RF-06.02 · Pi and T matching networks · Art of Nanoelectronics

01Scope, reference planes and circuit assumptions

An impedance transformer is not defined only by a component list. Begin by selecting the source-side reference plane, a strictly positive real source resistance, a positive real load resistance, a design frequency and a phasor sign convention. The network is passive, reciprocal and lossless only in the ideal synthesis model. We seek a real input impedance equal to the specified source resistance at the chosen frequency, not a broadband guarantee.

The standard three-reactance pi and tee networks add one independent tuning parameter compared with an elementary L network. We represent that freedom by an intermediate virtual resistance: it is a mathematical impedance in a two-stage synthesis, not a physical resistor to solder into the network. Validity of the resulting circuit must be checked by independently reducing the actual network from the load to the source.

Z_{\mathrm{in}}=R_S,\qquad R_S>0,\quad R_L>0

\Gamma_{\mathrm{in}}=\frac{Z_{\mathrm{in}}-R_S}{Z_{\mathrm{in}}+R_S}

Z_{\mathrm{in}}(\omega_0)=R_S\;\Longrightarrow\;\Gamma_{\mathrm{in}}(\omega_0)=0

02Pi synthesis: choosing the lower virtual resistance

A low-pass pi network has a shunt capacitor at the input node, one series inductor between the input and output nodes, and a second shunt capacitor across the load. To build it as two low-pass L sections, introduce a positive virtual resistance strictly smaller than both terminations. Each endpoint then represents the high-resistance side of its L section, so its capacitive shunt element is placed directly at that endpoint.

The two virtual-side series inductors meet without an intervening physical node that requires another component; they are combined into one series inductance. The source and load shunt susceptances remain distinct because they connect to different nodes. As the virtual resistance changes, the three values change although the specified input match at the design frequency remains exact in the ideal model.

0<R_v<\min(R_S,R_L)

Q_{\pi,S}=\sqrt{\frac{R_S}{R_v}-1},\qquad Q_{\pi,L}=\sqrt{\frac{R_L}{R_v}-1}

B_S=\frac{Q_{\pi,S}}{R_S},\quad B_L=\frac{Q_{\pi,L}}{R_L},\quad X_{\pi}=R_v(Q_{\pi,S}+Q_{\pi,L})

03Pi derivation by Kirchhoff and complex inversion

At the load node, the current through the load resistor adds to the capacitive current. Therefore their admittances add. Invert the resulting complex admittance to obtain the load-side impedance; at the design frequency its real part equals the chosen virtual resistance, and its imaginary part has the opposite sign from the series inductor contribution of the second L section.

Now add the full series reactance and invert again to find the admittance at the input node. The input shunt capacitance adds a positive imaginary current which cancels the remaining imaginary admittance. This last operation leaves the desired real source conductance. This sequence of sums and inversions is the physical circuit equation and serves as a test independent of the virtual-resistance construction.

Y_L=\frac1{R_L}+jB_L,\qquad Z_p=Y_L^{-1}

Z_p=R_v-jR_vQ_{\pi,L}

Z_{\mathrm{in}}=\left[jB_S+\left(jX_{\pi}+Z_p\right)^{-1}\right]^{-1}=R_S

04T synthesis: choosing the higher virtual resistance

A low-pass T network has one series inductor at the input, a shunt capacitor from its central junction to return, and a second series inductor next to the load. Choose the virtual resistance strictly greater than both endpoint resistances. The central junction is then the high-resistance side of each of the two conceptual L sections, so both required shunt capacitors occupy exactly the same electrical node.

The two center-node shunt susceptances add into one capacitor, but the series inductors remain separate because they occupy opposite sides of the common node. Replacing them by one series inductance would short-circuit the intended shunt branch and change the topology. No additional dissipative resistor is inserted at the virtual node; it represents the target impedance used to construct the two L transformations.

R_v>\max(R_S,R_L)

Q_{T,S}=\sqrt{\frac{R_v}{R_S}-1},\qquad Q_{T,L}=\sqrt{\frac{R_v}{R_L}-1}

X_S=R_SQ_{T,S},\quad X_L=R_LQ_{T,L},\quad B_T=\frac{Q_{T,S}+Q_{T,L}}{R_v}

05T derivation and the role of the common junction

Start from the load resistor and the inductor directly adjacent to it: their impedances add, since the same current flows through both. The reciprocal of that series impedance gives the branch admittance seen from the central node. Kirchhoff current balance adds the center capacitor susceptance to this branch admittance.

