Technical Article

# Using the Available Power Gain to Design a Bilateral Low-Noise Amplifier

December 17, 2023 by Dr. Steve Arar

## Learn how the available power gain concept can help us solve the problem of designing a bilateral RF amplifier for a specified gain and noise figure.

When designing a low-noise amplifier (LNA), we need to consider both gain and noise performance. As we learned in the preceding article, we can use an RF transistor’s constant noise figure (NF) contours to determine the appropriate source termination for a given noise performance. The constant NF contours are plotted in the ΓS plane; to design an amplifier for noise and gain, we need to plot the transistor’s gain contours in the ΓS plane as well.

We already covered how to accomplish this in the case of a unilateral device, where the gain of the input and output matching networks are independent of each other. This article explores the design of bilateral LNAs for a specified gain and noise figure, which can be a bit more complicated.

To design a bilateral amplifier for a specified gain other than the maximum, we can use either the operating power gain (GP) or available power gain (GA). However, constant GP circles are plotted in the ΓL plane, so they can’t be directly used to analyze the amplifier’s gain-noise trade-off. Constant GA circles, on the other hand, are specified in the ΓS plane. For that reason, we’ll use the available power gain concept to design our bilateral LNA.

### Available Power Gain

The available power gain (GA) is the ratio of the power available from the network (PAVN) to the power available from the source (PAVS):

$$G_A ~=~ \frac{P_{AVN}}{P_{AVS}}$$

##### Equation 1.

Figure 1 illustrates how the available power gain of a module is determined.

##### Figure 1. Determining the available power gain from the amplifier (a) and the maximum power available from the source (b).

In the GA equation, PAVN is normalized to the available power from the source, so our point of reference for this power gain measurement is the input voltage source (VS). As the RF power travels from the source to the amplifier output, it’s affected by the impedance mismatch at the transistor’s input (ZS and ZIN). As a result, PAVN depends on the source termination (ΓS).

However, as Figure 1(a) illustrates, we connect the module output to a conjugate-matched load to measure PAVN. That’s why PAVN doesn’t depend on the load termination (ΓL) that we’ll eventually connect to our amplifier. We can verify this by examining the available power gain equation below:

$$G_A ~=~ \frac{1~-~|\Gamma_S|^2}{|1~-~S_{11}\Gamma_S|^2} ~\times~ |S_{21}|^2 ~\times~ \frac{1}{1~-~|\Gamma_{OUT}|^2}$$

##### Equation 2.

where ΓOUT, the reflection coefficient seen at the output of the transistor, is given by:

$$\Gamma_{OUT}~=~S_{22} ~+~ \frac{S_{12}S_{21} \Gamma_S}{1~-~S_{11}\Gamma_S}$$

##### Equation 3.

Note that GA isn’t a function of ΓL, only of ΓS.

### Using the Available Power Gain in Amplifier Design

Since Equation 2 is only a function of ΓS and the S-parameters, we can use it to find the appropriate ΓS for a given GA. But does this completely solve our design problem? The actual gain exhibited by an amplifier is its transducer gain, which accounts for both ΓS and ΓL. The transducer gain is given by:

$$G_{T} ~=~ \frac{P_L}{P_{AVS}}$$

##### Equation 4.

If we want to design an amplifier using its available power gain, we need to find a relationship between GA and GT. Comparing Equations 1 and 4, we see that these two power gains become identical if we have PL = PAVN. Therefore, if we set ΓL equal to Γ*OUT, we’ll have GA = GT.

To summarize, in order to design for a specific gain through the available power gain concept, we use Equation 2 to find the ΓS value that produces the desired GA, then provide a conjugate match at the output so that the actual gain GT exhibited by the device becomes equal to the chosen GA.

#### Design Equations for an Unconditionally Stable Device

Now that we understand the overall design procedure, let’s have a look at the required equations for an unconditionally stable device. The values of ΓS that produce a given GA lie on the constant GA circle. The center (cA) of this circle is given by:

$$c_A ~=~ \frac{g_A c_1^*}{1~+~g_A(|S_{11}|^{2}~-~|\Delta|^2)}$$

##### Equation 5.

$$r_A ~=~ \frac{\Big ( 1~-~2K|S_{12}S_{21}|g_A ~+~ |S_{12}S_{21}|^2g_A^2 \Big )^{\frac{1}{2}}}{|1~+~g_A(|S_{11}|^2~-~|\Delta|^2)|}$$

##### Equation 6.

The parameters gA and c1 in the above equations are defined, respectively, by:

$$g_A ~=~ \frac{G_A}{|S_{21}|^2}$$

##### Equation 7.

and:

$$c_1~=~S_{11}~-~\Delta S_{22}^*$$

##### Equation 8.

$$K ~=~ \frac{1~-~|S_{11}|^2 ~-~ |S_{22}|^2 ~+~ |\Delta|^2}{2|S_{12}S_{21}|}$$

##### Equation 9.

and Δ is the determinant of the S-parameters matrix:

$$\Delta ~=~ S_{11}S_{22}~-~S_{12}S_{21}$$

##### Equation 10.

For an unconditionally stable device, the maximum of GA is given by:

$$G_{A, max} ~=~ \frac{|S_{21}|}{|S_{12}|} ~\times~ (K~-~\sqrt{K^{2}~-~1})$$

##### Equation 11.

As an aside, the equation for GA,max is also the equation for GP,max and GT,max.

