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# TRF370417 / output gain

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

• In reply to Mark Whittington:

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.

• In reply to Abdallah Obidat:

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

• In reply to Mark Whittington:

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.

• In reply to Abdallah Obidat:

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
• In reply to Mark Whittington:

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.

• In reply to Abdallah Obidat:

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
• In reply to Mark Whittington:

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.

• In reply to Abdallah Obidat:

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