Using The Decibel Formula
To Calculate Attenuation
The decibel:
“ a unit for expressing the ratio of two amounts of electric or acoustic signal power equal to 10 times the common logarithm of this ratio “
The formula for dB is:
dB = 10 LOG (P1/P2) ; where P1 = Power In , P2 =
Power Out
*** In Amateur Radio the impedance of modern equipment antenna input and output is 50 ohms ***
*** In Amateur Radio the impedance of modern equipment antenna input and output is 50 ohms ***
Aside from the use of
the decibel formula at work, I found myself in need to understand its use in my
hobby, Amateur Radio.
I have a HB1B YouKits CW
transceiver that has an average output of 5 watts across all the bands it
transmits , that is 15, 17, 20, 30 and 40 meters.
Unfortunately it does not have output
power adjustment. In order to reduce the output to lower levels for QRPp (less than 5 Watts) operation.
A simple PI or T Network Resistive Attenuator can be made to reduce the Power Output of the Tranceiver to desired levels. First I need to figure out how much attenuation is needed for different output levels less than 5 Watts.
A simple PI or T Network Resistive Attenuator can be made to reduce the Power Output of the Tranceiver to desired levels. First I need to figure out how much attenuation is needed for different output levels less than 5 Watts.
Here is where the dB formula comes in:
Expressed in dB , how many dB of attenuation is needed to reduce
5 Watts to 4, 3, 2.5, 2, 1 and .5 Watts?
5 Watts to 4, 3, 2.5, 2, 1 and .5 Watts?
dB = 10 LOG (P1/P2) ; where P1 = Power In , P2 =
Power Out
Power In = 5 W
Desired Power Out = 4 W
The Attenuation expressed in dB : dB =10*LOG(5/4) = .9691 we can round off to 1 dB
Power In = 5 W
Desired Power Out = 3 W
The Attenuation expressed in dB : dB =10*LOG(5/3) = 2.218 we can round off to 2 dB
Power In = 5 W
Power Out = 2.5 W
The Attenuation expressed in dB : dB =10*LOG(5/2.5) = 3.010 we can round off to 3 dB
*** A Change in 3 dB is = to half the Power! ***
Power In = 5 W
Power Out = 2 W
The Attenuation expressed in dB : dB =10*LOG(5/2) = 3.979 we can round off to 4 dB
Power In = 5 W
Power Out = 1 W
The Attenuation expressed in dB : dB =10*LOG(5/1) = 6.689 we can round off to 7 dB
Power In = 5 W
Power Out = .5 W
The Attenuation expressed in dB : dB =10*LOG(5/.5) = 10 dB
With this information I can build an Attenuator made of resistors to lower the power output of the Transceiver.
In Page # 22.44 of The ARRL Handbook For Radio Communications 2012
you can find a list of the Resistor Values for each level of attenuation from 1 to 25 dB in 1 dB increments and from 30 to 60 dB in 5 dB increments. Table 22.57 is for PI Network Resistive Attenuators and Table 22.58 is for T Network Resistive Attenuators. Both are assuming 50 Ohms impedance.
I will be adding more pages to this blog that will go into a bit more detail how to work with the decibel formula and my attempt to build my first Step Attenuator!
I couldn't have done this without the help of the following gentlemen:
I want to thank PA1B for his excellent QRPp Blog
PA1B's contribution to the Amateur Radio Community is priceless! Thank you! Please visit his blog for more information.
I also would like to thank Mr. Alan Wolke Callsign W2AEW. Alan has a excellent youtube channel where you can find useful informative videos on electronics.
https://www.youtube.com/user/w2aew
Thank you Alan! Your Instructional Videos have taught me LOTS!
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