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Consider a hypothetical star whose emitted specific intensity is exactly that of a uniform $5000\mathrm{K}$ blackbody. Assume transparent propagation with negligible Doppler shift and interstellar extinction, so geometric dilution multiplies every spectral interval by the same factor. The stipulated temperature givesUsing the converted frequency peak as though it maximized $B_{\lambda }$ would return $b/(c/{\nu }_{\max })\simeq 2841.26\mathrm{K}$, a substantial error caused solely by mixing spectral conventions.

Now stipulate a photodetector with constant quantum efficiency $\eta$ over the relevant wavelengths, collecting equal narrow wavelength bins. For fixed accepted projected area times solid angle $G$, with dilution included in $G$, its expected photon count rate density isThis follows by dividing the received power by the photon energy $hc/\lambda$. It peaks where $x=4(1-e^{-x})$, at $x=3.9206903949\dots$, giving $0.73394057\mu \mathrm{m}$ for the same model star. Constant quantum efficiency is not constant energy spectral responsivity: the latter includes the wavelength-dependent photon energy conversion. For example, $G=10^{-12}{\mathrm{m}}^2\mathrm{sr}$, $\eta =0.5$ and a $1\mathrm{nm}$ bin centered at $0.600\mu \mathrm{m}$ give approximately $1.93\times 10^{10}$ detected photons per second in the narrow-bin approximation. These are calculated model values, not observations of a named star.

A more general linear detector channel has outputwhere $S_{i,0}$ is the background reading and $R_i$ includes spectral selection, optical transmission and detector spectral responsivity. A raw channel maximum need not locate either Planck function maximum. Recovering a temperature requires an appropriate calibrated response and an emission model. For a real stellar atmosphere, opacity and radiative transfer can reshape the spectrum; interstellar extinction can reshape it again. A single peak does not establish a unique physical temperature without those assumptions.

The charge-transfer efficiency is the fraction of a packet delivered into its intended next site or time slot in one transfer, under stated conditions. It is distinct from quantum efficiency, which concerns conversion of incident photons into collected charge carriers. Define the complementary inefficiency as $ϵ=1-{\eta }_{\mathrm{tr}}$.

In an original model with constant $0\le {\eta }_{\mathrm{tr}}\le 1$, let $N_j$ denote the mean number of original packet electrons still delivered on time after $j$ transfers. Assuming no reinjection into that packet,Repeated substitution proves the result; for different efficiencies the fraction is ${\prod }_{j=1}^n{\eta }_j$. For $ϵ\ll 1$, a Taylor expansion of $\ln (1-ϵ)$ gives ${\eta }_{\mathrm{tr}}^n\simeq e^{-nϵ}$, while the simpler $1-nϵ$ requires $nϵ\ll 1$.

Choose $N_{\mathrm{in}}=10000$ and $n=2000$. For ${\eta }_{\mathrm{tr}}=0.9999$, the expected on-time output is $8187.23$ electrons; for ${\eta }_{\mathrm{tr}}=0.99999$, it is $9801.99$. These are model means, so fractional electron counts do not describe an individual packet. Even a small per-transfer imperfection can imprint distance-dependent image errors.

If every original electron independently stays on time with probability $p={\eta }_{\mathrm{tr}}^n$, a fixed initial count gives a binomial distribution with variance $N_{\mathrm{in}}p(1-p)$. An initial Poisson distribution with mean $\mu$ instead gives an on-time Poisson distribution with mean and variance $\mu p$, by Poisson thinning. A deterministic attenuation formula alone does not specify noise. Real trapping can correlate packets and depends on timing, illumination and temperature; the independent constant-efficiency model then fails.

