Phototransistor (HPT and Photo-FET)

1. Theory

• Definition (HPT): A phototransistor is a bipolar transistor, usually operated with a floating base, whose base-collector junction absorbs the light; the photogenerated carriers form the base current $I_B=I_{ph}$, which the transistor multiplies to $I_C=\beta I_B$, and a heterojunction phototransistor (HPT) has an emitter of wider band gap than the base.
• Definition (Photo-FET): A field-effect phototransistor has a semiconductor channel (a thin film, a nanowire or a two-dimensional layer) whose current changes with light by photoconductive gain and by photogating, the shift $\Delta V_{TH}$ of the threshold voltage by photogenerated carriers that are trapped.
• Current gain: In the HPT, $\beta ={\tau }_R/{\tau }_{tB}$ is the ratio of the recombination lifetime to the base transit time, and typical values are $10^2$ to $10^5$, with a measured $\beta \approx 200$ for a SiGe device and about 2000 for an InP/InGaAs device.
• Heterojunction advantage: The wide-gap emitter multiplies the gain of the homojunction by $e^{\Delta E_g/k_{B}T}$, which is $5\times 10^5$ for the 0.34 eV of InGaAs/InP (computed).
• Gain-speed trade-off: The optical gain raises the responsivity from 26 mA/W to 1.3 A/W in a SiGe device, at the price of a lower transit frequency, 25 GHz against 48 GHz of the control transistor, and of a $3\mathrm{dB}$ optical bandwidth of 550 MHz.
• Absorber position: Light absorbed in the base-collector region is amplified, whereas carriers generated in the substrate or far in the collector give a photocurrent without gain and a slow tail.
• Photo-FET gain: The internal gain $G_{int}={\eta }_{sep}{\tau }_{lifetime}/{\tau }_{transit}$ is that of the photoconductor, and photogating adds the term $I_{ph}=g_m\left|\Delta V_{TH}\right|$.
• Figures of merit: Responsivity, gain, gain-bandwidth product, dark current, noise and linearity, and for the photo-FET the gate dependence of the responsivity.

2. Working Principle

• Absorption and base current: Photons absorbed in the reverse-biased base-collector junction create pairs, the electrons go to the collector and the holes are held in the base, so that the photocurrent acts as the base current.
• Amplification: The accumulated holes lower the base-emitter barrier, many more electrons are injected from the emitter than the holes that were supplied, and these cross the base into the collector, which gives $I_C=\beta I_B$.
• Operating modes: In the photodiode mode the emitter is open and only the junction photocurrent flows, and in the transistor mode the base is open (or biased with a current) and the junction photocurrent is amplified; a base current raises the emitter current and with it the bandwidth.
• Power dependence: At low power the recombination through traps limits the gain, it rises as the traps saturate, and at high power the Kirk effect (base push-out) increases the transit time and lowers it, while the bandwidth rises with the power.
• Photoconductive channel: In the photo-FET the photogenerated carriers raise the conductance of the channel, and the gain is the ratio of the lifetime to the transit time $L^2/(V_{DS}{\mu }_{FE})$.
• Photogating: Carriers trapped in long-lived surface or interface states act as an extra gate, shift the threshold voltage and raise the channel current, and a fractional exponent $\gamma <1$ of the photocurrent against power indicates this effect.
• Gate control: The gate voltage sets the threshold, the carrier density and the occupancy of the traps, so that the responsivity of the same device can be tuned electrically.

3. Applications

• Low-voltage optical gain in SiGe BiCMOS receivers: A phototransistor made in the heterojunction bipolar process gave 1.3 A/W at 850 nm at 3 V and operates from a 1.8 V supply in a receiver of 3.5 Gb/s with the amplifier on the same chip.
• Gate-tunable photodetection: Photo-FETs of layered semiconductors, in which the gate voltage changes the responsivity of one device from 0.5 A/W to about 5 A/W.

