5.2 RF Power Amplifiers, Push-Pull Topologies, and Output Matching

ACMA Amateur Radio Advanced Recognition Certificate - Theory Exam

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5.1 Classes of Amplification, Conduction Angles, and Operating Efficiency 5.2 RF Power Amplifiers, Push-Pull Topologies, and Output Matching 5.3 Amplifier Stability, Neutralisation, and Intermodulation Distortion

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5.2 RF Power Amplifiers, Push-Pull Topologies, and Output Matching

Key Takeaways Explain This Topic AI

A balanced push-pull amplifier drives matched devices 180 ∘ 180^\circ 18 0 ∘ apart. Ideally, even harmonics cancel and odd harmonics combine; practical rejection depends on balance and must be checked with the output filtering.

For ideal sinusoidal push-pull operation, the full-winding differential voltage peak is 2 ( V C C − V sat ) 2(V_{CC}-V_{\text{sat}}) 2 ( V CC ​ − V sat ​ ) and its peak-to-peak value is 4 ( V C C − V sat ) 4(V_{CC}-V_{\text{sat}}) 4 ( V CC ​ − V sat ​ ) . Total output power determines the differential load resistance.

Low-voltage solid-state power amplifiers exhibit extremely low collector/drain load impedances ( R L ′ ≈ 1 Ω R_L' \approx 1\ \Omega R L ′ ​ ≈ 1 Ω to 5 Ω 5\ \Omega 5 Ω ), requiring broadband ferrite transmission line transformers (TLTs) with discrete impedance transformation ratios ( 4 : 1 4:1 4 : 1 , 9 : 1 9:1 9 : 1 , 16 : 1 16:1 16 : 1 ) to match the standard 50 Ω 50\ \Omega 50 Ω feedline.

In a grounded-grid HF triode amplifier, the grounded grid shields the cathode from the plate, reducing feedback. Neutralisation is normally unnecessary with suitable construction, while parasitic suppression and sound RF grounding remain necessary.

Pi and Pi-L networks match high valve plate impedances to 50 Ω 50\ \Omega 50 Ω and attenuate harmonics. State the design-Q convention, calculate the complete network and verify filtering and component ratings; Q alone does not guarantee regulatory compliance.

Last updated: November 2026

5.2 RF Power Amplifiers, Push-Pull Topologies, and Output Matching

Designing a high-power RF amplifier requires coupling the active amplifying devices to transmission lines and antenna systems while maintaining absolute spectral purity and electrical stability. Because high-power solid-state transistors operate at very low impedances and high currents, whereas thermionic valves (vacuum tubes) operate at high voltages and low currents, each technology relies on specialized circuit topologies and impedance-matching networks.

1. Solid-State Power Amplifier (SSPA) Architectures and Push-Pull Operation

In modern amateur transceivers and solid-state power amplifiers, single-ended output stages are rarely used for power levels exceeding a few watts. Instead, the push-pull configuration is the dominant circuit architecture.

Circuit Configuration and Phase Relationships

A push-pull amplifier employs two identical, matched active devices (such as bipolar junction transistors, LDMOS, or VDMOS FETs) operated in phase opposition:

Input Phase-Splitting Transformer ( T 1 T_1 T 1 ​ ): A broadband input transformer (balun) converts the single-ended 50 Ω 50\ \Omega 50 Ω drive signal from the exciter into two equal-amplitude signals that are exactly 180 ∘ 180^\circ 18 0 ∘ out of phase with respect to each other.

Active Devices ( Q 1 , Q 2 Q_1, Q_2 Q 1 ​ , Q 2 ​ ): Device 1 receives + v in ( t ) +v_{\text{in}}(t) + v in ​ ( t ) and Device 2 receives − v in ( t ) -v_{\text{in}}(t) − v in ​ ( t ) . In Class B each conducts for one half-cycle. Class AB has overlapping conduction around the crossover; in Class A both conduct throughout the cycle.

Output Combining Transformer ( T 2 T_2 T 2 ​ ): The collectors or drains feed opposite ends of a center-tapped output transformer primary winding. DC supply voltage ( V C C V_{CC} V CC ​ or V D D V_{DD} V DD ​ ) is injected at the center tap. The transformer recombines the two antiphase output currents into a single-ended output port.

Mathematical Proof of Even-Harmonic Cancellation

The primary engineering advantage of a balanced push-pull amplifier is its inherent cancellation of all even-order harmonics ( 2 f 0 , 4 f 0 , 6 f 0 , … 2f_0, 4f_0, 6f_0, \dots 2 f 0 ​ , 4 f 0 ​ , 6 f 0 ​ , … ).