For an isolated fluorescent molecule at displacement $\mathbf{r}$ from the scanning optical focus, write $E(\mathbf{r})$ for relative excitation light intensity and $D_a(\mathbf{r})$ for the fraction of its collected emission passed by a centred detection pinhole of radius $a$. The expected response, apart from a common brightness and collection factor, isHere $\mathbf{u}$ is position in the detection-pinhole plane and $p_{\mathrm{em}}$ is the emitted light intensity distribution there conditional on collection, with units of inverse area. This point spread function normalization retains throughput: $D_a$ is a fraction, not a curve rescaled to have peak one. The product expresses excitation followed by detection; it is an intensity/probability product, not multiplication of coherent wave amplitudes from different molecules. ^[ref]

Assume linear, unsaturated fluorescence, independent emitters whose light intensities add, a stationary specimen, fixed collection efficiency and quantum efficiency, and negligible changes from scattering, absorption or bleaching during a scan. Different excitation and emission wavelengths generally give different excitation and detection responses. For a sufficiently small pinhole, $D_a$ is approximately its area times $p_{\mathrm{em}}$ at the centre; this is the familiar excitation-times-detection point spread function. It is not universally the square of one point spread function. A spatially invariant response further allows the image to be expressed as a convolution with the emitter distribution; depth-dependent optical aberrations can invalidate that approximation.

For an explicitly synthetic axial example, put $t=z/z_0$ and choose an on-axis excitation $E(t)=1/(1+t^2)$, inspired by a Gaussian beam. Stipulate a circular Gaussian function for the detection-plane light intensity, with standard deviation $s(t)=s_0\sqrt{1+t^2}$. In dimensionless coordinates $\mathbf{v}=\mathbf{u}/s_0$, put $\alpha =a/s_0$. ThenThe expression for $D_{\alpha }$ follows by integrating $p$ over a disk using polar coordinates. The assumptions specify a toy model with matched axial scales; its parameters are not a calibration for a real objective optical lens. It omits Airy disk rings, vector diffraction, and unequal excitation and emission wavelengths.

For a vanishingly small pinhole, the peak-normalized synthetic response tends to $(1+t^2{)}^{-2}$, whose full width at half maximum is $2\sqrt{\sqrt{2}-1}{z}_0\simeq 1.2872z_0$. With all collected emission accepted it tends to $(1+t^2{)}^{-1}$, of width $2z_0$. Meanwhile $H_{\alpha }(0)=1-e^{-{\alpha }^2/2}\to 0$ as the pinhole closes. The narrower normalized curve therefore accompanies a vanishing detected signal; peak normalization must not conceal that loss.

Detector responsivity relates a detector's output to the incident radiant input under stated conditions. For a photodiode illuminated at a specified wavelength, current spectral responsivity iswhere $I_0$ is the background current and $P_{\lambda }$ is incident optical power. Its units are $\mathrm{A}{\mathrm{W}}^{-1}$. NIST's detector-calibration publication develops spectral radiant-power responsivity as the quantity transferred from optical standards to photodetectors.^[ref]

For an original example, $20\mu \mathrm{W}$ incident power produces $10\mu \mathrm{A}$ net photocurrent, giving $R_{\lambda }=0.5\mathrm{A}{\mathrm{W}}^{-1}$. If readout electronics convert that current to $1\mathrm{V}$, the complete system's voltage responsivity is $50,000\mathrm{V}{\mathrm{W}}^{-1}$. A value without its output quantity and units can therefore be misleading.

For a photodiode with no internal multiplication, let quantum efficiency $\eta$ be collected electrons per incident photon. A photon has energy $hc/\lambda$, with Planck constant $h$, speed of light $c$ and elementary charge $e$. Thus a photon arrival rate $\Phi$ gives $P=\Phi hc/\lambda$ and net current $I=\eta e\Phi$. ThereforeAt $800\mathrm{nm}$, the original $0.5\mathrm{A}{\mathrm{W}}^{-1}$ example corresponds to $\eta \simeq 0.775$. This relation assumes the stated charge-collection model; internal gain changes the interpretation.

Responsivity can depend on wavelength, irradiation geometry, bias and temperature. Detector linearity tests whether a response ratio stays constant as input level changes; detector saturation can make it strongly level-dependent. In a nonlinear response, the ratio of corrected output to input and the local transducer sensitivity are different quantities. Retain the wavelength, beam conditions, readout convention and measurement uncertainty when reporting a responsivity measurement.

Extreme ultraviolet denotes short-wavelength ultraviolet between longer-wavelength ultraviolet and X-rays; precise band boundaries depend on convention. Its photons have sufficient energy to cause strong optical absorption in many materials that transmit visible light.

