Photonic Communications Engineering Lecture Dr. Demetris Geddis Department of Engineering Norfolk State University
Light Detectors
How does this detector work? Image from visionweb.com Responds to range of light signals Very dim to very bright(10 5 ) Spectrum of colors Photoreceptors in the retina Rods scotopic vision (Low light) Cones photopic vision (High light and Color) The change in the visual pigment of the rods and cones is transformed into an electrical signal
Relative Response of the Eye
Photodetection Mechanisms Detectors convert light signals into electrical signals. Internal Photoelectric Effect Generation of mobile charge carriers in semiconductors by absorption of photons. External Photoelectric Effect Generation of free electrons when photons strike the surface of a metal. Electrons are emitted from the surface. 5
Photodetection Mechanisms Devices: Vacuum Photodiode pn photodiode PIN photodiode Metal Semiconductor Metal PD 6
Important Detector Properties 1. Responsivity: P ρ = Output Current Input Optical Power i Optical Power Electrical Current Photodetector ρ i P AW /
η = P hf η ( energy/second ) ( energy/photon ) P hf e η P hf i = number of emitted / generated electrons number of incident photons = eη P hf Coulomb electrons = photons second = electrons generated per electrons second And the responsivity is: i eη ρ = = = P hf second eηλ hc
2. Spectral Response: Range of optical wavelengths over which the detector is useful. It is often displayed as a curve of responsivity versus wavelength. Edmund Optics Silicon Photodiode
Edmund Optics InGaAs Photodiode
3. Speed of Response: Range of modulation frequencies over which the detector is useful. As before, if t r is the rise time, the bandwidth is (approximately) f 3 db = 0.35 t r P Input i 10% 90% Output t r
Vacuum Photodiode And Photomultiplier Cathode - hf Anode + Vacuum Photodiode Electrons - + i V v R L 12
The work function φ is defined to be: Energy required to liberate an electron from the metal cathode. Units of energy: Joules In order to free an electron: the photon energy must equal, or exceed, the work function. hf φ hc λ φ λ hc φ λ c = 1.24 φ
Semiconductor Devices
Compound Semiconductors Made out of elements from different columns of the periodic table III-V, II-VI, IV-VI, or IV-IV III-V is the most used compound GaAs and InP Same average number of valence electrons per silicon atom Column represent the number of valence electrons
Abbreviated Periodic Table
Indium Antimonide (InSb) the 1 st III-V Discovered in 1950 Easily synthesized Electron Mobility Low bandgap energy (E g =0.17 ev) Good for infrared detection
Other important III-V compounds Semiconductor Photodetectors GaAs Bandgap of E g = 1.43 ev Direct bandgap InP Bandgap of E g = 1.35 ev Direct bandgap GaP Bandgap of Eg = 2.1 ev Indirect bandgap
Elemental Versus III-V III-V has a higher mobilities and higher velocities III-V are more efficient High-Speed optoelectronic devices Higher Responsivity Many III-Vs are direct bandgap
Direct vs. Indirect Bandgap Semiconductors
Absorption Coefficient Direct bandgap semiconductors (GaAs, InAs, InP, GaSb, InGaAs, GaAsSb), the photon absorption does not require assistant from lattice vibrations. The photon is absorbed and the electron is excited directly from the VB to CB without a change in its k-vector (crystal momentum ħk), since photon momentum is very small. k k CB VB = photon momentum 0 Absorption coefficient α for direct bandgap semiconductors rise sharply with decreasing wavelength from λ g (GaAs and InP).
Absorption Coefficient Indirect bandgap semiconductors (Si and Ge), the photon absorption requires assistant from lattice vibrations (phonon). If K is wave vector of lattice wave, then ħk represents the momentum associated with lattice vibration ħk is phonon momentum. k k = phonon momentum = CB VB K Thus, the probability of photon absorption is not as high as in a direct transition.
