If the physical size of a circuit is much smaller then the wavelength of signals in that circuit, the wave-like nature and the finite speed of an EM-wave does not really have to be taken into account. In that case Kirchhoff’s voltage and current laws can be applied. For example, for an audio amplifier operating up to (let’s say) 20 kHz, assuming that the speed of an EM-wave in a metal is about 2/3 of its speed in vacuum, the wavelength of an EM-wave is about 10 km. Any audio amplifier that is orders of magnitude smaller than that can very well be described and analyzed using the (quasi static) Kirchhoff voltage and current law.

13.1 A single wire

At very low frequencies a wire simply behaves like a short or a low-ohmic resistor. At radio frequencies (RF) this is not the case any more, due to the sheer length of a wire and the finite speed of an EM-wave passing through the wire. There are many models that describe the series inductance associated with wires at RF; one of the more simple ones being the following which is valid for a round wire that is far away from any return ground path [18, 19]:

$$L=2\cdot l\left(\ln <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{4\cdot l}{d}\right)-0.75\right)\text{[nH]}$$

(13.1)

where $l$ is the wire length in cm, and $d$ is the wire diameter in cm.

This translates into — as a rule of thumb — 1 nH/mm wire length. Usually this amount of self inductance is not that relevant at low frequencies and/or for very short wires, but it may be very harmful already at circuits operating at 100 MHz with wires that are longer than a few mm. Wires with other shapes or wires that are relatively close to other conductors exhibit a different relation with wire length.

As example for the impact of a wire, let’s consider a capacitor including wires.

The total impedance of the capacitor including wires $Z_{\mathit{𝑡𝑜𝑡𝑎𝑙}}$ is then: $$\begin{array}{rcll}& Z_{\mathit{𝑡𝑜𝑡𝑎𝑙}}& =\frac{1}{\mathit{𝑗𝜔𝐶}}+2\cdot \mathit{𝑗𝜔}{L}_{\mathit{𝑤𝑖𝑟𝑒}}& \end{array}$$

Which can be used to derive an equation for the equivalent capacitance $\mathit{𝐶\prime }$for the original capacitor including wires. This equation is very much (angular) frequency dependent:

$$\begin{array}{rcll}& \frac{1}{\mathit{𝑗𝜔𝐶\prime }}& =\frac{1}{\mathit{𝑗𝜔𝐶}}+2\cdot \mathit{𝑗𝜔}{L}_{\mathit{𝑤𝑖𝑟𝑒}}\Rightarrow \mathit{𝐶\prime }=\frac{C}{1-2\cdot {\omega }^2{L}_{\mathit{𝑤𝑖𝑟𝑒}}C}& \end{array}$$

Aiming at (arbitrarily) an impedance of $Z_C=-j\cdot 50\mathrm{\Omega }$, to be used in an oscillator, ideally the capacitance value is

• at $f_{\mathit{𝑜𝑠𝑐}}=1\mathit{𝑀𝐻𝑧}$ the corresponding capacitive value is $C=1∕2\pi f_{\mathit{𝑜𝑠𝑐}}|Z_C|=3.18\mathit{𝑛𝐹}$, and 1 cm wires would add a total series inductance equal to $20\mathit{𝑛𝐻}$ which corresponds to a series impedance of about $Z_w\mathit{𝑖𝑟𝑒𝑠}=j0.12\mathrm{\Omega }$ which is negligible compared to the impedance of the capacitor.
• at $f_{\mathit{𝑜𝑠𝑐}}=10\mathit{𝑀𝐻𝑧}$ the corresponding capacitive value is $C=1∕2\pi f_{\mathit{𝑜𝑠𝑐}}|Z_C|=318\mathit{𝑝𝐹}$, and 1 cm wires add a series impedance of about $Z_w\mathit{𝑖𝑟𝑒𝑠}=j1.2\mathrm{\Omega }$ which is small compared to the impedance of the capacitor.
• at $f_{\mathit{𝑜𝑠𝑐}}=100\mathit{𝑀𝐻𝑧}$ the corresponding capacitive value is $C=1∕2\pi f_{\mathit{𝑜𝑠𝑐}}|Z_C|=31.8\mathit{𝑝𝐹}$. Having wires of 1 cm add a series impedance of about $Z_w\mathit{𝑖𝑟𝑒𝑠}=j12\mathrm{\Omega }$ which does impact on the net capacitor value. Now $\mathit{𝐶\prime }\approx 43\mathit{𝑝𝐹}$. Note that the wires serve to increase the (equivalent) capacitance.
• at $f_{\mathit{𝑜𝑠𝑐}}=500\mathit{𝑀𝐻𝑧}$ the corresponding capacitive value is $C=1∕2\pi f_{\mathit{𝑜𝑠𝑐}}|Z_C|=6.4\mathit{𝑝𝐹}$, and 1 cm wire adds about $Z_w\mathit{𝑖𝑟𝑒𝑠}=j63\mathrm{\Omega }$. Then the total impedance of the capacitor including 1 cm wires turns out to be inductive. Clearly using this capacitor including wires as ”C” in an LC-oscillator will not lead to oscillation.

