RS-485 termination and biasing calculator

Terminate an RS-485 bus when its one-way delay exceeds a quarter of the driver's rise time, per TI AN-903. On 5 ns/m twisted pair, a 100 ns edge allows 5 m; a 1 µs edge, 50 m. Past that, put one resistor matching the cable, typically 120 Ω, at each end. Stubs are held to a tenth of the rise time, 2 m here, per TI SLLA272.

Does this bus need terminators, how long can the stubs be, and what bias resistors keep the pair from floating when nobody is driving — per TIA-485 and the TI design guides.

Fig 1 — 100 m bus, 8 nodes, 120 Ω at each end; bias 680 Ω up and down.

Cable length (m)

Longest stub (m)

Driver rise time (ns)

Bit rate (bit/s)

Cable delay (ns/m)

Termination at each end (0 = none) (Ω)

Bias supply (V)

Required idle differential (V)

Transceivers on the bus

Unit load per transceiver

Termination: required — unterminated limit is 5.0 m
One-way cable delay: 500 ns · 1.0 round trips per bit
Stub limit: 2.00 m ✓
Failsafe bias, each resistor: 692 Ω → E24 680 Ω
Idle differential with 680 Ω: 203 mV
Bus load the bias sees: 57.7 Ω (inputs 1.50 kΩ)
Unit loads: 8.00 nodes + 8.82 bias = 16.8 of 32
Bias current, idle: 3.53 mA

How this is calculated

Standard: TIA/EIA-485-A; TI SNLA034 (AN-903, when a cable is a transmission line), SLLA272 (design guide, stubs), SLLA166 (unit loads and biasing)

$t_{delay} = L \cdot t_{pd}$: One way. Twisted pair runs at roughly 5 ns/m.
$L \le \frac{t_{rise}}{4 \cdot t_{pd}}$: Termination is optional below this length.
$L_{stub} \le \frac{t_{rise}}{10 \cdot t_{pd}}$: Stub limit.
$R_{in} = \frac{12\,\text{k}\Omega}{N \cdot \text{UL}}$: Every receiver input on the bus, in parallel.
$R_{bus} = \frac{R_T}{2} \parallel R_{in}$
$R_{bias} = \frac{R_{bus} \cdot (V_{cc} / V_{idle} - 1)}{2}$: One resistor to V_cc on A, one to ground on B.
$V_{idle} = \frac{V_{cc} \cdot R_{bus}}{2 \cdot R_{bias} + R_{bus}}$
$\text{UL}_{bias} = \frac{12\,\text{k}\Omega}{2 \cdot R_{bias}}$: Unit loads the bias network itself puts on the bus.

Assumptions

Rise time is the driver datasheet 10–90 % figure into the specified load; slew-limited parts have much longer edges and relax every limit here.
The 1/4 and 1/10 rules are rules of thumb from TI AN-903 and SLLA272, not the standard; a line that fails them is not guaranteed to fail, it is guaranteed to reflect.
Receiver inputs are taken as 12 kΩ per unit load minimum. Real parts are often higher, which only helps.
The bias network sits in one place. Two sets of bias resistors on one bus double the load and halve the idle voltage.
Common-mode range (−7 to +12 V) and ground shift are not checked; isolate if the ends do not share a ground.

What sets RS-485 termination and bias resistors

Two things go wrong on an RS-485 bus that has nothing to do with the protocol: the line rings because nobody terminated it, or the line floats because nobody biased it. The tool sizes both.

Termination first. A driver edge travels down the cable at about 5 ns/m (v = 0.66 c on typical twisted pair), hits the far end, and comes back. If it comes back while the edge is still rising, it is absorbed into the edge and the receiver never sees it. TI's AN-903 (SNLA034) draws the line at a quarter: when the rise time is more than four times the one-way delay, the cable is no longer a transmission line; beyond that, terminate. SLLA272 holds stubs tighter, a tenth of the rise time, so they never show up as a visible step. The delay, the unterminated length and the stub limit are the first three expressions under How this is calculated above.

Biasing second. With every driver tri-stated the A–B pair sits at 0 V and a classic receiver, whose threshold is anywhere in ±200 mV, reads noise. A pull-up on A and a pull-down on B force idle to a mark. They form a divider with everything across the pair: both terminators in parallel and the receiver inputs, a unit load being 12 kΩ. To a driver the bias pair looks like 2R_b between the rails, and SLLA166 counts that against the 32 unit loads the standard allows.

RS-485 bias resistor chart

The failsafe bias pair for an eight-node bus, computed by the calculator above, and what each pair costs from the 32 unit-load budget. Terminated means 120 Ω at each end, which is what the bias has to fight; the receivers barely matter next to it. Unterminated is the short, slow bus where the receivers are all there is.

