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An input bias current is a direct-current component drawn by an input terminal of a real operational amplifier. Through a source resistance it produces a voltage error, and through the feedback capacitor of an op amp integrator it can produce an output ramp. Its magnitude and sign depend on the amplifier technology and operating conditions.

An integrated circuit combines interconnected electrical circuit elements in a fabricated structure. A monolithic integrated circuit forms its elements in or on one piece of semiconductor, rather than merely putting separately manufactured components into one enclosure. Transistors supply control in electronic amplifiers; resistors and capacitors can supply bias and timing. Jack St. Clair Kilby's patent describes making such functions from a common semiconductor body with compatible fabrication steps.^[ref]

The useful change is to fabricate connections and functions together. Shrinking an enclosure still leaves individual parts to position and join. Integration makes their relative arrangement part of a repeatable fabrication pattern. The distinction is functional as well as geometric: one small piece can implement a whole electrical circuit, while its external terminals expose only the connections needed by its surroundings.

An ideal operational amplifier with a grounded noninverting input, input resistor $R_i$, and parallel feedback resistor $R_f$ and capacitor $C$ implements a low-pass filter. The virtual short and Kirchhoff's current law give $V_{\mathrm{out}}/V_{\mathrm{in}}=-Z_f/R_i$, where $Z_f=R_f/(1+i\omega R_{f}C)$. The DC gain magnitude is $R_f/R_i$ and the cutoff frequency is $1/(2\pi R_{f}C)$. The capacitance reduces high-frequency feedback electrical impedance, causing the gain magnitude to fall.

An ideal operational amplifier with a series input resistor and capacitor, resistive feedback and a grounded noninverting terminal realizes the displayed high-pass filter. Virtual ground and Kirchhoff's current law give $H=-R_f/[R_i+1/(i\omega C)]$. Its phase is $-\pi /2-{\tan }^{-1}(\omega R_{i}C)$ in the $e^{i\omega t}$ convention. Equal resistances do not make its phase frequency independent.

Joule heating is the dissipative conversion of electrical work into internal energy in an electrical conductor. In a metal, the electric field supplies energy to charge carriers, which transfer it to the material. This makes an immersed resistor a useful calibrated heater and makes electrical resistance a source of losses in wiring. Wind-driven electric currents are a proposed source of heating of an inflated giant planet.

For a stationary, purely resistive Ohmic conductor, use the passive sign convention: $V$ is the terminal electric potential difference from the entry to the exit of positive electric current $I$. The absorbed electrical power and Ohm's law giveThe last two expressions assume $V=IR$ at the relevant material temperature. For constant electrical resistance and electric current over a duration $\Delta t$, the dissipated energy is $E_{\mathrm{diss}}=I^{2}R\Delta t$; otherwise use $E_{\mathrm{diss}}=\int V(t)I(t)dt$. In SI, $\mathrm{V}\mathrm{A}=\mathrm{W}=J/s$.^[ref]

Absorbed electrical power $VI$ is more general than Joule heating: a capacitor can store energy, a battery can change its chemical energy, and an electrical circuit can deliver mechanical work or emit light. Even a heated resistor need not show an indefinitely rising temperature: in a thermal steady state, its dissipated power leaves by heat transfer. With the heater inside a calorimeter boundary, electrical input crosses as work; the subsequent heat transfer to the room is a different term in the first law of thermodynamics.

In his 2000 lecture, Jack St. Clair Kilby describes an August 1958 comparison circuit made from separate silicon elements. Resistors were cut and etched; capacitors came from diffused, metallized semiconductor wafers. It tested whether semiconductor elements could perform all the required functions, but was still a discrete circuit. For the September prototype he used already diffused germanium material with contacts, cut into bars, and hand-applied wax masks to define a transistor and a distributed resistor-capacitor region. Applying supply power produced an oscillation.^[ref]

An inspected eyewitness interview describes the germanium on a microscope slide with protruding wires; changing its connections changed the oscilloscope display from a straight line to a sinusoid. This supplied a qualitative connected/disconnected comparison, not a calibrated record of amplitude, loading or repeatability.^[ref]

The inference was feasibility: integrated semiconductor elements could cooperate as an operating electrical circuit. The hand-wired germanium prototype did not yet supply the deposited integrated circuit interconnects of Robert Noyce's silicon design. Nor does a working prototype establish manufacturing yield or long-term reliability.