Invert the total center admittance and then add the input-side series inductor impedance. At the design frequency the result becomes the specified source resistance, as required by zero input reflection. Notice the important contrast with the pi circuit: the pi calculation begins with a parallel combination at the load and ends with one at the input, whereas the T calculation begins and ends with series combinations.

Z_L=R_L+jX_L,\qquad Y_m=Z_L^{-1}+jB_T

Y_m^{-1}=R_S-jX_S

Z_{\mathrm{in}}=jX_S+\left[jB_T+(R_L+jX_L)^{-1}\right]^{-1}=R_S

06Pi numerical design at one gigahertz

Use a fifty-ohm real source, a two-hundred-ohm resistive load, and one gigahertz. Select twenty-five ohms as the pi virtual resistance: the two step ratios give quality parameters of one and square root of seven. The source-side shunt susceptance is twenty millisiemens, while the load-side value is smaller because its shunt branch spans the larger resistance.

Converting susceptances to capacitances and the combined positive reactance to an inductance produces a three-element low-pass pi circuit. Keep track of which capacitor is physically attached to the source-side input node. Recalculate the two internal complex impedances before claiming an exact match; the rounded nominal component values shown in the schematic are illustrative and will introduce a tiny numerical residual.

Low-pass π match, 50 Ω to 200 Ω at 1 GHz: end shunt capacitors at different nodes and one series inductor.

R_S=50\,\Omega,\quad R_L=200\,\Omega,\quad f_0=1\,\mathrm{GHz}

R_{v,\pi}=25\,\Omega,\quad Q_{\pi,S}=1,\quad Q_{\pi,L}=\sqrt7

C_S=\frac{0.02}{2\pi\,10^9}\approx3.18310\,\mathrm{pF},\quad C_L=\frac{\sqrt7/200}{2\pi\,10^9}\approx2.10542\,\mathrm{pF},\quad L_{\pi}\approx14.50598\,\mathrm{nH}

07T numerical design at one gigahertz

Use the same terminations and frequency but choose a four-hundred-ohm virtual resistance, now above the two endpoint resistances. The input-side step has the quality parameter square root of seven, and the load-side step has unit quality parameter. Both inductors are positive and distinct: the source series inductance is not equal to the load series inductance.

The two shunt susceptances at the intermediate junction add to give one physical capacitor. Verify the result by computing the complex impedance of the load and its adjacent inductor, adding the middle capacitive admittance, inverting, and finally adding the input inductor. The final imaginary part is zero at the design frequency and the real part is fifty ohms, up to rounding.

Low-pass T match, 50 Ω to 200 Ω at 1 GHz: two separate series inductors and one center shunt capacitor.

R_{v,T}=400\,\Omega,\quad Q_{T,S}=\sqrt7,\quad Q_{T,L}=1

L_S=\frac{50\sqrt7}{2\pi\,10^9}\approx21.05422\,\mathrm{nH},\quad L_L=\frac{200}{2\pi\,10^9}\approx31.83099\,\mathrm{nH}

C_T=\frac{(\sqrt7+1)/400}{2\pi\,10^9}\approx1.45060\,\mathrm{pF}

08Bandwidth, component losses and measured validation

Keeping each physical inductor and capacitor constant while sweeping frequency changes every signed reactance and susceptance, so both three-element networks generally reflect power away from the design frequency. Their virtual-resistance choices alter how sharply this mismatch develops. A high transformation quality parameter is not the same as the finite quality factor of a real inductor, and ideal matching alone proves nothing about dissipative insertion loss.

A realistic experiment includes series resistance, capacitor losses, pad capacitance, solder/trace inductance and the reference-plane positions. Calibrate the VNA to known planes, measure complex input reflection, and use a true two-port setup when reporting transmission or insertion loss. Preserve the measured S parameters and compare the measured bandwidth and uncertainty to the loss-aware model rather than only checking that a single frequency is matched.

Z_{\pi}(\omega)=\left[j\omega C_S+\left(j\omega L_{\pi}+\left(R_L^{-1}+j\omega C_L\right)^{-1}\right)^{-1}\right]^{-1}

Z_T(\omega)=j\omega L_S+\left[j\omega C_T+\left(R_L+j\omega L_L\right)^{-1}\right]^{-1}

\mathrm{RL}(\omega)=-20\log_{10}|\Gamma_{\mathrm{in}}(\omega)|,\qquad Q_{\mathrm{part}}\ne Q_{\pi,S},\,Q_{T,S}

