Trace capacitance, and whether the driver can charge it

A 50pF load takes 20ns to charge to 3.3V from an 8mA driver. Toggled at 50MHz it burns 27mW. Trace capacitance gets called "parasitic," but it has a clear, numeric effect on both timing and power.

Last updated: 2026-10-30 Trace capacitanceParasitic capacitanceDriverPower consumption

A trace runs about 1pF per centimeter

Capacitance per length is C = tpd / Z0. A 50 ohm microstrip (6.5ps/mm) gives 0.13pF/mm, or 1.3pF per cm. An inner-layer stripline (7ps/mm) gives 1.4pF/cm, and a narrow, high-impedance (100 ohm) trace far from the plane about 0.65pF/cm.

What the trace connects to adds more (table). A 10cm trace feeding three receiver ICs (5pF each), one ESD device (3pF) and one connector (2pF) adds up to 13 + 15 + 3 + 2 = 33pF. Another 30cm of flat cable (50-100pF/m) adds 15-30pF, landing at 50-60pF.

Typical capacitance of things a trace connects to
ElementCapacitanceNotes
50 ohm microstrip1.3pF/cmLower with higher Z0
CMOS input pin3-10pFDatasheet Cin
ESD protection diode1-5pFLow-cap, high-speed parts: 0.5pF or less
Via0.3-0.5pFDepends on board thickness and pad size
Connector pin1-3pF-
Flat cable50-100pF/mHigher if adjacent lines are grounded
Scope passive probe (10:1)10-15pFChanges the waveform you're measuring
A driver charging the capacitance of a PCB trace
A driver charging the capacitance of a PCB trace. AI-generated conceptual illustration; see the article’s diagrams and text for circuit details and conditions.
Breakdown of a signal line's load capacitance and charge timeDiagram of a trace from a driver feeding ESD protection, three CMOS inputs, and a connector, summing to 33pF total. On the right, a comparison of the time to charge 15pF and 50pF to 3.3V at a constant 8mA. Capacitance adds up. Charge time is C x delta-V / I Example: 10cm trace, 3 receiver ICs, ESD protection, connector Driver 8mA 10cm trace: 13pF ESD 3pF IC input 5pF IC input 5pF IC input 5pF Connector 2pF Total 33pF (over 50pF with cable added) 10ns 20ns 30ns 3.3V 50pF: 20ns 15pF: 6ns To 3.3V at 8mA t = C x 3.3V / 8mA 20ns assumes constant 8mA; it does not establish a clock limit. Power is C x V^2 x f. 50pF x 3.3^2 x 50MHz = 27mW/line.
Figure 1: capacitance adding up, and the charge time at 8mA.

Time to charge it

A constant-current approximation gives t = C × ΔV / I: about 20.6ns for 50pF, 3.3V and 8mA. IOH/IOL, however, are specified at particular output voltages. They do not guarantee constant current throughout an edge, so this calculation is not a guaranteed upper bound on rise time.

Another approximation treats the output as Ron. Assuming 50Ω from the single point VOL = 0.4V at 8mA gives τ = 2.5ns and a 10–90% edge of about 5.5ns into 50pF. Pull-up and pull-down behaviour can differ. Propagation delay measures arrival between defined reference points; rise time measures an edge between thresholds. Adding them does not give a rise time.

The half-period is 25ns at 20MHz and 10ns at 50MHz. A 5.5ns transition alone does not establish that 50MHz fails. Check receiver thresholds, setup and hold requirements, duty cycle and driver timing at the relevant load. For loads outside the timing specification or long traces, inspect the receiver waveform using suitable IBIS and interconnect models.

If drive strength is selectable, pick the lowest setting that meets the load; stronger means more ringing, crosstalk and radiated noise.

Two estimates of the time to charge 50pF to 3.3VGraph of voltage 0-3.3V against time 0-30ns. A constant 8mA reaches 3.3V in a straight line at 20ns. An RC with 50 ohm output resistance has tau = 2.5ns and takes 5.5ns from 10% to 90%. A dashed line marks the 25ns half-period of 20MHz. Charging 50pF: constant-current and RC estimates 8mA, 3.3V, R_on = 50 ohm (tau = 2.5ns) 0ns 10ns 20ns 30ns 3.3V 0V 20MHz half-period 25ns Constant current: 3.3V at 20ns 10% 90% RC: 10%-90% in 5.5ns time
Figure 2: two estimates of the 50pF charge time.