Bipolar and Heterojunction Phototransistor (HPT)

Name	Formula	Related Formulae	Variables/Constants
Base photocurrent, collector current and responsivity	$I_B=I_{ph}=\left(1-R\right)\left(1-e^{-\alpha d}\right)\frac{q\lambda }{hc}{P}_0,I_C=\beta I_B$	$I_E=\left(1+\beta \right)I_B$ $\mathfrak{R}=\beta \left(1-R\right)\left(1-e^{-\alpha d}\right)\frac{q\lambda }{hc}$	$I_B:basecurrentofthefloatingbase$ $I_{ph}:photocurrentofthebase-collectorjunction$ $R:reflectance$ $\alpha :absorptioncoefficient$ $d:thicknessoftheabsorbinglayer$ $P_0:incidentopticalpower$ $\beta :currentgain$ $I_C,I_E:collectorandemittercurrents$ $\begin{array}{r}(aSiGephototransistorgave4.6\mu \mathrm{A}for\\ 180\mu \mathrm{W}inthephotodiodemode(26mA/W)\\ and235\mu \mathrm{A}inthetransistormode(1.3\\ A/W),aratioof50,computed)\end{array}$
Current gain and heterojunction factor	$\beta =\frac{{\tau }_R}{{\tau }_{tB}},{\beta }_{het}={\beta }_{homo}{e}^{\Delta E_g/k_{B}T}$	${\tau }_R=\frac{\Delta n}{{\sum }_i{U}_i}$ ${\tau }_R=\beta {\tau }_{tB}$	${\tau }_R:recombinationlifetimeinthebase$ $\begin{array}{r}{\tau }_{tB}:transittimeofthemajoritycarriers\\ throughthebase\end{array}$ $\begin{array}{r}\Delta E_g:differenceofthebandgapsof\\ emitterandbase(about0.34eVfor\\ InGaAs/InP)\end{array}$ $k_B:Boltzmannconstant$ $T:temperature$ $\Delta n:excesscarrierconcentration$ $\begin{array}{r}{U}_i:recombinationratethroughtraplevel\\ i,(e^{\Delta E_g/k_{B}T}=5.1\times 10^{5}for0.34eVat300\\ K;\beta =2000and{\tau }_{tB}=2\times 10^{-10}sgive{\tau }_R=\\ 0.4\mu \mathrm{s},computed)\end{array}$
Dependence of the gain on the optical power	$\Delta n=\frac{\mathfrak{R}{P}_{opt}}{v_{D}Aq},{\tau }_R=\frac{\Delta n}{{\sum }_i{U}_i}$	$U_i=\frac{{\sigma }_n{\sigma }_p{v}_{th}{N}_t\left(np-n_i^2\right)}{{\sigma }_n\left(n+n_1\right)+{\sigma }_p\left(p+p_1\right)}$ $N_t=N_{tA}{\left(E_t-E_i\right)}^2+N_{tC}$	$v_D:diffusionvelocityofthecarriers$ $A:detectorarea$ $P_{opt}:opticalpower$ ${\sigma }_n,{\sigma }_p:capturecrosssections$ $v_{th}:thermalvelocity$ $N_t:trapdensity$ $\begin{array}{r}{n}_1,p_1:densitiesfortheFermilevelat\\ thetrapenergy\end{array}$ $\begin{array}{r}{N}_{tA},N_{tC}:fitparametersofaparabolictrap\\ distribution(Shockley-Read-Hallmodel),\\ thegainriseswithpowerasthetraps\\ saturateandfallsabovetheonsetofthe\\ Kirkeffect(about100nW,10\mu \mathrm{W}and100\\ \mu \mathrm{W}forthreestructures)\end{array}$
Frequency response and gain-bandwidth product	$f_{3dB}\approx \frac{1}{2\pi r_e\left(C_{BE}+C_{BC}\right)},r_e=\frac{k_{B}T}{q{I}_E}$	$\mathrm{GBP}=\frac{\beta }{{\tau }_{rise}}$ $f_{max}=\sqrt{\frac{f_t}{8\pi r_b{C}_c}}$	$r_e:small-signalemitterresistance$ $\begin{array}{r}{C}_{BE},C_{BC}:base-emitterandbase-collector\\ capacitances,thebandwidthriseswiththe\\ emittercurrentandsowiththeoptical\\ power(derived)\end{array}$ ${\tau }_{rise}:risetime$ $GBP:gain-bandwidthproduct$ $f_t:transitfrequency$ $r_b:baseresistance$ $\begin{array}{r}{C}_c:collectorcapacitance,(for{C}_{BE}+C_{BC}=50\\ fF(assumed)thebandwidthis0.12GHzat\\ I_E=1\mu \mathrm{A}and12GHzat100\mu \mathrm{A},computed,\\ withouttransit-timeandsubstrate\\ effects)\end{array}$ $k_B:Boltzmannconstant$
Noise of the phototransistor	$⟨i_n^2⟩=2q\beta \left(1+\beta \right)I_{ph}\Delta f,F=\frac{⟨G^2⟩}{{⟨G⟩}^2}$	$F\approx 1+\frac{1}{\beta }$ $\mathrm{SNR}=\frac{\beta }{1+\beta }\cdot \frac{I_{ph}}{2q\Delta f}$	$\begin{array}{r}⟨i_n^2⟩:collectornoisecurrentofthe\\ floatingbasefromtheamplifiedshot\\ noiseofthephotocurrentandtheshot\\ noiseofthecollectorcurrent(derived,\\ baserecombinationnoiseanddarkcurrent\\ neglected)\end{array}$ $\Delta f:noisebandwidth$ $F:excessnoisefactor$ $\begin{array}{r}G:gain,F=1foragainwithout\\ variance,(F=1.005for\beta =200,\\ computed)\end{array}$ $SNR:signal-to-noiseratiooftheoutputcurrent$

