# 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.*

*Figure 1.*

The transducer power gain (*G _{T}*) is defined as:

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

**Equation 1.**

**Equation 1.**

where:

*P _{L}* is the power delivered to the load

*P _{AVS}* is the power available from the source.

*P _{L}* 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

*Z*. What constitutes

_{S}*P*might be a bit more confusing.

_{AVS}*P*is the power that the source delivers to a conjugately matched load with impedance Z

_{AVS}_{s*}. The following diagram shows how we can determine

*P*.

_{AVS}

**Figure 2.** Simple RF power circuit with conjugately matched load.

**Figure 2.**Simple RF power circuit with conjugately matched load.

We can use the following, somewhat intimidating equation to express *G _{T}* in terms of the transistor’s S-parameters:

$$G_{T}~=~\frac{1~-~|\Gamma_S|^2}{|1~-~\Gamma_S \Gamma_{in}|^2}~\times~|S_{21}|^2~\times~\frac{1~-~|\Gamma_L|^2}{|1~-~S_{22} \Gamma_{L}|^2}$$

**Equation 2.**

**Equation 2.**

where Γ_{IN} is the reflection coefficient at the input of the transistor, given by:

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

**Equation 3.**

**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 (Γ_{S} = Γ_{L} = 0). In this case, the transducer power gain is equal to |*S*_{21}|^{2}. This is sometimes referred to as the “basic *Z*_{0}-based transducer power gain” of the transistor.

### Unilateral Transducer Power Gain

We can see that *G _{T}* depends on the mismatch of the source (Γ

_{S}) and load (Γ

_{L}). If we know Γ

_{S}and Γ

_{L}, 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 Γ

_{S}and Γ

_{L}for a given gain (G

_{T}). 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 *S*_{12} = 0. In this case, the device is called unilateral and Γ_{IN} is simply equal to *S*_{11} (see Equation 3). This simplifies *G _{T}* to:

$$G_{T}~=~\frac{1~-~|\Gamma_S|^2}{|1~-~S_{11} \Gamma_S|^2}~\times~|S_{21}|^{2}~ \times~ \frac{1~-~|\Gamma_L|^2}{|1~-~S_{22} \Gamma_{L}|^2}$$

**Equation 4.**

**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 |*S*_{21}| = 6.5. For this example, |*S*_{21}|^{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 Γ_{S} and Γ_{L}. Note that the two fractional terms have the same form, meaning that we can develop a single procedure to find both Γ_{S} and Γ_{L}.

### Physical Interpretation of Gain Terms

We can write the unilateral gain in Equation 4 as the product of three gain terms defined below:

$$G_{TU} ~=~G_{S}~ \times~ G_{0} ~\times~ G_{L}$$

**Equation 5.**

**Equation 5.**

where:

$$G_{S}~=~\frac{1~-~|\Gamma_S|^2}{|1~-~S_{11} \Gamma_S|^2} $$

**Equation 6.**

**Equation 6.**

$$G_{0}~=~|S_{21}|^{2} $$

**Equation 7.**

**Equation 7.**

$$G_{L}~=~\frac{1~-~|\Gamma_L|^{2}}{|1~-~S_{22} \Gamma_{L}|^{2}}$$

**Equation 8.**

**Equation 8.**

As you can see, *G _{S}* is only related to the input parameters (Γ

_{S}and

*S*

_{11}), and

*G*is only dependent on the output parameters (Γ

_{L}_{L}and

*S*

_{22}).

We can interpret *G _{S}* and

*G*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

_{L}*G*,

_{S}*G*

_{0}, and

*G*(Figure 3).

_{L}

**Figure 3.** Model of a single-stage amplifier using input and output matching networks.

**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 Γ_{S} and *S*_{11} (at the input) or Γ_{L} and *S*_{22} (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 *G _{S}* and

*G*can each provide, so that we can appropriately split the total gain between the two terms. Assuming that |

_{L}*S*

_{11}| < 1, the maximum of

*G*is:

_{S}$$G_{S, max}~=~\frac{1}{1~-~|S_{11}|^2}$$

**Equation 9.**

**Equation 9.**

which occurs at \(\Gamma_{S}~=~S^{*}_{11}\).

