THVD1419: Load Test of THVD1419

A unit load can be modelled as a 15 kΩ resistor connected to a −3 V or 5 V supply; see RS-485 Unit Load and Maximum Number of Bus Connections for details.

1/8 UL corresponds to 120 kΩ.

255/8 UL corresponds to about 470 Ω.

Hi Shuang,

The current RS-485 standard has 1 unit load ~ 12K Ohms (the older standard that Clemens mentioned is slightly outdated) and it technically specifies bias current versus input voltage but a 12K load line is usually assumed, where there can be 32 parallel unit loads on 1 bus (with two termination resistors at each end of the bus 120 ohms each). The 32 Parallel unit loads has an equivalent impedance of ~375 Ohms. The 375 Ohms is the standard common mode loading for differential voltage output testing with common mode voltage inputs - so a 1/8th unit load device can support 8 times as many devices on the bus (assuming no other passives besides termination resistors and all devices are also 1/8th unit load) and so have 8 times the input impedance (12K * 8 = 96K Ohms).

You do not need 256 devices to test the maximum nodes supported. Testing the bias current over the common mode voltage range (i.e. on each differential input sweep a DC voltage from -12V to 12V on this device (this device has an extended common mode rating compared to normal RS-485) and recording the input current) then from the data you will be able to determine the minimum input impedance over range - for a 1/8th unit load device this should be a minimum of ~96K - but it could be higher and that's ok. 

Please let me know if you have any other questions!

Best,

Parker Dodson

Hi Shuang,

So the output impedance isn't directly spec'd.

However we do have output current/Output voltage graph shown in figure 1:

From this you can find the effective output impedance on A/B when they are driving a low or a high signal.

The curves are pretty linear from 0mA to 60mA and the impedance of A or B when driving a high or low can be approximated by (|ΔY|) / (ΔX)

So A or B when driving high at a VCC of 5V, using the 50mA point as the 60mA is cut off of the graph, so the resistance is (|4.25V-5V|)/(50mA - 0mA) = (.75V/50mA) =  15 Ohms - so A or B pin when driving a high has an impedance of ~15 Ohms

For VCC = 3.3V its the same process but it would be (|3.3-2.2|)/(60mA - 0mA) = 18.33 Ohms - so ~18 Ohms 

When A or B are driving low you use the same process but with the VOL lines.

For VCC = 5V: (0.4V - 0V)/(60mA - 0mA) = 6.67 Ohms when A/B driving a Low at VCC = 5V

For VCC = 3.3V: (0.55V - 0V) / (60mA - 0mA)  = 9.167 Ohms when A/B are driving a Low at VCC = 3.3V

So at VCC = 5V, in a typical use case (T = 25C) :

A logic one is driven - the A pin has ~ 15 Ohms; the B pin has ~6.67 Ohms

A logic zero is driven - the B pin has ~15 Ohms; the A pin has ~6.67 Ohms

At VCC = 3.3V, in a typical use case (T = 25C):

A logic one is driven - the A pin has ~ 18 Ohms; the B pin has ~9.167 Ohms

A logic zero is driven - the B pin has ~18 Ohms; the A pin has ~9.167 Ohms

Those are the typical impedances you should see under most operating conditions - there will be some variance over temperature and with each device but this should be a decent approximation of a typical use case. 

Please let me know if you have any other questions!

Best,

Parker Dodson