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eHam Forums => Elmers => Topic started by: KF6YHK on October 19, 2012, 04:46:20 AM



Title: Calculating Power Output
Post by: KF6YHK on October 19, 2012, 04:46:20 AM
Hi. Can someone please help me understand and resolve the following?

In order to have an Effective Radiated Power of 500 watts, with 6dB of feed line loss and 12 dB antenna gain, what power output must you have?

When I plug the values into a db calculator I get 125.59 watts.

I know that Level in db=10Log(Power Ratio). So then 6=10Log(500/X)?  If this is correct, how would I simplify this (solve for X)? I was never very good with logarithms and it turns out that this deficiency has only increased with age.

Thanks in advance.





Title: RE: Calculating Power Output
Post by: KA4POL on October 19, 2012, 05:29:11 AM
If x = by, then y is the logarithm of x to base b, and is written y = logb(x). Usually the base b is 10.
In your case 500/x=106/10 and x= 500/106/10


Title: RE: Calculating Power Output
Post by: AC5UP on October 19, 2012, 05:30:44 AM
You can ballpark Decibels by remembering two values... 3 is either double or half (depending on whether it's gain or loss) and 10 dB is 10 times (or 1/10th) the original value.

In the example given, the 12 dB of gain in the antenna is reduced 6 dB by the loss in the feed line. Overall antenna system gain is now 6 dB which is 3dB + 3dB or double, then double again, for a multiplier of 4. If we want 500 watts ERP we'd need 125 watts from the transmitter. The first 3 dB of gain takes the ERP to 250 watts, the second 3 dB of gain takes the ERP to 500 watts. Or, working the reverse, 500 divided by 4 = 125.

Here's an example for you to work:  An amplifier is rated at 23 dB of gain. What's the ratio between the input and output level?




Title: RE: Calculating Power Output
Post by: KF6YHK on October 19, 2012, 06:33:48 AM
If x = by, then y is the logarithm of x to base b, and is written y = logb(x). Usually the base b is 10.
In your case 500/x=106/10 and x= 500/106/10

Thanks,

This is exactly what I was looking for.


Title: RE: Calculating Power Output
Post by: KF6YHK on October 19, 2012, 06:44:34 AM
You can ballpark Decibels by remembering two values... 3 is either double or half (depending on whether it's gain or loss) and 10 dB is 10 times (or 1/10th) the original value.

In the example given, the 12 dB of gain in the antenna is reduced 6 dB by the loss in the feed line. Overall antenna system gain is now 6 dB which is 3dB + 3dB or double, then double again, for a multiplier of 4. If we want 500 watts ERP we'd need 125 watts from the transmitter. The first 3 dB of gain takes the ERP to 250 watts, the second 3 dB of gain takes the ERP to 500 watts. Or, working the reverse, 500 divided by 4 = 125.

Here's an example for you to work:  An amplifier is rated at 23 dB of gain. What's the ratio between the input and output level?




Thanks,

This really helps. So from your example I divided 23 by 3 and came up with 7.6 (approx). Then I used an arbitrary output of 5 watts power. Doubling that 7.6 times would be 5X27.6 which gives me about 970 Watts. Correct?


Title: RE: Calculating Power Output
Post by: AC5UP on October 19, 2012, 07:08:33 AM
Very Close... Your answer of 5 watts in = 970 out should be 1000 watts out. You can do this in your head.

Let's take the number ' 23 ' and break it down as 10 + 10 + 3.  The first multiplier is 10. Second multiplier is 10 again. 10 x 10 = 100. Third multiplier is 2. 100 x 2 = 200, so 5 watts into the amplifier yields 1000 watts out. If you work it backwards, 2 x 10 = 20 x 10 = 200.

Same difference.

Run 23 dB through the calculator here (http://www.daycounter.com/Calculators/Decibels-Calculator.phtml) and you'll see the multiplier comes up as 200 for power but not voltage gain. (14.1) I don't have a shortcut for ballparking voltage gain, but where the 3 / 10 method helps is on multiple choice tests for questions like the one you asked. Almost invariably the numbers given will be multiples of 3 or 10 which makes it easy to double check the properly calculated result...


Title: RE: Calculating Power Output
Post by: KF6YHK on October 19, 2012, 07:39:54 AM
Very Close... Your answer of 5 watts in = 970 out should be 1000 watts out. You can do this in your head.

Let's take the number ' 23 ' and break it down as 10 + 10 + 3.  The first multiplier is 10. Second multiplier is 10 again. 10 x 10 = 100. Third multiplier is 2. 100 x 2 = 200, so 5 watts into the amplifier yields 1000 watts out. If you work it backwards, 2 x 10 = 20 x 10 = 200.

Same difference.

