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Re: [Discuss-gnuradio] Re: how much power can the BasicTX output?

From: Brian Padalino
Subject: Re: [Discuss-gnuradio] Re: how much power can the BasicTX output?
Date: Sat, 26 Jun 2010 00:07:03 -0400

On Fri, Jun 25, 2010 at 11:19 PM, Jeffrey Lambert <address@hidden> wrote:
> 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.

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


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