Power is C x V^2 x f

Charging a capacitor to V dissipates 1/2·CV^2 as heat in the driver's resistance, and discharging dissipates another 1/2·CV^2. That's CV^2 per cycle, so toggling at f gives P = C x V^2 x f: 50pF x 3.3^2 x 50MHz = 27mW. A 16-bit bus is 0.44W, possibly more than the microcontroller's core power. It heats the driving IC. Here f is the number of 0→1 charging events per second on each line. The 0.44W bus total assumes all 16 lines toggle every cycle; apply each line's activity factor for ordinary data traffic.

The levers: shrink C (shorter traces, fewer receivers, low-capacitance ESD parts), lower V (3.3V to 1.8V cuts it by 3.4x), or lower f (gate unused clocks). If a battery design draws more than predicted, a constantly toggling clock is often the culprit.

How C x V^2 x f comes about1/2·CV^2 is dissipated in the driver's resistance on charge and another 1/2·CV^2 on discharge, CV^2 per cycle. Multiplied by frequency f this is the power: 27mW for 50pF, 3.3V and 50MHz. Loss per cycle, CV^2, times frequency is the power Example: 50pF, 3.3V, 50MHz Charge ½CV² heat in driver R Discharge ½CV² heat in driver R One cycle CV² total Power per line 27mW C x V^2 x f + = ×f The power is dissipated in the IC that drives the I/O. About 0.44W only if all 16 lines toggle every cycle.
Figure 3: CV^2 per cycle times frequency is the power.

Connecting a probe changes the waveform

A 10:1 passive probe carries 10-15pF. On a 33pF trace the load becomes 45pF, slowing the rise by about 30%. You are measuring the waveform with the probe attached.

"It works with the probe on" in a marginal design means the probe's capacitance is making up a hold-time shortfall or damping ringing. Use an active probe with 1pF or less, count the probe's capacitance as part of the load, or verify without a probe (via an LED or another pin).

Only a short trace behaves as a capacitor

All this assumes a lumped view, where the whole trace sits at one voltage. With a fast driver and a long trace it is a transmission line: the driver sees characteristic impedance (50 ohm) instead of capacitance, and rise time is set by reflections. The dividing line is covered in how long before a trace becomes a transmission line.

The trace capacitance calculator computes capacitance from trace dimensions; the I/O buffer drive calculator computes rise time and power from load capacitance and driver strength.

Frequently asked questions

Can a GPIO drive a long trace (a 1m cable)?
A 1m cable is 50-100pF, so an 8mA GPIO rises in 20-40ns, fine around 100kHz. For faster or noisier use, use a line driver (RS-422/485, etc.).
Does making the trace narrower reduce its capacitance?
Yes (50 ohm to 100 ohm roughly halves it), but reflections and crosstalk grow. Shortening the trace or reducing receivers usually helps more.
What does a driver datasheet's "50pF load capacitance" figure mean?
The load used to measure the specified delay or transition time. A larger real load makes both worse; adjust from your actual load.
Is I/O power really that significant?
A 16-bit bus at 50MHz with 30pF per pin gives 16 x 30p x 3.3^2 x 50M = 0.26W. Slow GPIOs and rarely toggling lines are negligible; constantly running clocks and busy data buses dominate.

Standards and references

  • Logic IC and microcontroller manufacturers' datasheets — IOH/IOL, input capacitance, test load capacitance, I/O power
  • IPC-2141A — Transmission line characteristic impedance and per-length capacitance
  • Tektronix / Keysight probe application notes — Probe input capacitance and loading effects
  • Texas Instruments, Understanding and Interpreting Standard-Logic Data Sheets, SZZA036C (June 2016) — Sections 4.6, 4.9 and 4.12: output-current conditions, propagation/transition times and test loads.

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