ADC calibration

Hi,

I’m calibrating my USRP setup using a function generator. I remember
seeing that someone else had done this before in the past and written
down
a bunch of (input voltage, ADC output) pairs on the web somewhere, but
I’m
unable to find it on the wiki even with google site: search.

Does anyone have a pointer?

Thanks,
Dan

On 07/30/2010 11:32 AM, Daniel H. wrote:

Hi,

I’m calibrating my USRP setup using a function generator. I remember
seeing that someone else had done this before in the past and written
down a bunch of (input voltage, ADC output) pairs on the web somewhere,
but I’m unable to find it on the wiki even with google site: search.

Does anyone have a pointer?

When using a BasicRX or LFRX with the PGA set at 0dB gain (USRP1 only),
+10dBm is full scale on the ADC. On the USRP2 full scale on the ADC is
full scale when it gets back into gnu radio. On the USRP1 full scale on
the ADC is (1.67/2) times full scale due to the gain of the CORDIC.

+10dBm = 2Vpp = 1 Vpeak = 0.7 Vrms

With other daughterboards you will need to calibrate at the frequency of
interest.
Matt

On Tue, 10 Aug 2010, Matt E. wrote:

When using a BasicRX or LFRX with the PGA set at 0dB gain (USRP1 only),
+10dBm is full scale on the ADC. On the USRP2 full scale on the ADC is full
scale when it gets back into gnu radio. On the USRP1 full scale on the ADC
is (1.67/2) times full scale due to the gain of the CORDIC.

+10dBm = 2Vpp = 1 Vpeak = 0.7 Vrms

With other daughterboards you will need to calibrate at the frequency of
interest.

Thanks, that’s helpful. However, I can’t quite make the math work:

P = V^2/R ; 0.01 W = (1 V)^2/R ; R = 100 Ohms.

But we’re using SMA connectors which are 50 Ohms right? Or should I be
putting 1 Vpeak I and 1 Vpeak Q and getting the 100 Ohms from the
combination of the two 50 Ohm connectors?

Thanks,
Dan

(forgot to reply all first, sorry for the double-msg Matt)

On 08/13/2010 12:14 PM, Daniel H. wrote:

With other daughterboards you will need to calibrate at the frequency
of interest.

Thanks, that’s helpful. However, I can’t quite make the math work:

P = V^2/R ; 0.01 W = 1^2 / R ; R = 100 Ohms.

RMS power = Vpeak^2 / (2*R)
RMS power = Vrms^2 / R

Matt

On Fri, 13 Aug 2010, Matt E. wrote:

RMS power = Vrms^2 / R
… so as long as I don’t send a pure DC signal (which presumably I
can’t
b/c there will be filters so the device doesn’t support it anyway) I
should be using RMS power to account for the carrier wave variation?

Makes sense.

Thanks,
Dan

On Fri, Aug 13, 2010 at 3:17 PM, Daniel H.
[email protected] wrote:

interest.

Thanks, that’s helpful. However, I can’t quite make the math work:

P = V^2/R ; 0.01 W = (1 V)^2/R ; R = 100 Ohms.

But we’re using SMA connectors which are 50 Ohms right? Or should I be
putting 1 Vpeak I and 1 Vpeak Q and getting the 100 Ohms from the
combination of the two 50 Ohm connectors?

This may be helpful:

http://en.wikipedia.org/wiki/Alternating_current#Mathematics_of_AC_voltages

Note the usage of Vrms for power calculations and not Vpeak.

Thanks,
Dan

(forgot to reply all first, sorry for the double-msg Matt)

Good luck.

Brian

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