Question asked by Golan_X on May 30, 2018
Latest reply on Jun 13, 2018 by PVALAVAN

Hello,

The AD9375 will be part of a LTE10MHz system works on 2.4GHz Frequency.

I want to define the Dynamic range of the Transmitter:

By the DS the FS of the Transmitter is 7dBm for CW so it is the maximum power that can be transmitted for the LTE is 7dBm-10dB PAR -4dB Back off = -7dBm

IF the Noise of the transmitter is -155dBFS/Hz then -148dBm/HZ+10dB guard will get -138dBm/HZ

The conclusion is that the minimum signal that we can transmit is above -138dBm/Hz.

And the Maximum signal we can transmit is -7dBm.

Is that Maximum and minimum signal will give a good EVM? There is some graph that show the EVM vs the transmitted power of the AD9375?

Do I need to take under consideration other information according to the TX path of the Transceivers?

I want to understand a few things on the Receiver side also:

By the RX path the maximum signal that can be injected according to the DS is -14dBm for CW. If we will take back off it will be -28dBm (for maximum Gain).

But there is a note over the DS.

Does it affect my calculation for maximum input power for LTE signal?

There is no reference for the Noise level of the Receiver. I would like to know what is the minimum detected power.

Also, in the DS it written that the maximum gain is 30dB.

But by the Graph I don’t see that:

Is the 0dB gain will be achieved by setting the attenuator on 20dB And the maximum Gain will be 20dB and not 30dB?

By the NF/Attenuation graph it seems that the NF is linear to the Attenuation

For conclusion:

If I’ll use 0dB attenuation, it means that the Gain will be 18dB and the NF will be 15

Then: Minimum detected signal = -174+18dB+15dB= -141dBm/Hz

Then we will take 10dB back off = -141+10dB = -131dBm/Hz

Now with estimation of our receiver NF ~ 3.5dB

We can calculate the necessary Gain we need:

Gain = 174+3.5-131 = 46.5dB Is that correct?

I would like to know what is the meaning of the Gain that I peeked? What is the Best Gain value we should use?

By the graph you introduce in the DS the sample rate and the BW is 122.88MBPS.

What will change if I’ll use LTE10MHZ BW with another sample rate?

Also, the EVM graph show that I need to inject minimum -55dBm if I would like to have 1.4% EVM.

What is the Gain value in that Graph? Do I need to consider that as my minimum detected value?

How can I adapt this graph for 10MHZ?

In the DS it shows that the maximum spur is -95dBm.

Is that the SFDR of the Receiver?

Do I need to take this level + 10dB back off in order to define the minimum detected signal?

All those questions leading to define what will be the Gain in the RX path and what Gain should we define in the transvers.

Thank you !