An ideal electrical circuit exchanges energy between an inductor and a capacitor. With no electrical resistance or applied drive, the electric charge obeys $L{q}^{\prime \prime }+q/C=0$, giving angular frequency ${\omega }_0=1/\sqrt{LC}$. A series RLC circuit adds dissipation and can include an external drive.

A null electrical circuit for measuring a lossy inductor against a standard capacitor. If opposite arms are $R_x+i\omega L_x$ and $(1/R_1+i\omega C{)}^{-1}$, and the remaining arms are $R_2,R_3$, balance requires their products to agree. Consequently $R_x=R_2{R}_3/R_1$ and $L_x=C{R}_2{R}_3$. Two independent real adjustments are necessary; changing the source frequency does not change the ideal balance conditions.

Membrane capacitance is the capacitance associated with separated electric charge across cell membrane. A local ideal capacitor has charging electric current $I_C=C_{m}dV/dt$. If a known actual membrane voltage change is $\Delta V$, integrating the charging transient after subtracting sustained ionic and leak components gives $C_m=\Delta V^{-1}\int I_{C}dt$. Pipette capacitance, finite patch-clamp series resistance and imperfect transient subtraction can bias the estimate. A capacitive transient is not by itself an ion channel opening.

A membrane potential can coexist with near electroneutrality in the bulk cytosol and external chemical solution. The small imbalance producing the voltage lies near the cell membrane; most positive and negative electric charges balance in the bulk.^[ref]

For a separate original capacitor model, choose a total membrane capacitance $C=5\mathrm{pF}$ and $V_m=-70\mathrm{mV}$. These are stipulated parameters, not measured properties of a particular biological cell, and this voltage is not asserted to equal the preceding $E_{\mathrm{K}}$. Model the separated electric charge relative to an uncharged reference byUsing the elementary charge $e=1.602176634\times 10^{-19}\mathrm{C}$, this representsexcess monovalent negative-charge equivalents inside. These can represent a deficit of positive ions, rather than the arrival of that many negative ions.

Independently choose an internal reservoir volume $\mathcal{V}=1\mathrm{pL}=10^{-12}\mathrm{L}$ containing $140\mathrm{mM}=0.140\mathrm{mol}{\mathrm{L}}^{-1}$ potassium, with balancing negative electric charge. With the Avogadro constant $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$, its potassium inventory isThus, if charging were entirely due to outward monovalent potassium transport, about one in $38600$ of these potassium ions would suffice. This is a comparison with the chosen potassium inventory, not a fraction of all charged particles or a measurement of spatial charge density. No shape, surface area or thickness has been inferred from the chosen capacitance and volume. The calculation illustrates why substantial voltage does not require a strongly charged bulk cytosol.

A gate, insulating oxide and semiconductor form a capacitor. Applied electric potential difference changes the near-surface distribution of charge carriers.

For an ideal operational amplifier with grounded non-inverting input, connect an input capacitor $C$ to the inverting input and a feedback resistor $R$ from output to that input. The virtual ground and Kirchhoff's current law give $C{\dot{V}}_{\mathrm{in}}=-V_{\mathrm{out}}/R$. The ideal response amplifies high-frequency noise, so practical differentiators restrict its operating band and maintain feedback stability.