This makes beam transport an apparatus problem: an ordinary air path or visible-light window can absorb the beam, and suitable reflective optics can be more useful than an ordinary transmitting optical lens. Before interpreting a weak detector signal, compare a blocked-beam reading with the open-beam reading and account for wavelength-dependent detector quantum efficiency.

Fluorescence quantum yield counts emitted fluorescent photons per absorbed excitation photon. It differs from quantum efficiency, which describes the photodetector.^[ref]

For an original constant-rate model, let $k_{\mathrm{r}}$ be the radiative decay rate and $k_{\mathrm{nr}}$ the sum of competing rates without fluorescent photon emission. An initial population $N_0$ in the emitting excited state obeys the rate equationIntegrating $k_{\mathrm{r}}N(t)$ over all time givesHere every absorbed excitation is assumed to reach this emitting population; additional losses before that stage require another factor. Under steady weak excitation, absorbed photon rate $R_{\mathrm{abs}}$ and total detection fraction $\eta$ give $C=\eta \Phi R_{\mathrm{abs}}$. The count rate therefore mixes excitation, yield, and detection, whereas $\tau$ specifies a decay timescale.

Indium antimonide is a narrow-gap semiconductor made of indium and antimony, with chemical formula $\mathrm{InSb}$. Its small band gap permits absorption of comparatively low-energy infrared photons, making it useful in infrared detection.

For an ideal interband threshold, $hc/\lambda$ must reach the gap; actual detector response also depends on optical coupling, carrier collection and temperature. Thermal excitation across a small band gap can supply unwanted carriers, so an operating detector's dark current is an essential measurement alongside quantum efficiency. ^[ref]

Masayuki Senoh coauthored work on blue light-emitting diodes using an indium gallium nitride/aluminium gallium nitride double heterostructure. The devices emitted near $450$ nm and reached $2.7\%$ external quantum efficiency at $20$ mA electric current.^[ref]

A charge-coupled device converts absorbed photons into stored charge in semiconductor potential wells. An exposure accumulates a pixelated image; clocked electrodes transfer rows toward a serial register and output amplifier. Quantum efficiency, pixel sampling and optical throughput connect collected photons to the expected signal. In ordinary optical imaging it primarily counts photoelectrons rather than independently measuring every photon's energy.

The CCD noise budget combines source and sky shot noise, dark-current fluctuations and read noise. For an aperture containing $n$ pixels, known background expectation $N_b$, total dark expectation $N_d$, source expectation $N_s$ and per-pixel read noise $r$,Estimating the background from a finite separate region adds its own uncertainty. Cooling reduces dark current; longer integration reduces the relative importance of one read but can encounter saturation, cosmic rays or changing conditions. Full well capacity, amplifier gain, linearity, blooming and charge-transfer efficiency limit useful detector dynamic range.

Bias frames establish the electronic offset; matched dark frames characterize time- and temperature-dependent dark signal. Flat fields calibrate pixel-to-pixel response and illumination, with appropriate attention to wavelength. Bad-pixel masks, repeated exposures and dithering help reject defects and cosmic rays. A calibrated PSF and astrometric solution are needed for reliable photometry and positions. High quantum efficiency and stable linear integration are advantages, but an uncalibrated count array is not yet an astronomical flux map.

The detector is intended for arcsecond-scale astronomical seeing, rather than the telescope's diffraction limit. A circular aperture of this area has diameter $1.13\mathrm{m}$, giving a Rayleigh criterion angle about $0.11$ arcsec at the observing wavelength. One-arcsecond pixels conventionally sample an astronomical seeing profile of roughly two arcseconds across with two samples; a one-arcsecond profile would be undersampled. The pixel scale alone does not determine an exact astronomical seeing width.

Use the supplied surface brightness as radiative flux integrated over the photometric passband: its units contain no inverse hertz. Treat $h\nu$ as the representative photon energy, and assume unit throughput and quantum efficiency. A genuinely frequency-specific brightness would require a bandwidth. With angles measured consistently in arcseconds, an annulus has area $2\pi rdr$ square arcseconds, soThis is flux per collecting area; collected energy is $AtF$. There is no extra conversion to steradians when both brightness and angular area use square arcseconds.