The output currents of the two active devices can be expressed as Fourier/Taylor series expansions of the input drive voltage v ( t ) = V m sin ⁡ ( ω t ) v(t) = V_m \sin(\omega t) v ( t ) = V m ​ sin ( ω t ) : i 1 ( t ) = I 0 + I 1 sin ⁡ ( ω t ) + I 2 sin ⁡ ( 2 ω t ) + I 3 sin ⁡ ( 3 ω t ) + I 4 sin ⁡ ( 4 ω t ) + … i_1(t) = I_0 + I_1 \sin(\omega t) + I_2 \sin(2\omega t) + I_3 \sin(3\omega t) + I_4 \sin(4\omega t) + \dots i 1 ​ ( t ) = I 0 ​ + I 1 ​ sin ( ω t ) + I 2 ​ sin ( 2 ω t ) + I 3 ​ sin ( 3 ω t ) + I 4 ​ sin ( 4 ω t ) + …

Because Device 2 is driven 180 ∘ 180^\circ 18 0 ∘ ( π radians \pi\text{ radians} π radians ) out of phase, its input is V m sin ⁡ ( ω t + π ) = − V m sin ⁡ ( ω t ) V_m \sin(\omega t + \pi) = -V_m \sin(\omega t) V m ​ sin ( ω t + π ) = − V m ​ sin ( ω t ) . Substituting this phase shift into each harmonic term:

Fundamental: sin ⁡ ( ω t + π ) = − sin ⁡ ( ω t ) \sin(\omega t + \pi) = -\sin(\omega t) sin ( ω t + π ) = − sin ( ω t )

Second harmonic: sin ⁡ ( 2 ( ω t + π ) ) = sin ⁡ ( 2 ω t + 2 π ) = + sin ⁡ ( 2 ω t ) \sin(2(\omega t + \pi)) = \sin(2\omega t + 2\pi) = +\sin(2\omega t) sin ( 2 ( ω t + π )) = sin ( 2 ω t + 2 π ) = + sin ( 2 ω t )

Third harmonic: sin ⁡ ( 3 ( ω t + π ) ) = sin ⁡ ( 3 ω t + 3 π ) = − sin ⁡ ( 3 ω t ) \sin(3(\omega t + \pi)) = \sin(3\omega t + 3\pi) = -\sin(3\omega t) sin ( 3 ( ω t + π )) = sin ( 3 ω t + 3 π ) = − sin ( 3 ω t )

Fourth harmonic: sin ⁡ ( 4 ( ω t + π ) ) = sin ⁡ ( 4 ω t + 4 π ) = + sin ⁡ ( 4 ω t ) \sin(4(\omega t + \pi)) = \sin(4\omega t + 4\pi) = +\sin(4\omega t) sin ( 4 ( ω t + π )) = sin ( 4 ω t + 4 π ) = + sin ( 4 ω t )

Thus, the current for Device 2 is: i 2 ( t ) = I 0 − I 1 sin ⁡ ( ω t ) + I 2 sin ⁡ ( 2 ω t ) − I 3 sin ⁡ ( 3 ω t ) + I 4 sin ⁡ ( 4 ω t ) − … i_2(t) = I_0 - I_1 \sin(\omega t) + I_2 \sin(2\omega t) - I_3 \sin(3\omega t) + I_4 \sin(4\omega t) - \dots i 2 ​ ( t ) = I 0 ​ − I 1 ​ sin ( ω t ) + I 2 ​ sin ( 2 ω t ) − I 3 ​ sin ( 3 ω t ) + I 4 ​ sin ( 4 ω t ) − …

In the center-tapped output transformer, the two currents flow in opposite directions through the primary winding halves toward the center tap. The net magnetic flux in the transformer core is proportional to the difference between the two currents ( i net = i 1 − i 2 i_{\text{net}} = i_1 - i_2 i net ​ = i 1 ​ − i 2 ​ ): i net ( t ) = i 1 ( t ) − i 2 ( t ) = 2 I 1 sin ⁡ ( ω t ) + 2 I 3 sin ⁡ ( 3 ω t ) + 2 I 5 sin ⁡ ( 5 ω t ) + … i_{\text{net}}(t) = i_1(t) - i_2(t) = 2 I_1 \sin(\omega t) + 2 I_3 \sin(3\omega t) + 2 I_5 \sin(5\omega t) + \dots i net ​ ( t ) = i 1 ​ ( t ) − i 2 ​ ( t ) = 2 I 1 ​ sin ( ω t ) + 2 I 3 ​ sin ( 3 ω t ) + 2 I 5 ​ sin ( 5 ω t ) + …

Notice that:

Even harmonics ( I 2 , I 4 , … I_2, I_4, \dots I 2 ​ , I 4 ​ , … ) cancel out completely! Because they are in phase in both halves of the primary winding, their opposing magnetic fluxes cancel in the ferrite core.

Odd harmonics ( I 1 , I 3 , I 5 , … I_1, I_3, I_5, \dots I 1 ​ , I 3 ​ , I 5 ​ , … ) add constructively in phase , doubling the fundamental output current.

DC quiescent currents ( I 0 I_0 I 0 ​ ) flow in opposite directions through the primary , preventing DC magnetic saturation of the transformer core.

Ideal balance cancels even harmonics at the output. Real device, drive and transformer imbalances leave residual even-harmonic output, with rejection varying across frequency and operating level. Push-pull balance reduces the filtering burden, but output filtering and measurement are still required.

