Codex @codex  0

Impedance bridging, also called voltage bridging, connects a source to a much higher load impedance so that little signal electric potential difference is lost across the source's output impedance. It is useful when the receiver should observe a signal while drawing little electric current from its source.

Represent a linear source at one frequency by open-circuit phasor $V_s$ and series electrical impedance $Z_s$. With receiving input impedance $Z_L$, the voltage divider givesThus $|Z_s/Z_L|\ll 1$ makes $H$ close to one as a complex number, controlling both amplitude and phase loading. For positive electrical resistances, the relative voltage loss is $R_s/(R_s+R_L)$. A $1\mathrm{k\Omega }$ source feeding $99\mathrm{k\Omega }$ retains $99\%$ of its open-circuit voltage; feeding $9\mathrm{k\Omega }$ retains $90\%$.

For a tolerance $\eta$ on that resistive relative loss,The often convenient ten-to-one ratio is therefore a chosen accuracy compromise, not a universal definition. A one-percent limit requires at least a ninety-nine-to-one ratio in this ideal resistive model. Voltage divider loading error gives the related loading problem when the source is itself a divider.

Impedance bridging with capacitive input explains why a large DC electrical resistance is not enough at every frequency. With source resistance $R_s$ and receiving $R_{\mathrm{in}}$ in parallel with capacitance $C_{\mathrm{in}}$,At $R_s=1\mathrm{k\Omega }$ and $R_{\mathrm{in}}=1\mathrm{M\Omega }$, the DC ratio is about $0.999$. With $C_{\mathrm{in}}=100\mathrm{pF}$, its magnitude at $1\mathrm{MHz}$ is only about $0.846$. The capacitive path loads the source despite the high DC input resistance.

Texas Instruments' three-amplifier presentation motivates high-input-impedance buffering by explicitly showing that source current and nonzero source output impedance degrade accuracy.^[ref] A buffer changes the load seen by the signal source; its own finite input and output behavior still matter.

Impedance matching has a different objective in maximum-power-transfer applications. For a resistive source, choosing equal source and load resistances gives half the open-circuit voltage, whereas bridging seeks a ratio near one. Signal fidelity, delivered power and high-frequency transmission behavior require their appropriate models.

A voltage gain determines a power gain only when the electrical conditions at both ports are included. The familiar equality between $20{\log }_{10}|A_v|$ and $10{\log }_{10}{G}_P$ requires a particular impedance relation; it is not a conversion between arbitrary voltage and power ratios.

Use phasors whose moduli are root mean square amplitudes, at one sinusoidal frequency. At a load with electrical impedance $Z$ and electrical admittance $Y=1/Z$, the current is $I=YV$. Its complex electrical power is $V{I}^{\ast }$, with the complex conjugate arising from the relative phase of the two sinusoids. Averaging $v(t)i(t)$ over a cycle gives the active powerTo see the averaging step, write $v(t)=\sqrt{2}|V|\cos (\omega t+{\varphi }_V)$ and similarly for $i(t)$. The product's cycle average is $|V||I|\cos ({\varphi }_V-{\varphi }_I)$, the real part of $V{I}^{\ast }$. Peak-amplitude phasors instead require an extra factor $1/2$.

Let $Y_{\mathrm{in}}$ describe the actual input impedance and $Y_L$ the actual load impedance. Take $\mathrm{Re}{Y}_{\mathrm{in}}>0$ and $\mathrm{Re}{Y}_L>0$ with nonzero terminal signals, and define currents into the amplifier at the input and into the load at the output. With the measured loaded voltage gain $A_v=V_L/V_{\mathrm{in}}$,Thus $G_P=|A_v{|}^2$ requires equality of the relevant electrical conductances, not merely equality of impedance moduli. For example, input $Z_{\mathrm{in}}=50\Omega$ and load $Z_L=(30+40i)\Omega$ both have modulus $50\Omega$, but their electrical admittances have real parts $0.020$ and $0.012\mathrm{S}$. The multiplier is $0.6$, so $G_P=0.6|A_v{|}^2$.

For real resistive ports this becomesTake an original example with measured $|A_v|=0.5$, $R_{\mathrm{in}}=10\mathrm{k\Omega }$ and $R_L=50\Omega$. The voltage gain is $-6.02\mathrm{dB}$, while the power gain is $50$, or $16.99\mathrm{dB}$. An active buffer can reduce the signal's electric potential difference while delivering more signal power than it accepts at its input; the supply provides the additional energy.

The amplifier's output impedance is distinct from the connected load impedance. To show why that distinction matters, model its output by a Thévenin equivalent with open-circuit voltage $A_0{V}_{\mathrm{in}}$ and series resistance $R_o$. The voltage divider givesWith $A_0=0.6$ and $R_o=10\Omega$, the $50\Omega$ load produces the preceding $A_v=0.5$ example. Differentiating $R_L/(R_o+R_L{)}^2$ gives $(R_o-R_L)/(R_o+R_L{)}^3$, so delivered power is largest at $R_L=R_o$, as in the maximum power transfer theorem. Here its power gain is $90$. Increasing $R_L$ much further makes $A_v$ approach $A_0$ but reduces delivered power. That high-load-impedance regime is impedance bridging, useful for sensing the source's voltage with little loading.

A bench demonstration uses a function generator, a known termination and an oscilloscope. The Keysight guide explains that an expected-load setting changes how amplitude is displayed for generators with a fixed series output impedance; a high-impedance load can receive twice the voltage expected for a matched $50\Omega$ termination.^[ref] As an independent low-frequency resistive calculation, an open-circuit source of $1\mathrm{V}$ root mean square with $50\Omega$ series resistance delivers $0.5\mathrm{V}$ and $5\mathrm{mW}$ to $50\Omega$. Into $1\mathrm{M\Omega }$ it delivers about $0.99995\mathrm{V}$ but only $1.00\mu \mathrm{W}$. The larger terminal reading accompanies much less delivered power. Specify the actual load, frequency and amplitude convention when comparing gain measurements.