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RF power atteunation

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Hi Forum,

I plan to reduce the transmitted power of standard 433MHz transmitter .

Following is the Module and its parameters ;

Frequency Range:315 / 433.92 MHZ.
Supply Voltage: 3~12V.
Output Power : 4~16dBm
Circuit Shape: Saw

As it can be seen it is transmitting around 16dBm at 12V input would be good for 300meter distance or so . I want to employ Pi atteunator on this to reduce the power to such a level that is is able to transmit 5 meters approx.

I plan to do this;

Transmitter Ant------>Pi Atteunator----->loop antenna.

Can anyone help me with what amount of atteunation is required to achieve this distance ?

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Super Moderator
Most Helpful Member
WARNING - This post contains references to decibels and other serious radio content, viewer discretion is advised.:D:D:D

The accepted formula for calculating path loss is:

Pathloss = 32.45 +20Log(f) + 20Log(d)

Pathloss is in dB
f is in Mhz
d is in km

This formula is applicable to "long" paths, hence d in km.
I have no idea how well is scales to metre distances, but lets try it anyway.

The pathloss at 433Mhz over 300metres is 74.7dB
The pathloss at 433Mhz over 5metres is 39.2dB

The difference is 35.5dB.

So I suggest that you try a 30dB attenuator to start with.

On a practical note, if you are expecting this thing to work reliably and repeatedly at 4.9m, but never work at 5.1m, you are in for a surprise.
The range will be very much affected by surrounding objects, how well the transmitter and receiver are screened, how well the supply leads are filtered, etc, etc.



New Member
Just decrease the voltage supply to the module down to 3V.
I agree. If the module's power if variable by reducing it's DC input voltage as the specs claim, why run it at a higher power level just to then reduce the power by attenuation. Seems to me that is like driving your car with your feet pressing on both the gas and brakes at the same time. ;)

Long live the decibel

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