay manufacturers datasheets. - The output voltage of the clock. In the case of Tekron this is 8 Vpeak to peak
Now that you have this information, the first step is to calculate the total load on the IRIG-B
bus. This can be done by adding together all the input impedances of the slave devices. As
they are connected in parallel we would expect the equation to look like this:
Where:
RL is the total calculated load
R1 to Rn are the input impedances of the slave devices.
In our example, we have 5 protection relays each with an input impedance of 6 kΩ. This
makes our equation:
Solving for RL:
Now that we know what RL is, we can work out the required terminating resistor by using
the following equation:
Where:
Vreq is the minimum required voltage for the slave device to operate
Vout is the AM IRIG-B output voltage
Rs is the output Impedance of the AM IRIG-B output
RL is the total calculate load
Rterm is the value we are solving for, which is the terminating resistor
In this example, we will use the following values:
Vreq = 6 Vdc
Vout = 8 Vpeak to peak
Rs = 120 Ω
RL = 1,200 Ω
This gives us the following calculation:
From the E24 resistor range, the closest match is a 510 Ω resistor which will be sufficient to
achieve the required voltage level.
Extracted from (Almost) Everything you need to know about IRIG-B