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In December 1947, John Bardeen and Walter Houser Brattain obtained transistor action at Bell Telephone Laboratories. William Shockley proposed the distinct bipolar junction transistor within the following month.^[ref]

Their 1948 paper describes ground, etched germanium, initially also anodized and rinsed. Surface contacts separated by $50$–$250\mu \mathrm{m}$ share a broad base contact. Positive emitter biasing injects electron holes; negative collector biasing draws them toward the collector. An input signal changes emitter electric current, and a collector electrical load converts the resulting change into voltage.^[ref]

Early electric field control experiments gave unexpectedly weak responses. John Bardeen proposed that surface states screened the interior. Surface charge could create a barrier and a hole-rich semiconductor inversion layer. Replacing an electrolyte control with a gold contact reversed the response sign, suggesting electron hole carrier injection. Closely spaced contacts made from a gold-coated wedge split with a razor then enabled power gain. The relative importance of surface and bulk transport depended on preparation and collector forming.^[ref]

The 1948 authors found that collector-current increments could exceed emitter-current increments; their current gain depended on biasing.^[ref] A terminal-current ratio alone cannot identify the microscopic path. Surface states, the local barrier, carrier injection and contact preparation prevent a single ideal modern bipolar junction transistor model from explaining every point-contact transistor.

For the conventional electric current directions drawn in the figure, $I_E$ enters the emitter, $I_C$ leaves the collector and signed $I_B$ enters the base. Kirchhoff's current law requiresThus a local current gain $\alpha >1$ requires an accompanying base-current change. The collector increment is not simply a count of injected electron holes, and it does not violate electric charge conservation.

For an experiment following the functional arrangement, first establish each contact's separate current-voltage characteristic relative to the base. Then hold collector biasing fixed, vary a small emitter signal, and compare the collector electrical load voltage with the input. A decisive power gain check measures both input electric current and voltage and the output power in a known electrical load; voltage enlargement alone is insufficient. These are proposed measurement controls, not additional claims about the 1947 laboratory record.

Repeating a setting tests drift; reversing or removing emitter biasing tests whether the response requires carrier injection. Contact motion, surface contamination, temperature, direct electrical coupling and noise can all affect the inference. A small-signal model describes measured local terminal response without deciding every microscopic mechanism.

The ratio of output to input signal electric current, with specified reference directions and load conditions. A large current gain alone does not determine power gain.

Electronic gain expresses how an electronic circuit scales a specified input signal quantity into an output quantity. For a linearized voltage amplifier around its operating point, the voltage gain is $A_v=\delta v_{\mathrm{out}}/\delta v_{\mathrm{in}}$. At a sinusoidal frequency it can be a complex phasor ratio, specifying both amplitude ratio and phase shift. Power gain instead compares the average power delivered to the load with the average signal power entering the amplifier:State the input/output quantities, frequency, load and operating conditions when specifying gain. A transimpedance amplifier relates output electric potential difference to input current and has a dimensional transfer ratio in ohms.

For an original circuit example, use an inverting operational amplifier with input resistor $R_1=10\mathrm{k\Omega }$, feedback resistor $R_2=100\mathrm{k\Omega }$ and load $R_L=1\mathrm{k\Omega }$. In its ideal negative-feedback linear regime,An input of $50\mathrm{mV}$ root mean square therefore gives $0.50\mathrm{V}$ root mean square at the output, with a phase inversion. The manufacturer's primary explanation derives this resistance-ratio gain using the input and feedback currents.^[ref] Suitable supplies, output-current capability and amplifier output range are assumptions of this worked model, not specifications of a particular device.

Here the input resistance is $R_1$ in the ideal model. For resistive ports and root-mean-square signal amplitudes,The example input power is $0.25\mu \mathrm{W}$ and load power $0.25\mathrm{mW}$, giving $G_P=1000$. The additional output energy comes from the amplifier's supply, controlled by the input signal. The impedance condition for converting voltage gain to power gain is essential: voltage magnitude alone does not specify load power.