Also, from Equation 6, |Γ_{S}| = 1 produces the minimum value of *G _{S}* = 0. For other values of Γ

_{S}, we have \(0~ < ~G_{S}~ < ~G_{S,max}\). Note that when |

*S*

_{11}| > 1,

*G*can approach infinity for a passive input termination (|Γ

_{S}_{S}| < 1) having a value of \(\Gamma_{S}~=~\frac{1}{S_{11}}\).

We’ve been looking at the range of the *G _{S}* term, but a similar discussion is also applicable to the

*G*term—both gain equations take the same form.

_{L}In this article, we’re assuming that |*S*_{11}| < 1 and |*S*_{22}| < 1. This is commonly the case, but not always—for the potentially unstable case where |*S*_{11}| > 1 and |*S*_{22}| > 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 (Γ_{S} or Γ_{L}) for a given gain (*G _{S}* or

*G*)? In the interest of brevity, we’ll skip the derivation details and go straight to the final equations for this problem.

_{L}Let’s consider the *G _{S}* term. We want to find the appropriate Γ

_{S}to produce

*G*=

_{S}*G*. Interestingly, there are an infinite number of Γ

_{S}values that produce a certain G

_{S}. All these Γ

_{S}values lie on a circle known as the constant

*G*circle. The

_{S}*G*=

_{S}*G*circle has a center location of:

$$c_{S}~=~\frac{g_{S} S_{11}^{*}}{1~-~|S_{11}|^{2}(1~-~g_{S})}$$

**Equation 10.**

**Equation 10.**

and radius of:

$$r_{S}~=~\frac{\sqrt{1~-~g_{S}}~\times~(1~-~|S_{11}|^{2})}{1~-~|S_{11}|^{2}(1~-~g_S)}$$

**Equation 11.**

**Equation 11.**

where *g _{S}* =

*G*/

*G*is the normalized gain. Similarly, the center of the

_{s,max}*G*=

_{L}*G*circle is given by:

$$c_L~=~\frac{g_{L} S_{22}^{*}}{1~-~|S_{22}|^{2}(1~-~g_{L})}$$

**Equation 12.**

**Equation 12.**

and its radius by:

$$r_{L}~=~\frac{\sqrt{1~-~g_{L}}~\times~(1~-~|S_{22}|^{2})}{1~-~S_{22}|^{2}(1~-~g_{L})}$$

**Equation 13.**

**Equation 13.**

where *g _{L}* =

*G*/

*G*is the normalized gain.

_{L,max}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 *Z*_{0} = 50 Ω for a transistor with the S-parameters in Table 1.

**Table 1.** S-parameters for an example transistor.

**Table 1.**S-parameters for an example transistor.

f (GHz) |
S_{11} |
S_{21} |
S_{12} |
S_{22} |

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 *G _{S}* and

*G*at this frequency.

_{L}The basic *Z*_{0}-based transducer power gain of the transistor is:

$$G_{0}~ = ~|S_{11}|^{2}~ = ~7.84~ = ~8.94 ~\text{dB}$$

**Equation 14.**

**Equation 14.**

The maximum matching section gains are:

$$G_{S, max}~=~\frac{1}{1~-~|S_{11}|^2}~=~1.4~=~ 1.46~ \text{dB}$$

**Equation 15.**

**Equation 15.**

and

$$G_{L, max}~=~\frac{1}{1~-~|S_{22}|^2}~=~1.57~=~1.97~ \text{dB}$$

**Equation 16.**

**Equation 16.**

The maximum transducer gain works out to:

$$G_{TU,max}~=~G_{S, max}~+~G_0~+~G_{L,max}~=~ 1.46~+~8.94~+~1.97~=~12.37~\text{dB}$$

**Equation 17.**

**Equation 17.**

The centers and radii of some constant *G _{S}* circles are tabulated below.