#### Example 1: Plotting GA Circles for an RF Transistor

Now that we have the necessary equations, we’ll work through some examples. To start with, let’s plot some constant GA circles for a transistor with the S-parameters in Table 1 at f = 1.4 GHz.

##### Table 1. S-parameters for an example transistor. Z0 = 50 Ω.
 f (GHz) S11 S21 S12 S22 0.8 0.440 ∠ –157.6 degrees 4.725 ∠ 84.3 degrees 0.06 ∠ 55.4 degrees 0.339 ∠ –51.8 degrees 1.4 0.533 ∠ 176.6 degrees 2.800 ∠ 64.5 degrees 0.06 ∠ 58.4 degrees 0.604 ∠ –58.3 degrees 2.0 0.439 ∠ 159.6 degrees 2.057 ∠ 49.2 degrees 0.17 ∠ 58.1 degrees 0.294 ∠ –68.1 degrees

Properly, we should verify that this device is unconditionally stable at all three of the frequencies above. For the sake of brevity, though, I’m including a K-factor test only for f = 1.4 GHz.

$$K ~=~ \frac{1~-~|S_{11}|^2 ~-~ |S_{22}|^2 ~+~ |\Delta|^2}{2|S_{12}S_{21}|} ~=~ 1.12$$

##### Equation 12.

Since  K is greater than unity and |Δ| < 1, the device is unconditionally stable at our chosen frequency. We can therefore use the design procedure outlined above to find the appropriate source and load terminations. First, applying Equation 11, we find the maximum of GA:

$$G_{A, max} ~=~ \frac{|S_{21}|}{|S_{12}|} ~\times~ (K-\sqrt{K^{2}-1})~=~28.73$$

##### Equation 13.

This converts to GA,max = 14.58 dB.

I’ve arbitrarily chosen to plot the constant gain circles of GA = 11 dB, 12 dB, 13 dB, and 14 dB. Using the data and equations above, we know the following:

|S21|2 = 7.84

K = 1.12

Δ = 0.16 ∠ 113.32 degrees

c1 = 0.44 ∠ 177.66 degrees.

We can therefore apply Equations 5 and 6 to find the center and radius of the constant GA circles. Table 2 provides a summary of the calculations.

Note that the conversion of GA from dB to linear terms is the first set of calculations we have to do. We need to use Equation 7 to determine gA before we can calculate either cA or rA, and Equation 7 requires GA to be in linear terms.

##### Table 2. Summary of the calculations necessary to plot constant GA circles for an example transistor.
 GA (dB) GA (linear terms) gA cA rA 11 dB 12.59 1.61 0.50 ∠ –177.66 degrees 0.48 12 dB 15.85 2.00 0.58 ∠ –177.66 degrees 0.39 13 dB 19.95 2.55 0.67 ∠ –177.66 degrees 0.29 14 dB 25.12 3.20 0.77 ∠ –177.66 degrees 0.16

These constant GA circles are plotted in Figure 2.

##### Figure 2. Constant GA circles for an example transistor at GA = 11 dB (pink circle), 12 dB (red circle), 13 dB (green circle), and 14 dB (purple circle).

The centers of the constant GA circles are always on the line connecting c*1 to the origin of the Smith chart. To find the origin, see Equation 5.

#### Example 2: Designing an Amplifier With a Perfectly Matched Output for a Specific Gain

Let’s use the transistor from the previous example design an amplifier with a gain of 13 dB at f = 1.4 GHz, and ensure a perfect match on the output side of the transistor when we do so.

Since we want a perfect match on the output side, we’ll use the available power gain method. Any source termination residing on the green constant GA circle in Figure 2 can be used to achieve a 13 dB available gain—I’ve arbitrarily chosen ΓS = 0.38 ∠ –177.66 degrees, marked as point A in the above figure. With ΓS determined, we can find the reflection coefficient seen at the output of the transistor:

$$\Gamma_{OUT}~=~S_{22} ~+~ \frac{S_{12}S_{21} \Gamma_S}{1~-~S_{11}\Gamma_S}~=~0.68 ~\angle~ -57.92~\text{degrees}$$

##### Equation 14.

Now we only need to provide a complex conjugate match on the output side to have GT = GA = 13 dB:

$$\Gamma_L ~=~ \Gamma_{OUT}^{*}~=~0.68 ~\angle~ -57.92~\text{degrees}$$

##### Equation 15.

This choice of ΓL also ensures a perfect match at the output (VSWR = 1).

