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TRF370417 / output gain

Guru 20090 points
Other Parts Discussed in Thread: TRF370417

Hello,

In the datasheet figure 2 , TRF370417 output power @ 5.8GHz is about -7dBm.
And Vin 98mVrms is about -7.16dBm.(Please see the attached image as my calculation)
Thus, voltage gain is about 0dB.

However, Voltage gain @ fLO=5.8GHz is -5.5dB(typ).
Thus, there is discrepancy of gain.
Could you please let me know which gain should we use?
If my understanding is incorrect, please let me know.

Best Regards,
Ryuji Asaka

  • Ryuji:

    Please check out the following App Note: http://www.ti.com/lit/pdf/slwa056.  This App note specifically addressed the calculations used to determine gain.  I think the discrepancy above is related to SE vs. differential and/or the ideal combining power of the I/Q inputs in the modulator.

    --RJH

  • Hello RJH san,

    Thank you for your prompt reply.
    I understood.

    Best Regards,
    Ryuji Asaka

  • RJH,

    Looking at the app note and the example worked out in equations 1 through 3 the signal voltage at the input is listed as 0.1Vrms SE x 2. Is this indicating the differential voltage on a single I or Q input or is it meant to indicate that there is a 0.1Vrms SE voltage at each I and Q input? If the factor of 2 is to represent the differential signal level then is the presence of both I and Q signals implicit in the output power calculation?

    Thanks.

    Mark
  • I believe that the 0.1 Vrms SE voltage mentioned indicates that a 0.1 Vrms voltage is applied on both I and Q inputs and that the factor of 2 does indeed imply that the presence of I and Q are implicit in the output power calculation.

  • Abdallah,

    To be clear, in equation 1, the multiplication of the 0.1Vrms by a factor of 2 is to represent the fact that the 0.1Vrms single ended signal is present on both the I and the Q port and not to represent the actual differential voltage present on the I or Q (but not both) input, is that correct?

    If that is the case then i'm confused by the gain terms used in the TRF370417 datasheet.  As an example, at 2140MHz the gain is listed as -2.4dB and the gain is defined as the output rms voltage over the I (or Q) input rms voltage.  From that I would infer that the gain would be 3dB higher if I had both I and Q present.  But in the example worked from the app note, if my understanding per my first sentence is correct, then the gain when both I and Q are present is really -2.4 and there is a conflict between the app note and the data sheet.  

    Fundamentally I find the usage of the single ended voltage for differential inputs confusing since it isn't clear if I should truly be driving 0.2Vrms differential (equivalent of 0.1Vrms SE on each P/N input) or if I truly should drive the input with 0.1Vrms (either single ended or 0.05Vrms on each P/N input).  

    Thanks.

    Mark

  • Looking back at the block diagram in the datasheet, I see that both I and Q are provided as differential inputs. The datasheet and app note are consistent and correct. The factor of 2 is needed as 98mV is applied on the P and N inputs. 

    Mark Whittington said:

    Fundamentally I find the usage of the single ended voltage for differential inputs confusing since it isn't clear if I should truly be driving 0.2Vrms differential (equivalent of 0.1Vrms SE on each P/N input) or if I truly should drive the input with 0.1Vrms (either single ended or 0.05Vrms on each P/N input).  

    In this case you should drive the input with .1Vrms on each pin.

    Regards,

    Abdallah

  • Abdallah,

    I think you are now telling me that equation 1 in the app note is indicating a differential voltage of 0.2Vrms (0.1Vrms on the P input and 0.1Vrms on the N input). The voltage gain derived (-2.4dB) is the same as in the TRF 370417 datasheet at the same frequency where it explicitly states that the gain number is Output rms voltage over input I (or Q) rms voltage. But the figure used to find the output power, I think, is based on both I and Q inputs being loaded with the 0.1Vrms SE.

    I guess I'm just confused as to why the gain of the device defined as the dBV at a single I or Q input to the dBV of the combined I and Q power at the output. Or am I still confused?

