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RF amplifier gain

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Rémi:
Hello,

I'm a beginner in the RF field and I have a question regarding RF amplifiers.

I'm trying to calculate the output power level of a HMC414, based on the datasheet the gain is 20dB and input return loss 8dB.

If I apply a 10dBm signal at 2.4Ghz to the input, how can I calculate the output level of the amplifier?

Is it 10dBm-8dB+20dB = 22dBm ?

Thanks for your help.

Rémi

rfclown:
The input return loss isn't part of the output level equation in this amplifier. So the output it is just input+gain which would be 10dBm +20dB = +30 dBm BUT... this amplifier can only go up to +30 dBm maximum, so the output will actually be less than that. Look at the Pout vs. Input Power graph on the datasheet. At a low level input like -9 dBm, the output is +11 dBm (20 dB gain). But at +10 dBm input the output looks to be about +28 dBm (18dB gain). This is called gain compression. Input return loss is a measure of how close the input impedance is to 50 ohms. Since the manufacturer specifies the gain with a 50 ohm input connected, it wouldn't even matter if the input return loss was horrible, the gain would be what is specified on the datasheet.

hcglitte:
You also need to consider if you want to use the PA in its linear operating range, ie below the 1dB compression point. If your signal has variable input amplitude, it should operate in its linear range. Otherwise the output amplitude will not vary in correspondence to the input.

If it is FSK or similar, it can operate in compression. However, in compression you will generate a lot of harmonics, and these needs to be filtered in order to comply with most regulations.

Operating near compression will also ensure that the maximum efficiency is achieved. Otherwise a lot of power is wasted as heat and not pushed out to the load.

A lot of research tries to make RFPA operate in compression at varying input signals by changing the the bias current relative to the input signal. Just to achieve better efficiency.

Rémi:
Thanks for your answer so far.

Actually I'm trying to amplify the signal coming from a DJI Lightroom 2. The modulation used is based on OFDM from what I could find online.

The signal I apply to the amplifier is around 0dBm:


The signal I get on the output is around 10dBm, I get an amplification of ~10dB:


I have also measured the S21 with a VNA:

At 2.44GHz I have a S21 of 9,23 -> Gain=20*log(S21)=20*log(9,23)=19dB

Why is my input signal not amplified by 20 ?

Thanks

rfclown:
For a device with 50 ohm input and output, the VNA S21 reading is gain (no math needed). So you are measuring a gain of 9.23 dB, which according to the datasheet is not right, so something is wrong. Could be many things:
1. power supply is low voltage, current limiting
2. control voltage is not correct (Vctl on datasheet)
3. cables are bad
4. overdriving the amplifier (gain is compressing)
5. overdriving the instrument (display shows 0 dBm in, which would give +20 dBm out. Can the VNA handle that? if not you have to have an attenuator on the amplifier output. In general, when measuring an amplifier with a large output you ALWAYS want to put an attenuator at the instrument input in order to avoid damage.
6. more than one of the above (my guess)

With the spectrum analyzer you can't measure power this way. This is a very wide bandwidth signal (looks like 75 MHz). You have to add up the power in the entire spectrum, not just measure at one point. Some analyzers will have a band power measurement function. Another way is to read the spectrum into a computer, convert the points to a linear format (from dBm to watts), add up all the points.

OFDM has a very high peak to average ratio, so for a given average power the peaks are much greater. In order not to compress the gain your output has to stay in the linear range of the amplifier. With Vs=5v the P1dB is +27 dBm. With Vs=3.6 the P1dB is +25 dBm (datasheet numbers). The P1dB is the point where the gain is compressed 1 dB. You can often operate up to that point. Say your signal has a 10 dB peak-to-average ratio (I don't know, and this is hopefully a worse case guess) the maximum average input with a 5V supply would be P1dB-gain-PkAvg = +27-20-10 = -3 dBm.

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