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 2008-12-15 23:45

on 2008-12-16 00:05

On Mon, Dec 15, 2008 at 4:41 PM, Aadil V. <removed_email_address@domain.invalid> 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_s... 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. 10*log10( P / 1mW ) 10*log10( 40mW / 1 mW ) 16.02 dBm Hope this helped. Brian

on 2008-12-16 00:26

> "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? Hello Aadil --- 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

on 2008-12-16 00:29

> Since there are two inputs, we double that to be 40mW of power used up. > > 10*log10( 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

on 2008-12-16 00:46

Brian P. wrote: > On Mon, Dec 15, 2008 at 4:41 PM, Aadil V. <removed_email_address@domain.invalid> 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

on 2008-12-16 18:55

> 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