    Thanks.

    Mark
  • I apologize if my initial response caused confusion. I didn’t initially realize that the inputs to the TRF370417 were differential. I can confirm that the equation is for a single channel, I or Q, but not both. The datasheet specifically states that “VinBB = 98 mVrms single-ended sine wave in quadrature”, which leads me to believe that the figure used to find the output power is based on Q alone. 

  • Abdallah,

    Just to repeat back to you so that I can confirm my understanding:

    The datasheet indicates ~0.1Vrms single ended voltage which is present on both P and N inputs applied to only one of the I or Q ports.  For an analysis that uses both I and Q inputs  the output power would be 3dB higher.

    Do I have it right?

    Thanks.

    Mark

  • Hi Mark,

    I was looking over your last post and I would like to make a correction: For an analysis that uses both I and Q inputs the output power would be 6dB higher.

  • Abdallah,

    Why 6dB instead of 3?  Given that in an I or Q (not both) situation I have a defined output power why wouldn't the addition of the Q or I port simply double the power at the output?  The two voltage sources are non-coherent so even if there is a voltage summing before the output wouldn't the summed power be the RSS of the voltages instead of the coherent adding?

    Thanks.

    Mark

  • The output voltage would double, but the output power is proportional to the voltage squared. Assuming one input would result in: P = V2/R . When the second input is also used, the voltage is doubled: P = (2*V) 2/R, It can be seen that power increases by a factor of 4, therefore 10*Log(4) = 6dB.

  • Abdallah,

    I find it very surprising that the output voltage would double. I would expect that since the two sources are non-coherent their voltage sum would be the RSS of the two sources and not linear addition of amplitudes. Can you explain how the voltage sources in quadrature are coherently adding to the output?

    Thanks.
    Mark
  • The sources are indeed non-coherent as they are in quadrature. However, I is multiplied by the LO and Q is multiplied by a “rotated” LO and both are then finally summed. A total of four terms are produced, but two of them cancel (ideally). The remaining two terms are coherent and equal, causing the output voltage to double.

    The doubling of the amplitudes is not a result of a direct linear addition of the amplitudes. 

  • Abdallah,

    I think I understand why you are saying 6dB now. You are assuming that the I and Q inputs are at the same frequency and in quadrature such that you are adding voltage at the summer coherently. However for non-quadrature inputs or for inputs with a different frequency the LO up-conversion wouldn't cause the signals to be coherent and the power summing to the output would be 3dB per doubling instead of 6dB. Do I have that right? My usage case doesn't have the I and Q inputs entering the modulator already in quadrature.

    Thanks.
    Mark
  • Yes, I had assumed that the I and Q inputs were at the same frequency and in quadrature.

    I think the answer can be explained best with an example. For this example, let’s assume that the output is 0 dBm with both I and Q at the same frequency. When one of the frequencies are changed, they will no longer sum as coherent signals and you will now see two outputs, each at – 6 dBm. It is also worth noting that each output now has an unsuppressed sideband.

    So in this example,  if you only had I or Q, the total output power would be – 6 dBm, but if you had I and Q (at different frequencies), then they wouldn’t sum as coherent signals and your total output power would be -6 dBm + -6 dBm = -3 dBm. So yes, you would see a 3dBm increase if you have I and Q at different frequencies.

  • Abdallah,

    I was able to get the TRF370417 evaluation board working and measured the increase in power from having both I and Q inputs present and it is only a 3dB increase.  I think it is easier to think about in the frequency domain where, if only I is present you would have two tones modulated offset from the LO with half the total power in each of the tones.  If you were then to put a signal on the Q that is in quadrature with the tone on the I it will add with one of the two tones and cancel with the other.  Since half the power was in each tone and you lost one of the tones you will only get a net increase in power of 3dB.  This is backed up by the spectrum analyzer for all cases (I only, Q only, I&Q no phase delta, I&Q in quadrature)

    Thanks.

    Mark