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AGC dynamic range in AD9371

Hi

i wanted to know that what is the dynamic range of AGC in AD9371?

regards

js Hyanki

  • As you can see from the above graph that max and min gain for receiver chain is 17dB and -12dB approximately.

    You can use the ADIsimRF tool to calculate the SFDR form the max and minimum Rx gain and Rx input.

  • Hi srimoyi

    I need AGC dynamic range of 70dB  for my application but as per your reply it is 29dB. How can i achieve 70dB AGC dynamic range?

    Regards

    J S Hyanki

  • Gain range of AD9371 is 30  dB out of which 18 dB gain and -12 dB attenuation.

    Noise Floor  = -174+Noise Figure of AD9371 = -174+13.5 = -160.5 dBm / Hz @ 2.4 GHz with Max gain of AD9371

    Max input = -14+30= 16 dBm / Hz with max attenuation setting. 

    For 10 MHz analysis bandwidth

    Noise Floor  = -160.5+79=-90.5 dBm  /10  MHz

    Max input = -14+30= 16 dBm / 10 MHz with max attenuation setting. 

    Note: When you keep AD9371 at maximum setting the NF would degrade to 43.5 dB.

    Dynamic range with AGC enabled would be 16 - (-90.5) = 106.5 dB

    Same can be verifed using ADISimRF tool. ADIsimRF Signal Chain Calculator | Analog Devices 

     

  • Hi PVALAVAN

    asper you calculation Noise Floor  should be  = -160.5+79=-81.5 dBm. In data sheet Gain range is given 30dB and maximum power is 23dBm(Peak). 

    I wanted to know that what is the minimum power at RF port so that signal should come in dynamic range of ADC for demodulation?

    For my application maximum input will be +5dBm and minimum input will be -60dBm at RF ports. and signal bandwidth is 20MHz centered at 4000 MHz . For this scenario will the part work?

    Regards

    J S Hyanki

  • Thanks for pointing out the typo error.

    For my application maximum input will be +5dBm and minimum input will be -60dBm at RF ports. and signal bandwidth is 20MHz centered at 4000 MHz . For this scenario will the part work?

    For -60 dBm input use case 

    Output noise floor  = -174+10log(20MHz)+14(NF)+17 (Gain)= -70 dBm

    Signal output power = -60+17(Gain) = -43 dBm.

    AD9371 output SNR  = 27 dB. 

    For +5 dBm input use case

    Max input for AD9371 is -14 dBm with 17 dB gain, You have to reduce the gain to -1 dB to avoid ADC saturation for 5 dBm input,

    Output noise floor  = -174+10log(20MHz)+32 (NFat -1 dB gain setting) + (-1)(Gain)= -70 dBm

    Signal output power = 5+ (-1)(Gain) = 4 dBm.

    AD9371 output SNR  = 74 dB. 

    If your baseband SNR requirement for demodulation is within this range AD9371 should work for your requirement.

    From datasheet,

  • Hi PVALAVAN

    I did not understood

    For +5 dBm input use case

    Max input for AD9371 is -14 dBm with 17 dB gain, You have to reduce the gain to -1 dB to avoid ADC saturation for 5 dBm input,

    Question: How?

    Output noise floor  = -174+10log(20MHz)+32 (NFat -1 dB gain setting) + (-1)(Gain)= -70 dBm

    Question: If NF is 32 why noise floor is still same as taken in -60dBm case?

    Signal output power = 5+ (-1)(Gain) = 4 dBm.

    AD9371 output SNR  = 74 dB. 

    I have below question also.

    Does AGC is maintaining constant input to the ADC of the part?

    or we have to implement agc in baseband processor for demodulation ?

    regards

    j s hyanki

  • -14 dBm is allowed with max gain, when the input power increases more than -14 dBm the internal gain should be reduced by same value. 

    Noise Figure is increased and Gain also decreased to the same proposition. Hence there is no change in noise floor.

    AGC will not maintain constant input to the ADC. AGC is designed in such a way not to saturate the ADC at large signal input.

    From UG-992,

    Automatic gain control (AGC) allows the receivers to autonomously adjust the receiver gain depending on variations of the input signal, such as the onset of a strong interferer overloading the receiver datapath.

  • Hi PVALAVAN

     Please see the comment below.

    -14 dBm is allowed with max gain, when the input power increases more than -14 dBm the internal gain should be reduced by same value.

    Does it mean that if i feed +5dBm signal it will be attenuated by 19 dB and actual signal will be -14dBm to the ADC?

     

    What is the saturation point of ADC?

    Regards

    J S Hyanki

  • Does it mean that if i feed +5dBm signal it will be attenuated by 19 dB and actual signal will be -14dBm to the ADC?

    ADC input level = input power + gain. With 5 dBm input signal will be attenuated by 1 dB and input to ADC will be +4 dBm.

    What is the saturation point of ADC?

    Please refer to UG-992,

    The full-scale voltage of the Rx/ORx ADC is 707 mV peak. 

     

    Please refer to below post for voltage to dBm conversion.

    https://ez.analog.com/message/277760-re-calculate-dbfs-for-ad9371?commentID=277760#comment-277760 

     

  • Hi  

    For -60 dBm input use case 

    Output noise floor  = -174+10log(20MHz)+14(NF)+17 (Gain)= -70 dBm

    In the above calculation, you have considered noise figure of 14 based on the receiver noise figure vs receiver attenuation graph blue curve. 17 is the noise figure when receiver attenuation=0dB which implies a gain of 17dB.

    Now in case of +5dBm input power, gain should be -1dB. Receiver attenuation for gain of -1dB is around 18dB.  Noise figure at receiver attenuation of 18dB is around 22dB.  But in the below calculation, you have considered NF of 32. 

    Output noise floor  = -174+10log(20MHz)+32 (NFat -1 dB gain setting) + (-1)(Gain)= -70 dBm

    Is it a typo mistake or am i missing something. Please clarify.

    Thanks