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Simultaneous Conjugate Match for Maximum Amplifier Gain

By John · Updated · 5 min read

When S12 is not zero, the input and output of an amplifier must be matched together, not one at a time. For an unconditionally stable device, Gmax = |S21|/|S12| · (K − sqrt(K² − 1)), which rises to the MSG at K = 1.

Engineer’s note: I don't trust a K of 1.5 from a single frequency. Compute it at every point in the file, out of band too, because a stage that's clean at the design point can still oscillate somewhere else. And don't quote 10 dB of MSG as your gain when K = 1.5 gives you 5.82 dB.

For practitionersNew to this? Plain-language version

For practitioners

A transistor gives its most gain when both ends are conjugate matched. The source sees its own conjugate, and so does the load. That sounds like two separate matching problems. It is one problem, and this page shows why. It also gives the closed-form solution and the gain you get at the end.

Why the two ends cannot be matched independently

A two-port with S-parameters S11, S21, S12 and S22 sits between an input matching network and an output matching network. The input network turns the 50 Ω source (Z0) into a source reflection coefficient Γ_S at the device input. The output network turns the Z0 load into a load reflection coefficient Γ_L at the device output.

Looking into the device, the input reflection coefficient is:

Γin=S11+S12S21ΓL1−S22ΓL\Gamma_{in} = S_{11} + \frac{S_{12}S_{21}\Gamma_L}{1 - S_{22}\Gamma_L}

Looking back into the output, the reflection coefficient is:

Γout=S22+S12S21ΓS1−S11ΓS\Gamma_{out} = S_{22} + \frac{S_{12}S_{21}\Gamma_S}{1 - S_{11}\Gamma_S}

The second term in each expression carries S12. It is the feedback path through the device. Because of it, Γ_in depends on the load you pick, and Γ_out depends on the source you pick.

Γin depends on ΓL,Γout depends on ΓSinput sideoutput sideSourceZ0InputmatchingnetworkTransistorS11 S12S21 S22OutputmatchingnetworkLoadZ0ΓinΓoutΓMS(ΓS)ΓML(ΓL)
Amplifier with input and output matching networks, showing the reference planes for Γ_MS and Γ_ML and how S12 feedback couples the input and output matches. Concept drawing.

The figure shows the loop. Suppose you set Γ_S = S11* to match the input, then design the output for Γ_L = Γ_out*. The new Γ_L changes Γ_in, so the input is no longer matched. Fix the input again and Γ_out moves. Iterating by hand may converge, but it is slow and gives no insight. If S12 = 0, the feedback terms vanish, Γ_in = S11 and Γ_out = S22, and each end can be matched alone. That is the unilateral case. For a real transistor, S12 is small but rarely negligible at high gain.

The bilateral match needs both conditions to hold at the same time:

  • Γ_S = Γ_in*
  • Γ_L = Γ_out*

Substituting the Γ_in and Γ_out expressions gives two coupled equations in Γ_S and Γ_L. They can be solved in closed form.

The B and C terms

Start with the determinant of the S-matrix:

Δ=S11S22−S12S21\Delta = S_{11}S_{22} - S_{12}S_{21}

Then define two real numbers and two complex numbers:

B1=1+∣S11∣2−∣S22∣2−∣Δ∣2B_1 = 1 + |S_{11}|^2 - |S_{22}|^2 - |\Delta|^2
B2=1+∣S22∣2−∣S11∣2−∣Δ∣2B_2 = 1 + |S_{22}|^2 - |S_{11}|^2 - |\Delta|^2
C1=S11−ΔS22∗C_1 = S_{11} - \Delta S_{22}^{*}
C2=S22−ΔS11∗C_2 = S_{22} - \Delta S_{11}^{*}

B1 and B2 are always real. C1 and C2 are complex, and their phases set the phases of the matching reflection coefficients. B1 and C1 belong to the input side, B2 and C2 to the output side. Each one still contains all four S-parameters through Δ. That is how the coupling between the ends enters the solution.

