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Signals & Wireless (RF)

Impedance matching explained

A load takes the most power from a source when its resistance equals the source's own resistance. That's the maximum power transfer theorem, and making a load look like its source is impedance matching. It's why radio transmitters, cables, connectors and antennas are all built for the same impedance, usually 50 Ω.

Every source has resistance inside

Seen from its terminals, any source is a voltage behind a resistance, its Thevenin equivalent. A radio transmitter has 50 Ω inside, by convention. The load gets whatever that internal resistance leaves.

  • A huge load gets nearly all the volts, but hardly any current: little power.
  • A tiny load gets plenty of current, but almost no volts: little power again.
  • In between there's a best, and it's a load equal to the source's resistance.
Pmax = V² over 4 × RS when RL = RS

The sign of a match: the load gets exactly half the source's voltage. More than half and the load is too big; less and it's too small.

Live experiment

Find the load that takes the most

A 1 V generator with 50 Ω inside drives Pip's load. Step it from 5 Ω to 1 kΩ: a big load gets the volts but little current, a small one the current but no volts. The power peaks at 2.5 mW when the load is 50 Ω and gets exactly half the volts.

Try it yourself: free lesson

“Match the load” is lesson 4 of 6 in Carrier Cove, Signals & Wireless (RF) World 1. Free with an account. Lessons unlock in order, so you’ll do the 3 before it first.

A worked example

A 50 Ω transmitter drives a 200 Ω antenna. The antenna gets 200 ÷ 250 = 80% of the voltage, and its power compared with a matched load is:

4 × RS × RL over (RS + RL)² = 4 × 50 × 200 over 250² = 64%

Over a third of the power that could reach it doesn't.

Matching with an L-network

Two reactive parts can make one impedance look like another at a chosen frequency. For 200 Ω to 50 Ω, put a capacitor across the 200 Ω antenna and a coil in series towards the transmitter. The capacitor pulls the antenna's impedance down towards 50 Ω, leaving some reactance; the coil cancels it.

At 10 MHz, about 140 pF and 1.4 µH do it. From a 1 V source, the antenna now gets a full 1 V instead of 0.8 V: the network steps the voltage up as it steps the impedance. Coils and capacitors store energy rather than burning it, so the network itself wastes almost nothing. But it only works near one frequency.

Matching with reactance

When the source and load have reactance too, the best match is the complex conjugate: the same resistance, with equal and opposite reactance, so the reactances cancel. A mismatch on a long cable also reflects part of the signal back, measured as SWR (standing wave ratio). Radio uses 50 Ω throughout, and TV aerials 75 Ω, so every link in the chain matches the next.

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.

Try it yourself: free lesson

“Match the load” is lesson 4 of 6 in Carrier Cove, Signals & Wireless (RF) World 1. Free with an account. Lessons unlock in order, so you’ll do the 3 before it first.

Or start from the very first lesson, no account needed

Practice questions

Three quick questions from the lessons. Pick an answer to check it.

  1. Question 1 of 3The load is matched. It gets…
    Show the answer

    Answer: Half the source's volts. 50 Ω inside and 50 Ω outside split the volts equally. Half the volts, and the maximum power.

    • Why not “A tenth”? A tenth would mean the load was far smaller than the source, which delivers less power.
    • Why not “Double”? A passive load can never get more than the source's own voltage.
    • Why not “All the volts”? All the volts would mean no current flowing, which only an open circuit gets. That delivers no power.
  2. Question 2 of 3A source with 50 Ω inside drives a 1 kΩ load. The load gets 95 % of the volts, so it gets…
    Show the answer

    Answer: Little power: lots of volts but hardly any current. Power is volts times amps. 1 kΩ takes almost no current, so almost no power, however many volts it has. The most power goes to 50 Ω.

    • Why not “The most power possible”? The most power goes to a load equal to the source's 50 Ω. A 1 kΩ load gets the volts but hardly any current.
    • Why not “No volts”? The question says the load gets 95% of the volts. It's the current, not the voltage, that it's short of.
    • Why not “Half the power”? Half the power goes to a matched 50 Ω load. At 1 kΩ the load is twenty times too big and takes far less.
  3. Question 3 of 3A load takes the most power from a source when…
    Show the answer

    Answer: Its resistance equals the source's. Matched: half the volts, the most power.

    • Why not “Its resistance is zero”? A short circuit draws plenty of current, but with no voltage across it, it takes no power.
    • Why not “Its resistance is huge”? A huge load gets nearly all the volts but almost no current, so its power is small.
    • Why not “It's an open circuit”? An open circuit draws no current at all, so it takes no power.
Keep practising: free lesson

“Match the load” is lesson 4 of 6 in Carrier Cove, Signals & Wireless (RF) World 1. Free with an account. Lessons unlock in order, so you’ll do the 3 before it first.

Questions people ask

What is the maximum power transfer theorem?

A source with internal resistance R_S delivers the most power to a load when the load's resistance equals R_S. The load then gets half the source's voltage and V² ÷ (4 × R_S) of power. Any bigger or smaller load takes less.

Why is impedance matching important in RF?

Radio signals are weak, so every bit of power counts, and a mismatched transmission line reflects part of the signal back towards the source. Matching the transmitter, cable, connectors and antenna to the same impedance, usually 50 Ω, delivers the most power with no reflection.

Does matching mean 100% efficiency?

No, only 50%. At a match the source's own resistance heats up exactly as much as the load gets. That's why power grids don't match: their loads are far bigger than the grid's resistance, so very little is lost on the way.

How do you match a 200 Ω antenna to 50 Ω?

With an L-network: a capacitor across the higher-impedance side and a coil in series towards the lower. At 10 MHz, about 140 pF across the 200 Ω antenna and 1.4 µH in series make it look like 50 Ω. The values change with frequency.

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