For Poisson photon counting, a pixel with expected count $N$ has energy $e=Nh\nu$ and standard deviation ${\sigma }_e=h\nu \sqrt{N}$. Thus the local four-sigma rule gives $N=4\sqrt{N}$, or 16 photons at the boundary. With pixel area $\Delta \Omega =1{\mathrm{arcsec}}^2$,Consequently $r_m=(1-16/32.6991)/k=10.2138$ arcsec andIntegrating to the model's outer radius $R=1/k=20$ arcsec instead givesThe pixel cut retains 51.6% of this total. It is an imposed isophotal photometry rule, not a fundamental limit: averaging annuli or fitting a justified profile can recover signal below the per-pixel cut.

In practice, sky background photons add shot noise even after subtraction of the mean sky. read noise, dark current, imperfect flat-field correction, throughput calibration and atmospheric extinction also matter. At large radius the dominant uncertainty can be a tiny error in the sky level, scattered light or the wings of the point spread function, rather than source photon statistics. Mask foreground stars and background objects, compare blank-sky regions and dithered exposures, and check profile recovery by inserting artificial extended sources. astronomical seeing and finite pixels convolve the intrinsic profile; model that response when extrapolating the total.

The luminosity convergence of an outer surface-brightness profile follows directly: $I\propto r^{-p}$ contributes $dF\propto r^{1-p}dr$, hence $F(<R)\propto R^{2-p}$ at large radius for $p<2$. For either shallow index considered, ever larger annuli dominate: extending the law to infinity gives an unphysical divergent luminosity. Such a law must eventually steepen or truncate. Over a finite range, however, substantial light can lie below a surface-brightness selection threshold and be mistaken for sky. This biases galaxy luminosities, the luminosity function and the inferred diffuse light budget; distant diffuse galaxies become especially incomplete because bolometric cosmological surface-brightness dimming scales as $(1+z{)}^{-4}$. An assumed finite-aperture total is therefore a model-dependent quantity.

In a CCD, absorbed optical photons create photoelectrons collected in electrostatic wells beneath gate electrodes. During readout, clock voltages move the charge packets row by row into a serial register; serial clocks move them one pixel at a time to an output node. The node converts charge to a voltage, approximately $\Delta V=Q/C$, which is amplified and digitized. pixel addresses are recovered from the row and serial-clock sequence. Resetting and sampling the output between packets establishes the baseline. Efficient transfer preserves charge; trapping or incomplete transfer produces loss and trailing. This is CCD clocked charge transfer, not independent wiring of an amplifier to every imaging pixel.^[ref]

dark current is thermally generated charge accumulated even without illumination. If its mean rate is $d$ electrons per pixel per second, a single exposure adds mean $dt$ and Poisson variance $dt$. Cooling the detector and using low-defect material suppresses this generation. A dark calibration removes the reproducible mean pattern, but cannot remove the dark electrons' random shot noise in the science exposure. read noise comes from measuring the output packet and resetting the sense node, amplifiers and electronics; write its rms as $r$ electrons per read. correlated double sampling reduces reset and correlated offsets by subtracting the baseline measurement from the signal measurement. Low-noise electronics and an appropriate slower readout reduce the remaining contribution. The CCD noise budget for $M$ separate exposures combined over total time $t$ contains independent dark and read contributions $dt+M{r}^2$, not $(dt{)}^2+M^2{r}^2$. Long exposures reduce repeated read noise, but saturation, cosmic rays and tracking requirements constrain the exposure length. On-chip binning can reduce read-noise penalties for a fixed binned area at the cost of angular sampling.

For the source and sky calculation, define the count conversion factor $K=At\Delta \nu /(h\nu )$ and the source flux per pixel $s_{\nu }=F_{\nu }/N_{\mathrm{pix}}$. The ideal source and background electron counts are $S=K{s}_{\nu }$ and $B=K{b}_{\nu }{\Omega }_{\mathrm{pix}}$. These assume unit throughput and quantum efficiency, approximately constant quantities across the band and a well-determined mean sky level from many background pixels. Source and sky photons have independent Poisson fluctuations, so the variance after subtracting that mean sky is $S+B$. Subtracting a background removes its mean, not its photon noise. ThereforeThus the required expression follows directly from adding photon-count variances. Finite sky-estimation noise, dark current and read noise would add further variance. A nonunit common efficiency replaces $A$ by the effective collecting area; a nonflat spectrum requires the corresponding frequency integral. For example, estimating the sky from one equally exposed background pixel would add an extra $B$, whereas a large sky region makes that extra term small.