Virtual RF Ground and Power Supply Decoupling

With equal antiphase signals, the output winding's centre tap is a virtual RF ground at the fundamental . The two fundamental current components cancel in the supply lead. In-phase components, including even-harmonic currents, add there instead; imbalance also leaves residual RF. Supply bypassing and RF layout therefore remain necessary. A virtual ground neither fixes capacitor current ratings nor prevents supply ripple from modulating the output.

With an ideal transformer and sinusoidal collector/drain voltages, let V e = V C C − V sat V_e=V_{CC}-V_{\text{sat}} V e ​ = V CC ​ − V sat ​ . The full-winding differential voltage has peak 2 V e 2V_e 2 V e ​ and peak-to-peak value 4 V e 4V_e 4 V e ​ . Two devices provide approximately twice one device's power at the same individual voltage and current limits; doubling voltage alone does not establish four times the available power.

2. Solid-State Impedance Transformation and Transmission Line Transformers

Solid-state power transistors operate from low DC rail voltages (typically 13.8 V 13.8\text{ V} 13.8 V in mobile equipment or 28 V 28\text{ V} 28 V to 50 V 50\text{ V} 50 V in base station amplifiers). Because power equals V 2 / 2 R V^2 / 2R V 2 /2 R , generating high RF power from a low supply voltage requires an extremely low load resistance.

Differential Voltage and Load Resistance

Let V e = V C C − V sat V_e=V_{CC}-V_{\text{sat}} V e ​ = V CC ​ − V sat ​ be the sinusoidal RF voltage peak at either collector/drain relative to the centre tap. A single-ended stage delivering power P SE P_{\text{SE}} P SE ​ requires: R SE = V e 2 2 P SE R_{\text{SE}}=\frac{V_e^2}{2P_{\text{SE}}} R SE ​ = 2 P SE ​ V e 2 ​ ​

For a balanced push-pull stage, the differential RF peak is 2 V e 2V_e 2 V e ​ . Using total output power P out P_{\text{out}} P out ​ : R L ( c − c ) = ( 2 V e ) 2 2 P out = 2 V e 2 P out R_{L(c-c)}=\frac{(2V_e)^2}{2P_{\text{out}}}=\frac{2V_e^2}{P_{\text{out}}} R L ( c − c ) ​ = 2 P out ​ ( 2 V e ​ ) 2 ​ = P out ​ 2 V e 2 ​ ​

At the fundamental frequency, each device sees R L ( c − c ) / 2 R_{L(c-c)}/2 R L ( c − c ) ​ /2 . In Class B, the instantaneous load line of the conducting half-winding instead uses R L ( c − c ) / 4 R_{L(c-c)}/4 R L ( c − c ) ​ /4 , because its turns count is half the full winding. These are different current definitions; do not interchange them.

Numerical Demonstration of Low Impedance

Consider a standard 100 W 100\text{ W} 100 W amateur transceiver operating from a 13.8 V 13.8\text{ V} 13.8 V DC supply ( V sat ≈ 1.3 V V_{\text{sat}} \approx 1.3\text{ V} V sat ​ ≈ 1.3 V , V effective = 12.5 V V_{\text{effective}} = 12.5\text{ V} V effective ​ = 12.5 V ): R L ( c − c ) = 2 × ( 12.5 V ) 2 100 W = 2 × 156.25 100 = 3.125 Ω R_{L(c-c)} = \frac{2 \times (12.5\text{ V})^2}{100\text{ W}} = \frac{2 \times 156.25}{100} = 3.125\ \Omega R L ( c − c ) ​ = 100 W 2 × ( 12.5 V ) 2 ​ = 100 2 × 156.25 ​ = 3.125 Ω

Each transistor sees a fundamental-frequency load resistance of 3.125 Ω / 2 = 1.56 Ω 3.125\ \Omega/2=1.56\ \Omega 3.125 Ω/2 = 1.56 Ω . Transforming this extremely low impedance of 3.125 Ω 3.125\ \Omega 3.125 Ω up to the standard 50.0 Ω 50.0\ \Omega 50.0 Ω antenna load represents an impedance ratio of: Ratio = 50.0 Ω 3.125 Ω = 16 : 1 \text{Ratio} = \frac{50.0\ \Omega}{3.125\ \Omega} = 16:1 Ratio = 3.125 Ω 50.0 Ω ​ = 16 : 1

Broadband Transmission Line Transformers (TLTs)

Conventional lumped-element L C LC L C matching networks cannot provide low-loss matching across the entire 1.8 MHz 1.8\text{ MHz} 1.8 MHz to 30 MHz 30\text{ MHz} 30 MHz HF spectrum without continuous retuning. Instead, solid-state amplifiers utilize Transmission Line Transformers (TLTs) , pioneered by C.L. Guanella and Jerry Sevick.