An op amp integrator places a resistor $R$ between the input source and the inverting input of an operational amplifier, and a feedback capacitor $C$ between that input and the output, with the non-inverting input grounded. For an ideal operational amplifier, Kirchhoff's current law gives $v_{\mathrm{in}}/R=-Cd{v}_o/dt$, henceThe initial capacitor charge supplies the integration constant. A parallel feedback resistor $R_f$ replaces the transfer by $H(s)=-(R_f/R)/(1+s{R}_{f}C)$: the direct-current gain becomes finite while integration is recovered when $|s{R}_{f}C|\gg 1$. This limits offset-driven saturation. ^[ref]

An ideal inductor and capacitor in parallel have electrical admittance $1/Z=1/(j\omega L)+j\omega C$ for phasors with time dependence $e^{j\omega t}$. The branch currents cancel at angular frequency ${\omega }_0=1/\sqrt{LC}$, giving zero total electrical admittance and an infinite limiting electrical impedance. Below this resonance its electrical reactance is positive; above it the electrical reactance is negative. Losses make the peak finite.

A parallel-plate capacitor consists of two parallel electrical conductors with overlapping area $A$ and separation $d$. If the gap is small compared with both plate widths and is filled by a uniform material of positive permittivity $ϵ$, its capacitance is approximatelyThis neglects the fringing field of a capacitor near the plate edges. The interior electric field points from the plate of positive electric charge to the plate of negative electric charge, and is approximately uniform. ^[ref].

For a worked vacuum example, stipulate $A=0.01{\mathrm{m}}^2$, $d=0.001\mathrm{m}$, and vacuum permittivity ${ϵ}_0\approx 8.854\times 10^{-12}F/m$. This rounds the value in Fundamental Physical Constants — Complete Listing by National Institute of Standards and Technology and others. Within the approximation,At terminal electric potential difference $V=10\mathrm{V}$, the electric field magnitude is $E\simeq V/d=10^{4}V/m$, the plate electric charge magnitude is $Q=CV=0.8854\mathrm{nC}$, and the capacitor energy is $U=C{V}^2/2=4.427\mathrm{nJ}$. These numbers use the stipulated rounded vacuum permittivity; finite plates also have corrections from the fringing field of a capacitor.

The formula makes design changes easy to predict: doubling the overlapping area doubles the capacitance, and halving the gap doubles the capacitance, with permittivity fixed and the fringing field of a capacitor neglected. Doubling permittivity also doubles capacitance if the material uniformly fills the gap and remains linear. This lets a compact device store more separated electric charge at the same terminal electric potential difference.

The short derivation uses surface charge density $\sigma =Q/A$. In the interior, Gauss's law gives electric field magnitude $E\simeq \sigma /ϵ=Q/(ϵA)$. The electric potential difference is $V\simeq Ed=Qd/(ϵA)$, so dividing $Q$ by $V$ gives $C\simeq ϵA/d$. A narrow curved capacitor gap has the same leading local approximation when its width is much less than the curvature radius.

An ideal resistor, inductor and capacitor connected across the same voltage have admittance $Y=1/R+i(\omega C-1/(\omega L))$ in the $e^{i\omega t}$ convention. At ${\omega }_0=1/\sqrt{LC}$ the reactive branch currents cancel, giving minimum source current $V_0/R$ and maximum impedance $R$. This differs from a lossy series coil placed across a capacitor.

Model a spatially uniform cell membrane as a capacitor of membrane capacitance $C_m$ in parallel with membrane electrical resistance $R_m$ and a fixed leak reversal potential $E_L$. Exclude active ion channels, seal leak current and pipette capacitance. For electric current $I$ injected into the biological cell, electric charge conservation yieldsStarting at $V(0)=E_L$ and applying constant $I$ gives the ordinary differential equation solutionThe steady state rise is $I{R}_m$; the time constant is $R_m{C}_m$. With $I=50\mathrm{pA}$, $R_m=200\mathrm{M\Omega }$, $C_m=50\mathrm{pF}$ and $E_L=-70\mathrm{mV}$, the rise is $10\mathrm{mV}$ and ${\tau }_m=10\mathrm{ms}$. At one time constant the rise is $10(1-e^{-1})\mathrm{mV}\approx 6.321\mathrm{mV}$. The figure's uncompensated pipette voltage adds $I{R}_s=1\mathrm{mV}$ for $R_s=20\mathrm{M\Omega }$. The membrane voltage is continuous at step onset in this model, while the pipette voltage immediately acquires that access drop.

