### Loadings for DC IRIG-B

Loading Recommendations for DC IRIG-B (IRIG-B00X)

A question that is commonly asked is “how many Intelligent Electronic Devices (IEDs) can a single DC IRIG-B output drive”? The answer is almost always “well it depends…”

First you need the following information:

What is the drive power of the clock’s output? For Tekron this is commonly 150mA

What is the input impedance of the IED? Or what is the current drain of the IED? (These parameters should be available in the vendors datasheets under the timing section.)

The distance between the 1^{st} IED and the last IED that you want to synchronise.

Once you have this information the calculation for how many IEDs you can drive is quite simple.

To demonstrate this calculation, we'll work through an example and apply the following equation:

Where:

I_{L }is the total current load

I_{1 }to I_{n} is the current drain of each IED on the IRIG-B bus

V_{s} is the supply voltage from the clock (typically 5 Vdc)

R_{term} is the terminating resistor which matches the cable impedance (120 Ω for shielded twisted pair cabling or 50 Ω for RG58 Coax)

The first step to starting this calculation is to know what load each IED is going to put on the IRIG-B line. This is different for each manufacturer.

To find this data, you will need to look in the datasheet of the IED for the IRIG-B or time sync section. Here you will generally find the input voltage range (5 Vdc) and either the input impedance (kΩs) or the current load (mA).

If the datasheet is nice enough to have the load current, this is your I_{1} value. If it only gives you the input impedance you can calculate the current load using:

Where

V is the source voltage (5 Vdc)

R is the input impedance of the IED

For this example we will use 25 protection relays with an input impedance of 5 kΩ.

This would mean that each IED has a current burden of:

Across 25 relays this would come to a total of 25 mA loading.

This then brings us to the main equation:

Great! Now we know the total loading of the relays on the IRIG-B output. The next point is to check that this I_{L }is not larger than the clocks drive power. As Tekron’s devices supply a 150 mA drive power, this leaves 73 mA to spare!

So, does this mean I can add another 70 relays to this IRIG-B line?

Yes, technically you can add a further 70 relays to this output, but first you need to consider the total cable length between the clock and the last relay. If the cable length is getting greater than 50 m it’s recommended that you either split the remaining relays off onto another output, or use a signal repeater to regenerate this signal.

There are several reasons for this suggestion. The first is that after 50 m of transmission, the square IRIG-B signal may start to show rounded edges as the cable capacitance starts to degrade the signal quality. It may even start to degrade to a point where IEDs will reject it as a valid signal, or the signal accuracy will decrease due to the rounded rising and falling signal edges.

To correct this issue you can install a simple signal repeater to regenerate the signal, giving much sharper rising and falling edges, filtering out the noise, and adding in an isolation barrier.

The second point to think about is the accumulated propagation delay of the signal as it passes down a long piece of wire. For Belden 9841 shielded twisted pair cabling the propagation delay is 5.25 ns/ m. Over 50 m this adds to 262.5 ns of delay. For most applications this is minimal, especially when your target accuracy is only 1 ms. But in an application where you are aiming for a < 1 µs accuracy, this is important as you could lose 26% of your overhead just in the cable transmission delay.