Thanks to rclocher3 for helping edit more to English

why closed so my struggle to get thought to English thank rclocher3 I round back too best I put in avuenue what I was asking

why closed i tryed so

SDR's are low power output, as I know it

can you control the watts out on an SDR

is if you like more output power use an amplifier

as if you respect max power watt output to fcc band plan,

By the way I'm good in math not writing(dyslexia dysgraphia, that not excess, I do try to do better next time)

[sdr out put] watts * [amplifier gain]db = [to antenna]watts 
  • 1
    $\begingroup$ is there a question? $\endgroup$ Feb 3 '20 at 17:53
  • 2
    $\begingroup$ if you're having a hard time putting your question to words or drawings, often formulas help! $\endgroup$ Feb 3 '20 at 18:39
  • 2
    $\begingroup$ Hello, I took a stab at editing your question. Please feel free to edit it further or reject my edit if I changed what you meant. I'm trying to help! $\endgroup$
    – rclocher3
    Feb 3 '20 at 19:54
  • 1
    $\begingroup$ I Googled dysgraphia, and found that many folks that have it are quite articulate and coherent when talking. If you are, then perhaps you might benefit greatly from speech-to-text software. The resulting text could then be pasted into your question. Might that work? Just trying to help, as your question has now has three close votes. $\endgroup$
    – Mike Waters
    Feb 7 '20 at 15:26
  • 1
    $\begingroup$ Closed because the question was rolled back and re-edited by the original author so many times, that it had lost all meaning. $\endgroup$
    – Scott Earle
    Feb 8 '20 at 7:57

You have forgotten that decibels are a way of expressing the difference of two things logarithmically.

20dB is a gain of 100.

So with 1W in you get 100W out, and with 2W in you get 200W.

Also remember that many commercial amplifiers talk a good talk about what gain they can provide, but they won't always walk the walk, if you see what I mean. That is to say that an amplifier sold as having a 20dB gain might only give 80W out for 1W in.

Also, since you are thinking in watts, make sure you don't overload the amplifier. Just because it says it gives a 20dB gain, you won't get 200W out for a 2W input if the amplifier itself is rated at 100W maximum output.

  • $\begingroup$ cam up set watt output on sdr? i lack english duh i try so ok then stackexchange be prefix and have it have it , do not try it does not count $\endgroup$
    – kc0hwa
    Feb 9 '20 at 6:46

I know you want watts, and in this case since you are adding just one stage (the amplifier) it is a little bit annoying that the amplifier's gain is in decibels. The reason for that is because when you have a lot of things to add, for example…

SDR + amplifier - coax loss + antenna gain = Effective radiated power

…then decibels becomes very handy. The trick I use is to start with my power (watts) in decibels too!

Think of a plain decibel dB unit like an increase/decrease ratio. But there are also decibel-with-reference units, like dBm (power relative to a milliwatt) and dBuV (voltage relative to a microvolt).

You can think of numbers with these "reference decibels" as absolute values — e.g. 0 dBm is the same as 1 milliwatt. That's because it's the reference milliwatt (the …m part) with no increase or decrease (the 0 dB… part).

Let's say your SDR is 0.1W. That is 20 dBm, because 100 mW (0.1W) is a 20 dB increase from the …m reference unit of 1 mW. I made myself a little "dBm ↔︎ watts converter" that's available here https://utils.ipcalf.com/#dbm and you can type in your actual SDR watt in the right hand (watts) box and the dBM will show up in the left box.

I'll continue with a 100 mW transmitter for example, into an amplifier that has 35 dB gain for example. You can replace with your actual numbers. We know 100 mW is the same as 20 dBm. So now we can do your math:

SDR output (20 dBm) + amplifier gain (35 dB) = to antenna (55 dBm)

So you get 55 dBm to the antenna. But the FCC safety limits and some band restrictions are given in watts on most charts. So we don't want dBm but W.

I put that back into the left box (dBm) of my calculator and it shows 316.2277660168379 on the watts side. So our example SDR plus example amplifier together output about 316 watts total!

Or to get Effective Radiated Power, we keep it in dBm for a little longer:

20 dBm (SDR) + 35 dB (amplifier) - 5 dB (coax loss) + 3 dB (antenna gain) = 53 dBm (ERP)

Converting to watts again, 53 dBm is approximately 200 W (199.5…). But it was very handy to add in more steps while it was still in decibels. You can even add more dB things, like your predicted path loss and a receiver antenna gain to predict whether communication is possible


Not the answer you're looking for? Browse other questions tagged or ask your own question.