Checking the units is instructive: electrical resistance times capacitance has units $\Omega \mathrm{F}=\mathrm{s}$, and electric current times electrical resistance has units of voltage. Unlike the passive voltage clamp transient under patch-clamp series resistance, this ideal current clamp membrane time constant is set by $R_m{C}_m$, not $R_s{C}_m$. Strong current steps can recruit voltage-gated ion channels; the passive model then fails and cannot generate an action potential. Small pulses of both signs, stable access and repeated baseline measurements help test whether the passive approximation is appropriate.

Treat the membrane as a capacitor, and count the additional positive charge on its inner surface. With radius $r=10^{-6}$ m and length $\ell =10^{-4}$ m, including both end caps givesThe membrane capacitance is $C=C_{m}A=6.3460$ pF. Thus $\Delta Q=C\Delta V=5.3941\times 10^{-13}$ C, corresponding to $\Delta N=\Delta Q/e=3.3667\times 10^6$ monovalent sodium ions. Converting charge to amount of substance using the Faraday constant,Omitting the ends changes the answer to $0.01762$ mM, a one-percent correction. A large membrane potential change therefore requires only about $0.12\%$ of the intracellular sodium population to enter. The bulk liquid remains nearly electrically neutral; excess charge resides chiefly near the cell membrane. This explains why an action potential need not exhaust the concentration difference that drives it, and why fixed reversal potentials are a useful approximation during one spike. This membrane capacitive ion budget assumes sodium supplies the minimum net charging requirement. Simultaneous opposing electric currents can require more sodium entry; repeated spikes eventually require the sodium-potassium ATPase to maintain the gradients.

Suppose initially $c_1>c_2$. Positive ions initially move by diffusion from side 1 to side 2. Since negative counterions cannot cross, side 1 acquires a negative interfacial charge and side 2 a positive one. The resulting electrical force opposes further positive-ion transfer. The two contributions balance when the electrochemical potentials are equal; individual ions still cross in both directions, but there is no net electric current of the permeant chemical species.

A simple capacitor model makes the feedback quantitative. If $n$ moles of monovalent positive ions move from initially neutral side 1 to side 2, the charge on side 1 is $-Fn$, hence $V=-Fn/C$. The free-energy stationarity condition isAs $n$ increases, both the concentration imbalance and the voltage drive toward reversal diminish the original diffusion tendency. The equilibrium lies before the bulk concentrations become equal. This demonstrates why a nonzero Nernst potential can coexist with very little change in bulk composition and approximate bulk electroneutrality. A concentration difference alone does not maintain an electrical voltage if both counterions can freely follow each other; selective permeability is essential to this model.

The original Panel A is readable: the baseline is roughly $-45\mathrm{mV}$ and the peak about $+40\mathrm{mV}$, giving a voltage excursion near $85\mathrm{mV}$. Using the specific membrane capacitance and surface area,A coarser $100\mathrm{mV}$ spike estimate gives $6.2\times 10^6$ ions, illustrating the precision appropriate to the graph. These are the net monovalent charges needed to charge the capacitor if sodium supplies the inward charge; simultaneous opposing currents can make the actual sodium influx larger.

The cell contains $0.015\mathrm{mol}{\mathrm{L}}^{-1}\times 3\times 10^{-13}\mathrm{L}\times N_A\simeq 2.71\times 10^9$ intracellular sodium ions. The capacitive estimate adds only about $0.20\%$, or $0.029\mathrm{mM}$. Thus a spike does not equilibrate bulk concentrations. The sodium-potassium ATPase maintains the gradients over many spikes, exporting three sodium ions per two potassium ions imported per ATP; the estimated sodium load alone corresponds to at least about $1.8\times 10^6$ pump cycles. Rapid repolarization itself comes mainly from changing channel conductances, while sustained fast firing requires adequate energy supply and gradient restoration. The small stated volume and large area are used as supplied, without assuming a particular cell geometry.