Worked problems with answers

For the given terminations, prove that the selected virtual resistance is admissible for a low-pass pi circuit and calculate both section-quality parameters.
Starting from Kirchhoff current law, construct the full complex input impedance of the pi circuit without using an intermediate resistor in the actual schematic.
At one gigahertz, synthesize all three components of the pi match between fifty and two hundred ohms and check the units.
Verify the exact pi input match using two independent complex inversions at the design frequency, without relying on the virtual-resistance diagram.
Choose the virtual resistance for a low-pass T network and derive its two quality parameters from the endpoint resistance ratios.
At one gigahertz, compute the source inductor, the load inductor and the shared center capacitor of the T circuit.
Starting at the load, evaluate the T circuit admittance at its middle node and recover the real input resistance by complex inversion.
Increase operating frequency by twenty percent without changing any physical component values. Compare both input reflections with the design-frequency values.
Include a finite series resistance in the pi inductor. Derive its correct placement in the input-impedance equation and distinguish transformation Q from part Q.
Write a reproducible VNA measurement protocol that distinguishes the one-port match check from a two-port transmission and uncertainty measurement.

Show detailed solutions

1. For the given terminations, prove that the selected virtual resistance is admissible for a low-pass pi circuit and calculate both section-quality parameters.

A pi transformation through two ideal low-pass L cells requires the intermediate resistance to lie below both real terminations. Twenty-five ohms satisfies this because it is strictly below fifty and two hundred. The source step has a resistance ratio of two, while the load step has a ratio of eight.

Taking the square root of each ratio minus unity yields respectively one and the square root of seven. Both are positive real quantities. These are transformation parameters used for element synthesis, not statements about the finite unloaded quality factor of an inductor.

Q_{\pi,S}=\sqrt{50/25-1}=1

Q_{\pi,L}=\sqrt{200/25-1}=\sqrt7

2. Starting from Kirchhoff current law, construct the full complex input impedance of the pi circuit without using an intermediate resistor in the actual schematic.

At the load node, current divides between the real load resistor and its shunt capacitor. Their admittances sum, so invert that sum to recover the complex load-node impedance. The central inductor and this equivalent load are then in series, and their impedances add.

Looking into the source node, the source capacitor current is in parallel with the series branch. Add that capacitive admittance to the reciprocal of the series-branch impedance, then invert one last time. The resulting expression contains only the physically installed two capacitors, one inductor and load resistor.

Z_p=\left(R_L^{-1}+jB_L\right)^{-1}

Z_{\mathrm{in}}=\left[jB_S+\left(jX_\pi+Z_p\right)^{-1}\right]^{-1}

3. At one gigahertz, synthesize all three components of the pi match between fifty and two hundred ohms and check the units.

The virtual-resistance construction gives a source-end susceptance of twenty millisiemens, a load-end susceptance of square root of seven over two hundred siemens, and a total series reactance of about ninety-one point fourteen ohms. Both susceptances are positive, so each end branch must be a capacitor.

Divide the susceptances by angular frequency to obtain approximately three point one eight three picofarads at the input and two point one zero five picofarads across the load. Divide the positive series reactance by angular frequency to obtain about fourteen point five zero six nanohenries. These are ideal nominal values at one gigahertz.

B_S=\frac1{50}=0.02\,\mathrm S,\quad B_L=\frac{\sqrt7}{200}\,\mathrm S

X_\pi=25(1+\sqrt7)\,\Omega\approx91.14378\,\Omega

4. Verify the exact pi input match using two independent complex inversions at the design frequency, without relying on the virtual-resistance diagram.

First invert the load node admittance. At the design frequency it becomes a virtual resistance of twenty-five ohms with a negative imaginary part of twenty-five times square root of seven ohms. Adding the two virtual-side inductive reactances cancels the second section contribution and leaves a real and imaginary series impedance each of twenty-five ohms.

Invert that impedance: its admittance becomes twenty millisiemens with a negative imaginary part of twenty millisiemens. The input capacitor adds the opposite positive susceptance, so the input admittance is purely twenty millisiemens. Its reciprocal is exactly fifty ohms; the input reflection is zero.

Z_p=25-j25\sqrt7\,\Omega

Z_p+j25(1+\sqrt7)\,\Omega=25+j25\,\Omega,\quad Z_{\mathrm{in}}=50\,\Omega

5. Choose the virtual resistance for a low-pass T network and derive its two quality parameters from the endpoint resistance ratios.

For a low-pass T circuit the central node must be the high-resistance side of both conceptual L transformations. Therefore the virtual resistance must exceed both fifty and two hundred ohms. Four hundred ohms is admissible and gives resistance ratios of eight on the source side and two on the load side.

Taking the square root of each ratio minus one gives the square root of seven and unity. These determine two unequal positive series reactances and the sum of two positive shunt susceptances at one physical central node. Do not connect a physical four-hundred-ohm resistor to that node.