Solving for Γ_MS and Γ_ML

Solving the two conjugate conditions together gives a quadratic for each end. The roots are:

ΓMS=B1±B12−4∣C1∣22C1\Gamma_{MS} = \frac{B_1 \pm \sqrt{B_1^2 - 4|C_1|^2}}{2C_1}
ΓML=B2±B22−4∣C2∣22C2\Gamma_{ML} = \frac{B_2 \pm \sqrt{B_2^2 - 4|C_2|^2}}{2C_2}

Γ_MS is the reflection coefficient the input network must present, looking toward the source. Γ_ML is what the output network must present, looking toward the load. Each quadratic has two roots whose magnitudes multiply to one. One root is inside the Smith chart and one is outside. When B1 is positive, the minus sign gives the root with |Γ_MS| < 1. The same rule applies to B2 and Γ_ML.

A passive network can only present |Γ| < 1. So the question is when both roots inside the chart exist. The answer is unconditional stability: K > 1 and |Δ| < 1, where

K=1−∣S11∣2−∣S22∣2+∣Δ∣22∣S12S21∣K = \frac{1 - |S_{11}|^2 - |S_{22}|^2 + |\Delta|^2}{2|S_{12}S_{21}|}

Under those conditions B1² − 4|C1|² is positive, the square root is real, and both |Γ_MS| and |Γ_ML| are below one. At the solution, Γ_in = Γ_MS* and Γ_out = Γ_ML*. Both ports are conjugate matched together.

If K is less than 1, do not force the formulas. The square root goes imaginary or the result lands outside the unit circle. Stabilise the device first, with resistive loading or feedback, then recompute the S-parameters and K. See the Rollett K factor page.

With both ends conjugate matched, the transducer gain reaches its maximum. For K > 1 it can be written directly from the S-parameters:

Gmax=∣S21∣∣S12∣(K−K2−1)G_{max} = \frac{|S_{21}|}{|S_{12}|}\left(K - \sqrt{K^2 - 1}\right)

The factor (K − sqrt(K² − 1)) is always at most one, and it shrinks as K grows. A device with a large stability margin gives up gain. At K = 1 the factor is exactly one, and Gmax reduces to |S21|/|S12|. This is the maximum stable gain (MSG). It marks the boundary of the unconditionally stable region. It is not a gain you can reach with a conjugate match at K < 1.

Worked example. Take |S21|/|S12| = 10 and K = 1.5.

  • sqrt(K² − 1) = sqrt(2.25 − 1) = sqrt(1.25) = 1.118
  • K − sqrt(K² − 1) = 1.5 − 1.118 = 0.382
  • Gmax = 10 × 0.382 = 3.82 linear
  • 10 log10(3.82) = 5.82 dB
  • MSG = 10 log10(10) = 10 dB
11.522.530246810Stability factor KGmax (dB)K = 1: MSG = 10 dBunconditional-stability edgeK = 1.5: Gmax = 5.82 dB(3.82 linear)Gmax, |S21|/|S12| = 10
Maximum transducer gain Gmax versus stability factor K for |S21|/|S12| = 10, falling from the 10 dB MSG at K = 1 to 5.82 dB at K = 1.5. Computed from the formula.

The plot shows Gmax in dB against K for |S21|/|S12| = 10. The curve starts at 10 dB (the MSG) at K = 1 and falls as K rises. At K = 1.5 it sits at 5.82 dB. The gap of about 4.2 dB is the gain price of that stability margin.

Design procedure

  1. Measure or simulate the S-parameters at the design frequency, with the bias you will use.
  2. Compute Δ and K. Confirm K > 1 and |Δ| < 1. If not, stabilise first.
  3. Compute B1, B2, C1 and C2.
  4. Solve for Γ_MS and Γ_ML, using the minus sign when B1 and B2 are positive. Check that both magnitudes are below one.
  5. Compute Gmax from K as a check on the final gain.
  6. Design the input network to turn Z0 into Γ_MS, and the output network to turn Z0 into Γ_ML. L, Pi or T sections all work; see RF Matching Networks: L, Pi and T Compared. Plot both points on a Smith chart to pick the topology.

