Impedance matching for LNA having infinite input impedance

Let say there are 50ohm receiving antenna, 50ohm transmission line, and common source LNA. (or in high speed wire-line link, TX's output driver having 50ohm output impedance, 50ohm transmission line, and a common source amp in RX)
If you design the LNA so that only the small input gate capacitance becomes high input impedance without input matching, there is an advantage that the input signal voltage at the gate is doubled by the total reflection, so why not use this design?

Although the voltage is doubled, the power transferred to the LNA is not necessarily maximized because of the reflection losses. This may not be effective for applications where power transfer efficiency is critical.

aliarifat794 said:

Although the voltage is doubled, the power transferred to the LNA is not necessarily maximized because of the reflection losses. This may not be effective for applications where power transfer efficiency is critical.

Click to expand...

thanks for the answer, aliarifat794.

But the CS LNA amplifies the gate-source voltage not the power, doesn't it?

It's useless because waves return back to the source and then they are dissipated by source impedance.

BigBoss said:

It's useless because waves return back to the source and then they are dissipated by source impedance.

Click to expand...

Thanks for the answer, I understand that the power transfered by the antenna cannot be delivered into the transistor at all. However, I still have questions.
Let me rephrase the question:

Why does the input of an RF CS LNA need to be matched to a 50-ohm antenna or a 50-ohm transmission line (longer enough than the wavelength of interest) for maximum power transfer, even though the CS LNA amplifies voltage, not power, between the gate and the source of the input transistor by 'gm*Rload'? According to the principle of the ideally full reflective condition without input matching, the input voltage signal becomes doubled by constructive interference at the gate of the transistor, even though the incident RF power is not delivered into the LNA at all. So, isn't it better to design an RF AFE without input matching for a 6 dB passive voltage pre-gain before the LNA if the input gate-source capacitance, Cgs, of the LNA is small enough to be considered high input impedance? Why isn't the fully power reflective input condition showing the standing wave voltage doubled employed to LNA? Am I missing something about how an amplifier operates or RF signal propagation theory here?

Hello, I'm new here and currently studying RF and high-speed analog front-end design.

I understand the principles of phenomena such as reflection, maximum power transfer conditions(conjugate matching), transmission lines, VSWR, resonance, noise matching, and other concepts in RF analog front-end design.
I know that input impedance must be optimized in some way for the best noise performance. Additionally, I understand that using source degeneration inductance along with an additional gate series inductance can induce a Q-boosted gate-source voltage and provide controllability of the resonant frequency.

However, I am confused about the logic in the attached slide.

A common-source transistor transduces the voltage between the gate and the source into output current, scaled by the proportional constant gm.
According to this principle, the input form is voltage, not power. So why do we need the maximum power transfer condition for the voltage gain of an LNA?
Can't we simply amplify the voltage standing wave, which is twice the magnitude of the signal voltage source due to perfect reflection?

What am I missing? please let me know.

You assume that LNA input impedance is a negligible small capacitance so that it achieves +1 reflection factor if no load resistor is intentionally added.

Looking at s11 of typical low noise RF FETs shows that the assumption isn't right, you see a complex frequency dependant impedance.

Secondly the source (e.g. antenna) impedance is rarely exactly matched. Allowing a high reflection factor at the load results in waves traveling forth and back and unwanted gain ripple. Therefore both side impedance matching is the standard solution at least for wideband systems.

IF it is low frequency, AND it has a very high input impedance so that it is truly a voltage controlled device, YES the voltage at an effective open circuit is doubled.

but your gain flatness will be very poor, as you have deliberately created standing waves at the input of your amplifier. Maybe if you only had a single frequency you could pull it off.

You ALSO could use L-C networks to boost the rf voltage at the gate input

Hello all.
I'm studying about impedance matching for RF amplifiers.

● I'd like to ask a conceptual question. I understand that a transistor with an actual 0F capacitance doesn't exist, but let's make an exception for the sake of this discussion.
Let's assume there is an RF front-end system shown in the attached slide, and it is under the following four conditions:

1. The input impedance of the common source amplifier is infinite over all frequencies.
2. The signal source is a single-tone continuous wave with a single wavelength (λ).
3. Between the signal source and the input of the common source amplifier, there is a 50-ohm transmission line that is much longer than the signal wavelength (λ « l).
4. All components are noiseless. Noise is not an issue to deal with here.

● Under the circuit conditions described above, I have reached the following five conclusions. Are these interpretations theoretically correct?

1. All signal power is reflected at the transistor gate and then dissipated entirely in the source impedance (Rsrc). Therefore, no signal distortion occurs due to cumulative and successive bouncing.
2. Since the input impedance of the common source amplifier is infinite, the behavior can be analyzed using the voltage transfer principle, not the power transfer principle.
3. With the infinite input impedance of the common source amplifier, a voltage standing wave is formed due to perfect reflection, doubling the voltage. As a result, vgs equals vrf.
4. There is no need to add a noiseless 50-ohm shunt resistor to the gate for impedance matching. If impedance matching is done in this manner, vgs would be halved, reducing the final output voltage (vout).
5. If the load impedance is ideally infinite over all frequencies, the length of the transmission line does not matter at all. Whether it is long or short, it makes no difference.

Thanks.

biff44 said:

IF it is low frequency, AND it has a very high input impedance so that it is truly a voltage controlled device, YES the voltage at an effective open circuit is doubled.

but your gain flatness will be very poor, as you have deliberately created standing waves at the input of your amplifier. Maybe if you only had a single frequency you could pull it off.

You ALSO could use L-C networks to boost the rf voltage at the gate input

Click to expand...

thanks biff44,
would you check if my conclusions in the following post are right?

Maximum power transfer \[\neq\] Maximum voltage present at the gate. For audio and other low-frequency applications you will usually strive for the latter in most situations, since, in audio, you can get away with impedance bridging, which is what you described. For RF/MW you have to keep impedances matched for reasons that others here have pointed out.