Q_{T,S}=\sqrt{400/50-1}=\sqrt7

Q_{T,L}=\sqrt{400/200-1}=1

6. At one gigahertz, compute the source inductor, the load inductor and the shared center capacitor of the T circuit.

On the source side, multiply fifty ohms by the square root of seven to obtain roughly one hundred thirty-two point twenty-nine ohms of positive series reactance. Dividing by angular frequency gives approximately twenty-one point zero five four nanohenries. On the load side the series reactance is two hundred ohms and the inductance is about thirty-one point eight three one nanohenries.

At the central node the shunt susceptance is the sum of the two L-section values. Converting that value to capacitance at one gigahertz gives about one point four five zero six picofarads. The parts form a genuine T: one input series inductor, a grounded center capacitor, and one output series inductor.

X_S=50\sqrt7\,\Omega,\quad X_L=200\,\Omega

B_T=\frac{1+\sqrt7}{400}\,\mathrm S

7. Starting at the load, evaluate the T circuit admittance at its middle node and recover the real input resistance by complex inversion.

The load resistor and the adjacent output inductor present a series impedance with equal positive real and imaginary parts, each two hundred ohms. Its reciprocal admittance has real part one over four hundred siemens and an equal negative imaginary part. Kirchhoff current law adds the positive susceptance of the central capacitor.

After cancellation and addition, the total central admittance has real part one over four hundred siemens and imaginary part square root of seven over four hundred siemens. Complex inversion gives fifty ohms with negative input-side inductive reactance; adding the input inductor cancels the imaginary part and leaves fifty ohms.

\left(200+j200\right)^{-1}=\frac{1-j}{400}\,\mathrm S

jB_T+\left(200+j200\right)^{-1}=\frac{1+j\sqrt7}{400}\,\mathrm S

8. Increase operating frequency by twenty percent without changing any physical component values. Compare both input reflections with the design-frequency values.

Hold the designed capacitances and inductances constant and evaluate their susceptances and reactances at one point two times the original angular frequency. The pi input impedance is approximately seventy-two point two two minus j eighty-three point twenty-seven ohms. The T input impedance is about twenty-six point nine three plus j forty-seven point three one ohms.

Both produce a nonzero input reflection, with magnitude approximately zero point five eight three in this particular dual example. Hence an exact center-frequency match does not mean broadband matching, and equality of reflection magnitude here does not imply identical internal voltages or loss under real component models.

\omega_1=1.2\omega_0,\qquad \Gamma_\pi=\frac{Z_\pi(\omega_1)-50\,\Omega}{Z_\pi(\omega_1)+50\,\Omega}

\Gamma_T=\frac{Z_T(\omega_1)-50\,\Omega}{Z_T(\omega_1)+50\,\Omega}

9. Include a finite series resistance in the pi inductor. Derive its correct placement in the input-impedance equation and distinguish transformation Q from part Q.

A real series inductor may be modeled by a small positive resistance in series with its ideal reactance. In a pi circuit this extra resistance lies between the two shunt nodes, inside the source-side parallel combination. Placing it artificially ahead of the input capacitor would change the measured network and give the wrong reflection.

The part quality factor compares reactance magnitude to series resistance at the frequency considered; the virtual-resistance transformation parameter is instead fixed by the resistance ratio. Power dissipated in the winding depends on the actual RMS branch current. Both return loss and insertion loss must be recomputed using the corrected circuit.

Z_{\pi,\mathrm{loss}}=\left[jB_S+\left(r_L+jX_\pi+Z_p\right)^{-1}\right]^{-1}

Q_{\mathrm{part}}=\frac{\omega L}{r_s},\qquad P_{\mathrm{loss},L}=I_{\mathrm{rms}}^2r_s

10. Write a reproducible VNA measurement protocol that distinguishes the one-port match check from a two-port transmission and uncertainty measurement.

Calibrate a vector network analyzer to documented physical reference planes with standards appropriate to the connectors and substrate. Terminate the network consistently, sweep across the design frequency and save complex input reflection together with calibration records and uncertainty assumptions.

To study dissipative insertion loss, construct a two-port fixture, calibrate both ports, record complex transmission and compare with a circuit model including element resistance, parasitic capacitance and connection inductance. Repeat with realistic component tolerances; a one-port reflection dip by itself cannot quantify energy lost in the network.

S_{11}=\frac{b_1}{a_1}\quad(a_2=0)

\mathrm{RL}=-20\log_{10}|S_{11}|,\qquad \mathrm{IL}=-20\log_{10}|S_{21}|

Next: RF-06.03, single- and double-stub matching.