Measuring receiver gain...

IMHO you should just use dBuV, and make it clear that you are doing so. I take it that your IF is low enough that you don't _have_ to terminate everything precisely -- given that dBuV is probably "best", whatever that means.

The key to your question lies in the fact that the term "gain" by itself is ambiguous.

You first need to decide if you want to measure the VOLTAGE GAIN or the CURRENT gain or the POWER GAIN.

Then your solution should be clear.

Mark

In message , dated Tue, 22 Aug 2006, billcalley writes

The simple answer is that you can choose what to do, provided you give all the relevant metadata with your gain figure (i.e. exactly how you measured it).

You can use dBV in the first case, if you say what load you actually did measure into. The op-amp output source impedance is probably quite low, so you may see a difference in output voltage with 1 kohm and 10 kohm loads (you can use that to calculate the source impedance, of course).

You can inject a known signal and measure the output voltage, but what about AGC? If you can't disable it, you will need to plot the output/input characteristic and look below the AGC threshold to see the real gain.

If you can disable the AGC, use a low input voltage so as not to overload the amplifier.

You can quote a power gain between 50 ohm loads if you pad out the op-amp output with, say, a 1 kohm and 50 ohm in series and measure across the 50 ohm. Remember that puts 25 ohms, not 50 ohms, in series with the 1 kohm.

Its not the output impedance of the op-amp that is important but the impedance wich it is designed to drive. If it is happy driving a 50ohm load then use 50ohms to make it easy as then voltage gain in db = power gain in db /2

Colin =^.^=

Thanks John, Tim, Colin, and Mark -- great stuff, and a real clarifier on testing the gain of a receiver!

-Bill

In message , dated Wed, 23 Aug

Power gain = 80 dB, voltage gain = 40 dB?

Please show working.

Voltage gain = V2/V1

Power gain = P2/P1 = V2^2/V1^2

dB = 20lg(V2/V1)

dB = 10lg(P2/P1)

Change your mind?

not sure exactly what I was thinking of with the db/2 ... as its already included in the conversion of db->volt but its a lot easier to work out the voltage gain if power gain is calculated with zin=zout, voltage gain is then sqrt of power gain (wich if u divide a log by 2 it takes the sqrt of it.)

Colin =^.^=