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I understand that by introducing the series resistance to the NMOS in the amplifier circuit the gain becomes: $$A_{v}=-\frac{R_D}{\left(\frac{1}{g_m}+R_s\right)}$$ Which means that higher values of \$R_{s}\$ correspond to lower gain, but perhaps this in return makes it more "linear". I can't quite comprehend what does linear gain mean, how does it make the amplifier better even though the gain goes down and how can I see that it was achieved by looking at the equation.

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  • \$\begingroup\$ \$\$\text{thing you want to write}\$\$ \$\endgroup\$
    – Andy aka
    Nov 27 '20 at 15:33
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\$ \frac{1}{gm}\$ can be modeled as a resistor in the drain of the transistor, now, the important thing to consider is that \$gm\$ depends on the current through the FET, so if you input a signal, the current of the transistor will vary, and thus, the gain of the FET will also vary.

If the gain is not constant but varies with input signal level, there will be distortion in the output waveform, this is a non-linearity, a linear amplifier is such that if you plot the input VS output it is a line, meaning that the output is proportional to the input, the proportionality constant or slope of the line is the gain of the amp, if the gain is changing, then the in/out relationship will no longer be a line, hence non-linear.

When you add a degeneration resistor, the small variations in \$\frac{1}{gm}\$ are swamped by the larger value of \$R_D\$ so the gain will not change that much, hence the amp will be more linear, the larger the value of \$R_D\$ with respect to \$\frac{1}{gm}\$ the less impact \$\frac{1}{gm}\$ will have in the gain, so gain will remain relatively constant, however, gain will also be reduced the larger \$R_D\$ gets.

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  • \$\begingroup\$ This is so insightful. Thank you. \$\endgroup\$
    – Essam
    Nov 27 '20 at 16:38
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As you have shown in your question, the formula for gain involves \$g_m\$ and we should like, for stability and distortion reduction reasons, for \$g_m\$ not to change but, it does: -

$$g_m \approx \dfrac{2\cdot I_D}{V_{GS} - V_{TH}}$$

To reduce changes in \$g_m\$ causing changes to \$A_v\$, we increase \$R_S\$ and, the \$A_v\$ formula evolves into this simplified version: -

$$A_v = -\dfrac{R_D}{R_S}$$

In effect \$R_S\$ now swamps \$\frac{1}{g_m}\$.

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Typically, we model distortion by examining the coefficients of the Taylor Series that described the polynomial of the transconductance.

With FETs having several degrees of freedom (gate width, gate length, density of well ties, back_biasing), I will give you some bipolar numbers instead.

For both FETs and bipolars (and vacuum tubes, and other power RF devices), we use the INTERCEPT POINTS as basis for our math.

By using logarithms, specifically log of input power and of output power, we convert all the math to this

  • input_to_output_fundamental transfer_function is a straight line, with 1:1 slope

  • input_to_output_2nd_harmonic conversion_function is a straight line, with 2:1 slope

  • input_to_output_3rd_harmonic conversion_function is a straight line, with 3:1 slope

Since these are non_parallel lines (the slopes are different), we know there are intercept points.

The IP2 is where the energy of 2nd harmonic is the same as fundamental energy.

The IP3 is similar.

As I mentioned earlier, these numbers come from coefficients in the Taylor Series descriptions of polynomials.

Very very usefully, give the 1:1, 2:1 and 3:1 slopes, you have overt presentation of the degrees of freedom in performing low_distortion design.

  • reducing the input RF signal (or audio) 20 dB will reduce the 2nd harmonic by 40 db; given the fundamental also is reduced by 20dB, we must subtract (40 - 20) and observe the 2nd harmonic power_ratio improves by 20dB.

  • for the 3rd harmonic, a 20 dB reduction of input RF signal results in a 40dB improvement of power/distortion

The IP3 is particularly important for rigidly_channelized systems, as are many communications_public_utility provides (you AM and FM and Cellphones).

Example:

  • you are a weak signal at 990 MHz

  • there are two strong signals at 992 and 994 MHz

  • problem: the 3rd order products result in distortion at 996 and 990

  • that distortion at 990 sits directly atop your own signal, and what with having the same modulation properties, may seriously impair your own channel SNR and recovery of modulation

Now for the bipolar (remember the FET properties have 3+ degrees of freedom, whereas bipolars have NO DEGREES OF FREEDOM).

  • bipolar IP2 and IP3 are approximately the same (within 10dB), and are about -10 dBv.

  • -10 dBv is 0.316 volts; expect equal distortion and fundamental with that base drive (notice I've not mentioned peak, peakpeak or RMS, which does matter)

  • a rule of thumb for bipolars is 4 milliVolts PeakPeak base drive causes 10% distortion; how can that be?

Given the math of

  • Iout ~~ e^[ q * Vbe / K * T * nu]

which is approximately

  • Iout = e^ [ Vbe / 0.026 volts] at room temp

We know 0.026 volts causes e^1 ratio_change of collector current

We know 0.018 volts (often used in bandgap design) causes 2:1 change of collector current.

Now you can evaluate that --- e^[Vbe / 0.026] --- to learn what Vbe produces a 10% change.

By the way, Willy Sansen has papers that describe the derivation of the IP2 and IP3 numbers; so does Barry Gilbert.

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