AMPLIFIER DESIGN FOR SPECIFIED GAIN Design an amplifier to have a gain of 11 dB at 4.0 GHz. Plot constant-gain circles for GS = 2 and 3 dB, and GL = 0 and 1 dB. Calculate and plot the input return loss and overall amplifier gain from 3 to 5 GHz. The transistor has the following scattering

Question

AMPLIFIER DESIGN FOR SPECIFIED GAIN

Design an amplifier to have a gain of 11 dB at 4.0 GHz. Plot constant-gain circles for [latex]G_{S} = 2[/latex] and 3 dB, and [latex]G_{L} = 0[/latex] and 1 dB. Calculate and plot the input return loss and overall amplifier gain from 3 to 5 GHz. The transistor has the following scattering parameters ([latex]Z_{0} = 50 Ω[/latex]) :

f(GHz)	[latex]S_{11}[/latex]	[latex]S_{12}[/latex]	[latex]S_{21}[/latex]	[latex]S_{22}[/latex]
3	0.80 ∠-90°	0	2.8 ∠100°	0.66 ∠-50°
4	0.75∠ -120°	0	2.5 ∠180°	0.60 ∠-70°
5	0.71 ∠-140°	0	2.3 ∠60°	0.58 ∠-85°

Accepted Answer

Since [latex]S_{12} = 0[/latex] and [latex]|S_{11}|[/latex] < 1 and [latex]|S_{22}|[/latex] < 1, the transistor is unilateral and unconditionally stable at each frequency in the above table. From (12.47) we calculate the maximum matching section gains as

[latex]G_{S_{max}}=\frac{1}{1-\left|S_{11} ight|^{2} }=2.29=3.6 dB[/latex],

[latex]G_{L_{max}}=\frac{1}{1-\left|S_{22} ight|^{2} }=1.56=1.9 dB[/latex],

The gain of the mismatched transistor is

[latex]G_0 =|S_{21}|^2 = 6.25 = 8.0 dB[/latex],

so the maximum unilateral transducer gain is

[latex]G_{TU_{max}}=3.6+1.9+8.0=13.5 dB[/latex]

We therefore have 2.5 dB more available gain than is required by the specifications. Next, use (12.48), (12.51), and (12.52) to calculate the following data for the constant-gain circles:

[latex]g_{s}=\frac{G_{s}}{G_{S_{max}}}=\frac{1-\left|\Gamma _{s} ight|^{2} }{\left|1-S_{11}\Gamma _{S} ight|^{2} }(1-\left|S_{11} ight|^{2} )[/latex],

[latex]g_{L}=\frac{G_{L}}{G_{L_{max}}}=\frac{1-\left|\Gamma _{L} ight|^{2} }{\left|1-S_{22}\Gamma _{L} ight|^{2} }(1-\left|S_{22} ight|^{2} )[/latex],

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

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

[latex]C_{L}=\frac{g_{L}S_{22}^{*}}{1-(1-g_{L}\left|S_{22} ight|^{2} )}[/latex],

[latex]R_{L}=\frac{\sqrt{1-g_{L}}(1-\left|S_{22} ight|^{2} ) }{1-(1-g_{L}\left|S_{22} ight|^{2} )}[/latex],

[latex] \begin{matrix} G_{S}=3 dB & g_{S} =0.875 & C_{S}=0.706\angle 120^{\circ }&R_{S}=0.166 \ G_{S}=2 dB & g_{S}=0.691 & C_{S}=0.627\angle 120^{\circ } & R_{S}=0.294 \ G_{L}=1 dB & g_{L}=0.806 & C_{L}=0.520\angle 70^{\circ} & R_{L}=0.303 \ G_{L}=0 dB & g_{L}=0.640 & C_{L}=0.440\angle 70^{\circ } & R_{L}=0.440 \end{matrix} [/latex]

The constant-gain circles are shown in Figure 12.8a. We choose [latex]G_{S} = 2 dB[/latex] and [latex]G_{L} = 1 dB[/latex], for an overall amplifier gain of 11 dB. Then we select [latex]Γ_{S}[/latex] and [latex]Γ_{L}[/latex] along these circles as shown, to minimize the distance from the center of the chart (this places [latex]Γ_{S}[/latex] and [latex]Γ_{L}[/latex] along the radial lines at 120° and 70°, respectively). Thus, [latex]Γ_{S} = 0.33∠ 120°[/latex] and [latex]Γ_{L} = 0.22∠70°[/latex], and the matching networks can be designed using shunt stubs as in Example 12.3.

The final amplifier circuit is shown in Figure 12.8b. The response was calculated using CAD software, with interpolation of the given scattering parameter data. The results are shown in Figure 12.8c, where it is seen the desired gain of 11 dB is achieved at 4.0 GHz. The bandwidth over which the gain varies by ±1 dB or less is about 25%, which is considerably better than the bandwidth of the maximum gain design in Example 12.3. The return loss, however, is not very good, being only about 5 dB at the design frequency. This is due to the deliberate mismatch introduced into the matching sections to achieve the specified gain.

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