Note that the impact of wires, for a capacitor, is very dependent on both the capacitive value and on the frequency. For example, for a 1 nF capacitor with 2x1.5 mm wires at 100 MHz $\mathit{𝐶\prime }\approx 80\mathit{𝑛𝐹}$ while somewhat longer wires result in inductive behavior at that frequency.

Example As example of the impact of wire length on circuit performance, a quite straight forward common emitter circuit is used. Assuming 5 mm long wires for /textitevery connection, the following circuit is obtained.

After some simplifications — merging inductances in series — this simplifies to the following circuit. Note that for biasing, the presence of the (inductances associated with the) wires does not change anything. However, for sufficiently high signal frequencies the small signl properties of the circuit may be changed quite a lot. Analyzing/deriving the small signal properties this full circuit is a lot more work that analyzing the original circuit because of the increase component count and it leads to hard-to-read relations.

Analyzing the impact of each wire-inductance individually is however straight forward, and the resulting relations are readable and interpretable The impact of the inductances that are in series with the capacitors, $L_{\mathit{𝐶𝐼𝑁}}$ and $L_{\mathit{𝑜𝑢𝑡}}$ is clearly an impedance in series with the input node and output node.. Assuming that $L_{\mathit{𝐶𝐼𝑁}}$ is the only (significant) wire, and assuming that the impedance of the capacitors is small at signal frequencies, the (magnitude of the) voltage gain is lowered and the input impedance is increased.

$$\begin{array}{rcll}& z_{\mathit{𝑖𝑛}}& =R_B∕∕\frac{{\beta }_{\mathit{𝑓𝑒}}}{g_m}+\mathit{𝑗𝜔}{L}_{\mathit{𝑐𝑖𝑛}}& \\ & A_V& =-g_m\cdot R_C\cdot \frac{{\beta }_{\mathit{𝑓𝑒}}∕g_m}{\mathit{𝑗𝜔}(L_{\mathit{𝐶𝐼𝑁}}+L_B)+{\beta }_{\mathit{𝑓𝑒}}∕g_m}& \end{array}$$

Similarly, assuming that only $L_E$ has significant inductance, again the input impedance is increased and the (magnitude of the) voltage gain is reduced:

$$\begin{array}{rcll}& z_{\mathit{𝑖𝑛}}& =R_B∕∕\frac{{\beta }_{\mathit{𝑓𝑒}}}{g_m}\cdot (1+g_m\cdot \mathit{𝑗𝜔}{L}_E)& \\ & A_V& =-\frac{g_m}{1+g_m\cdot \mathit{𝑗𝜔}{L}_E}\cdot R_C& \end{array}$$

The impact of $L_{\mathit{𝑅𝐵}}$ on small signal properties is mainly an increase in the amplifier’s input impedance. $L_C$ mainly causes an increase of the (magnitude of the unloaded) voltage gain and an increase of the output impedance of the amplifier:

$$\begin{array}{rcll}& A_V& =-g_m\cdot (R_C+\mathit{𝑗𝜔}{L}_C)& \\ & z_{\mathit{𝑜𝑢𝑡}}& =R_C+\mathit{𝑗𝜔}{L}_C& \end{array}$$

13.2 Transistor capacitances

Also note that whereas capacitive effects in transistors — such as junction capacitances in BJTs, see e.g. $2.3.2, §2.3.3 and gate-source and gate-drain capacitances in MOS transistors, see e.g. §2.5 are usually ignored, this may not be allowed towards higher frequencies. Typically, at RF these capacitances must be taken into account, which makes circuit design or circuit optimization quite a bit more more complex.

