I’m getting conflicting info on how much power the BasicTX can transmit.
Ettus FAQ says, “Yes, the BasicTX will put out about 1mW up to about 50
MHz.”
This post says it’s, “0.5mW”:
http://old.nabble.com/question-about-BasicTX-td28010044.html
When I use “usrp_siggen.py” with full amplitude, my sineway has a
peak-to-peak voltage of nearly 500mV → (0.707*0.5V)^2/50ohm = 2.5mW
So, which should I go by? I’m trying to find an amplifier that will take
my
signal up to 200mW to go into a laser bias-T.
Thanks.
-William
Sorry for the bump, but I’m confused about the output power of the
BasicTX,
especially in regards to 1 channel or both.
Any thoughts?
William
It is worth noting that your calculation of power is missing a divide by
2, since you specify a peak-to-peak voltage, and not a peak voltage.
The equation for peak-to-peak voltage to power (given a 50 ohm
impedance) is P (watts) = (.707*V/2)^2/50, which when I calculated
yields something like .61 mw, so this is correct.
Also, as far as I am aware, the BasicTX board directly accessing the TX
output of the AD9862 chip, so perhaps your best bet is to check the
datasheet for this IC. This should yield a power output versus
frequency. The alternative is to measure yourself, as you have done.
Remember, the datasheet is only a reference. Actuall power levels can
vary within a tolerance (which the datasheet should also provide) due to
variations in gain and/or biasing, connector loss, transmission line
loss, impedance mismatch, and so on.
~Jeff
On Fri, Jun 25, 2010 at 11:19 PM, Jeffrey L. [email protected]
wrote:
datasheet is only a reference. Actuall power levels can vary within a
tolerance (which the datasheet should also provide) due to variations in
gain and/or biasing, connector loss, transmission line loss, impedance
mismatch, and so on.
I read this thread and saw the conversation revolved around voltage.
I then remembered that the AD9862 is actually a current mode output
DAC. The datasheet defines 20mA as the maximum current that the
differential pair to drive and all the parameters assume a 50Ohm load.
I am guessing the 50Ohm load is for optimal power transfer, but I
still wonder what would happen if the resistance were increased while
keeping Rset at 4k to try to keep driving 20mA. Could you increase
the power? Would you be stressing the internal driver of that pin too
much? Could you burn it out? Could you get a little more power out
of it if you drove 75Ohms instead?
Either way, depending on what Rset is on the USRP, and if you are
actually driving a 50Ohm load, it looks like the AD9862 itself can
drive 20mA into a 50Ohm load for each differential pair.
I’d also be interested to hear if what I assumed here is correct at
all, or if I am completely mistaken.
~Jeff
Brian