If we substitute ΓS = 0.38 ∠ –177.66 degrees and ΓL = 0.68 ∠ 57.92 degrees into the transducer gain equation, we obtain:

$$G_{T}~=~\frac{1~-~|\Gamma_{S}|^{2}}{|1~-~S_{11} \Gamma_{S}|^{2}}~\times~|S_{21}|^{2}~\times~\frac{1~-~| \Gamma_{L}|^{2}}{|1~-~\Gamma_{OUT} \Gamma_{L}|^{2}}~=~19.82$$

##### Equation 16.

Converted from linear terms, this becomes 12.97 dB. This is very close to the target value of GT = 13 dB.

We can now use a Z Smith chart to design the input and output matching networks. To design the input matching network, we first locate ΓS on the Smith chart and find its associated normalized admittance (yS) through a 180 degree rotation along the constant |ΓS| circle (Figure 3).

##### Figure 3. Smith chart for the design of an example RF amplifier's input matching network.

We interpret the Smith chart as a Y Smith chart from now on. We want a circuit that takes us from the center of the chart (the 50 Ω termination) to yS. The intersection point of the constant |ΓS| circle with the 1 + jb circle, marked as point A, has a susceptance of j0.84.

To create a susceptance of j0.84, we add a parallel open-circuited stub of length l1 = 0.109λ to the 50 Ω termination. We then add a series line of length l2 = 0.091λ to travel along the constant |ΓS| circle to yS. That takes care of the input matching section; the output matching section can be designed in a similar way, as we see in Figure 4.

##### Figure 4. Smith chart for the design of the example amplifier's output matching network.

We can see that the output matching network needs an open-circuited stub of length l3 = 0.171λ and a series line of length l4 = 0.236λ.

Figure 5 shows the AC schematic of the final design, including both input and output matching networks.

##### Figure 5. The final design of our example amplifier.

Figure 6 shows the simulated gain of the amplifier, which is very close to the desired value GT = 13 dB.

##### Figure 6. Simulated gain of the amplifier we designed above. At 1400 MHz, gain is 12.96 dB.

Figure 7 shows the input and output reflection coefficients of the amplifier. We can see from this figure that using the available power gain approach produces amplifiers with well-matched output and mismatched input.

#### Example 3: Designing a Bilateral Amplifier for a Specific Gain and Noise Performance

Assume that, at f = 1.4 GHz, the noise parameters of the transistor we’ve been investigating are:

Fmin = 1.6 dB

Γopt = 0.62 ∠ 100 degrees

RN = 20 Ω.

Using this transistor, let’s design an amplifier with a noise figure of 3 dB and the maximum gain that’s compatible with that noise figure.

First, we find the center and radius of the NF = 3 dB constant noise circle. Table 3 provides a summary of the required calculations. Note that we’re plugging the noise factor, not the noise figure, into our equations.

##### Table 3. Summarized calculations for the NF = 3 dB constant noise circle.
 NF (Noise Figure) F (Noise Factor) N cF rF 3 dB 2 0.4 0.44 ∠ 100 degrees 0.46

Figure 8 plots the constant noise circle of NF = 3 dB alongside the constant GA circles from Example 1.

##### Figure 8. Example 1’s constant GA circles plotted alongside the 3 dB constant NF circle.

The NF = 3 dB noise circle intersects the GA = 13 dB gain circle at point A, which corresponds to ΓS = 0.46 ∠ 161.4 degrees. With ΓS determined, we can now find the reflection coefficient seen at the output of the transistor:

$$\Gamma_{OUT}~=~S_{22} ~+~ \frac{S_{12}S_{21} \Gamma_S}{1~-~S_{11}\Gamma_S}~=~0.7 ~\angle~ -61.65~\text{degrees}$$

##### Equation 17.

Finally, to produce GT = GA = 13 dB, we provide a complex conjugate match on the output side:

$$\Gamma_L ~=~ \Gamma_{OUT}^{*}~=~0.7 ~\angle~ 61.65~\text{degrees}$$

##### Equation 18.

We’ll once again use a Z Smith chart to design the matching networks. Figures 9 and 10 show the design details for the input and output matching networks, respectively.

##### Figure 10. Smith chart for the design of the amplifier’s output matching network.

Figure 11 shows the final AC schematic.

##### Figure 11. AC schematic for a bilateral LNA intended to have 13 dB of gain and 3 dB of noise at f = 1.4 GHz.

A computer analysis of the above design gives a gain of 12.94 dB and noise figure of 3 dB at f = 1.4 GHz.

### Wrapping Up

RF LNAs are a critical component of our increasingly connected world. I hope that this article, taken together with the previous one, has helped improve your understanding of low-noise amplifier design. If there is more about LNA design that you’d like me to cover, please consider leaving a note for me in the comments section.

Featured image used courtesy of Adobe Stock; all other images used courtesy of Steve Arar