With $A=\pi (1\mathrm{m}{)}^2$ and the supplied photon frequency, the sky flux per pixel is $b_{\nu }{\Omega }_{\mathrm{pix}}=1.75\times 10^{-32}\mathrm{W}{\mathrm{m}}^{-2}\mathrm{H}{\mathrm{z}}^{-1}$. Multiplying arcsecond-squared pixel area by brightness per arcsecond squared already gives a flux: no additional steradian conversion belongs in this product. Rearranging givesUsing $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ yields

The first case is mainly source-shot-noise limited, while the last is strongly sky limited. At high source brightness $t/R^2\propto F_{\nu }^{-1}$; at low source brightness it approaches $F_{\nu }^{-2}$. Reducing source flux by ten can therefore require nearly a hundred times the exposure once sky noise dominates. These are per-pixel values for the ideal model, not guaranteed integration times including calibration, cosmic rays or read noise.

Increasing the diameter increases both source and sky photon rates as $D^2$ for fixed angular pixels. In this ideal limit the required exposure decreases as $D^{-2}$, whereas the stated telescope cost increases as $D^{2.5}$: collecting area grows only as cost to the power $0.8$. A larger ground telescope does not remove atmospheric photons or recover arbitrary wavefront information lost in a long seeing-limited exposure. Space observations avoid atmospheric airglow, weather and atmospheric turbulence, and offer stable high-resolution imaging and access to very faint structure. They still have zodiacal light, instrumental backgrounds and their own calibration limits. adaptive optics can provide excellent ground-based resolution in suitable fields, but its guide-star, wavelength and field-dependent limitations do not remove the need for space-based continuum imaging. The need is for complementary low-background, stable and high-resolution measurements; the cost exponent alone is not a proof that every imaging programme is cheaper in space.

For X rays, one fully absorbed photon produces many electron-hole pairs in silicon, rather than one optical photoelectron count. At moderate surface brightness, the probability of two photons landing in the same event island during one CCD frame is small. The charge packet or sum over neighbouring split-event pixels then supplies a pulse height for each photon, while its pixel position supplies the image coordinate. Calibrating pulse height against energy produces an event list from which a spectrum can be accumulated for any spatial region. photon pile-up at larger occupancies merges photon energies and corrupts both spectra and event grades, so fast enough frames and event-pattern screening are essential.^[ref]

A measured silicon pair-creation energy $w\simeq 3.65\mathrm{eV}$ gives $N_e=E/w\simeq 2.74\times 10^3$ electrons for a $10\mathrm{keV}$ photon. Let $F$ be the Fano factor of pair production, $n_r$ the number of independently read pixels in the event sum, and $r$ the rms read noise per pixel. ThenThe Fano-limited X-ray CCD energy resolution using the representative silicon value $F\simeq 0.12$ gives an ideal Fano-limited $\Delta E_{\mathrm{FWHM}}\simeq 0.16\mathrm{keV}$ at $10\mathrm{keV}$, or resolving power $E/\Delta E\simeq 64$. Finite read noise, charge loss and event summing broaden this; an order-of-magnitude practical estimate is a few tenths of a keV, but the particular detector's value is not determined by the given optical-pixel parameters. These $w$ and $F$ values are external detector properties, not supplied question data. Treating every pair as independent Poisson production instead would give $0.19\mathrm{keV}$ rms, or $0.45\mathrm{keV}$ FWHM; stating which width is meant avoids conflating those estimates. Radiation-induced charge-transfer inefficiency requires additional gain and spectral-response calibration.^[ref]

With constant detected rates, $Q=R_{Q}T$, $B=R_{B}T$. Squaring the previous signal-to-noise ratio in photon counting givesFor $R_Q>0$, nonnegative background and target $Z>0$, choose the positive root:The other root is nonpositive. With no read noise, this reduces to $T=Z^2(R_Q+2R_B)/R_Q^2$; in a read-dominated short exposure it approaches $Z\sqrt{2n}r/R_Q$. If $R_Q=0$, no positive target stellar signal-to-noise is attainable. A zero target can be assigned zero exposure, with its limiting interpretation when all noise terms also vanish. Real quantum efficiency or multiple reads modifies the rates or read-variance term before solving the quadratic.