The match is exact at one frequency. Away from it, both Γ_MS and Γ_ML move, and so does the gain.

In plain words · for students and beginners

Simultaneous conjugate match in plain words

An everyday picture

Think of an amplifier as a box between two tuning knobs. The input knob sets what the signal source sees. The output knob sets what the load sees. You want both ends "matched": no signal bounces back, and the most power gets through. (Matched means the impedances fit each other, so nothing reflects.)

With an ideal amplifier, signal only flows forward, so each knob only affects its own end. Real transistors leak a little signal backward. Engineers call that leak S12. Because of it, turning the output knob changes what the input looks like. Turning the input knob changes what the output looks like.

turn one and theother end changesinput knob(what the source sees)output knob(what the load sees)amplifierS12 = backward leakTune both together,not one at a time
Because of the backward leak S12, the input and output settings of an amplifier depend on each other, so both matches have to be tuned together. Concept drawing.

So you have to tune both together, not one at a time. The B and C formulas work out both knob settings in one go.

Where the analogy breaks down: real knobs turn on a scale. Here each "setting" is a complex number, a reflection coefficient with a size and an angle. It is a point on the Smith chart, not one number on a dial.

The same example, step by step

Say the forward gain is 10 times the backward leak, so |S21|/|S12| = 10. Say the stability number K is 1.5. (K above 1 means the amplifier will not oscillate, whatever passive source and load you connect.)

  1. Square K: 1.5 × 1.5 = 2.25.
  2. Subtract 1: 1.25.
  3. Take the square root: 1.118.
  4. Subtract that from K: 1.5 − 1.118 = 0.382.
  5. Multiply by 10: 3.82. The amplifier multiplies power by 3.82.
  6. In decibels, that is 5.82 dB.

If K were exactly 1, step 4 would give 1, and the gain would be 10, or 10 dB. That top value is called the maximum stable gain. More stability margin costs gain.

Mistakes beginners make

  • Matching one end, then the other, and stopping. The second match spoils the first. You need the joint solution.
  • Using the wrong sign. The formula has a plus-or-minus. When B1 or B2 is positive, take the minus sign. The other root gives a reflection bigger than one, which a passive circuit cannot make.
  • Skipping the stability check. If K is below 1, the formulas give nonsense or a design that oscillates.
  • Treating 10 dB (MSG) as the gain you will get. At K = 1.5 you get 5.82 dB, not 10 dB.

Remember

  • The backward leak S12 ties the two ends together, so match them at the same time.
  • A clean solution needs K > 1 and |Δ| < 1.
  • Gmax = |S21|/|S12| × (K − sqrt(K² − 1)), and it falls as K rises.

Frequently asked questions

Why can't I conjugate match the input and output of a transistor separately?

The reverse transmission S12 makes the input reflection coefficient depend on the load and the output reflection coefficient depend on the source. If you change one matching network, the other end moves. Only a unilateral device with S12 = 0 can be matched one end at a time.

When does a simultaneous conjugate match exist?

A match with passive terminations, |Γ_MS| < 1 and |Γ_ML| < 1, exists when the device is unconditionally stable, which means K > 1 and |Δ| < 1. If the device is only conditionally stable, the solution falls outside the Smith chart or makes the amplifier oscillate.

Which sign do I use in the Γ_MS and Γ_ML formulas?

Use the minus sign in front of the square root when B1 (or B2) is positive. That root gives the solution with magnitude less than one, which a passive matching network can realise.

What is the difference between Gmax and MSG?

Gmax is the maximum transducer gain with a simultaneous conjugate match, valid for K > 1. At K = 1 the Gmax expression reduces to |S21|/|S12|, the maximum stable gain, which marks the edge of the unconditionally stable region.

References and further reading

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