Photo Field-Effect Transistor (Photo-FET)

Name	Formula	Related Formulae	Variables/Constants
Photoconductive gain of the channel	$G_{int}=\frac{{\eta }_{sep}{\tau }_{lifetime}}{{\tau }_{transit}},{\tau }_{transit}=\frac{L^2}{V_{DS}{\mu }_{FE}}$	$G_{int}=\frac{{\mu }_{FE}{V}_{DS}{\eta }_{sep}{\tau }_{lifetime}}{L^2}$ $G_{int}=\frac{\mathfrak{R}hc}{q\lambda }$	$G_{int}:internalgain$ ${\eta }_{sep}:efficiencyofthechargeseparation$ ${\tau }_{lifetime}:lifetimeofthephotogeneratedcarriers$ ${\tau }_{transit}:transittime$ $L:channellength$ $V_{DS}:drain-sourcevoltage$ ${\mu }_{FE}:field-effectmobility$ $\begin{array}{r}(thelastformholdsatlowpower;5A/W\\ at658nmgives{G}_{int}=9.4and0.5A/W\\ gives0.94,computed)\end{array}$
Photogating	$\Delta V_{TH}=V_{TH}-V_{TH,dark},I_{ph}=g_m\left|\Delta V_{TH}\right|$	$g_m=\frac{W{\mu }_{FE}{C}_{ox}{V}_{DS}}{L},{\mu }_{FE}=\frac{L{g}_m}{W{C}_{ox}{V}_{DS}}$ $I_{photo}\propto P_{eff}^{\gamma }$	$\begin{array}{r}{V}_{TH}:thresholdvoltageextractedfromthe\\ linearregionofthetransfercurve(it\\ shiftstonegativegatevoltageunder\\ light)\end{array}$ $\begin{array}{r}{g}_m:transconductance(theformforthe\\ photocurrentisderived)\end{array}$ $W:channelwidth$ $C_{ox}:gatecapacitanceperarea$ $P_{eff}:poweronthedevice$ $\begin{array}{r}\gamma :exponent(\gamma \approx 1forphotoconductionand\\ afractionalvaluefortrap-mediated\\ photogating:1at{V}_{GS}=-10Vandabout0.3\\ at{V}_{GS}>0foroneInSedevice)\end{array}$
Responsivity and mobility of the photo-FET	$\mathfrak{R}=\frac{I_{photo}}{P_{in}},P_{eff}=P_{in}\frac{A_{dev}}{A_{spot}}$	${\mu }_{FE}\propto \frac{\mathfrak{R}{L}^2}{V_{DS}}$ $\sigma \propto {\left(V_{GS}-V_{TH}\right)}^{\chi }$	$I_{photo}:illuminatedminusdarkcurrent$ $P_{in}:powerofthebeam$ $\begin{array}{r}{A}_{dev},A_{spot}:areasofthedeviceandofthe\\ spot,theproportionalitytothemobility\\ assumes{\eta }_{sep}\approx 1andaconstantlifetime\end{array}$ $\begin{array}{r}\chi :screeningexponent(\chi =2for\\ unscreenedand1forfullyscreenedcharge\\ traps)\end{array}$ $\begin{array}{r}{V}_{GS}:gate-sourcevoltage,(R/V_{DS}fell9.3\\ timesfora130-foldfallofthemobility\\ infiveInSedevices,anexponentof0.46,\\ computed)\end{array}$

