Designing a Unilateral RF Amplifier for a Specified Gain
Learn about the role of gain analysis in the design of unilateral RF amplifiers, first by reviewing the basic concepts and then by working through a pair of design examples.
Like many other RF design topics, RF amplifier design can be intimidatingly math-intensive. With today’s RF design software tools, though, we don’t need to spend all our time plugging numbers into equations. Instead, we need to understand the design principles underlying those equations, so that we can use our software tools confidently. To that end, the previous article in this series discussed the basics of stability analysis for RF amplifiers. Now it’s time to introduce another important piece of the design process: assessing the amplifier’s gain.
Analyzing the gain of a low-frequency amplifier is relatively straightforward, since we don’t need to worry about signal reflections with a low frequency. At very low frequencies, we can simplify our analysis even further by assuming that the transistor is unilateral, meaning that there’s no internal feedback from its output to its input.
Transducer Power Gain
When designing amplifiers, we use the transducer power gain for an RF two-port network. To understand the transducer power gain, consider the RF amplifier schematic in Figure 1.
Figure 1. RF single-stage amplifier with arbitrary source and load impedances.
The transducer power gain (GT) is defined as:
Equation 1.
where:
PL is the power delivered to the load
PAVS is the power available from the source.
PL is easy to understand—it’s simply the power that the circuit delivers to the load when the transistor is driven by the source, which has an impedance of ZS. What constitutes PAVS might be a bit more confusing. PAVS is the power that the source delivers to a conjugately matched load with impedance Zs*. The following diagram shows how we can determine PAVS.
Figure 2. Simple RF power circuit with conjugately matched load.
We can use the following, somewhat intimidating equation to express GT in terms of the transistor’s S-parameters:
Equation 2.
where ΓIN is the reflection coefficient at the input of the transistor, given by:
Equation 3.
One special case of Equation 2 happens when both input and output terminations are reflectionless, meaning that they are matched for zero reflection (ΓS = ΓL = 0). In this case, the transducer power gain is equal to |S21|2. This is sometimes referred to as the “basic Z0-based transducer power gain” of the transistor.
Unilateral Transducer Power Gain
We can see that GT depends on the mismatch of the source (ΓS) and load (ΓL). If we know ΓS and ΓL, we can use Equations 2 and 3 to find the gain of the circuit. However, most design problems require us to determine the value of ΓS and ΓL for a given gain (GT). In that case, directly working with Equation 2 can be a bit tricky, since—as we can see in Equation 3—there’s an interaction between input and output reflection coefficient values.
In some practical amplifier circuits, we can neglect the influence of the transistor’s internal feedback and assume that S12 = 0. In this case, the device is called unilateral and ΓIN is simply equal to S11 (see Equation 3). This simplifies GT to:
Equation 4.
Note that the unilateral transducer power gain consists of three distinct, independent gain terms.
How does Equation 4 simplify the design process? Assume that we want to design an 18 dB single-stage amplifier using a unilateral device with |S21| = 6.5. For this example, |S21|2 = 16.26 dB. Therefore, 1.74 dB of gain should be provided by the fractional terms in the above equation.
Since these two fractional terms are independent of each other, we can easily split the remaining gain between them and find appropriate values of ΓS and ΓL. Note that the two fractional terms have the same form, meaning that we can develop a single procedure to find both ΓS and ΓL.
Physical Interpretation of Gain Terms
We can write the unilateral gain in Equation 4 as the product of three gain terms defined below:
Equation 5.
where:
Equation 6.
Equation 7.
Equation 8.
As you can see, GS is only related to the input parameters (ΓS and S11), and GL is only dependent on the output parameters (ΓL and S22).
We can interpret GS and GL as the gain or loss produced by the input or output matching network. Based on this, we can model a single-stage amplifier as the cascade of three blocks with gains GS, G0, and GL (Figure 3).
Figure 3. Model of a single-stage amplifier using input and output matching networks.
But how can a matching network made up of passive components produce a gain? To understand this seemingly contradictory behavior, note that we can have a significant mismatch loss when either ΓS and S11 (at the input) or ΓL and S22 (at the output) aren’t matched. A better impedance match provided by the matching networks can reduce the mismatch losses, effectively increasing gain.
Minimum and Maximum Values of Gain Terms
When designing an amplifier, it’s important to know how much gain GS and GL can each provide, so that we can appropriately split the total gain between the two terms. Assuming that |S11| < 1, the maximum of GS is:
Equation 9.
which occurs at
.
Also, from Equation 6, |ΓS| = 1 produces the minimum value of GS = 0. For other values of ΓS, we have
. Note that when |S11| > 1, GS can approach infinity for a passive input termination (|ΓS| < 1) having a value of.
We’ve been looking at the range of the GS term, but a similar discussion is also applicable to the GL term—both gain equations take the same form.
In this article, we’re assuming that |S11| < 1 and |S22| < 1. This is commonly the case, but not always—for the potentially unstable case where |S11| > 1 and |S22| > 1, please refer to “Microwave Transistor Amplifiers: Analysis and Design” by G. Gonzalez.
Determining Appropriate Terminations: Constant Gain Circles
At this point, another question arises: how can we determine the appropriate reflection coefficient (ΓS or ΓL) for a given gain (GS or GL)? In the interest of brevity, we’ll skip the derivation details and go straight to the final equations for this problem.
Let’s consider the GS term. We want to find the appropriate ΓS to produce GS = G. Interestingly, there are an infinite number of ΓS values that produce a certain GS. All these ΓS values lie on a circle known as the constant GS circle. The GS = G circle has a center location of:
Equation 10.
and radius of:
Equation 11.
where gS = G / Gs,max is the normalized gain. Similarly, the center of the GL = G circle is given by:
Equation 12.
and its radius by:
Equation 13.
where gL = G / GL,max is the normalized gain.
