ADC readings does not correspond with input power

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Djaferbey

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Hello im currently working with an high speed ADC (LTC2387-18) and i use DC2622A driver circuit to drive the differential inputs of the ADC. I have made a poor block diagram of the setup.


The input is a 0dBm single tone signal the ADC full scale voltage is 4.096V. So i calculate the full scale voltage given in dBm to dBm(fullscale)= 10*log10((Vrms^2/R)/1mW)= 22dBm
So if the input power is 0dBm i should expect the power in dBFS to -22dBFS but i get -17dBFS. This was also the case when the input signal was 10dBm i got -7dBm. To calculate the input power i calculate the RMS voltage by writing RMS= sqrt(mean(abs(adcdata.^2))), and then convert it to dBFS by dBFS= 20*log10(RMS*sqrt(2)/2^17).

 

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Hi,

what is your signal frequency?

Here you see a 3 dB change of your power (dBm), which means that there is twice the power present. What is the input impedance of your circuitry? If it is HIGH-Z, the signal gernerator will simply provide twice the power, as it is supposed to deliver it into a 50 Ohm load/impedance.

BR
 

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There is a 50 Ohm termination in the attenuator, so i measured the voltage at the input of the single-ended to differential converter. So when the signal generator delivered 10dBm i measured 2Vpp at the input which corresponds to 10dBm. The schematic for the driver is shown below


The driver deliver differential signal centered on a common mode voltage of 2.048 and the ADC quantizes the difference between the differential signals. I am trying to figure out how common mode voltage works in a differential signal, i look at it as a biasing voltage that is common to both the in phase and out of phase signals. And im trying to figure out what is the input signal if the ADC takes the difference between these two signals i made a plot to illustrate it

I have no idea but does this have something to do with what I'm measuring?
 

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You mean "50 Ohm termination", not attenuator?
The input of the DC2622A is AC-coupled, so the max AC-signal is 4.096V p-p with mean = 0.
The maximum RMS value is Vpeak 2.048V / sqrt(2) = 1.448V
So your full-scale power is 10*log10(1.448^2/50/0.001) = 16.2 dBm.
 
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An attenuator is a series element with a certain attenuation e.g. 10 dB. A termination will be connected towards GND. Please clarify what are you using, or better show us what are you using.

BR
 

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"The LTC2387-18 has a fully differential ±4.096V input range. The IN+ and IN– pins should be driven 180 degrees out-of-phase with respect to each other, centered around a common mode voltage (IN+ + IN–)/2 that is restricted to (VREFBUF/2 ± 0.1V). The ADC samples and digitizes the voltage difference between the two analog input pins (IN+− IN–), and any unwanted signal that is common to both inputs is reduced by the common mode rejection ratio(CMRR) of the ADC".

I calculated the maximum RMS value based on the text above. If I have misunderstood something, please inform me.
 

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Hi,

I agree with std_match.

The result is independent of common mode voltage.
But for sure the common mode voltage needs to be at mid level in a way that
* the signals are not distorted (clipping)
* both signals match the absolute input voltage range of the ADC.

So the 16.2dBm is a rather theoretical value that never can be achieved. Still this value is used to calculate the actual dBm value.

Klaus
 

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stenzer said:
An attenuator is a series element with a certain attenuation e.g. 10 dB. A termination will be connected towards GND. Please clarify what are you using, or better show us what are you using.

BR
Click to expand...

Im sorry for any misunderstanding. Im using this attenuator to have impedance match so the generator has 50 and is delivering to 50 Ohm no reflections and max power transfer is achieved. The voltage i measure at the input always corresponds to the voltage i calculate. If i have misunderstood something you are always welcomed to correct me
--- Updated ---

Can you explain how, when the Max input voltage of the adc is 4.096V i need more clarification and Thanks
 
Last edited:

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Have you set the attenuator to 0 dB? If so have a look here, to see how the attenuator works for a 0 dB switch position.
 

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DC2622A has unity gain for both the + and - outputs, which means that the differential output voltage is double the input single ended voltage.
 
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@std_match of course you are right. I was suspecting the doubled power by the wrong use of the attenuator, and it looks to me it is set to 0 dB. But in the picture a a 50 Ohm terminator can be seen on a BNC-T connector as well.
 
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std_match said:
DC2622A has unity gain for both the + and - outputs, which means that the differential output voltage is double the input single ended voltage.
Click to expand...
Okay, I need to sum it all up. The full-scale voltage is 22 dBm, and the reason I get a value that is 6 dB higher is because the differential signal is double the single-ended signal.
--- Updated ---

By the way the entire circuit schematic is shown below, i tried to model the entire front end but it didn't go well since i cant open the ltspice file i get this error

 

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  • DC2290A.zip
    2.2 MB · Views: 86
Last edited:

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Don't know why? A PCB file is a PCB file. Who told it should be something different?

Obviously there's no circuit simulation provided for DC2622A.
 

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