Voltage gain is conventionally expressed as $20{\log }_{10}|A_v|$ decibels, while power gain uses $10{\log }_{10}{G}_P$. This example has voltage gain $20\mathrm{dB}$ and power gain $30\mathrm{dB}$; the two expressions agree when the resistive input and load impedances are equal. A negative voltage gain indicates phase inversion; its magnitude enters the logarithm.

A direct test drives a small sine wave into the input and measures the actual terminal signals with an oscilloscope. Sweeping frequency reveals the frequency response; increasing amplitude reveals amplifier clipping or other departure from the assumed linear gain. Keep the load and measurement impedances specified. Gain can amplify input noise along with signal, so the signal-to-noise ratio requires its own measurement.

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.

For series resistance $R$, inductance $L$ and capacitance $C$, with output measured across the inductor, the $e^{i\omega t}$ transfer function is $H_L=i\omega L/[R+i(\omega L-1/(\omega C))]$. At current resonance ${\omega }_0=1/\sqrt{LC}$ its modulus is ${\omega }_{0}L/R=Q$. Large and oppositely phased inductor and capacitor voltages nearly cancel in the loop sum. This is voltage magnification in a passive circuit, not power gain. The maximum of $|H_L|$ need not occur exactly at the current-resonance frequency.

For sinusoidal steady state with convention $e^{i\omega t}$, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage-current phase difference. Since $V_L=LdI/dt$,The original circuit is a series RLC circuit with output across the inductor. Its transfer function isFor the original capacitance, the outputs are $10.15\mathrm{mV}$ at $9\mathrm{MHz}$ and $4.65\mathrm{mV}$ at $10\mathrm{MHz}$, a ratio $0.458$. The current resonance is $f_0=1/(2\pi \sqrt{LC})=8.967\mathrm{MHz}$, so the circuit favors the first transmitter.

The resistor voltage is in phase with the current; the inductor voltage leads it by $\pi /2$ and the capacitor voltage lags by $\pi /2$. Large reactive voltages therefore nearly cancel in their phasor sum. A large inductor voltage is consistent with a small supply voltage: it is inductor voltage magnification, not power gain in a passive circuit.

Reducing the capacitance raises $f_0$ to $9.452\mathrm{MHz}$, between the two transmitters. Direct evaluation gives $7.02\mathrm{mV}$ and $7.22\mathrm{mV}$, respectively. Their similar detunings give approximately equal amplitudes; they are not exactly equal, partly because the numerator of $H$ also depends on frequency.

The two-tone output of a linear resonant circuit, for coherent monochromatic sources of equal input amplitude and arbitrary fixed input phases, isWith $A\simeq B=V_o$, this is approximatelyThus a $9.5\mathrm{MHz}$ rapid oscillation has a beat envelope repeating in magnitude at $1\mathrm{MHz}$. The signed envelope cosine has frequency $0.5\mathrm{MHz}$. Unequal amplitudes leave a small nonzero envelope minimum; independently drifting transmitter phases make the beating drift. No new frequency is generated by this ideal linear circuit.

The ratio of signal power delivered to a specified output load to signal power accepted at a specified input, with $P_{\mathrm{in}}>0$. An amplifier draws additional energy from its supply, so a power gain above one does not create energy.

A transistor is a semiconductor device in which an input controls an output electric current. Used with biasing and an electrical load, it can form an electronic amplifier or a switch. Its power gain comes from an external supply: the input signal controls the conversion of supplied energy into output signal power.

The point-contact transistor makes the idea tangible: two nearby contacts interact through one piece of semiconductor. A bipolar junction transistor instead uses two p-n junctions and a thin intervening region. The distinction matters both for the history and for choosing a physical model.