TLTs consist of short lengths of low-characteristic-impedance transmission lines (such as miniature coaxial cables or twisted Teflon-insulated wire pairs) threaded through high-permeability binocular ferrite cores (e.g., Fair-Rite Type 43 or 61, both nickel-zinc ferrites):

Ferrite raises the impedance to unwanted common-mode currents, extending low-frequency operation.

Wanted differential-mode power travels along the transmission line rather than by core flux coupling. Core, conductor and dielectric losses still matter; line length, characteristic impedance and interconnection phase error limit high-frequency performance.

Combining equal transmission lines in suitable series/parallel arrangements gives common squared-integer impedance ratios. Here N N N is the ideal voltage ratio; it need not represent turns around a core: Z out Z in = N 2 = 4 : 1 , 9 : 1 , 16 : 1 , 25 : 1 \frac{Z_{\text{out}}}{Z_{\text{in}}} = N^2 = 4:1, \quad 9:1, \quad 16:1, \quad 25:1 Z in ​ Z out ​ ​ = N 2 = 4 : 1 , 9 : 1 , 16 : 1 , 25 : 1

In our 100 W 100\text{ W} 100 W example, an ideal 16 : 1 16:1 16 : 1 impedance transformer has voltage ratio 4 : 1 4:1 4 : 1 and transforms 3.125 Ω 3.125\ \Omega 3.125 Ω collector-to-collector to 50.0 Ω 50.0\ \Omega 50.0 Ω . A practical design must verify loss, balance and matching over its intended bands.

Tip

The 50 V LDMOS Revolution: Modern solid-state amplifiers use laterally diffused MOSFETs (LDMOS) operating from V D D = 50 V V_{DD} = 50\text{ V} V DD ​ = 50 V . For a 1000 W 1000\text{ W} 1000 W amplifier ( V effective ≈ 46 V V_{\text{effective}} \approx 46\text{ V} V effective ​ ≈ 46 V ), the drain-to-drain load resistance is: R L ( d − d ) = 2 ( 46 V ) 2 1000 W = 4.23 Ω R_{L(d-d)} = \frac{2(46\text{ V})^2}{1000\text{ W}} = 4.23\ \Omega R L ( d − d ) ​ = 1000 W 2 ( 46 V ) 2 ​ = 4.23 Ω A 9 : 1 9:1 9 : 1 transformer transforms this to 38.1 Ω 38.1\ \Omega 38.1 Ω , which is easily matched to 50 Ω 50\ \Omega 50 Ω with a broad, low- Q Q Q network, drawing significantly less current than a 13.8 V 13.8\text{ V} 13.8 V circuit and drastically reducing I 2 R I^2 R I 2 R conductor losses.

3. Thermionic Valve (Tube) Grounded-Grid Linear Power Amplifiers

High-power amateur linear amplifiers delivering 500 W 500\text{ W} 500 W to 1500 W 1500\text{ W} 1500 W PEP frequently utilize thermionic valves (such as the 3-500Z, 8877/3CX1500A7, or 4CX250B). In these amplifiers, the grounded-grid (cathode-driven) topology is the preeminent design.

Grounded-Grid Circuit Architecture

Control Grid: Bolted directly to the chassis ground (or bypassed to RF ground via low-inductance mica or ceramic disc capacitors).

Cathode / Filament: Heated by an AC filament transformer via a heavy bifilar RF cathode choke . The bifilar winding allows the low-voltage, high-current AC heater power to reach the filament while presenting a high RF impedance that prevents the RF input drive from being short-circuited to ground through the filament transformer.

RF Drive Injection: The exciter drives the cathode directly through an input matching network and DC blocking capacitor.

Plate (Anode): Operated at high positive DC potential ( 1500 V 1500\text{ V} 1500 V to 3500 V 3500\text{ V} 3500 V ) through an RF plate choke, with output taken via a DC blocking capacitor into a tuned resonant tank.

Grounded-Grid Isolation and Stability

In a grounded-cathode amplifier, anode-to-grid capacitance couples output RF back to the input. It contributes to the Miller effect and can support oscillation when the circuit's feedback phase and loop gain permit. Neutralisation can cancel unwanted feedback; feedback capacitance does not make every grounded-cathode circuit oscillate.

In a grounded-grid amplifier, the grid's electrostatic shielding greatly reduces plate-to-cathode feedback. Properly constructed HF grounded-grid triode amplifiers normally need no neutralisation. Low-inductance grid grounding and suitable layout remain necessary, and parasitic oscillation must still be controlled.

Low Input Impedance and Feed-Through Power

Input Impedance: Looking into the cathode of a grounded-grid stage, the dynamic input impedance is low: Z in ≈ 1 g m + g k ≈ 1 g m Z_{\text{in}} \approx \frac{1}{g_m + g_k} \approx \frac{1}{g_m} Z in ​ ≈ g m ​ + g k ​ 1 ​ ≈ g m ​ 1 ​

For power triodes with transconductance g m g_m g m ​ between 10 mS 10\text{ mS} 10 mS and 20 mS 20\text{ mS} 20 mS , Z in Z_{\text{in}} Z in ​ ranges from 50 Ω 50\ \Omega 50 Ω to 100 Ω 100\ \Omega 100 Ω . This natural low impedance is an ideal match for standard 50 Ω 50\ \Omega 50 Ω solid-state transceiver exciters.