To learn about important features of these constant gain circles, let’s work through some example calculations.
Example 1: Calculating Gain for a Given Frequency
Assume that Z0 = 50 Ω for a transistor with the S-parameters in Table 1.
Table 1. S-parameters for an example transistor.
| f (GHz) | S11 | S21 | S12 | S22 |
| 0.8 | 0.44 ∠ –157.6 | 4.725 ∠ 84.3 | 0 | 0.339 ∠ –51.8 |
| 1.4 | 0.533 ∠ 176.6 | 2.800 ∠ 64.5 | 0 | 0.604 ∠ –58.3 |
| 2.0 | 0.439 ∠ 159.6 | 2.057 ∠ 49.2 | 0 | 0.294 ∠ –68.1 |
First, we’ll find the maximum input and output matching section gains at f = 1.4 GHz and the maximum unilateral transducer power gain. After that, we’ll plot the constant gain circles for various values of GS and GL at this frequency.
The basic Z0-based transducer power gain of the transistor is:
Equation 14.
The maximum matching section gains are:
Equation 15.
and
Equation 16.
The maximum transducer gain works out to:
Equation 17.
The centers and radii of some constant GS circles are tabulated below.
Table 2. Centers and radii of constant GS circles.
| Gain | Normalized Gain | Center | Radius |
| GS = 0.0 dB | gS = 0.71 | cS1 = 0.41 ∠ –176.6 | rS1 = 0.41 |
| GS = 0.5 dB | gS = 0.80 | cS1 = 0.45 ∠ –176.6 | rS1 = 0.34 |
| GS = 1.0 dB | gS = 0.90 | cS2 = 0.49 ∠ –176.6 | rS2 = 0.23 |
| GS = 1.4 dB | gS = 0.99 | cS2 = 0.53 ∠ –176.6 | rS2 = 0.07 |
Figure 4 plots these circles in the ΓS plane. The blue dot shows the location of ΓS = S11*.
Figure 4. Constant GS circles.
We now have the maximum value of GS. Observe that the centers of the constant GS circles all lie on a line connecting the origin to S11*. As the gain increases, the gain circles become smaller, and the centers of the circles get closer to S11*. Finally, for the maximum value of gain, the gain circle reduces to a single point at S11*.
Note that the GS = 1 (0 dB) circle passes through the origin. This isn’t a coincidence—it always happens for both constant GS and constant GL circles. It’s also worth mentioning that we can choose ΓS to produce a GS less than unity, such as GS = –1 dB.
We can find the constant GL circles similarly. Table 3 lists the centers and radii of some example constant GL circles.
Table 3. Centers and radii of constant GL circles.
| Gain | Normalized Gain | Center | Radius |
| GL = 0.00 dB | gL = 0.64 | cL1 = 0.44 ∠ 58.3 | rL1 = 0.44 |
| GL = 0.66 dB | gL = 0.74 | cL1 = 0.49 ∠ 58.3 | rL1 = 0.36 |
| GL = 1.00 dB | gL = 0.80 | cL2 = 0.52 ∠ 58.3 | rL2 = 0.31 |
| GL = 1.56 dB | gL = 0.91 | cL1 = 0.57 ∠ 58.3 | rL1 = 0.20 |
Figure 5 plots these circles in the ΓL plane.
Figure 5. Constant GL circles.
Features similar to those mentioned for the constant GS circles are also observable in the constant GL circles.
Example 2: Designing Based on Total Gain
Using the results of the previous example, let’s design an amplifier with a gain of 11 dB at 1.4 GHz.
The basic Z0-based transducer power gain of the transistor is G0 = 8.94 dB. Therefore, to have an 11 dB amplifier, the input and output matching sections should provide a total gain of 2.06 dB. One possible solution is to choose GS = 1 dB and GL = 1 dB. The gain circles for this design are shown in Figure 6.
Figure 6. Constant gain circles for an example amplifier. GS and GL both equal 1 dB.
Any point on these constant gain circles can be used to achieve the desired total gain. We’ll use the reflection coefficient values (ΓS and ΓL) shown in the above figure. Using the Smith chart, we obtain:
ΓS = 0.26 ∠ –176.6
ΓL = 0.22 ∠ 58.3
Figure 7 shows the final design with the input and output matching sections added.
Figure 7. RF amplifier with input and output matching sections added.
Though the above diagram shows only the AC amplifier schematic, the amplifier will also require some bias circuitry. Additionally, I’ve assumed that the active device is a bipolar transistor.
Need to brush up on impedance matching techniques? This article on single stub impedance matching can help.
Figure 8 plots the simulated gain of the amplifier.
Figure 8. Gain (y-axis) vs. frequency (x-axis) of an example RF amplifier.
The gain at 1.4 GHz is 10.95 dB, which is almost equal to the desired value. In the above simulation, the software was provided with the S-parameters at 0.8, 1.4, and 2 GHz. The S-parameters for any other required frequency points are obtained through interpolation. The input reflection coefficient of the amplifier is also shown below, in Figure 9.
Figure 9. Input reflection coefficient (y-axis) vs. frequency (x-axis) of a simulated amplifier.
We can see that the input isn’t very well matched to the 50 Ω source impedance. In this example, we’re actually deliberately reducing the impedance matching at the input and output to reduce the gain to the desired value.
Up Next
The gain analysis of an RF amplifier should account for signal reflections at both the input and output of the transistor, as well as the transistor’s bilateral response. Because we’ve specifically been looking at unilateral amplifiers, we’ve largely ignored this aspect. We’ll learn more about this next time, when we explore bilateral amplifier design.
