Decibel Formula To Calculate Attenuation

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 ***

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.

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?

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

http://pa1b-qrp.blogspot.com/.  

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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