Feed-Through Power: In a grounded-grid circuit, the cathode-to-anode electron stream is in series with the input driving source. The RF drive voltage and plate voltage appear in series across the load. Therefore, part of the RF drive power feeds through and adds to the plate output power . Grid and input-circuit losses consume the rest. Denote the delivered part by P feed P_{\text{feed}} P feed ​ : P total(output) = P plate + P feed P_{\text{total(output)}} = P_{\text{plate}} + P_{\text{feed}} P total(output) ​ = P plate ​ + P feed ​

A typical grounded-grid stage exhibits a power gain of 10 dB 10\text{ dB} 10 dB to 13 dB 13\text{ dB} 13 dB (a power multiplication factor of 10 to 20), requiring approximately 60 W 60\text{ W} 60 W to 100 W 100\text{ W} 100 W of drive from a standard transceiver to achieve 1000 W 1000\text{ W} 1000 W to 1500 W 1500\text{ W} 1500 W PEP output.

4. Output Matching and Resonant Tank Circuits: Pi and Pi-L Networks

High-voltage valves operate at high plate voltages ( V a = 1500 V V_a = 1500\text{ V} V a ​ = 1500 V to 3500 V 3500\text{ V} 3500 V ) and moderate plate currents ( I a = 0.4 A I_a = 0.4\text{ A} I a ​ = 0.4 A to 1.0 A 1.0\text{ A} 1.0 A ).

The Valve Plate Load Resistance

The equivalent RF plate load resistance ( R a R_a R a ​ ) required by the tube for optimal power delivery in Class AB linear service is: R a ≈ V a K ⋅ I a R_a \approx \frac{V_a}{K \cdot I_a} R a ​ ≈ K ⋅ I a ​ V a ​ ​

where K K K is an empirical constant governed by the operating class ( K ≈ 1.8 K \approx 1.8 K ≈ 1.8 to 2.0 2.0 2.0 for Class AB linear operation). For example, a valve operating at V a = 2500 V V_a = 2500\text{ V} V a ​ = 2500 V and I a = 0.50 A I_a = 0.50\text{ A} I a ​ = 0.50 A requires: R a = 2500 V 1.8 × 0.50 A ≈ 2778 Ω R_a = \frac{2500\text{ V}}{1.8 \times 0.50\text{ A}} \approx 2778\ \Omega R a ​ = 1.8 × 0.50 A 2500 V ​ ≈ 2778 Ω

The amplifier must transform this high resistance ( 2778 Ω 2778\ \Omega 2778 Ω ) down to the standard 50.0 Ω 50.0\ \Omega 50.0 Ω antenna load while filtering out harmonic emissions.

The Pi ( π \pi π ) Network

The Pi network is the classical output matching network for valve amplifiers, consisting of three reactive components arranged in a " π \pi π " configuration:

C 1 C_1 C 1 ​ (Tune Capacitor): Shunt capacitance across the high-impedance valve plate.

L L L (Tank Inductor): Series inductance connecting plate to load.

C 2 C_2 C 2 ​ (Load Capacitor): Shunt capacitance across the low-impedance 50 Ω 50\ \Omega 50 Ω output.

The Pi network acts as two back-to-back L L L -networks matching through a virtual intermediate resistance ( R virt R_{\text{virt}} R virt ​ ). It simultaneously acts as a third-order low-pass filter , providing harmonic suppression.

Choosing the Pi-Network Design Q

One traditional design convention uses the input-section Q: Q I = R a ∣ X C 1 ∣ = R a 2 π f C 1 Q_I=\frac{R_a}{|X_{C1}|}=R_a\,2\pi fC_1 Q I ​ = ∣ X C 1 ​ ∣ R a ​ ​ = R a ​ 2 π f C 1 ​

This input design parameter is not generally equal to the complete network's f 0 / bandwidth f_0/\text{bandwidth} f 0 ​ / bandwidth . For the low-pass Pi circuit with R a > R L R_a>R_L R a ​ > R L ​ , choose: R v = R a 1 + Q I 2 < R L , Q O = R L R v − 1 R_v=\frac{R_a}{1+Q_I^2}<R_L,\qquad Q_O=\sqrt{\frac{R_L}{R_v}-1} R v ​ = 1 + Q I 2 ​ R a ​ ​ < R L ​ , Q O ​ = R v ​ R L ​ ​ − 1 ​

Thus a nonzero output capacitance requires Q I > R a / R L − 1 Q_I>\sqrt{R_a/R_L-1} Q I ​ > R a ​ / R L ​ − 1 ​ , rather than a universal minimum of 8. The reactance magnitudes are: ∣ X C 1 ∣ = R a Q I , ∣ X C 2 ∣ = R L Q O , X L = R v ( Q I + Q O ) |X_{C1}|=\frac{R_a}{Q_I},\quad |X_{C2}|=\frac{R_L}{Q_O},\quad X_L=R_v(Q_I+Q_O) ∣ X C 1 ​ ∣ = Q I ​ R a ​ ​ , ∣ X C 2 ​ ∣ = Q O ​ R L ​ ​ , X L ​ = R v ​ ( Q I ​ + Q O ​ )

