Electronics 101 // impedance lab
reflection · matching · bandwidth

Make the reflection
disappear.

Every RF link fights the same enemy: impedance mismatch. When source and load disagree, energy bounces back as standing waves. This is a hands-on lab for the Smith chart, L / Pi / T networks, stub matching, and the Q–bandwidth tradeoff — drag, tune, and watch the physics respond.

Γ reflection coefficient VSWR standing waves Q = F / BW λ stub lengths
An Electronics 101 initiative by Sanu · theory from Silicon Labs AN1275
01

Why reactance costs you power

A purely real load keeps voltage and current in phase — all power reaches the load. Add reactance and current lags voltage; the phase gap generates reflections and standing waves.

Voltage Current
PHASE Δ: 0° IN PHASE: YES STATE: matched — no standing wave
02

The interactive Smith chart

Constant-resistance circles nest toward the right; constant-reactance arcs curve into the top half (inductive, +jX) and bottom half (capacitive, −jX). The center is a perfect 50Ω match. Drag the marker — Γ, VSWR, return loss and delivered power recompute live.

drag the marker · or pick a preset →

Live measurement

z (normalized)1.00 + j0.00
Z (Ω, Z₀=50)50 + j0 Ω
|Γ|0.000
VSWR1.00 : 1
Return loss∞ dB
Power to load100.0%

Jump to a case

03

Q vs bandwidth — the core tradeoff

BW = F / Q. A low-Q network matches over a wide band but lets harmonics through; a high-Q network is selective but touchy — component tolerance can shift the whole notch. Slide Q and watch the S11 reflection curve narrow.

Network Q

Q factor1.2
2.04 GHz −3dB BW @ 2.445 GHz

Reference points

L network (ex. 2)Q = 1.17
Pi network (ex. 2)Q = 4.89
Target BW500 MHz
04

Four ways to move the point

Series parts slide the impedance along resistance circles; shunt parts slide it along admittance circles. More elements buy control over Q. Coil = inductor, plates = capacitor.

2 ELEMENTS

L Network

Series L + shunt C (low-pass) or series C + shunt L (high-pass). Lowest loss; Q is fixed by the impedances.

L series C shunt
bandwidth WIDE · Q not adjustable
3 ELEMENTS

Pi Network

Shunt C — series L — shunt C. Two L networks back-to-back with a virtual R below both terminations. Q is yours to choose.

between HIGH impedances (>50Ω) · one inductor only
3 ELEMENTS

T Network

Series — shunt — series. Same Q formula as Pi; virtual R is larger than either termination.

between LOW impedances (<50Ω) · needs two series L
TRANSMISSION LINE

Single Stub

No lumped parts at all — a line length d rotates the load to the unit conductance circle, then a shorted stub of length ℓ cancels the susceptance.

ℓ stub ← d → ZL
microstrip friendly · free at PCB level
05

Single-stub matching calculator

Enter a load and get the two lengths that match it: the distance d from the load where the line admittance hits Re(y)=1, and the shunt stub length ℓ whose susceptance cancels what remains. Lengths are in wavelengths (λ) on a Z₀ = 50Ω line.

Load & stub type

Solution (first of two)

Distance to stub d— λ
Stub length ℓ— λ
d physical (εr=1)— mm
ℓ physical (εr=1)— mm
Two mathematical solutions exist; the shorter-d one is shown. On real PCB, divide physical lengths by √εeff of your substrate. Short stubs are preferred at RF — an open end radiates and its fringing makes the true length uncertain.
06

The matching move, in four beats

STEP 01
Read the miss

Plot the load. Distance from center is your reflection; top or bottom half tells you inductive vs capacitive.

STEP 02
Shunt to the circle

A parallel part slides you along a conductance circle onto the unit resistance circle.

STEP 03
Series to center

A series part with equal-and-opposite reactance cancels the leftover and lands on 50Ω.

STEP 04
Absorb the strays

Fold the load's own parasitics into the network — fewer parts, smaller values, cheaper board.

07

Worked example: RFIC → 50Ω

From AN1275: match a 2.4 GHz radio whose optimum load is 23 + j11.5 Ω to a 50Ω trace at 2445 MHz, low-pass. Source conjugate is 23 − j11.5Ω; the load side is higher, so the shunt part goes toward the load.

Low-pass L network · series L, shunt C

Q = √(50/23 − 1) = 1.08

Q gives both reactances; then resonate out the source capacitance with an extra series inductor and merge it into the final value.

SERIES REACTANCE Xₛ
24.84 Ω
SHUNT REACTANCE Xₚ
46.29 Ω
SERIES INDUCTOR L
2.35 nH
SHUNT CAPACITOR C
1.40 pF