Important Questions

Ques: Why does a phototransistor have optical gain, and how large is it?

The photocurrent of the base-collector junction cannot leave through the floating base, so the holes accumulate there and lower the base-emitter barrier. The emitter then injects electrons in a number set by the transistor action, $I_C=\beta I_B$ with $\beta ={\tau }_R/{\tau }_{tB}$, and the emitter current is $(1+\beta )I_{ph}$ (derived). A SiGe device with a multiple-quantum-well absorber gave 4.6 μA for 180 μW in the photodiode mode and 235 μA in the transistor mode, i.e. 26 mA/W and 1.3 A/W, an amplification of 50 (computed), whereas its $\beta$ was about 200. The source does not explain why the optical gain is below $\beta$; a gain that depends on the power, which is the subject of a later question, is a possible reason. Reference: SiGe-Si heterojunction phototransistor with multiple-quantum-well absorber (scholars.lib.ntu.edu.tw, file 37.pdf; the title is descriptive, the tool gave no title or authors); Ulm University, Optoelectronics Annual Report 2014 (uni-ulm.de), InGaAs/GaAs heterojunction phototransistor; Bianconi et al., Engineering the gain-bandwidth product of phototransistor diodes, Appl. Phys. Lett. 115, 051104 (2019).

Ques: Is the gain of an InGaAs/InP phototransistor about $4\times 10^6$ or about $10^3$?

The patent application estimates ${\beta }_{dc}\approx 4\times 10^6$ for an abrupt InGaAs/InP NPN junction and lists about 200 for commercial devices, and the heterojunction factor $e^{\Delta E_g/k_{B}T}=5\times 10^5$ for 0.34 eV (computed) shows how a large number arises. Measured values are far lower: 90 to 2000 for three structures at 1550 nm. With ${\tau }_{tB}=2\times 10^{-10}$ s, $\beta =2000$ needs ${\tau }_R=0.4$ μs, whereas $4\times 10^6$ needs 0.8 ms (computed), which is not credible for a base with traps of $10^{10}$ to $10^{13}$ cm${}^{-3}$. The value of order $10^2$ to $10^3$ is therefore the more likely, and the theoretical number is an upper limit that ignores the recombination in the base and at the interfaces. Reference: US patent application 11430905, Hetero-junction phototransistor; Bianconi et al., Engineering the gain-bandwidth product of phototransistor diodes, Appl. Phys. Lett. 115, 051104 (2019).