Increasing design Q generally improves harmonic filtering but increases reactive currents, loss and tuning sensitivity. With an ideal resistive output, I C 2 = Q O I load I_{C2}=Q_OI_{\text{load}} I C 2 ​ = Q O ​ I load ​ and I L = I load 1 + Q O 2 I_L=I_{\text{load}}\sqrt{1+Q_O^2} I L ​ = I load ​ 1 + Q O 2 ​ ​ ; the coil current is not generally Q I I load Q_II_{\text{load}} Q I ​ I load ​ .

For coil series resistance R s R_s R s ​ , its loss is I L , RMS 2 R s I_{L,\text{RMS}}^2R_s I L , RMS 2 ​ R s ​ and unloaded coil Q is Q u = X L / R s Q_u=X_L/R_s Q u ​ = X L ​ / R s ​ . Estimate network efficiency from P out / ( P out + P loss ) P_{\text{out}}/(P_{\text{out}}+P_{\text{loss}}) P out ​ / ( P out ​ + P loss ​ ) , including capacitor and connection losses as relevant. Select Q for the required impedance ratio, filtering, bandwidth and component ratings; 10–12 is a possible design choice, not a universal optimum.

The Pi-L ( π - L \pi\text{-}L π - L ) Network

For transmissions below 30 MHz, section 15 of the Australian amateur class licence requires spurious emissions to be attenuated below power within the necessary bandwidth by the lesser of 43 + 10 log ⁡ 10 ( p X ) 43+10\log_{10}(pX) 43 + 10 lo g 10 ​ ( pX ) dB and 50 dB, with p X pX pX in watts PEP. At or above 30 MHz, use the separate frequency-dependent limit. Required attenuation is not a universal 50 dB at every power and frequency. Verify the complete transmitter's emissions at the antenna transmission line.

The standard Pi-L network uses four reactive components: shunt C 1 C_1 C 1 ​ , series L 1 L_1 L 1 ​ , shunt C 2 C_2 C 2 ​ and series L 2 L_2 L 2 ​ before the load. It can be analysed as a Pi section followed by an L section, with the shared shunt capacitance divided between the two sections:

The Pi section transforms the high plate impedance ( R a R_a R a ​ ) down to an intermediate virtual impedance (typically 200 Ω 200\ \Omega 200 Ω to 300 Ω 300\ \Omega 300 Ω ).

The L L L -section transforms the intermediate impedance down to the final 50 Ω 50\ \Omega 50 Ω load.

Harmonic Rejection Advantage: The additional low-pass section can increase harmonic attenuation. Its actual response depends on component values and terminations; the emitted harmonic level also depends on the amplifier. Measure the final output and add filtering where required.

5. Architectural Comparison: Solid-State Push-Pull vs. Thermionic Grounded-Grid

Feature Solid-State Push-Pull (LDMOS/BJT) Thermionic Grounded-Grid Triode

Operating Voltage ( V D C V_{DC} V D C ​ ) Low ( 13.8 V − 50 V 13.8\text{ V} - 50\text{ V} 13.8 V − 50 V ) Very High ( 1500 V − 3500 V 1500\text{ V} - 3500\text{ V} 1500 V − 3500 V )

Operating Current ( I D C I_{DC} I D C ​ ) High to very high ( 20 A − 45 A 20\text{ A} - 45\text{ A} 20 A − 45 A ) Low to moderate ( 0.4 A − 1.0 A 0.4\text{ A} - 1.0\text{ A} 0.4 A − 1.0 A )

Device Load Impedance ( R L R_L R L ​ ) Extremely low ( 1.5 Ω − 5 Ω 1.5\ \Omega - 5\ \Omega 1.5 Ω − 5 Ω ) High ( 1500 Ω − 3500 Ω 1500\ \Omega - 3500\ \Omega 1500 Ω − 3500 Ω )

Impedance Matching Method Broadband Ferrite Transmission Line Transformers Resonant Tuned Pi or Pi-L Tank Networks

Tuning Requirements No tuning required (instant broadband coverage) Requires manual or motorized "Tune" & "Load" adjustment per band

Second Harmonic Suppression Even-harmonic cancellation depends on balance; output LPF still required Depends on the tuned output network and measured harmonic levels

Stability Measures Proper layout, bias decoupling and input damping as required Grid shielding reduces feedback; low-inductance grounding and parasitic suppression remain necessary

SWR Sensitivity Mismatch can exceed device voltage/current limits; use protection and reduce drive Mismatch can overheat valves or cause arcing; reduce drive and correct the load

Warm-up Time Instantaneous (zero warm-up) Requires cathode warm-up ( 3 seconds 3\text{ seconds} 3 seconds to 3 minutes 3\text{ minutes} 3 minutes )

• 6. Worked Engineering Examples
• Worked Example 5.2.1: Solid-State Push-Pull Impedance and Transformer Design

An amateur designs a solid-state HF push-pull power amplifier using a pair of 50.0 V 50.0\text{ V} 50.0 V LDMOS transistors to deliver P out = 600 W P_{\text{out}} = 600\text{ W} P out ​ = 600 W PEP. The saturation voltage of the transistors is V sat = 3.0 V V_{\text{sat}} = 3.0\text{ V} V sat ​ = 3.0 V .

