Take a minute to learn more about you! Who consumes all my DBs

Set the signal generator to output the CW tone at a specific power. According to my mathematical equation, the ADC generates a –1 dBFS signal. However, I saw the –15 dBFS signal! Who consumes all my dB?

Many times, ADCs (Analog to Digital Converters) have nominal performance at –1 dBFS. Some data sheets give distortion of 0.5 dB below full scale. Whether it is 1 dB or 0.5 dB below full scale, this will prevent clipping of the signal if the ADC input is operating at full scale (0 dBFS). Benchtop RF signal generators typically output signals in units of dBm. To achieve –1 dBFS in a 1.7 V pp full-scale ADC range, the signal level is only 7.6 dBm (based on a 50 Ω reference impedance). However, in doing so, the ADC's single-tone FFT output shows –6.7 dBFS. What consumes all the dB?

The obvious answer is the ADC component. It is the ADC's front-end network. Let's take a closer look at the default front-end network for the ADCAD9680.

Take a minute to learn more about you! Who consumes all my DBs

Figure 1. Default front-end network for the AD9680 evaluation board.

Converting single-ended to differential mode can be achieved using the broadband balun BAL-0006SMG. A quick look at the BAL-0006SMG data sheet shows that it has 6 dB insertion loss. Another 6 dB has been added to the matching networks (Rs and RSH) after Barron. The matching network needs to provide broadband matching to the balun output. The series resistance in front of the ADC (RkB) exhibits a small amount of insertion loss. This resistor improves third-harmonic performance by reducing the kickback from the ADC sampling stage to the holding stage.

Therefore, let us overcome the related ADC challenges to understand the power required by the signal generator to obtain a –1 dBFS ADC signal. The 50 Ω reference resistor is used for the following mathematical equations. For the default full-scale level of 1.7 V pp, the –1 dBFS signal is 1.515 V pp. Since the 10 Ω resistor losses are quite small, we can assume that the loss voltage is the termination network voltage. The balun termination has 6 dB loss, so the balun swings about twice as much as 1.515 V on each side. This results in a single-ended input of approximately 3.03 V pp. Therefore, the signal generator must provide a signal corresponding to approximately 3.03 V pp or 14 dBm. Please note that this does not include the insertion loss of the bandpass filter or connector cable. Therefore, look at Figure 1 again. This time we obtain Figure 2 with some comments.

Take a minute to learn more about you! Who consumes all my DBs

Figure 2. Front-end network with integrated band-pass filter and signal generator.

Returning to our question again, if there is no interference next to the signal generator in front of the ADC, the required power of 7.6 dBm to obtain the -1 dBFS ADC signal is correct. Barron may need to be considered. There are other factors (broadband baluns, matching networks, backflush control, etc.) that can affect the insertion loss, resulting in an attenuated signal at -6.7 dBFS. Therefore, you can rest assured that "My front-end consumes all the dB". As you can see, mathematical equations can never be wrong.

Refer to the following formula:

Where VIN is the input voltage and VFS is the full-scale voltage

Where Vrms is the rms voltage and V pp is the peak-to-peak voltage

Where PdBm is the signal generator power (in dBm), Vrms is the rms voltage, R is the system impedance (in this case, 50 Ω), and P0 is 1 mW.

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