Licensed amateurs must know how to calculate power in terms of the decibel watt since this is the unit used in the Amateur Radio License (A) or (B) Terms, Provisions and Limitations Booklet BR68. Column 4 of the schedule in the middle of the booklet gives the Maximum Power level (in dB relative to one watt) PEP.

**Logarithms** (a simple explanation of logs)

**dBW** (the decibel watt)

**Convert Watts to dBW** (some examples of how to)

**Convert dBW to Watts** (some examples of how to)

**Calculations** (examples)

The dBW is a curious but very useful unit. Once you understand it you will find it very easy to use. In fact you should be able to do the calculations in your head without using an electronic calculator. I know that this is difficult for people who were brought up with a calculator fitted to their push chair or pram; nevertheless, it is possible to do mental arithmetic.

The basis of calculating using dBW is related to how we use logarithms to multiply numbers. Come back to this section when you have read the rest of the page.

Here is a very simple example:

**2 x 2 = 4** (you can do this sum in your head)

**Log 2 + Log 2 = Log 4** (using logs we add the them instead of multiplying)

The value of **Log 2 is 0.3** (you will find that in your Log tables or calculator)

The value of **Log 4 is 0.6** (this is **Log 2 **plus **Log 2**).

Here is another example:

**2 x 4 = 8** (another easy one).

**Log 2 + Log 4 = Log 8** (add to two logs together)

The value of **Log 2 is 0.3** (try to remember this value)

The value of **Log 4 is 0.6** (the answer in the last example)

The value of **Log 8 is 0.9** (this is **Log 2 **plus **Log 4**)

When we want to multiply power values we actually add the dBW values. So far this has been straightforward, we have been using single digits. So here is an example using tens:

**20 x 40 = 800** (another easy one).

**Log 20 + Log 40 = Log 800** (count the number of zeros in each number)

The value of **Log 20 is 1.3** (the figure **1** in front of the .3 tells you there is 1 zero)

The value of **Log 40 is 1.6** (again there is 1 zero)

The value of **Log 800 is 2.9** (the figure **2** shows that there are 2 zeros).

Lets try another:

**20 x 8 = 160** (another easy one).

**Log 20 + Log 8 = Log 160** (count the number of zeros in each number)

The value of **Log 20 is 1.3** (just as before)

**Log 8 is 0.9** (the figure **0** says there are no zeros)

The value of **Log 160 is 2.2** (think about this one).

If you found this easy, you will not be surprised to know that dividing us just as simple: all we do is subtract on log from the other. Here is one example:

400 / 5 = 80 (another easy one).

**Log 400 – Log 5 = Log 80** (subtract log 5)

The value of **Log 400 is 2.6** (just as before)

The value of **Log 5 is 0.7** (the figure **0** says there are no zeros)

The value of **Log 80 is 1.9** (**2.6** minus **0.7**).

Decibel watts are calculated just like logarithms except that there is a “deci-” prefix which you must take account of when converting from watts into dBW. When converting from Watts to dBW you must multiply by ten. So here are some values to remember:

**Watts dBW
10 10
32 15
40 16
100 20
160 22
400 26
1000 30**

**Convert Watts to dBW**

Look up the log of **32** (use log tables or calculator): **1.5 **Now we must multiply by 10 to make it dBW so we get **15**

Look up the log of **400** (use log tables or calculator): **2.6 **Now we must multiply by 10 to make it dBW so we get **26**

**Convert dBW to Watts**

First we must divide by 10; so ** 30**dBW/10 = 3.0** **Now we look up the anti-log and get **1000 **(there are three zeros)

First we must divide by 10, ** 26**dBW/10 = 2.6** **Now we look up the anti-log and get **400 **(there are two zeros)

**Back to logs**

**Calculations**

Once we have our power values in dBW we can add and subtract them. The schedule says that I must only use 15dBW between 1.85 and 2.00MHz (this is the bottom end of “top band”). My rig can generate 50w power using telephony on this band. I have 20m of feeder between the rig and the antenna. The loss in this feeder is 0.1dBW per metre of cable. The total loss in the feeder will therefore be 2dBW. 50w is 17dBW subtract 2dBW (loss in the feeder) to find the power delivered to the antenna is 15dBW which is what I am allowed. When I replace my feeder with a shorter one or one with lower losses, I must reduce power or I will be in breach of the terms and conditions of my licence.

Here is a simplification from the “schedule”:

**Watts ** **dBW ** **Bands MHz**

**32 ** **15 ** **1.85 – 2.00**

**40 ** **16 ** **430.0 – 432.0**

**100 ****20 ****51.00 – 52.00**

**160 ** **22 ** **70.00 – 70.05**

**400 ** **26 ** **All Other Bands**

400 watts +26dBW is the maximum permitted power in the UK. The values in the table above only refer to attended operation; if you use automatic methods you should look at the terms and conditions booklet more carefully.

You should also bear in mind that it is not necessary to use maximum power. Even without a linear amplifier I have made contacts with the USA and the Middle East, that is with just 50w. You do NOT need 50w let alone (400w) to contact your neighbour! Some amateurs take great delight in using QRP **see the Q codes page** they make interesting contacts using less than 1w.

This page was written many years ago: things change in the world of Amateur Radio, so don’t rely on the comments on Amateur radio here. However the content on Logs and decibelwatts is otherwies correct.