Ques: Why does the gain depend on the optical power, rising first and then falling?

At very low power the carriers are lost to recombination through traps, and the lifetime is ${\tau }_R=\Delta n/{\sum }_i{U}_i$. As the excess density grows the trap levels saturate, the lifetime rises and with it $\beta ={\tau }_R/{\tau }_{tB}$. At high power the base is charged and pushed out (Kirk effect) and the transit time grows, so the gain falls. The onset of the Kirk effect was about 100 nW for an InP/InGaAs structure with $7.5\times 10^9$ cm${}^{-3}$ traps, 10 μW and 100 μW for two structures with about $6\times 10^{12}$ cm${}^{-3}$. The gain-bandwidth product $\beta /{\tau }_{rise}$ is therefore highest at an intermediate power, and its peak was similar for all structures, since more traps lower the gain but speed up the response. Reference: Bianconi et al., Engineering the gain-bandwidth product of phototransistor diodes, Appl. Phys. Lett. 115, 051104 (2019).

Ques: Why does the bandwidth of a phototransistor rise with the optical power and with a base current?

With a floating base the base-emitter and base-collector capacitances are charged by the emitter current through the small-signal resistance $r_e=k_{B}T/(q{I}_E)$, so $f_{3dB}\approx 1/[2\pi r_e(C_{BE}+C_{BC})]$ (derived). With 50 fF (assumed) this is 0.12 GHz at 1 μA and 12 GHz at 100 μA (computed, ignoring the transit time and the substrate). The power itself sets the emitter current, and a base current bias raises it independently: a 10 μm by 16 μm silicon device gave about 500 MHz with the base open, and above 5 GHz is reported with a base current bias. The optical $3\mathrm{dB}$ bandwidth of the SiGe device was 550 MHz while its $f_T$ was 25 GHz. In the travelling-wave waveguide devices of the Surrey thesis the $3\mathrm{dB}$ bandwidth was set by the base discharge time constant, and a scaled 1 μm by 10 μm model reached 59.4 GHz. Reference: US patent 8062919, Monolithic silicon-based photonic receiver; SiGe-Si heterojunction phototransistor with multiple-quantum-well absorber (scholars.lib.ntu.edu.tw, file 37.pdf; the title is descriptive, the tool gave no title or authors); H. N. Ang, Modelling and characterisation of 2-terminal heterojunction phototransistors, PhD thesis, University of Surrey (2005), abstract.

Ques: Why should the emitter of a heterojunction phototransistor be smaller than the absorber?

The abstract of a study of five devices states that a smaller emitter-to-absorber area ratio $A_e/A_a$ gives a higher optical gain: at $A_e/A_a=0.0625$ the gain was about 100 times, and the dark current about 40 times, that at a ratio of 1, measured at 3 V, with a maximum signal-to-noise ratio at about 0.16. If the shot noise of the dark current dominated, a 40-fold larger dark current would raise the noise by $\sqrt{40}=6.3$ and the improvement would be 16 times at best (computed), so a maximum at a larger ratio indicates that the dark current and the noise rise faster at small ratios. The mechanism and the analytical model are not given in the abstract. Reference: Chin. Phys. Lett. 36, 108501 (2019) (heterojunction phototransistors with scaled emitters; abstract only).

Ques: Why does the SiGe phototransistor show a slow tail and two poles, and why is it better at 850 nm than at 1060 nm?

Carriers generated in the collector and in the substrate reach the junction by diffusion and give photocurrent without gain, so the response has a low-frequency pole from the collector-substrate junction and a high-frequency pole from the base-amplified carriers. The 50 ps laser pulse gave a rise time of 64 ps and a fall time of 442 ps, with an optical $3\mathrm{dB}$ bandwidth of 550 MHz limited by the tail, and a capacitor in series with the amplifier removed the low-frequency pole. At 850 nm a 43 μm by 51 μm detector gave 2.4 A/W, against 0.12 A/W at 1060 nm, where silicon absorbs weakly and the SiGe base supplies a larger part. The source suggests a silicon-on-insulator substrate to reduce the tail. Reference: SiGe-Si heterojunction phototransistor with multiple-quantum-well absorber (scholars.lib.ntu.edu.tw, file 37.pdf; the title is descriptive, the tool gave no title or authors); US patent 8062919, Monolithic silicon-based photonic receiver.

