7

This is an attempt to get a 'second opinion' using a different account (no history) and a much more specific prompt.

Prompt: Give me a circuit to match a 1 millivolt p-p 50 ohm signal generator source down to a 1 microvolt 400 ohm antenna input. Show each stage as a separate entity. Give me a factual, highly direct answer. Do not use creative language or stylistic flourishes. Stick strictly to the most probable data.

The source¹ is here.

To match a 1 mV p-p 50-ohm source to a 1 µV 400-ohm load, the circuit requires an impedance transformation and a 1000:1 voltage reduction (60 dB attenuation).

Attempting this in a single matching stage results in mathematically valid but physically impractical resistor values (e.g., < 0.2 ohms). The standard engineering solution is to cascade the required attenuation into manageable 50-ohm blocks, followed by a final impedance matching pad.

Here is the four-stage cascaded circuit to achieve exact matching and attenuation.

Stage 1: 20 dB Attenuator (50-ohm T-Pad)

This stage reduces the 1 mV p-p input to 100 µV p-p while maintaining a 50-ohm impedance on both sides.

        **Input Series Resistor:** 40.91 ohms
        **Output Series Resistor:** 40.91 ohms
        **Center Shunt Resistor:** 10.10 ohms (Connects from the junction of the two series resistors to ground)
    

Stage 2: 20 dB Attenuator (50-ohm T-Pad)

This stage reduces the 100 µV p-p signal to 10 µV p-p.

        **Input Series Resistor:** 40.91 ohms
        **Output Series Resistor:** 40.91 ohms
        **Center Shunt Resistor:** 10.10 ohms (Connects from the junction of the two series resistors to ground)
    

Stage 3: 14.27 dB Attenuator (50-ohm T-Pad)

This stage provides a specific 5.1669:1 voltage reduction, dropping the 10 µV p-p signal to 1.935 µV p-p.

        **Input Series Resistor:** 33.78 ohms
        **Output Series Resistor:** 33.78 ohms
        **Center Shunt Resistor:** 20.11 ohms (Connects from the junction of the two series resistors to ground)
    

Stage 4: Minimum Loss Matching Pad (50-ohm to 400-ohm L-Pad)

This final stage matches the 50-ohm line from Stage 3 to the 400-ohm antenna input. Because the 400-ohm load is higher than the 50-ohm source, the shunt resistor must be placed on the 50-ohm side. The inherent voltage division of this network reduces the 1.935 µV p-p signal down to exactly 1.00 µV p-p across the 400-ohm load.

        **Input Shunt Resistor:** 53.45 ohms (Connects directly from the 50-ohm input line to ground)
        **Series Resistor:** 374.17 ohms (Connects inline from the 50-ohm input directly to the 400-ohm antenna load)
    

Verification Data

        **Total Voltage Attenuation Factor:** $10 X 10 Xs 5.1669 Xs 1.9354 = 1000
        **Stage 4 Input Impedance:** $53.45 || 374.17 + 400 = 50.00 ohm
        **Stage 4 Output Impedance:** $374.17 + (53.45 || 50) = 400.00 ohm
    

Footnotes:

1. No idea how long Google will keep this 'pinned' result available. If it goes away, it's not much different than what is presented above.

Page: /Antenna_matching/7/7.shtml

Last modified: Wednesday, 1 Jul 2026