An original teaching model exposes the roles of input loading, output loading and the supply. Choose a small-signal model with positive input electrical resistance $r_i$, transconductance $g_m>0$, and output electrical resistance $r_o>0$. Its controlled output current source draws $g_m{v}_i$ from the output node toward the reference node. This chosen sign gives an inverting voltage gain; it is not the historical contact circuit's sign convention.

Assume a fixed operating point, negligible reverse coupling, a frequency range where coupling capacitors pass the signal, and negligible other capacitances and transit delays. A collector resistor $R_D$ connects the output to a steady ideal supply $V_{DD}$; an external electrical load $R_L$ is coupled through a capacitor, so it draws no direct current. The input source has series electrical resistance $R_s$. All voltages below are small departures from the operating point. An ideal fixed supply has zero incremental voltage and therefore becomes the reference node in the small-signal model.

Define $R_O=R_D\parallel r_o$ and $R_P=R_O\parallel R_L$, where $a\parallel b=ab/(a+b)$. Input voltage division and output Kirchhoff's current law giveConsequently the port voltage gain, overall source-to-output voltage gain, and load current gain magnitudes areThis current gain uses load electric current, not all the controlled-source electric current. It is also a different port definition from the historical emitter-to-collector current gain. Larger $R_s$ attenuates the input; smaller $R_L$ attenuates the output.

For a sinusoid input signal, use root mean square voltages. The power gain from accepted input signal power $P_i=v_i^2/r_i$ to useful load power $P_L=v_o^2/R_L$ isUnlike $|A_i|$, this tends to zero as either $R_L\to 0$ or $R_L\to \infty$. Differentiation givesThus matching the output electrical load maximizes delivered power in this unilateral resistive model. For fixed source voltage, input loading multiplies $P_L$ by $[r_i/(R_s+r_i){]}^2$; the accepted-input power gain above excludes the loss in $R_s$.

Choose $r_i=R_s=1\mathrm{k\Omega }$, $g_m=2\mathrm{mS}$, $r_o=20\mathrm{k\Omega }$, and $R_D=2\mathrm{k\Omega }$. Then $R_O=20/11\mathrm{k\Omega }$. With $R_L=2\mathrm{k\Omega }$ and $v_s=20\mathrm{mV}$ root mean square, the input is $10\mathrm{mV}$ and the inverted output has magnitude $19.0476\mathrm{mV}$ root mean square. Hence $P_i=0.100\mu \mathrm{W}$ and $P_L=0.181406\mu \mathrm{W}$: $G_P=1.81406$. These are calculated values of the chosen model, not historical measurements.

With the same parameters but $R_L=100\Omega$, $|A_i|=1.89573$ while $G_P=0.359381<1$. At $R_L=10\mathrm{k\Omega }$, $|A_v|=3.07692$ while $G_P=0.946746<1$. Neither current gain nor voltage gain alone establishes power gain.

For the direct current load line in the figure, choose $V_{DD}=6\mathrm{V}$, $I_Q=1\mathrm{mA}$ and $V_Q=4\mathrm{V}$, consistent with $V_Q=V_{DD}-R_D{I}_Q$. Locally extend the model as $I=I_Q+g_m{v}_i+(V-V_Q)/r_o$. With the external electrical load isolated at direct current, its intersection with $I=(V_{DD}-V)/R_D$ is $V=V_Q-g_m{R}_O{v}_i$. Keeping $|v_o|$ small compared with the available voltage headroom is essential; extrapolation to arbitrarily large signals would be unphysical.

At the chosen operating point the collector supply delivers $V_{DD}{I}_Q=6\mathrm{mW}$: $2\mathrm{mW}$ is dissipated in $R_D$ and $4\mathrm{mW}$ in the output device. Under a small signal, that energy flow is redistributed to provide load power. The incremental controlled current source hides the supply's steady energy flow; it does not create energy. A real complete amplifier also needs its input biasing network and has further losses. This calculation establishes neither its total efficiency nor a universal law for point-contact transistors.