13.3 Two parallel wires - transmission line

Similar to the single wire case described above, two parallel wires also have self inductance. Because these two wires “see” each other the relation between self inductance and geometric parameters is a little different [18, 19, 20]. Denoting the distance between the wires as $\mathrm{\Delta }x$ and again denoting the wire thickness as $d$, for $\mathrm{\Delta }x>>d$

$$\begin{array}{rcll}L& & \approx \frac{{\mu }_0{\mu }_r}{100\pi }\cdot l\cdot \mathit{𝑙𝑛}\left(\frac{2\mathrm{\Delta }x}{d}\right)& \\ & & \approx 4\cdot l\cdot \mathit{𝑙𝑛}\left(\frac{2\mathrm{\Delta }x}{d}\right)\text{[nH]}& \\ & & & \end{array}$$

Being two parallel conductors, there is also a capacitance between these two wires. Again assuming for $\mathrm{\Delta }x>>d$, the relation between capacitance and geometric parameters is

$$\begin{array}{rcll}C& & =\frac{\pi {𝜖}_0{𝜖}_r}{100}\cdot l\cdot \frac{1}{\mathit{𝑙𝑛}\left(\frac{2\mathrm{\Delta }x}{d}\right)}\text{[F]}& \\ & & \approx 0.3\cdot l\cdot \frac{1}{\mathit{𝑙𝑛}\left(\frac{2\mathrm{\Delta }x}{d}\right)}\text{[pF]}& \end{array}$$

From this, two parallel wires can be modelled using (many physically short) sections that have series inductance and parallel capacitance.

Looking into the pair of wires, this gives rise to an impedance

$$\begin{array}{rcll}{Z}_{2\mathit{𝑤𝑖𝑟𝑒𝑠}}=\sqrt{\frac{L}{C}}\approx \frac{1}{\pi }\cdot \sqrt{\frac{\mu }{{𝜖}_0{𝜖}_r}}\cdot \mathit{𝑙𝑛}\left(\frac{2\mathrm{\Delta }x}{d}\right)& & & \end{array}$$

This means that a signal passing these two parallel wires experience an impedance $Z_{2\mathit{𝑤𝑖𝑟𝑒𝑠}}$ while there is no (NO!) dissipation as there are only inductances and capacitances involved. The impedance only means that there is a ratio between voltage and current when the signal passes through the two parallel wires; this impedance is typically denoted as the characteristic impedance $Z_0$ of the pair of wires. Two-parallel-wire configurations are denoted as transmission lines as signals are transmitted through these.

One well known variant of two-wires transmission lines is the “wire over plane” construction. Using symmetry, is can readily be derived (just slide a ground plane in the symmetry plane between the two wires) that then the capacitance (per unit length) is doubled and that inductance (per unit length) is halved,, still using the (now imaginary) distance between the original 2 wires. Denoting the distance between the wire and the plane as $\mathrm{\Delta }s$ this leads to

$$\begin{array}{rcll}L& & \approx \frac{{\mu }_0{\mu }_r}{200\pi }\cdot l\cdot \mathit{𝑙𝑛}\left(\frac{4\mathrm{\Delta }s}{d}\right)& \\ & & \approx 2\cdot l\cdot \mathit{𝑙𝑛}\left(\frac{4\mathrm{\Delta }x}{d}\right)\text{[nH]}& \\ C& & =\frac{2\pi {𝜖}_0{𝜖}_r}{100}\cdot l\cdot \frac{1}{\mathit{𝑙𝑛}\left(\frac{4\mathrm{\Delta }s}{d}\right)}\text{[F]}& \\ & & \approx 0.6\cdot l\cdot \frac{1}{\mathit{𝑙𝑛}\left(\frac{4\mathrm{\Delta }s}{d}\right)}\text{[pF]}& \end{array}$$

Another well known variant is the coaxial cable.

13.4 Reflections

If the wavelength of EM-signals is not much bigger than the physical size of components, wires, ..., then reflections occur at any impedance step experienced by a signal. For electrical signals this might at first seem quite strange, but it is simply the same behavior underlying reflections, transmission and more in visible light that happens at a discontinuity in refractive index - you see and experience it every day. This refractive index is nothing more or less than an impedance change for the part of the EM-spectrum that we call light.

