Hi, I have read this information regarding the USRP ADC

“The full range on the ADCs is 2V peak to peak, and the input is 50 ohms
differential. This is equates to 40mW, or 16dBm.”

Please may someone enlighten me, how the 40mW and 16dBm values were
calculated?

thanks

On Mon, Dec 15, 2008 at 4:41 PM, Aadil V. [email protected] wrote:

Hi, I have read this information regarding the USRP ADC

“The full range on the ADCs is 2V peak to peak, and the input is 50 ohms
differential. This is equates to 40mW, or 16dBm.”

Please may someone enlighten me, how the 40mW and 16dBm values were
calculated?

I am not sure where those numbers came from directly, but the 50 Ohms
differential might be wrong. From the datasheet here:

``````http://www.analog.com/static/imported-files/data_sheets/AD9860_9862.pdf
``````

Page 2 shows the differential input resistance to be 200 Ohms with the
maximum differential input being 2Vp-p.

Regardless (and assuming the 50 Ohms), the equations are only Ohms Law
and conversion into dBm.

P = V^2 / R
P = 1.0V^2 / 50 Ohms
P = 20mW

Since there are two inputs, we double that to be 40mW of power used up.

10log10( P / 1mW )
10
log10( 40mW / 1 mW )
16.02 dBm

Hope this helped.

Brian

“The full range on the ADCs is 2V peak to peak, and the input is 50 ohms
differential. This is equates to 40mW, or 16dBm.”

Please may someone enlighten me, how the 40mW and 16dBm values were
calculated?

The impedance (effective resistance to ground) seen from the input pin
on the USRP coax, is 50 ohms. The calculation you quote assumes a
sinusoidal input at Vpp = 2 Volts peak-to-peak, which means the root
mean square amplitude is Vrms = (Vpp/sqrt(2)) = (2/sqrt(2)) = sqrt(2).
This RMS voltage is going through a 50 ohm resistor, so the power
dissipated is Vrms^2/R = (sqrt(2)^2)/50 = 2/50 = 0.04 Watts, or 40
milliWatts. To convert linear mW to logarithmic (decibel) mW, or dBm,
we calculate 10*log(40) = 16dBm.

That’s the short answer, please feel free to ask for more detail.

–Scott McDermott

Since there are two inputs, we double that to be 40mW of power used up.

10log10( P / 1mW )
10
log10( 40mW / 1 mW )
16.02 dBm

Ah, and at this point I encourage Aadil to ignore my response, as I was
making a different (not very bright) assumption about why my numbers had
come out a factor of two off (that the 2V was supposed to have been
peak, not peak-to-peak). Sorry about that, and thanks Brian for getting
the right answer out first.

–Scott

I’m afraid I dont understand. Are you saying I should be calculating:

P = (0.707V)^2/50 Ohms
P = 10mW

or

P = (0.707V)^2/200 Ohms
P = 2.5mW

The first one, 10mW, is what a signal source would be providing to the
BasicRX or LFRX input. So if you set your signal generator to 10 dBm,
you should see a signal at roughly full-scale.

The second one is the power that would be consumed at the input of the
ADC, but that is a completely meaningless number.

Matt

Brian P. wrote:

On Mon, Dec 15, 2008 at 4:41 PM, Aadil V. [email protected] wrote:

“The full range on the ADCs is 2V peak to peak, and the input is 50 ohms
differential. This is equates to 40mW, or 16dBm.”

I am not sure where that sentence came from, but it is incorrect. The
correct statement would be:

“The full-scale range of the ADCs is 2V peak to peak. The BasicRX and
LFRX have unity voltage gain and an input impedance of 50 ohms. Thus,
their full-scale input is 2Vpp, or 10dBm.”

2Vpp is equal to 1 Volt peak, or .707 Vrms. At 50 ohms, this comes to
10 mW or 10dBm. Also, while the ADC has a nominally 200 ohm input
impedance, the BasicRX and LFRX terminate in 50 ohms.

Matt