Run 23 dB through the calculator here (http://www.daycounter.com/Calculators/Decibels-Calculator.phtml) and you'll see the multiplier comes up as 200 for power but not voltage gain. (14.1) I don't have a shortcut for ballparking voltage gain, but where the 3 / 10 method helps is on multiple choice tests for questions like the one you asked. Almost invariably the numbers given will be multiples of 3 or 10 which makes it easy to double check the properly calculated result...

Yes, I see. It's much easier to use the 10X multiplier (10db) accompanied with the 2X (3db). Also, after playing around with the #s a bit I realized that 23/3 is closer to 7.7 than 7.6 which almost gives me another me an additional 70 Watts on top of my 970, so the method you show above with the 10X is definitely more accurate if I'm trying to get an estimate in my head. Especially when I get up into the higher db gain or loss.






Title: RE: Calculating Power Output
Post by: STAYVERTICAL on October 19, 2012, 04:21:10 PM
Decibels are logarithm based.
Logarithms are the powers of a number.
When you multiply numbers with a power (exponent), you add the powers.

So 10^2 times 10^3 equals 10^5
In English ten to the power of two TIMES ten to the power of three is ten to the power of five.
You see, you add the powers.
Since powers are the equivalent of logarithms, you simply add logarithms.
This is why they are used - it makes it simple to do multiplications.
In fact this is how the old slide rules worked - they added lengths which were logarithms - so allowing you to multiply by adding.

Logarithms can be powers of any base - but in our case the most common one is base 10.

So what is the power to which 10 has to be raised to give 100?
10 to the power of 2 is 100 :  So the logarithm of 100 to the base of 10 is 2.

A very commonly used one is: what is the power to which 10 must be raised to give 2 ?
The answer must lie between 0 and 1, because 10 to the power of 0 is 1 and 10 to the power of 1 is 10.
The answer is 10 to the power of 0.3 gives 2 (approximately).

Since we are using Decibels which are 1/10 (one tenth) of a Bel, we multiply the logarithm by 10 to give 3 (0.3 x 10).
This is why a 3dB increase in power is a doubling of power.

The good thing about logarithms is that they can simply be added together to give a final result.
So in your case you have 12dB positive and  6dB negative which gives 6dB total positive.

Since 3dB is a doubling, then 6dB is 2 times 2 or 4 times.
If you had 9db you would have 2 x 2 x 2 or 8 times.

So to get a 500W ERP with a  6dB positive gain you simply need 500/4 watts from the transmitter.
This is of course 125W.
Looking at it another way - if you have a gain of 4 times (6dB) you need 125 watts to get 500W ERP.

For a general way to convert dB into an actual number what you need to do is solve for the power to base 10.

So for example:
Lets say you want to convert 15.7dB into a number showing how many times something has been multiplied.
First convert Decibels into Bels by dividing by 10 giving 1.57 as the power of 10.
Next, solve for 10 to the power of 1.57 (windows calculator has a 10^x button in scientific mode).
10 to the power of 1.57 equals  37
So the if you were running 100W with a 15.7dB gain antenna you would have an ERP of 100 x 37 = 3700W

The main thing to remember is that decibels are just powers of 10 (which have been multiplied by 10 to convert from Bels)


Hope this helps,

73 - Rob


Title: RE: Calculating Power Output
Post by: W6OU on October 19, 2012, 04:49:34 PM
If you misplace your calculator, another handy thing to remember is there is roughly a linear relationship in dB vs voltage factor from 0 to -3 dB. This is not true beyond the 0 to -3dB range.

0 dB = 1.0 X
-1 dB = 0.9 X
-2 dB = 0.8 X
-3 dB = 0.7 X

If you square the voltage factors you get the power factors:

0 dB = 1.0 X
-1 dB = 0.81 X
-2 dB = 0.64 X
-3 dB = 0.49 X


Title: RE: Calculating Power Output
Post by: W8JI on October 19, 2012, 07:16:01 PM
Another interesting thing about this is power is log 10, while voltage or curent are log 20.

This means 3 dB change in voltage or current is the same as 3 dB change in power, or any other ratios.

Say we double voltage or double current. We have  log^20 of 2, or 6.02 dB. 
This is 4x the power level (since voltage and current both double), so we have log^10 of 4 = 6.02 dB. 

6 dB, or any dB, power change is the same as 6 dB, or any dB, power change. This means we don't have to worry if a change is dB volts or dB power as long as impedance is constant. 


Title: RE: Calculating Power Output
Post by: KA4POL on October 19, 2012, 10:08:49 PM
Another interesting thing about this is power is log 10, while voltage or curent are log 20.

Voltage and current: 20*logV1/V2 and power: 10*logP1/P2

Quote
6 dB, or any dB, power change is the same as 6 dB, or any dB, power change.

Absolutely true  ;D

No offense meant  ;)


Title: RE: Calculating Power Output
Post by: KF6YHK on October 20, 2012, 09:44:58 AM
 ;D Thanks everyone for taking the time to help me resolve this. It's good to know you guys are out there & willing to help.