A GaAs MESFET is biased for minimum noise figure, with the following scattering parameters and noise parameters at 4 GHz (Z0 = 50 Ω): S11 = 0.6̸−60◦, S12 = 0.05∠ 26°, S21 = 1.9∠81°, S22 = 0.5∠−60°, Fmin = 1.6 dB, Γopt = 0.62∠100°, and RN = 20 Ω. For design purposes, assume the device is unilateral,

Question

LOW-NOISE AMPLIFIER DESIGN

A GaAs MESFET is biased for minimum noise figure, with the following scattering parameters and noise parameters at 4 GHz ([latex]Z_{0} = 50 Ω)[/latex]: [latex]S_{11} = 0.6∠−60◦,S_{12} = 0.05∠ 26°,S_{21} = 1.9∠81°,S_{22} = 0.5∠−6°◦,F_{min} = 1.6 dB,Γ_{opt} =0.62∠100°, [/latex] and [latex]R_{N} = 20 Ω.[/latex] For design purposes, assume the device is unilateral, and calculate the maximum error in [latex]G_{T}[/latex] resulting from this assumption. Then design an amplifier having a 2.0 dB noise figure with the maximum gain that is compatible with this noise figure.

Accepted Answer

We first calculate that K = 2.78 and Δ = 0.37, so the device is unconditionally stable even without the approximation of a unilateral device. Next, compute the unilateral figure of merit from (12.46):

[latex]U=\frac{\left|S_{12}S_{21}S_{11}S_{22}\right| }{(1-\left|S_{11}\right|^{2} )(1-\left|S_{22}\right|^{2} )}=0.059,[/latex]

From (12.45) the ratio [latex]G_{T}/G_{TU}[/latex] is bounded as

[latex]\frac{1}{(1+U)^{2}}\lt \frac{G_{T}}{G_{TU}} \lt \frac{1}{(1-U)^{2}} ,[/latex]

or

[latex]0.891 \lt \frac{G_{T}}{G_{TU}} \lt 1.130 ,[/latex]

In dB, this is

[latex]-0.50 \lt G_{T}=G_{TU}\lt0.53 dB,[/latex]

where [latex]G_{T} [/latex] and [latex]G_{TU}[/latex] are now in dB. Thus, we should expect less than about ±0.5 dB error in gain.
Now use (12.58) and (12.60) to compute the center and radius of the 2 dB noise figure circle:

[latex]N=\frac{F-F_{min}}{4R_{N}/Z_{0}}\left|1+\Gamma _{opt}\right|^{2}=\frac{1.58-1.445}{4(20/50)}\left|1+0.62\angle 100^{\circ }\right|^{2} =0.0986[/latex],

[latex]C_{F}=\frac{\Gamma _{opt}}{N+1}=0.56 \angle 100^{\circ }[/latex],

[latex]R_{F}=\frac{\sqrt{N(N+1-\left|\Gamma _{opt}\right| ^{2})} }{N+1}=0.24 [/latex].

This noise figure circle is plotted in Figure 12.9a. Minimum noise figure ([latex]F_{min} =1.6 dB[/latex]) occurs for [latex]Γ_{S} = Γ_{opt} = 0.62∠ 100°.[/latex]Next we calculate data for several input section constant-gain circles. From (12.51), we have the following results:

[latex]C_{S}=\frac{g_{S}S_{11}^{*}}{1-(1-g_{S}\left|S_{11}\right|^{2} )}[/latex],

[latex]R_{S}=\frac{\sqrt{1-g_{S}}(1-\left|S_{11}\right|^{2} )} {1-(1-g_{S})\left|S_{11}\right|^{2} }[/latex]

[latex]G_{S}(dB)[/latex]	[latex]g_{S}[/latex]	[latex]C_{S}[/latex]	[latex]R_{S}[/latex]
1.0	0.805	0.52∠ 60°	0.300
1.5	0.904	0.56 ∠60°	0.205
1.7	0.946	0.58∠ 60°	0.150

These circles are plotted in Figure 12.9a. We see that the [latex]G_{S} = 1.7 dB[/latex] gain circle just intersects the F = 2 dB noise figure circle, and that any higher gain will result in a worse noise figure. From the Smith chart the optimum solution is [latex]Γ_{S} = 0.53∠75°[/latex], yielding [latex]G_{S} = 1.7 dB[/latex] and F = 2.0 dB.

For the output section we choose [latex]Γ_{L} = S^{∗}_{22} = 0.5∠60°[/latex] for a maximum [latex]G_{L}[/latex] of

[latex]G_{L}=\frac{1}{1-\left|S_{22}\right|^{2} }=1.33=1.25 dB,[/latex]

The transistor gain is

[latex]G_{0}=\left|S_{21}\right|^{2}=3.61 = 5.58 dB,[/latex]

so the overall transducer gain will be

[latex]G_{TU}=G_{S}+G_{0}+G_{L}=8.53[/latex]

A complete AC circuit for the amplifier, using open-circuited shunt stubs in the matching sections, is shown in Figure 12.9b. A computer analysis of the circuit gives a gain of 8.36 dB.

Question 12.5: A GaAs MESFET is biased for minimum noise figure, with the f...

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