Signals & Wireless (RF)
Build a receiver front end: trap, amplifier, band-pass
A receiver's front end is everything between the antenna and the rest of the radio. Its job is to keep what you want and lose what you don't. This guide builds one for a station at 1 MHz in three stages: an LC trap that shorts a loud interferer to ground, an op-amp that doubles the station, and an LC band-pass that passes the station and blocks harmonics far above it.
What arrives at the antenna
- Your station at 1 MHz
- A loud interferer at 1.5 MHz
- Harmonics from something nearby at 3 MHz
The antenna is a 50 Ω source and the receiver a 50 Ω load. The targets: the station arrives at ×1.2 to ×2.5, and the interferer and the harmonics each at 10 % or less.
Stage 1: trap the interferer
A 10 µH coil and a capacitor in series, from the line to ground. At their resonance they cancel into a near short:
At 1 MHz the same trap is about 78 Ω of reactance, a light load that the station mostly survives.
Stage 2: amplify by 2
A non-inverting op-amp stage with Rf = Rg = 1 kΩ:
With 10 MHz of gain-bandwidth, a ×2 stage is flat to about 5 MHz, so 1 MHz gets its full gain.
Stage 3: band-pass the station
A 100 µH coil and a capacitor in series between the amplifier and the receiver, tuned to 1 MHz:
At 3 MHz the pair is about 1.7 kΩ of reactance in series with the 50 Ω receiver, so the harmonics are cut to a few per cent.
Live experiment
Choose the three parts
The antenna brings your 1 MHz station, a 1.5 MHz interferer and 3 MHz harmonics. Choose the trap capacitor, the amplifier's feedback resistor and the band-pass capacitor, then test: the station at ×1.2 to ×2.5, the interferer and 3 MHz each at 10 % or less. The right build reads ×1.51, 0.12 % and 4.7 %.
“Project: Receiver front end” is the project that ends Oscillator Outcrop, Signals & Wireless (RF) World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.
The results
Built right, the front end gives the station ×1.51 (the source's 50 Ω and the trap take a share before the ×2), the interferer 0.12 %, and 3 MHz 4.7 %.
| Mistake | What goes wrong |
|---|---|
| No trap capacitor | Nothing trapped: the interferer reaches the amplifier |
| No feedback resistor | Open loop: far too much gain, set by the op-amp, not you |
| Band-pass left at 1.13 nF | Tuned to about 470 kHz: the station is mostly blocked |
Why this order
The trap comes first so the loud interferer never reaches the amplifier, where it could overload it and mix with your station. The band-pass comes last to clean up whatever the amplifier adds. Real receivers follow the same pattern: a preselector filter, a low-noise amplifier, then the mixer.
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.
“Project: Receiver front end” is the project that ends Oscillator Outcrop, Signals & Wireless (RF) World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 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: 470 kHz: the 1 MHz station is mostly blocked. 1 ÷ (2π√(100 µH × 1.13 nF)) is about 470 kHz. For 1 MHz the capacitor must be 253 pF.
- Why not “1.5 MHz”? 1.13 nF tunes 1.5 MHz only with the trap's 10 µH coil. With 100 µH it resonates near 470 kHz.
- Why not “1 MHz”? 1 MHz with 100 µH needs 253 pF. With 1.13 nF the resonance is much lower.
- Why not “3 MHz”? Adding capacitance or inductance lowers the resonance. 100 µH with 1.13 nF sits well below 1 MHz, not above it.
Show the answer
Answer: Lags the input by about 45°: it's at its corner. 1 MHz is this stage's corner, 10 MHz ÷ 10, where the gain is down to 70 % and the output trails by 45°, like an RC leg at its corner. ×2 puts the corner at 5 MHz and the 1 MHz output is nearly in phase.
- Why not “Leads the input”? An op-amp running out of gain behaves like an RC low-pass, so its output lags and never leads.
- Why not “Is exactly in phase with the input”? 1 MHz is this stage's corner, 10 MHz ÷ 10, and at a corner the output always trails by 45°.
- Why not “Is ×10 and in phase”? At its corner the gain is down to about 70 % of ×10, and the output trails by 45°.
Show the answer
Answer: ×8.2: each a little under ×3. Each ×3 stage has its corner at 3.3 MHz and delivers ×2.87 at 1 MHz. 2.87 × 2.87 is about 8.2.
- Why not “×9 exactly”? Each stage's corner is 10 MHz ÷ 3, about 3.3 MHz, so at 1 MHz each gives a little under ×3.
- Why not “×3”? Two stages in a row multiply their gains, so the total is near ×9, not ×3.
- Why not “×6”? Stage gains multiply rather than add: 2.87 × 2.87 is about 8.2.
“Project: Receiver front end” is the project that ends Oscillator Outcrop, Signals & Wireless (RF) World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.
Questions people ask
What is a receiver front end?
Everything between the antenna and the receiver's mixer or detector: the filters that keep unwanted signals out and the amplifier that lifts the wanted one above the noise. It decides how sensitive the receiver is and how well it ignores strong neighbours.
Why filter before amplifying?
A strong unwanted signal can push the amplifier towards its limits, and an overloaded amplifier mixes signals together into new frequencies that no later filter can remove. Trapping the worst offender first keeps the amplifier in its clean, linear range.
Why only ×2 gain from the op-amp?
Gain-bandwidth. A 10 MHz op-amp set for ×2 stays flat to about 5 MHz, so 1 MHz passes cleanly. Set for ×10, its corner would land right on 1 MHz, where it would give only about ×7 and shift the phase by 45°.
Related topics
- LC resonant frequency explainedA coil and a capacitor swapping energy at one frequency: f₀ = 1 ÷ (2π√(LC)), tanks and tuning.
- Non-inverting amplifier explainedAn op-amp, two resistors and gain G = 1 + R_f ÷ R_g: bigger, in step, and clipped at the rails.
- Op-amp gain-bandwidth product explainedGain times bandwidth stays fixed: ask an op-amp for more gain and its corner frequency falls.
- Impedance matching explainedA load takes the most power when its resistance equals the source's: why radio is 50 Ω, and how to match.