**Table 2. **Centers and radii of constant G_{S}* circles.*

**Table 2.**Centers and radii of constant

Gain |
Normalized Gain |
Center |
Radius |

G = 0.0 dB_{S} |
g = 0.71_{S} |
c_{S}_{1} = 0.41 ∠ –176.6 |
r_{S}_{1} = 0.41 |

G = 0.5 dB_{S} |
g = 0.80_{S} |
c_{S}_{1} = 0.45 ∠ –176.6 |
r_{S}_{1} = 0.34 |

G = 1.0 dB_{S} |
g = 0.90_{S} |
c_{S}_{2} = 0.49 ∠ –176.6 |
r_{S}_{2} = 0.23 |

G = 1.4 dB_{S} |
g = 0.99_{S} |
c_{S}_{2} = 0.53 ∠ –176.6 |
r_{S}_{2} = 0.07 |

Figure 4 plots these circles in the Γ_{S} plane. The blue dot shows the location of Γ_{S} = *S*_{11}*.

**Figure 4.** Constant G_{S}* circles.*

**Figure 4.**Constant

We now have the maximum value of *G _{S}*. Observe that the centers of the constant

*G*circles all lie on a line connecting the origin to

_{S}*S*

_{11}*. As the gain increases, the gain circles become smaller, and the centers of the circles get closer to

*S*

_{11}*. Finally, for the maximum value of gain, the gain circle reduces to a single point at

*S*

_{11}*.

Note that the *G _{S}* = 1 (0 dB) circle passes through the origin. This isn’t a coincidence—it always happens for both constant

*G*and constant

_{S}*G*circles. It’s also worth mentioning that we can choose Γ

_{L}_{S}to produce a

*G*less than unity, such as

_{S}*G*= –1 dB.

_{S}We can find the constant *G _{L}* circles similarly. Table 3 lists the centers and radii of some example constant

*G*circles.

_{L}

**Table 3.** Centers and radii of constant G_{L}* circles.*

**Table 3.**Centers and radii of constant

Gain |
Normalized Gain |
Center |
Radius |

G = 0.00 dB_{L} |
g = 0.64_{L} |
c_{L}_{1} = 0.44 ∠ 58.3 |
r_{L}_{1} = 0.44 |

G = 0.66 dB_{L} |
g = 0.74_{L} |
c_{L}_{1} = 0.49 ∠ 58.3 |
r_{L}_{1} = 0.36 |

G_{L} = 1.00 dB |
g = 0.80_{L} |
c_{L}_{2} = 0.52 ∠ 58.3 |
r_{L}_{2} = 0.31 |

G = 1.56 dB_{L} |
g = 0.91_{L} |
c_{L}_{1} = 0.57 ∠ 58.3 |
r_{L}_{1} = 0.20 |

Figure 5 plots these circles in the Γ_{L} plane.

**Figure 5.** Constant G_{L}* circles.*

**Figure 5.**Constant

Features similar to those mentioned for the constant *G _{S}* circles are also observable in the constant

*G*circles.

_{L}### 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 *Z*_{0}-based transducer power gain of the transistor is *G*_{0} = 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 *G _{S}* = 1 dB and

*G*= 1 dB. The gain circles for this design are shown in Figure 6.

_{L}

**Figure 6.** Constant gain circles for an example amplifier. G_{S}* and *G_{L}* both equal 1 dB.*

**Figure 6.**Constant gain circles for an example amplifier.

Any point on these constant gain circles can be used to achieve the desired total gain. We’ll use the reflection coefficient values (Γ_{S} and Γ_{L}) shown in the above figure. Using the Smith chart, we obtain:

Γ_{S }= 0.26 ∠ –176.6

Γ_{L }= 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.

**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.

**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.

**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.

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

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