Without digging into details, part of an incident (voltage) wave on an impedance discontinuity is reflected. The reflected fraction is denoted by the (complex) reflection coefficient $\mathrm{\Gamma }$ that depends on the source impedance $Z_0$ and the load impedance $Z_L$ as

$$\begin{array}{rcll}\mathrm{\Gamma }& & =\frac{Z_L-Z_0}{Z_L+Z_0}.& \end{array}$$

The part of the (voltage) wave that is transmitted across the impedance discontinuity can similarly be described by a transmission coefficient $T$ as described below. This has similarities with the theorem on maximum power transfer but it is fundamentally different as this $\mathrm{\Gamma }$ and $T$ deal with the reflected wave behavior of EM-waves.

$$\begin{array}{rcll}T& & =1-\mathrm{\Gamma }=\frac{2Z_0}{Z_L+Z_0}& \end{array}$$

13.5 Maximum power versus maximum power transfer

A recap of maximum power transfer and the associated conjungate impedance matching was presented in section 0.21. This assumes an equation for power ending up in a load impedance $R_{\mathit{𝑙𝑜𝑎𝑑}}$, for a source with source impedance $R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ and a voltage amplitude $V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ for the configuration in the figure below.

$$\begin{array}{rcll}{P}_{\mathit{𝑙𝑜𝑎𝑑}}& =& I_{\mathit{𝑙𝑜𝑎𝑑}}^2\cdot R_{\mathit{𝑙𝑜𝑎𝑑}}& \\ & =& {\left(\frac{V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}}{R_{\mathit{𝑙𝑜𝑎𝑑}}+R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}}\right)}^2\cdot R_{\mathit{𝑙𝑜𝑎𝑑}}& \end{array}$$

The maximum power as a function of $R_{\mathit{𝑙𝑜𝑎𝑑}}$ can be obtained via differentiation:

$$\begin{array}{rcll}\frac{\partial P_{\mathit{𝑙𝑜𝑎𝑑}}}{\partial R_{\mathit{𝑙𝑜𝑎𝑑}}}& =& V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}^2\cdot \frac{R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}-R_{\mathit{𝑙𝑜𝑎𝑑}}}{{(R_{\mathit{𝑙𝑜𝑎𝑑}}+R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}})}^3}& \end{array}$$

This directly shows that the power in the load is at its maximum if the load resistance is equal to the source resistance: $R_{\mathit{𝑙𝑜𝑎𝑑}}=R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$. In a similar fashion, we find that the power transfer with complex impedances is highest for $Z_{\mathit{𝑙𝑜𝑎𝑑}}=Z_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}^{\ast }$. A fairly simple result that you can use to design a load. At the same time, you should not use this when designing a power amplifier, or when having voltage or current limitations in the driving power source.

The optimum above holds for the case when you assume an ideal source with a certain — fixed — source impedance, which does not apply if you design your own power amplifier. If you design an amplifier, then you have to deal with limitations in output voltage and output current, and you have a degree of freedom in the output impedance of your amplifier. If you want a maximum output power, then you have to take these conditions into account.

Assuming an ideal voltage source, without any voltage limitation and without current limitation, the power into the load not only depends on $R_{\mathit{𝑙𝑜𝑎𝑑}}$ but also depends on $R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ and $V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$. Two additional ways to maximize the power $P_{\mathit{𝑙𝑜𝑎𝑑}}$ into $R_{\mathit{𝑙𝑜𝑎𝑑}}$ follows from the following two (partial) derivatives:

$$\begin{array}{rcll}\frac{\partial P_{\mathit{𝑙𝑜𝑎𝑑}}}{\partial R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}}& =& V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}^2\cdot \frac{-2\cdot R_{\mathit{𝑙𝑜𝑎𝑑}}}{{(R_{\mathit{𝑙𝑜𝑎𝑑}}+R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}})}^3}& \\ \frac{\partial P_{\mathit{𝑙𝑜𝑎𝑑}}}{\partial V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}}& =& V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}\cdot \frac{2\cdot R_{\mathit{𝑙𝑜𝑎𝑑}}}{{(R_{\mathit{𝑙𝑜𝑎𝑑}}+R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}})}^2}& \end{array}$$

