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I am trying to characterize a power amplifier fitted to transmitter board. I applied a sinewave at the required carrier frequency and I collected the input samples and the output samples by using a digital scope. I have noticed a strange behaviour which I can not explain.

when the input sample is zero the output is not zero but a significant value.

Should I expect when the input sample is zero, the output sample from the power amplifier should be zero ?

May the input and the output samples are not in sync. I tried to sync them, but still the zero input samples do not correspond to very small signal or zero sample.

When I plotted the output/input I received this plot.

Is this an expected plot ?

How I can get a smooth plot for the ratio of the output over the input ?

I appreciate any comments.

enter image description here

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Yes, this is (kind of) expected:

No semiconductor device in itself is truly linear over its full damage-free range; nor is it memoryless.

You're very likely seeing spikes whenever your input approaches zero, so that your division becomes unstable (mathematically), considering you'd still be amplifying noise and an offset current.

Also, you're likely seeing sub-linear amplification when your instantaneous signal becomes large. For a proper discussion I'd just refer you to a research on "intermodulation products", which you'll find in every real (non-perfect) amplifier.

To give you a short glimpse: Let's assume your amplifier is basically a transistor with support circuitry.

A transistor's output current (assume that's emitter current) over input voltage (assume that's base-emitter voltage) curve is (simplifiedly, not considering temporal effects etc):

$$I_\text{E} = I_\text{ES} \left(e^\frac{V_\text{BE}}{V_\text{T}} - 1\right) \tag1\label 1$$

Where $V_\text{T}$ and $I_\text{ES}$ are temperature- and semiconductor dictated constants.

You'll notice this is nothing close to a linear function. Now, the job of an amplifier designer is to operate the components (mainly, transistors) in the amplifier in a way that looks as linear as possible (for power amplifiers, that is a competing requirement to efficiency and large power capability, so these aren't typically all to linear or low-noise).

Imagine you do a Taylor series expansion of $\eqref 1$. You'll notice that the exponential function has the series representation

$$e^x = \sum\limits_{n = 0}^{\infty} \frac{x^n}{n!}\text.\tag2\label2$$

Transistor amplifiers aren't really perfectly represented by that exponential function, as you can do things like influence "weights" on individual series elements within bounds by careful semiconductor design:

$$f(x)=\sum\limits_{n = 0}^{\infty} \tilde a_n \frac{x^n}{n!}=\sum\limits_{n = 0}^{\infty} a_n {x^n}\label 3$$

What you don't want is the $x^0$ term, because that's just a constant bias current ($x^0 \equiv 1$), so you'd typically try to have a very small $a_0$.

What you do want is the $x^1=x$ term, because that's your linear amplification – and typically, you'd want that to be large, so if you can, you see towards that $a_1$ becomes large.

$a_2$ and higher lead to intermodulation, and you want them to be small – but compensating them technically leads to limited $a_1$, and that means your amplifier doesn't amplify much.

Now, your amplifier manufacturer won't offer you a specific input-output curve, often, but other things more important for people that work their amplifier with signals of different frequencies:

Second and Third order intercept points. I can only recommend reading these articles closely, and possibly get a book on semiconductor practice (Tietze-Schenk would be the go-to bible where I'm from), as you're working with amplifiers and need to know what you're doing

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  • $\begingroup$ Thanks a lot for the detailed explanation. It helps a lot. Just one point. $\endgroup$ – Adam Apr 29 '18 at 8:50
  • $\begingroup$ Should I expect when the input sample is zero, the output sample from the power amplifier should be zero ? But I do not get that !!! $\endgroup$ – Adam Apr 29 '18 at 8:51
  • $\begingroup$ No!! Please reread my answer if you thought that. $\endgroup$ – Marcus Müller Apr 29 '18 at 8:54

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