PCB Design
How a series termination resistor stops ringing
A fast logic edge driven down a PCB trace can ring: the trace's inductance and the receiving pin's capacitance form an LC circuit, and the edge sets it swinging. The pin overshoots its rail, dives below ground, and can see false clock edges. A series termination resistor, placed right at the driver's output, damps the swing. Choose it to roughly match the trace, about √(L ÷ C).
Why the edge rings
Model the trace as a coil and the input pin as a capacitor to ground. The numbers below are scaled up so the ringing is slow enough to watch; real traces have nanohenries and picofarads and ring far faster, but the maths is the same. With 10 µH and 1 nF:
With almost no resistance in the loop, a 0 to 5 V step overshoots to nearly 10 V, twice the step, and keeps swinging for microseconds. A 5 V chip sees almost double its rail on every edge, and the undershoot below ground can flip its input too.
The characteristic resistance
How much it overshoots depends on the damping ratio, ζ = (R ÷ 2) × √(C ÷ L):
| R at the driver | ζ | What the pin does |
|---|---|---|
| 1 Ω | 0.005 | Rings to 9.9 V and keeps going |
| 47 Ω | 0.24 | Overshoots to 7.3 V |
| 100 Ω | 0.5 | One small overshoot, 16 %, to 5.8 V |
| 200 Ω | 1 | No overshoot at all |
Live experiment
Damp a ringing clock edge
A 0 to 5 V clock at 100 kHz drives a 10 µH trace into a 1 nF pin. Pick the resistor at the driver: 1 Ω rings to 9.9 V, 47 Ω still reaches 7.3 V, 100 Ω overshoots once to 5.8 V and settles, 200 Ω doesn't overshoot, and 1 kΩ turns the edge into a slow ramp.
“Damp the edge” is lesson 4 of 6 in Solder Strait, PCB Design World 2. Free with an account. Lessons unlock in order, so you’ll do the 10 before it first.
Not too much
More resistance kills the ringing but slows the edge, because the resistor now charges the pin's capacitance: R × C is roughly the rise time. 100 Ω with 1 nF is 0.1 µs. 1 kΩ is 1 µs, and 0.6 µs after a 100 kHz clock's edge the pin has only reached 2.3 V. The characteristic value is the sweet spot: fast, with one small bump.
Why at the driver
The resistor goes as close to the driving pin as possible. There it limits the current flowing into the trace, so the LC never gets the energy to ring. At the far end, the whole trace between the driver and the resistor would still swing.
On a real board
A long, fast trace behaves as a transmission line with its own impedance, often 50 Ω. Source termination works by making the driver's output resistance plus the series resistor equal that impedance, so the reflection from the far end is absorbed when it returns. CMOS outputs are typically 10 to 30 Ω on their own, which is why 22 and 33 Ω resistors sit beside so many clock and data pins. Slow edges on short traces often need nothing; fast edges on long ones almost always do.
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.
“Damp the edge” is lesson 4 of 6 in Solder Strait, PCB Design World 2. Free with an account. Lessons unlock in order, so you’ll do the 10 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: Doesn't ring but takes a microsecond to rise: too slow for the clock. R × C is 1 kΩ × 1 nF = 1 µs of rise time. The ringing is gone, and so is the edge.
- Why not “Arrives faster”? A bigger resistor charges the pin's capacitance more slowly: R × C grows from 0.1 µs to 1 µs.
- Why not “Is the same”? The resistor sets both the damping and the rise time, so a tenfold change shows clearly.
- Why not “Rings more”? More series resistance means more damping, so the edge rings less, not more.
Show the answer
Answer: Still overshoots past 6 V, to about 7.3 V. Less damping than √(L ÷ C): the first overshoot is smaller than 1 Ω's 9.9 V but still over the check's 6 V, and the fall undershoots to −2.3 V. 100 Ω gets it down to one small bump.
- Why not “Is perfectly clean”? 47 Ω is below √(L ÷ C) = 100 Ω, so the circuit is still underdamped and overshoots.
- Why not “Rings exactly as much as with 1 Ω”? 47 Ω soaks up a lot of each swing's energy, so the overshoot is 7.3 V instead of 9.9 V.
- Why not “Never reaches 5 V”? An underdamped edge shoots past 5 V. Only a much bigger resistor makes it creep up slowly.
Show the answer
Answer: The same 100 Ω: √(L ÷ C) doesn't depend on the clock. The ringing is the trace's own, set by L and C. The terminator matches the trace, not the signal.
- Why not “Less resistance”? √(L ÷ C) doesn't involve the clock frequency, so the right value stays 100 Ω.
- Why not “No resistor”? A faster clock has the same fast edges, and each edge still rings without damping.
- Why not “More resistance”? The ringing is set by the trace's L and the pin's C, which don't change with the clock.
“Damp the edge” is lesson 4 of 6 in Solder Strait, PCB Design World 2. Free with an account. Lessons unlock in order, so you’ll do the 10 before it first.
Questions people ask
What value series termination resistor should I use?
Enough that the driver's own output resistance plus the resistor roughly equals the trace's impedance. For 50 Ω traces and typical CMOS outputs that usually means 22 to 33 Ω. In a lumped model like the one here, start at the square root of L over C.
Why does the termination resistor go at the driver end?
It has to limit the current going into the trace before the trace's inductance and the pin's capacitance can store it and swing. Put at the far end, the whole trace between the driver and the resistor still rings.
Do I need to change the resistor if the clock gets faster?
No. The ringing comes from the trace's inductance and the pin's capacitance, which don't depend on the clock rate. What matters is how fast each edge rises, not how often edges come.
Related topics
- LC resonant frequency explainedA coil and a capacitor swapping energy at one frequency: f₀ = 1 ÷ (2π√(LC)), tanks and tuning.
- Impedance matching explainedA load takes the most power when its resistance equals the source's: why radio is 50 Ω, and how to match.
- RC time constant explainedτ = R × C: how long a capacitor takes to charge through a resistor, and the 63% rule.
- What is a decoupling capacitor?Why every chip has a capacitor beside its power pins, how big, and why it must be close.