I am using AD8363 in a receiver chain. I would like to know what should be the SNR for the AD8363 so that it distinguishes (detect) the signal from the noise.
The AD8363 integrates all of the noise that it sees in its roughly 6 GHz of input bandwidth. Let's say for example that the integrated noise is -50 dBm (i.e. due to the noise of the components thar are driving the AD8363). If your carrier level sweeps from -55 dBm to -30 dBm, you will initially see the output voltage of the detector rise slowly (i.e. not at a rate of around 50 mV/dB) as your carrier level goes from -50 to -48 to -48 and so on. This is because the net power present at the input is equal to the sum of the carrier power and the -50 dBm of noise. However as the carrier level continues to increase, the influence of the noise quickly diminishes because of the log function (ie. Lot 10 = 1, Log 100 = 2, Log (100 + 10) = 2.04) and the nominal 50 mV/dB slope will kick in.
So the AD8363 will register the presence of the carrier as soon as it is similar in level to the integrated noise but the transfer function will initially be non linear as the carrier comes out of the noise.
You can see a good example of this phenomenon in Figure 43 of the AD8307 Rev. D datasheet. This plot shows the transfer function of an application circuit with and without a noise-limiting filter.
Thanks for a very good explanation. But is the level from the integrated noise when the device behaves linear characterized for 8363. This would be my required SNR.
No, this has not been characterized. I think that you are the question "If my SNR is 5 dB and my carrier power increasees by 1 dB, will the output voltage increase by 50 mV? If my SNR is 10 dB and my carrier power increases by 1 dB, will the output voltage increase by 50 mV?" I can give you a quantative answer to these questions except to say that higher SNR will bring you closer to linear operation and the log function that tends to ignore small signals when big signals are present, is helping things.
If you know the power level of your integrated noise, I think that you can do an estimate. using the equation
Vout = Slope x ((Psig+Pnoise) - Intercept).
The power levels in this equation are in dBm so to sumPsig and Pnoise, you have to first convert them back to Watts. If you set Pnoise to its fixed known value, then you can plot Vout vs. Pin and. This I believe will yield a plot that starts out flat (at low power levels), will gradually rise and move towards having the specified slope.
One thing to clarify is that when we talk about noise, we are talking about the noise of the signals that you apply to the device. I believe that the internal noise of the AD8363 does not really play into this. Notice how the Vout vs Pin transfer functions go almost all of the way to ground at low input power levels. If internal noise was limiting the dynamic range, you would expect the see the Vout trace flattening out to some offset voltage at low power levels. (e.g. Figure 7 in the AD8309 datasheet or the AD8307 plot that I mentioned in the previous posting.)
Thanks for your answer. My actual application is to measure signal as low as -90 dBm with less than 0.25dB accuracy. AD8363 was choosen since the error is very less at the output. The idea was to have sufficient amplification at the front to bring the signal into the linear range of AD8363. With a front end having 40dB gain, the input power which can be measured was only -70dBm. Under the assumption that the input is hitting the noise floor a filter (1MHz|)was used to narrow down the effective bandwidth. But still the minimum signal measured is only -75dBm whereas I expected an increase in the dynamic range. Can you tell me what am I missing,
The amplifier should increase your sensitivity not your dynamic range. But I think that you know that if your scheme works, you will just shift the effective range of the detector downwards.
So you are seeing an input referred sensitivity (i.e. at the input of the RF Amplifier) of either -70 dBm or -75 dBm, that is, you go below -75 dBm and the detector otuput does not move.
You should sum up your noise levels to see if that is the source of the problem. Let's assume that your amplifier has a noise figure of 3 dB. So the noise at the amplifier output is thermal (integrated over 1 MHz) plus gain plus noise figure . If you amplify -173.8 dBm/Hz and integrate it over 1 MHz (i.e. the pass-band of the fitler) and add in the noise figure of the 40 dB amp you get -173.8 + 40+3 + 10log(1000000) = -70.8 dBm (well below the minimum input sensitivity of the AD8363). This is the noise level at the input of the detector. If you refer this to the input of the amp, it comes out to about -110 dBm/Hz.
Is your 40 dB amplifier a single block? If it's two cascaded blocks, it might be worth trying to add just a single block of amplification to see how the circuit responds.
Thanks for the calculation. I had done the same way to find out the noise level at the input of the detector. I have a cascaded LNAs of 20 dB each here. In this set up my signal and noise levels at the input of detector are - -35dBm ( -75 + 40 dB gain) and -74 dBm (as you have shown above) respectively. The desired signal is well above the noise floor at the input of detector.
The lower limit of Detector being -56dBm, it should measure till -90dBm of input referred power as the noise floor at the detector is still 24 dB below than the desired
Signal : -90 dBm at LNA input -> -50 dBm input at detector,
Noise : --114 dBm at LNA input -> -74 dBm at Detector.
Am I missing something, or is there anything to do with the detector.
I'm going to suggest the following experiments (you may have tried these already) where you start with a bare bones set up (source & detector) and progressively build up your signal chain.
1. Sweep power to AD8363 with source connected directly.
2. Sweep power to AD8363 with your 1 MHz filter.
3. Sweep power to LNA&Filter&AD8363
4. Sweep power to LNA&LNA&Filter&AD8363
I'm not sure how feasible this is from a wiring point of view (e.g. your components may be soldered on to a single board) but anything you can do to measure the performance with the composition of the elements re-arranged will hopefully point to the source of the problem.
You can use the attached Excel file to plot your data.
As I have said above, I have performed the 1 and 4 of the above mentioned steps. But what are we looking into from these experiments ( mainly from 2 an 3)
When you do measurement 1, do you get the results you expect (i.e. a plot similar to what is in the datasheet). By progressively adding the other elements, I'm hoping that the source of the problem will become apparent.
For the detector alone, the minimum power measurable was -55dBm. With addition of two LNA alone it was -70dBm and then when we added the filter infront of LNA unfortunately there was no change.
What is the carrier frequency that your are trying to detect, and what is the BW of your LNA? If your LNA's have similar BW as the 8363, then the total noise from LNA is
N = No + NF + G + 10 log (BW) = -174 + 3 + 40 + 98 = -33 dBm
even when no signal is applied, regardless of your input filter. This corresponds to an equivalent input noise of -33 - 40 = -73 dBm at the input to the LNA, which may be what you have observed.
If the 1 MHz filter is used between the LNA & the 8363 (I assume this is the BW of a BPF around your carrier, or an LPF and your signal is inside this BW), noise from LNA (assuming a 3 dB noise figure) is
N = No + NF + G + 10 log (BW) = -174 + 3 + 40 + 60 = -71 dBm
which is lower than the -55 dBm due to the 8363, so the -55 dBm will dominate, and the noise referred to the LNA input will be -55 - 40 = -95 dBm ("sensitivity" at LNA input).
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