6 dB in-line 50 ohm attenuator

Question

Recently someone posted a question regarding an inline device to attenuate their voltage on a 50 \$\Omega\$ coax by half (i.e. 6 dB). Because the question asked for commercial recommendations, it was closed. However, the question of how to create an attenuator that has the same input and output impedance is a reasonable one, I was not able to find a similar question, although my search was perhaps superficial. If this is a duplicate, and provides no new value, I will gladly delete it.

The easy part of the answer is a circuit that gives 6 dB attenuation with both input and output impedances matching 50 ohms. I will give that answer shortly. A harder part, is how to physically construct such a circuit such that the transition between coaxial cable and circuit and back to coax is not afflicted by reflections due to changing impedance. This difficulty is perhaps part of the reason why commercial in-line attenuators tend to be expensive. Any advice in this area will be a welcome addition.

how to create a 6 dB attenuator with 50 \$\Omega\$ input and output impedance. — Math Keeps Me Busy, Oct 14 '21 at 11:22

Math Keeps Me Busy · Accepted Answer · 2021-10-14T13:24:12.217

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The following simple circuit provides "6" dB attenuation and 50\$\Omega\$ input impedance, assuming that the load at the output is also 50\$\Omega\$. (To be precise, it is an attenuator with a attenuation factor of 0.5 or 6.021 dB)

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The circuit can easily be adjusted for 75\$\Omega\$ lines simply by scaling each resistor by 75/50 = 1.5

Edit:

AndyAka has kindly given an equation which contrains the resistor values for a T-pad attenuator with equal input and output impedances:

$$R_{in} = R_{out} = R_1\sqrt{1 + \frac{2 \cdot R_3}{R_1}}$$

(I have modified it to match the schematic in my answer. In this equation \$R_1\$ is equal to \$R_2\$).

edited Oct 14 '21 at 13:24

answered Oct 13 '21 at 16:06

Math Keeps Me Busy

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The output voltage will depend on the load so it is wrong. – Jun Seo-He Oct 13 '21 at 16:12
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@JunSeo-He The load impedance is expected to be a matched 50 ohm load; all passive attenuators and likely many RF devices in general will have a load-dependent effect if there's a mismatch. – nanofarad Oct 13 '21 at 16:14
@JunSeo-He I added wording to make explicit the assumption that the load is 50 ohms. – Math Keeps Me Busy Oct 13 '21 at 16:19
Specifically it's a 6.021 dB attenuator. – Andy aka Oct 14 '21 at 11:35
@Andyaka yes... – Math Keeps Me Busy Oct 14 '21 at 12:54
\$R_{IN}=R_{OUT} = R_1\sqrt{1+\dfrac{2\cdot R_2}{R_1}}\$ – Andy aka Oct 14 '21 at 13:05

score 1 · Answer 2 · answered Apr 29 '22 at 15:16

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Of course you could just use a calculator specifically designed for that application: Everything RF T-attenuator

answered Apr 29 '22 at 15:16

analog

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6 dB in-line 50 ohm attenuator

2 Answers2