Use the ideal detection convention of one collected electron per detected photon; otherwise replace photon totals by detected electron totals using the quantum efficiency. Let $A$ be the summed measurement on the star patch and $B^{\prime }$ the independent background-only patch. After additive offsets are removed, their means are $Q+B$ and $B$. Independent Poisson distribution counts have variance equal to their mean, and a single read of $n$ detector pixels contributes $n{r}^2$ to each patch's variance. ThereforeThe two factors of two arise from the independently measured background and the second patch's read noise, not from doubling the stellar signal. The formula assumes equal background means, independent detector pixels and patch measurements, one read per patch, and negligible dark current or other noise. Several exposure reads or correlated detector pixels require their own variance and covariance terms.

Let $N$ be the input photoelectron count and $\mu =\eta Q$ its mean, with quantum efficiency $\eta$ and mean incident photon count $Q$. Independent arrivals give a Poisson distribution, so $\mathrm{Var}N=\mathbb{E}N=\mu$. Conditional on $N$, the preceding gamma distribution gives $\mathbb{E}[X\mid N]=gN$ and $\mathrm{Var}(X\mid N)=g^{2}N$. The law of total variance therefore yieldsOne contribution is the ordinary photon shot noise; the other is multiplication excess noise. Thus, neglecting read noise and backgrounds,This is an excess-noise factor $\sqrt{2}$ in analog operation: actual photon conversion efficiency has not been halved. At low occupancy, thresholding each pixel as zero or one event can avoid most multiplication noise, but high arrival rates produce coincident events that cannot be counted separately. That is why the high-rate result concerns charge measurement rather than ideal binary photon counting.

Treat $Q$ and $B$ as mean detected counts, or as photon counts with unit quantum efficiency. Let the independent patch measurements be $X\sim \mathrm{Poisson}(Q+B)$ and $Y\sim \mathrm{Poisson}(fB)$. For the specified unweighted background subtraction $\hat{Q}=X-Y$,The first error is a fixed bias of an estimator; the last expression is the sum of independent photon shot noise variances. Since $Q,B$ are already patch totals, no extra factor of the pixel count $n$ is needed, and read noise is neglected.

To obtain the systematic-error ceiling requested in the following clause, define the accuracy measure $Z$ using total root-mean-square error relative to the true source count. By the bias-variance decomposition of mean squared error,For $f\simeq 1$, the shot-noise term is approximately $Q+2B$, but the mismatch term must be retained.

There is a terminology qualification: the usual variance-based signal-to-noise ratio in photon counting is $Q/\sqrt{Q+(1+f)B}$ and does not include a fixed bias as noise. The printed next-part limit requires the root-mean-square accuracy convention above. A deterministic background mismatch contributes $b^2$ to mean squared error, not to the statistical variance.

The optical telescope supplies collecting area and spatial selection, the spectral selector supplies a passband, and the photodetector converts received radiant flux into a measured signal. For an extended source, a useful schematic photon count iswhere $A\Omega$ is the etendue, $\tau$ the integration time, $\eta$ the net transmission and quantum efficiency, and $L_{\lambda }$ the spectral radiance. An illustrative independent-noise model givesHere $N_b$ and $N_d$ denote background and dark counts, and ${\sigma }_{\mathrm{read}}$ is readout standard deviation. This makes the roles of the three components explicit.

A larger optical telescope collects more photons at fixed angular pixel size and improves the diffraction-limited resolution. Its limits include optical aberrations, alignment, launch size, stray light and pointing. Conservation of etendue constrains simultaneous collecting area and angular acceptance. Increasing spatial resolution by narrowing a pixel's solid angle reduces its photons.