that show that

• the partial derivative $\partial P_{\mathit{𝑙𝑜𝑎𝑑}}∕\partial R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}\le 0$ for every source voltage (amplitude) and combination of resistors. Hence, to obtain maximum $P_{\mathit{𝑙𝑜𝑎𝑑}}$, the output impedance of the amplifier $R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ must be as low as possible.
• the partial derivative $\partial P_{\mathit{𝑙𝑜𝑎𝑑}}∕\partial V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}\ge 0$, from which it follows that for maximum $P_{\mathit{𝑙𝑜𝑎𝑑}}$ the voltage (amplitude) $V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ must be as large as possible,

In case you design an amplifier, the output impedance $R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ is designed by you, the circuit designer. The second item is usually limited by things like (clipping to) a given supply voltage. As long as no clipping occurs — in voltage or current — the system is perfectly linear and for maximum power into the load, you can minimize the output impedance of the driver/amplifier, maximize the signal swing and use (conjungate) matched load.

Actual drivers/amplifiers always have limitations in output voltage (amplitude) and in output current (amplitude). This is straight forward implication of using non-linear components such as transistors. The current limitation and voltage limitation are properties of (the design of) your circuit and these limitations are usually fairly independent of the load impedance $Z_{\mathit{𝑙𝑜𝑎𝑑}}$. With current or voltage limitations, (13.2) does not hold and consequently (13.2) (13.2) nor (13.2) are not valid.

Maximum power into $Z_{\mathit{𝑙𝑜𝑎𝑑}}$ is now obtained:

• achieving maximum $V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ when driving $Z_{\mathit{𝑙𝑜𝑎𝑑}}$
• having minimum $R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$, for maximum $V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ and driving $Z_{\mathit{𝑙𝑜𝑎𝑑}}$

Note that to achieve maximum $P_{\mathit{𝑙𝑜𝑎𝑑}}$:

• when designing a driver/amplifier and having a fixed $R_{\mathit{𝑙𝑜𝑎𝑑}}$, the output voltage amplitude $V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ must be maximum and $R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}$ should be minimum and the maximum output current amplitude must be (at least)

  $$I_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒},\mathit{𝑚𝑎𝑥}}\ge \frac{V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒},\mathit{𝑚𝑎𝑥}}}{R_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒}}+R_{\mathit{𝑙𝑜𝑎𝑑}}}\approx \frac{V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒},\mathit{𝑚𝑎𝑥}}}{R_{\mathit{𝑙𝑜𝑎𝑑}}}$$

• when designing the load $R_{\mathit{𝑙𝑜𝑎𝑑}}$, the amplifier should be loaded in such a way that maximum output voltage amplitude $V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒},\mathit{𝑚𝑎𝑥}}$ and maximum output current amplitude $I_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒},\mathit{𝑚𝑎𝑥}}$ are achieved at the same time. Then [21],

  $$R_{\mathit{𝑙𝑜𝑎𝑑},\mathit{𝑜𝑝𝑡}}=\frac{V_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒},\mathit{𝑚𝑎𝑥}}}{I_{\mathit{𝑠𝑜𝑢𝑟𝑐𝑒},\mathit{𝑚𝑎𝑥}}}$$

  Note that this is quite different from the condition of maximum power transfer following from (13.2) and recapped in section 0.21. This is because the derivations above are for maximum power. And more power is more better.

In actual RF power amplifiers, the load impedance that results in maximum power is more complex. The conventional way to estimate this $Z_{\mathit{𝐿𝑂𝐴𝐷},\mathit{𝑜𝑝𝑡}}$ is to apply a variable load impedance (variable real and variable reactive part) and sweep both while monitoring the real power. This approach is called a load-pull measurement or simulation; a good intro is described in [21]. This also shows that conjungate matching is not applied to get maximum power which should be obvious when assuming ideal (but current limited) voltage sources or ideal (but voltage limited) current sources.

Wave-behavior of the RF-signals only kick in if lengths (of e.g. wires, cables, transmission lines, components) are significant compared to the wavelength of the signals. Then reflections as described in §13.4 become relevant and impedance matching should be done to get maximum power in your load. However, do not impedance match directly at the output of an RF amplifier if you want maximum power.