Ques: What separates photoconduction from photogating in a photo-FET?

In photoconduction the photogenerated carriers themselves raise the conductance, and the photocurrent is proportional to the power, $\gamma \approx 1$, with the gain $G_{int}={\eta }_{sep}{\tau }_{lifetime}/{\tau }_{transit}$. In photogating one carrier type is trapped in long-lived surface or interface states and acts as an extra gate, so that the threshold voltage shifts to negative gate voltage, and the photocurrent is $g_m\left|\Delta V_{TH}\right|$ (derived). A fractional exponent signals this: an InSe device gave $\gamma \approx 1$ at $V_{GS}=-10$ V and about 0.3 at positive gate voltage. In the same device the responsivity was about 0.5 A/W at zero gate voltage and about 5 A/W at 22 nW with 60 V, i.e. a gain of 0.94 and 9.4 (computed); the source gives no explanation for the rise. Reference: Role of Layer Thickness and Field-Effect Mobility on Photoresponsivity of Indium Selenide (InSe) Based Phototransistors, arXiv:2102.00365.

Ques: Is the responsivity of a photo-FET proportional to the mobility?

The InSe study assumes a separation efficiency of 1 and a constant lifetime, so that $G_{int}={\mu }_{FE}{V}_{DS}{\eta }_{sep}{\tau }_{lifetime}/L^2$ gives ${\mu }_{FE}\propto \mathfrak{R}{L}^2/V_{DS}$. Its five devices do not follow this: the mobility falls from 60 to 0.46 $\mathrm{c}{\mathrm{m}}^2/Vs$, a factor of 130, while $\mathfrak{R}/V_{DS}$ falls from 2.95 to 0.318 A/(W V), a factor of 9.3, which is an exponent of 0.46 (computed). Since the lifetime of one device (75 nm) was estimated at 20 μs, and the thickness changes the screening and the trapping, a lifetime that varies with the thickness is the more likely explanation, and the proportionality holds only for a fixed lifetime and separation efficiency. Reference: Role of Layer Thickness and Field-Effect Mobility on Photoresponsivity of Indium Selenide (InSe) Based Phototransistors, arXiv:2102.00365.

Ques: Does a phototransistor add excess noise, as an avalanche photodiode does?

The patent application sets the excess noise factor $F=⟨G^2⟩/⟨G{⟩}^2$ to 1, because it assumes no variance of the gain, and the Ulm report calls high gain without excess noise a potential advantage over the avalanche photodiode. For a floating base the amplified shot noise of the photocurrent, ${\beta }^2\cdot 2q{I}_{ph}$, and the shot noise of the collector current, $2q\beta I_{ph}$, add to $2q\beta (1+\beta )I_{ph}\Delta f$ (derived), so that $F\approx 1+1/\beta$, i.e. 1.005 for $\beta =200$ (computed). The base recombination, the dark current and the extra noise of a gain that varies with power are neglected in this estimate, so the true noise is somewhat higher than the estimate, although well below that of an avalanche gain of the same size. Reference: US patent application 11430905, Hetero-junction phototransistor; Ulm University, Optoelectronics Annual Report 2014 (uni-ulm.de), InGaAs/GaAs heterojunction phototransistor.

Related pages in this vault: Graphene and 2D-Material Photodetector, Photoconductor, Avalanche Photodiode (APD), PN Photodiode, PIN Photodiode, Bipolar Junction Transistor (BJT), Bipolar Transistors (BJT and SiGe HBT).