A narrower spectral passband or higher spectral resolving power separates diagnostic features and rejects unwanted background, but usually reduces photons per channel and hence signal-to-noise ratio. Selector transmission, leakage, slit width, detector format and the required time cadence limit the trade. A wider band helps only insofar as it adds useful signal rather than mostly background.

The photodetector's quantum efficiency, dark current, read noise, dynamic range and response time determine conversion losses and added noise. Cooling can suppress thermal background and dark current, but cannot remove photon shot noise. Longer integration improves a stationary signal's signal-to-noise ratio until motion, saturation, drift or required cadence becomes limiting.

The incident photon rate is ${\dot{N}}_{\gamma }=P/(hc/\lambda )\simeq 2.30\times 10^{15}{\mathrm{s}}^{-1}$. The stated quantum efficiency is $\eta =0.005$, so the emitted electron rate is $\eta {\dot{N}}_{\gamma }\simeq 1.15\times 10^{13}{\mathrm{s}}^{-1}$. If all emitted electrons are collected, the photocurrent magnitude isThe light power controls the number of events; the work function already determined the kinetic energy per emitted electron and should not be subtracted from the photon energy when counting incident photons. Incomplete collection would reduce the measured photocurrent below this value.

Photocurrent is the electric current attributable to illumination. Here $I_{\mathrm{ph}}\ge 0$ denotes its magnitude after background subtraction. For a photodiode without internal multiplication, let $P$ be incident radiant flux concentrated at wavelength $\lambda$, and let $\eta$ be external quantum efficiency, including losses before collection. An incident photon has energy $E_{\gamma }=hc/\lambda$, with Planck constant $h$ and speed of light $c$. Dividing energy per second by energy per photon gives the incident photon rate ${\Phi }_{\gamma }$. Multiplying by $\eta$ gives the mean collected pair rate ${\dot{N}}_{\mathrm{pair}}$; multiplying by elementary charge $q$ givesThe factor is $q$: each collected pair replenishes an electron at one contact and removes an electron at the other. A cross-section of the external wire therefore counts one transferred electron per completed pair, consistent with charge conservation. The electron and electron hole are two internal charge carriers, not two independent transfers through that wire.

Choose an illustrative photodiode model with $\lambda =650\mathrm{nm}$, $\eta =0.75$ and $P=1.00\mu \mathrm{W}$. Using the exact SI values $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ and $q=1.602176634\times 10^{-19}\mathrm{C}$ givesA mean fraction $0.75$ does not require each fourth photon to be lost in a repeating pattern. It specifies a mean conversion rate. Nor does $0.75$ mean $0.75\mathrm{A}{\mathrm{W}}^{-1}$: quantum efficiency is dimensionless, whereas spectral responsivity compares electric current with power.

At fixed radiant flux, longer-wavelength photons have less energy, so more arrive per second. This explains the rising straight line when quantum efficiency is held constant. A real photodiode's quantum efficiency varies with wavelength; the plotted assumption cannot be extended past its band gap absorption cutoff. At fixed incident photon rate instead, $I_{\mathrm{ph}}=\eta q{\Phi }_{\gamma }$ has no additional factor of $\lambda$.

A photodetector converts incident light into a measurable signal, often an electric current or stored electric charge. A photodiode collects carriers produced by absorbed photons; a photomultiplier first releases photoelectrons and then amplifies their number.

A basic characterization measures the output with the beam blocked and with several known incident radiant powers. The slope gives spectral responsivity at the chosen wavelength; the blocked signal reveals offsets and dark current. Quantum efficiency, response speed and noise-equivalent power describe different limitations. Detecting a weak signal requires separating its contribution from both background photons and the detector's own fluctuations.

A photodiode is a semiconductor diode producing photocurrent from photon absorption. Exciting an electron across the band gap leaves an electron hole. The depletion region's electric field separates the charge carriers; others can arrive by diffusion before electron-hole recombination. Contacts complete the electrical circuit.^[ref]

The useful bridge is from light to a measurable electric current: spectral responsivity converts incident radiant flux into photocurrent, while quantum efficiency counts collected signal charge carriers per incident photon. These answer different questions: how much electric current one watt produces, and how many incident photons contribute to the signal. The worked calculation below makes their relationship explicit.