Supply and idle target	R_b, terminated	E12	UL cost	R_b, unterminated	UL cost
5 V, 200 mV idle	692 Ω	680 Ω	8.67	18.0 kΩ	0.33
3.3 V, 200 mV idle	447 Ω	470 Ω	13.42	11.6 kΩ	0.52
5 V, 50 mV idle (failsafe receiver)	2.86 kΩ	2.7 kΩ	2.10	74.3 kΩ	0.08
3.3 V, 50 mV idle (failsafe receiver)	1.87 kΩ	1.8 kΩ	3.20	48.7 kΩ	0.12

The two halves of the table are the two arguments the page makes. Across a row, termination is what makes bias expensive: the same 200 mV costs 8.67 unit loads on a terminated bus and a third of one without. Down a column, the idle target is the lever: a receiver that reads a mark from 50 mV gives back six and a half of those unit loads at 5 V, and ten at 3.3 V, where the lower supply had made the bias even heavier.

Worked example: 100 m of cable with eight nodes

100 m of cable, 0.3 m longest stub, 100 ns driver rise time, 1 Mbit/s, 5 ns/m, 5 V supply, 200 mV idle target, 120 Ω at each end, eight nodes at 1 UL.

t_delay  = 100 m × 5 ns/m           = 500 ns
L_unterm = 100 ns / (4 × 5 ns/m)    = 5 m      100 m ≫ 5 m  →  terminate
L_stub   = 100 ns / (10 × 5 ns/m)   = 2 m      0.3 m, OK
t_bit    = 1 / 1 Mbit/s             = 1 µs     1 round trip per bit

R_in     = 12 kΩ / (8 × 1)          = 1500 Ω
R_T / 2  = 120 / 2                  = 60 Ω
R_bus    = 60 ∥ 1500                = 57.69 Ω

R_b      = 57.69 × (5 / 0.2 − 1) / 2
         = 57.69 × 24 / 2           = 692 Ω    →  680 Ω E12

check    V_idle = 5 × 57.69 / (2 × 692 + 57.69) = 0.200 V

UL_bias  = 12 kΩ / (2 × 692)        = 8.67
UL_total = 8 + 8.67                 = 16.67 of 32

The receivers are 1500 Ω; the terminators are 60 Ω. The bias is fighting the terminators, not the receivers, which is why it costs 8.67 unit loads.

Same bus, no termination (short cable, slow edge):

R_bus    = 1500 Ω
R_b      = 1500 × 24 / 2            = 18 kΩ
UL_bias  = 12 kΩ / 36 kΩ            = 0.33

Same terminated bus, but modern failsafe receivers (THVD1450, MAX3485 class) that read a mark from 50 mV of idle:

R_b      = 57.69 × (5 / 0.05 − 1) / 2
         = 57.69 × 99 / 2           = 2.86 kΩ
UL_bias  = 12 kΩ / (2 × 2856)       = 2.10
UL_total = 8 + 2.10                 = 10.10 of 32

That is the argument for a failsafe part: six unit loads come back.

Where the termination and bias model stops being valid

The 120 Ω is the cable's characteristic impedance, not a constant of nature. Cat 5 is 100 Ω, some instrumentation cable is 150 Ω; terminate with what the cable is. There are exactly two terminators, one at each physical end of the trunk, and never one per node.

The quarter-rise-time rule needs the driver's real edge. Slew-rate-limited transceivers (250 kbit/s parts with 500 ns to 1 µs edges) push L_unterm to 25–50 m, let short buses go unterminated, and take the terminator load out of the bias sum. Where EMI matters, split termination, two 60 Ω with a capacitor from the midpoint to ground, is the same differential load with a common-mode path; the tool treats it as 120 Ω.

The model assumes one ground. Two buildings or two power feeds and the receiver sees the signal riding on a ground shift. The standard tolerates −7 V to +12 V of common mode; past that the receiver clamps or dies. Run a ground wire with the pair, or isolate the transceiver (ISO1410, ADM2582E) when the shift cannot be bounded. The bias then lives on the isolated side.

Modern 1/8 UL receivers allow 256 nodes, and the tool accepts fractional UL, but 256 nodes plus 8.67 UL of bias is still 40.67 UL. Fix the bias first.

Topology is not in the model. The stub rule assumes a daisy chain, trunk running node to node with each transceiver on a short stub. A star is several long stubs meeting at a point, none terminated correctly, and no resistor fixes it. Rewire it, or put a repeater on each leg.

Common RS-485 mistakes

Termination on every board. Every designer added "the 120 Ω" and the bus is 30 Ω; drivers current-limit and the far end never reaches threshold. Two, at the ends, jumpered off everywhere else.
Bias at every node. Eight sets of 692 Ω in parallel is 87 Ω to each rail and no driver can swing the pair. Bias lives in one place, usually the master, sized for the whole bus.
Terminating a 4 m bus with a 100 ns driver. Under the 5 m limit the terminators do nothing for the signal and cost 25 mA of driver current plus 8.67 UL of bias. Leave them off and bias with 18 kΩ.
Testing with a driver always on. The bias only matters between packets; the bench works, the field throws framing errors every time the master goes quiet. Check idle with everything tri-stated.
Long stubs to a "nearby" node. 2 m is the limit for a 100 ns edge; the 3 m drop to the panel PC is a reflection source. Loop the trunk through the node instead.