PCB Design
How to terminate a trace at the receiving end
End termination stops a fast edge ringing at a chip's input with a resistor from the end of the trace to ground, right at the receiving pin, matched to the trace: about √(L ÷ C). The edge's energy goes into the resistor instead of bouncing back and forth. Put a capacitor in series with it and it becomes AC termination, which damps the edges without wasting current while the line sits still.
Why edges ring
A trace has inductance, and the pin at its end has capacitance. A fast edge excites the pair like a struck bell, and with nothing to absorb the energy it sloshes back and forth. A 0 to 5 V clock arriving through 10 µH into 1 nF overshoots to about 9 V and dips to about −5 V. That can trigger a pin twice per edge, or push current through its protection diodes.
Match the trace
At the match, the resistor soaks up each edge: it lands once, with only a small bump. A 10 kΩ pull-down, a hundred times too high, barely damps it. On a long trace treated as a transmission line, the same idea uses its characteristic impedance, often 50 to 100 Ω.
Live experiment
Catch the edge at the pin
A 0 to 5 V clock arrives on a cable through 10 µH into a pin's 1 nF. Pick the end resistor: 10 kΩ rings past 8 V, 100 Ω lands every edge cleanly but pulls 50 mA while the clock is high, and 100 Ω with 47 nF in series keeps the clean edge without the waste.
“Catch it at the end” is lesson 5 of 6 in Gerber Gulf, PCB Design World 3. Free with an account. Lessons unlock in order, so you’ll do the 18 before it first.
The cost: current
While the clock sits at 5 V, a plain 100 Ω terminator pulls 5 ÷ 100 = 50 mA and burns 0.25 W. For a clock that's high half the time, that's a lot of heat for nothing.
AC termination
Put a capacitor in series with the resistor. On an edge the empty capacitor is a short, so the edge still sees 100 Ω. Then it charges, and the current stops. Its time constant has to fit between the edges:
Too small a capacitor, 1 nF, fills during the edge itself and lets go just when it's needed: the pin rings again. Too big, and it never settles between edges, so the line keeps drawing current.
End or series termination?
Series termination puts the resistor at the driver, and only works if you can reach it. End termination works at the receiver, so it suits a clock arriving on a cable from another board, or a bus with a receiver at each end. Every car's CAN bus has a 120 Ω resistor at each end of its cable, and DDR memory terminates its data lines at the receiver, often with resistors switched on inside the chip.
Free lesson
Now build it yourself
Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.
“Catch it at the end” is lesson 5 of 6 in Gerber Gulf, PCB Design World 3. Free with an account. Lessons unlock in order, so you’ll do the 18 before it first.
Or start from the very first lesson, no account neededPractice questions
Three quick questions from the lessons. Pick an answer to check it.
Show the answer
Answer: At the receiving pin, from the trace's end to ground. It catches the edge where it arrives, so it works even when the driver is on another board.
- Why not “In the middle of the trace”? In the middle, the edge would still bounce off the open end beyond it.
- Why not “At the driver too”? That's series termination. End termination is for when the driver is out of reach.
- Why not “On the supply”? The terminator goes where the edge arrives, at the end of the trace, not at the supply.
Show the answer
Answer: Barely damps it: the edge still rings. 10 kΩ is a hundred times too high: hardly any of the edge's energy goes into it. The match is about √(L ÷ C).
- Why not “Doubles the voltage”? A resistor to ground can't raise the voltage. The overshoot comes from the ringing it fails to damp.
- Why not “Stops the signal arriving”? 10 kΩ to ground hardly loads the line, so the signal still arrives in full.
- Why not “Matches it perfectly”? A match is about √(L ÷ C), which is 100 Ω here. 10 kΩ is a hundred times too high.
Show the answer
Answer: Charges up during the edge and stops damping: the pin rings again. 1 nF fills in a fraction of a microsecond, about as long as one ring. It has to stay nearly empty through the edge, and charge up only afterwards.
- Why not “Is the same”? The time constant falls from 4.7 µs to 0.1 µs, about as long as one ring: far too short to damp it.
- Why not “Works better”? A smaller capacitor charges faster, so it stops absorbing the edge before the ringing is over.
- Why not “Wastes more current”? 1 nF charges sooner, so it draws current for less time, not more.
“Catch it at the end” is lesson 5 of 6 in Gerber Gulf, PCB Design World 3. Free with an account. Lessons unlock in order, so you’ll do the 18 before it first.
Questions people ask
What is a termination resistor?
A resistor on a signal line, usually matched to the line's characteristic impedance, that absorbs a fast edge's energy instead of letting it reflect. Without one, long or fast lines ring and can trigger falsely.
Why does a CAN bus need 120 Ω resistors?
Its twisted pair has a characteristic impedance of about 120 Ω, so a 120 Ω resistor at each end absorbs the signal instead of reflecting it. The two in parallel also make the 60 Ω load the transceivers expect.
When does a PCB trace need termination?
When it's long compared with the edge: a common rule says when the rise time is shorter than about twice the trace's one-way delay, roughly 6 to 7 ps per mm on FR-4. Short traces with slow edges don't need it.
Related topics
- How a series termination resistor stops ringingOne resistor at the driver turns a ringing, overshooting clock edge into a clean one.
- Impedance matching explainedA load takes the most power when its resistance equals the source's: why radio is 50 Ω, and how to match.
- LC resonant frequency explainedA coil and a capacitor swapping energy at one frequency: f₀ = 1 ÷ (2π√(LC)), tanks and tuning.
- RC time constant explainedτ = R × C: how long a capacitor takes to charge through a resistor, and the 63% rule.