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AD9208-3000EBZ, ACE tool input signal level question

I am using AD9208-3000EBZ, ADS7-V2EBZ with ACE software for seeing FFT analysis.

I have some question.

please help me.

Here's the thing

the default setting was kept (input Vp-p setting is 1.7V)

When i input +16dBm level(225MHz) to input A(J101) of AD9208 by Signal generator, the register 0x0563 keeps 0 (Not over-range).

But with input +17dBm level(225MHz), it starts over-range(0x0563 was set as 1)


in the FFT of ACE program, when over-range occurs, the FUND power was -0.478dBFS (225MHz, with +17dBm input). But Over-range indicator bit of 0x0563 was 1(indicates over-range).

I guess Only when Fund power exceeds 0dBFS, the over-range indicator bit can be set as 1.

Is this right? If my guess were correct, what is the problem?.

2. At that time,  the input differential voltage of AD9208 is approximately 1.06V.

the output signal generator is +17dBm. but real input differential is similar to +13dBm(50ohm system) by insertion loss of Balun of AD9208-3000EBZ A channel input circuit.

However the input configuration Vp-p(0x1910) was set to 1.7 Vp-p.

Although real input Vp-p is 1.06Vp-p, the over-range indicator bit of 0x0563 was 1....

I could not understand this. please help me.

3. And also. In the first picture(ACE's FFT), The parameter of fund power is -0.478dBFS. But the FFT graph seems like different(-3dBFS..). What is the real data;;;?

Thanks for helping me!!