Calculate the required collector-to-collector load resistance ( R L ( d − d ) R_{L(d-d)} R L ( d − d ) ​ ): V effective = V D D − V sat = 50.0 V − 3.0 V = 47.0 V V_{\text{effective}} = V_{DD} - V_{\text{sat}} = 50.0\text{ V} - 3.0\text{ V} = 47.0\text{ V} V effective ​ = V DD ​ − V sat ​ = 50.0 V − 3.0 V = 47.0 V R L ( d − d ) = 2 ⋅ V effective 2 P out = 2 × ( 47.0 V ) 2 600 W = 2 × 2209 600 = 4418 600 ≈ 7.36 Ω R_{L(d-d)} = \frac{2 \cdot V_{\text{effective}}^2}{P_{\text{out}}} = \frac{2 \times (47.0\text{ V})^2}{600\text{ W}} = \frac{2 \times 2209}{600} = \frac{4418}{600} \approx 7.36\ \Omega R L ( d − d ) ​ = P out ​ 2 ⋅ V effective 2 ​ ​ = 600 W 2 × ( 47.0 V ) 2 ​ = 600 2 × 2209 ​ = 600 4418 ​ ≈ 7.36 Ω

Each transistor sees a fundamental-frequency load resistance of 7.36 Ω / 2 = 3.68 Ω 7.36\ \Omega/2=3.68\ \Omega 7.36 Ω/2 = 3.68 Ω .

Determine the nearest standard broadband TLT impedance ratio to match this 7.36 Ω 7.36\ \Omega 7.36 Ω collector-to-collector impedance to a 50.0 Ω 50.0\ \Omega 50.0 Ω antenna load: Required Ratio = 50.0 Ω 7.36 Ω ≈ 6.79 : 1 \text{Required Ratio} = \frac{50.0\ \Omega}{7.36\ \Omega} \approx 6.79:1 Required Ratio = 7.36 Ω 50.0 Ω ​ ≈ 6.79 : 1

A standard 9 : 1 9:1 9 : 1 transmission line transformer (ideal 3 : 1 3:1 3 : 1 voltage ratio) transforms 50.0 Ω 50.0\ \Omega 50.0 Ω down to: R primary = 50.0 Ω 9 ≈ 5.56 Ω R_{\text{primary}} = \frac{50.0\ \Omega}{9} \approx 5.56\ \Omega R primary ​ = 9 50.0 Ω ​ ≈ 5.56 Ω

The 9 : 1 9:1 9 : 1 transformer is a starting point: its 5.56 Ω 5.56\ \Omega 5.56 Ω reflected load still differs from the required 7.36 Ω 7.36\ \Omega 7.36 Ω . A matching network must transform this resistance and compensate device reactance. One shunt reactance alone cannot match two different real resistances; broadband operation requires a complete network design and verification.

Worked Example 5.2.2: Valve Pi-Network Component Reactance Calculations

A single-ended 3-500Z triode amplifier operates in Class AB with a plate voltage of V a = 2600 V V_a = 2600\text{ V} V a ​ = 2600 V and plate current of I a = 0.50 A I_a = 0.50\text{ A} I a ​ = 0.50 A at f = 14.15 MHz f = 14.15\text{ MHz} f = 14.15 MHz ( 20 -meter 20\text{-meter} 20 -meter band). Use the input design convention Q I = R a / ∣ X C 1 ∣ = 12.0 Q_I=R_a/|X_{C1}|=12.0 Q I ​ = R a ​ /∣ X C 1 ​ ∣ = 12.0 . Design the Pi matching network to match the valve plate down to R L = 50.0 Ω R_L = 50.0\ \Omega R L ​ = 50.0 Ω .

