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I was reading the chapter about Low Noise Amplifiers from the book "The design of cmos radio-frequency integrated circuits" by Lee, and I have seen this example:

enter image description here

I do not understand the part inside the red circle. It is used to set the correct bias point for M1. The book says that:

Transistor M3 essentially forms a current mirror with M1, and its width is some small fraction of M1's width to minimize the power overhead of the bias circuit. The current through M3 is set by the supply voltage and Rref in conjunction with the Vgs of M3. The resistor RBIAS is chosen large eniugh that its equivalent noise current is small enough to be ignored. In a 50 Ohm system, values of several hundred ohms to a kilohm or so are adequate.

Precisely, I have the following questions:

1) Why do we need a current mirror? The current that flows on M3 is not given to any other circuit, what is its aim? I'd say that M1 needs a biasing voltage (and not current) and that may be achieved through for instance a simple voltage divider with two resistances.

2) I do not understand its considerations about RBIAS and noise: higher resistance means higher thermal noise, so I do not understand that statement.

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The clue is in the title... "The design of ... integrated circuits".

You are correct that you could set the bias voltage with a resistive divider ... IF you knew what the correct voltage was.

But you don't ...

Look at some MOSFET datasheets : note that there isn't a specific value given for Vgs(at some current) e.g. Vgs(th) often refers to a threshold current of 0.1 mA - but a range, say 1V to 3V.

All you can be reasonably sure of is that M1 and M3 require (reasonably closely) the same voltage, as they are made in the same process, and at the same temperature, by virtue of being close to each other on the same die.

If it helps to think of M3 as an analog computer calculating the correct voltage for Vbias, that's pretty much what it is.

EDIT : the point of Rbias is to isolate the RF path from all the parasitic capacitances around M3, and to attenuate any RF signals reaching M3 (which would attempt to amplify them onto Rref...) M3 is supposed to be DC only, and you can make R3 as high as you like because Ibias is practically 0. (If the book recommends kilohms, that is about keeping the physical size down)

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  • \$\begingroup\$ What is the problem if RF signals reach M3? \$\endgroup\$ Commented Jan 3, 2020 at 11:10
  • \$\begingroup\$ M3 can amplify, and it would be an inverting amplifier. What good purpose would a high level antiphase signal near your low level low noise input serve? None I can think of, but I can foresee plenty of bad consequences. \$\endgroup\$ Commented Jan 3, 2020 at 12:01
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Why do we need a current mirror? The current that flows on M3 is not given to any other circuit, what is its aim? I'd say that M1 needs a biasing voltage (and not current) and that may be achieved through for instance a simple voltage divider with two resistances.

Remember that resistors on an IC are often much larger than MOSFETs. Also they often have very large variation in value, maybe \$\pm\$ 20 or 30 per cent. Furthermore, the threshold voltage of the MOSFET is also subject to variability, and thus so is the required bias voltage.

Using a current mirror allows you to set the bias point much more accurately than a resistor divider would.

Although this design shows two resistors, so you'd expect it to take about the same area as a voltage divider, it's possible that having the additional free parameter of the W/L ratio between the two FETs allows you to choose minimum sized resistors. Also in a practical design \$R_{ref}\$ might actually be replaced by some other structure, or be shared between multiple circuits on the IC.

I do not understand its considerations about RBIAS and noise: higher resistance means higher thermal noise, so I do not understand that statement.

Higher resistance means a larger voltage noise, but a smaller current noise. Since this is driving a high-impedance node, we'd rather minimize current noise than voltage noise.

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  • \$\begingroup\$ What I continue not to understand on point one, is how can that current mirror set a voltage biasing point. I have never seen a current mirror used for this aim, instead for providing current for other circuits \$\endgroup\$ Commented Jan 2, 2020 at 21:27
  • \$\begingroup\$ @Kinka-Byo, in this circuit you are essentially using M1 as a current source to drive the common-gate M2. \$\endgroup\$ Commented Jan 3, 2020 at 1:41
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This is a std. sink current mirror with high gate resistance to reduce RF noise current.

Only the drain resistance and threshold conduction voltage with Vdd affect the current bias.

enter image description here

Current source - Gate to Gate noise attenuation occurs due to source impedance.

enter image description here

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If you want good thermal tracking, embed the one-strip bias-generator FET inside the multi-strip LNA FET

This answers Question #1.

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