SN75LBC180: How to calculate the maximum number of connections

You do not need to care about current or resistance; you just need to count unit loads. (The RS-485 specification defines it for this purpose.)

All RS-485 drivers support 32 unit loads. A 'normal' receiver like the SN75LBC180 is one unit load. Many modern receivers are better; e.g., the THVD1500's receiver is ⅛ of a unit load, and you could use 32 × 8 = 256 such receivers on a bus.

Hi Conor,

Generally speaking - all you need to do to see how many nodes on a bus a device can support if it isn't explicitly mentioned on the first page of the datasheet (many times it is) is look to see what type of unit loading the device has - this is also typically on the very first page of the datasheet.

1 Unit Load = 32 Devices on Bus 

1/2 Unit Load = 64 Devices on Bus

1/4 Unit Load = 128 Devices on Bus

1/8 Unit Load  = 256 Devices on bus.

Where 1 Unit load refers to the following figure:

Where the x axis represents input voltage on bus (either of the differential pins) from the -7V to 12V common mode range required by RS-485. The Y axis is the corresponding sink or source current - anything in the shaded region is at least a 1 Unit load device. The Y-Axis value is also referred to as input bus current - which will be defined in the datasheet as Ib - input bus current. 

On some older devices - we may only include the Ib spec and not explicitly mention the other versions of the specification. So you can look at the Datasheet and check the Ib value to determine what the unit load is for the device.

For the SN75LBC180 specifically 

We see that the bus input current at Vi = 12V has a max of 1ma and the Vi = -7V input has a minimum of -0.8mA which if we look back at the unit load chart conforms to a 1 unit load device - so this device can support 32 nodes on a bus (assuming all other nodes are also 1 Unit Load)

For other unit loadings such as 1/8th - you'd just divide the 12V max and the -7V min currents by 8 - so a 1/8th unit load device would have a max of 125uA of current at 12V and a min of -100uA at -7V. 

If you only have the IC's and 2 termination resistors at each end of the bus (with a minimum impedance between the differential pins of 54 Ohms - generally both termination resistors are 120 Ohms with up to a +/-10% tolerance) then you can support up to the 32 nodes for this device.

That being said - it does get a little bit more complicated if you have other passive devices on the bus that connect either of the differential pins to Ground or a source voltage - as this will also load the bus and diminish the amount of nodes that you can have on a bus. If this is a concern in your application - then there is one more step you can do to help calculate the maximum loading of the bus - which  is to look at the common mode loading.

To do this - you will analyze each differential line separately - and for the calculations there will be no termination resistor - but the actual system will still require one. The first step is to take all sources except the driver and short them to ground via superposition - so all other transceivers and other passives on the lines are connected to ground. Next calculate the approximate impedance from differential line to ground - this value should be >= 375 Ohms to prevent overloading of the bus.

To put an impedance value on the transceiver it is common for the following impedances to be used in calculations w.r.t. their unit loading - its a short-hand approximation that doesn't really cause any  issues to use for these types of applications

1 Unit Load ~ 12K Ohms

1/2 Unit Load ~ 24K Ohms

1/4 Unit Load ~ 48K Ohms

1/8 Unit Load ~ 96K Ohms

So lets say you have 24 SN75LBC180 on the bus + 1 pull up resistor at 10K on the Y pin and a pull down resistor of 10K on the Z pin. Since both pull-up and pull-down values are the same we only solve for one line.

Looking at the 'Y' Line we have 23 Receivers (24 devices in total - but one is actively driving so isn't counted) we can convert the 23 Recievers to a 12K resistor to ground. We short the pull-up on Y to ground so that the pull-up and other transceivers are in parallel. We then calaculate the loading to check if it is >= 375 Ohms.

Using the formula: 1/(N/12K + 1/10k) - where N is the number of transceivers - and the loading is 1/(23/12K + 1/10k) ~ 495 Ohms which is > = 375 ohms so its okay.

This is only necessary for passive components added to the bus that will add additional loading - if you only have the transceivers and termination resistors on the bus then it is just the bias current/unit load/ number of nodes supported and there really isn't any additional calculations.

Please let me know if you have any questions. 

Best,

Parker Dodson