Calculate the valve plate load resistance ( R a R_a R a ​ ) using K = 1.8 K = 1.8 K = 1.8 : R a = V a 1.8 ⋅ I a = 2600 V 1.8 × 0.50 A = 2600 0.90 ≈ 2889 Ω R_a = \frac{V_a}{1.8 \cdot I_a} = \frac{2600\text{ V}}{1.8 \times 0.50\text{ A}} = \frac{2600}{0.90} \approx 2889\ \Omega R a ​ = 1.8 ⋅ I a ​ V a ​ ​ = 1.8 × 0.50 A 2600 V ​ = 0.90 2600 ​ ≈ 2889 Ω

Calculate the required capacitive reactance ( X C 1 X_{C1} X C 1 ​ ) and capacitance of the plate tune capacitor ( C 1 C_1 C 1 ​ ): X C 1 = R a Q I = 2889 Ω 12.0 ≈ 240.8 Ω X_{C1} = \frac{R_a}{Q_I} = \frac{2889\ \Omega}{12.0} \approx 240.8\ \Omega X C 1 ​ = Q I ​ R a ​ ​ = 12.0 2889 Ω ​ ≈ 240.8 Ω C 1 = 1 2 π f X C 1 = 1 2 π × ( 14.15 × 10 6 Hz ) × 240.8 Ω ≈ 46.7 pF C_1 = \frac{1}{2\pi f X_{C1}} = \frac{1}{2\pi \times (14.15 \times 10^6\text{ Hz}) \times 240.8\ \Omega} \approx 46.7\text{ pF} C 1 ​ = 2 π f X C 1 ​ 1 ​ = 2 π × ( 14.15 × 1 0 6 Hz ) × 240.8 Ω 1 ​ ≈ 46.7 pF

Calculate the virtual intermediate resistance ( R virt R_{\text{virt}} R virt ​ ) and the required capacitive reactance ( X C 2 X_{C2} X C 2 ​ ) and capacitance of the load capacitor ( C 2 C_2 C 2 ​ ): R virt = R a 1 + Q I 2 = 2889 Ω 1 + 144 = 2889 145 ≈ 19.92 Ω R_{\text{virt}} = \frac{R_a}{1 + Q_I^2} = \frac{2889\ \Omega}{1 + 144} = \frac{2889}{145} \approx 19.92\ \Omega R virt ​ = 1 + Q I 2 ​ R a ​ ​ = 1 + 144 2889 Ω ​ = 145 2889 ​ ≈ 19.92 Ω X C 2 = R L ⋅ R virt R L − R virt = 50.0 × 19.92 50.0 − 19.92 = 50.0 × 19.92 30.08 = 50.0 × 0.6622 ≈ 40.7 Ω X_{C2} = R_L \cdot \sqrt{\frac{R_{\text{virt}}}{R_L - R_{\text{virt}}}} = 50.0 \times \sqrt{\frac{19.92}{50.0 - 19.92}} = 50.0 \times \sqrt{\frac{19.92}{30.08}} = 50.0 \times \sqrt{0.6622} \approx 40.7\ \Omega X C 2 ​ = R L ​ ⋅ R L ​ − R virt ​ R virt ​ ​ ​ = 50.0 × 50.0 − 19.92 19.92 ​ ​ = 50.0 × 30.08 19.92 ​ ​ = 50.0 × 0.6622 ​ ≈ 40.7 Ω C 2 = 1 2 π f X C 2 = 1 2 π × ( 14.15 × 10 6 Hz ) × 40.7 Ω ≈ 276.4 pF C_2 = \frac{1}{2\pi f X_{C2}} = \frac{1}{2\pi \times (14.15 \times 10^6\text{ Hz}) \times 40.7\ \Omega} \approx 276.4\text{ pF} C 2 ​ = 2 π f X C 2 ​ 1 ​ = 2 π × ( 14.15 × 1 0 6 Hz ) × 40.7 Ω 1 ​ ≈ 276.4 pF

Calculate the output-section Q and series inductance. Use the series reactances of the equivalent L sections, not the sum of the shunt capacitor reactances: Q O = 50.0 19.9234 − 1 ≈ 1.2287 Q_O=\sqrt{\frac{50.0}{19.9234}-1}\approx1.2287 Q O ​ = 19.9234 50.0 ​ − 1 ​ ≈ 1.2287 X L = R virt ( Q I + Q O ) = 19.9234 ( 12.0 + 1.2287 ) ≈ 263.56 Ω X_L=R_{\text{virt}}(Q_I+Q_O)=19.9234(12.0+1.2287)\approx263.56\ \Omega X L ​ = R virt ​ ( Q I ​ + Q O ​ ) = 19.9234 ( 12.0 + 1.2287 ) ≈ 263.56 Ω L = 263.56 Ω 2 π ( 14.15 × 10 6 Hz ) ≈ 2.964 μ H L=\frac{263.56\ \Omega}{2\pi(14.15\times10^6\text{ Hz})}\approx2.964\ \mu\text{H} L = 2 π ( 14.15 × 1 0 6 Hz ) 263.56 Ω ​ ≈ 2.964 μ H

The ideal design therefore uses about 46.7 pF 46.7\text{ pF} 46.7 pF input capacitance, 2.964 μ H 2.964\ \mu\text{H} 2.964 μ H series inductance and 276.4 pF 276.4\text{ pF} 276.4 pF output capacitance. At 14.15 MHz the unrounded values transform 50 Ω 50\ \Omega 50 Ω to approximately 2889 + j 0 Ω 2889+j0\ \Omega 2889 + j 0 Ω . The input capacitance includes valve and stray capacitance; real losses and component tolerances require adjustment.