Absorption Coefficient Photon absorption in a direct bandgap semiconductor: E ph =hv>e g E Photon absorption in an indirect bandgap semiconductor: E ph =hv>e g ± E phonon E C B E C Direct Bandgap E g E V Photon Photon C B Indirect Bandgap E C E g V B V B E V Phonons k k k k
Quantum Efficiency and Responsivity Quantum efficiency, η, is the number of carriers (EHP) collected to produce the photocurrent, I ph, divided by the number of incident photons (where P inc is incident optical power): I ph / e I ph hv η = = * Pinc / hv e Pinc η depends on absorption coefficient and thickness of absorbing region. Responsivity of a detector, R, is a performance parameter that indicates the magnitude of the electrical signal output from a photodetector in response to a particular light power: R = I ph /P inc = (ηe)/hν = (ηλ)/1.24 (µm)
Photodiodes Energy Level Diagram for V = 0 (zero bias) p n Electron Energy Conduction Band - - - - - - - + + + + + + Hole E g Valence Band Energy
SEMICONDUCTOR JUNCTION PHOTODIODE Reverse Biased Photodiode p n hf Electron Energy hf + + + + + + + + + - + E - - - - - - - - - - - E g Junction Region
SEMICONDUCTOR JUNCTION PHOTODIODE An incident photon absorbed in the junction gives up its energy, creating a free electron and a free hole in the junction. The generated free charges move due to the strong electric field E in the junction. Recall that the electric field is generated by the change in electric potential V given by: E = V
SEMICONDUCTOR JUNCTION As before PHOTODIODE A condition for detection is that the photon energy be greater than the bandgap energy: hf E g f E h g λ c = hc E g λ c = 1.24 E g Example: Compute the cutoff wavelength for Silicon. The bandgap energy for silicon is E g = 1.1 ev 1.24 λ c = = 1. 1µ m 1.1
Light Transmission
SEMICONDUCTOR JUNCTION PHOTODIODE Response Time The pn photodiode responds slowly because many photons are absorbed in the n and p regions close to the junction. hf Diffusion p - E n Electron Energy +
SEMICONDUCTOR JUNCTION PHOTODIODE The free charge carriers diffuse slowly (gold arrow) into the junction where they are accelerated (green arrow) by the large electric field there. This produces a delayed current in the external circuit, causing pulse spreading, lowering the data rates that can be received, and reducing the analog 3-dB frequency bandwidth of the receiver.
SEMICONDUCTOR JUNCTION PHOTODIODE The analog 3-dB frequency bandwidth is typically about t r = 1 µs for pn junction photodetectors. f 0.35 0.35 = = = = t 10 3 db 6 r 6 0.35 10 350 khz The solution to this problem is the pin diode.
PIN Photodiode PIN Photodiode hf Intrinsic Layer i p - + E n V - Thin Layer R L v +
PIN PHOTODIODE Materials Material Wavelength Range (µm) Peak Response λ (µm) Peak Responsivity ρ (A/W) Silicon 0.3 1.1 0.8 0.5 Germanium 0.5 1.8 1.55 0.7 InGaAs 1.0 1.7 1.7 1.1
PIN PHOTODIODE The intrinsic layer is on an insulator. Most of the photons are absorbed in that layer because it is long. Most of the voltage drop is across the intrinsic layer. This creates a high electric field in the intrinsic layer. Now there is no delay caused by diffusion and the response time is much faster that that of a pn photodiode. t tr t tr = 1 2 = τ ( modulation period) RC
Photoresponse Bandwidth Tradeoff The advantage of a PIN photodiode is that it has a controlled depletion region or width, which can be tailored to meet the requirements of photoresponse and bandwidth. By increasing the thickness of the intrinsic region, the photoresponse can be enhanced. However, increasing the thickness of the active region reduces the bandwidth. This photoresponse-bandwidth tradeoff limits the amount of tailoring that can be done.
Speed of Response Transit time: Time for free charges to move across the depletion region (the intrinsic layer in a pin photodiode). The speed of response is ultimately limited by the transit time. Example: d = depletion width = 50 µm v = carrier velocity = 5 x 10 4 m/s Then the transit time is: d v 50 10 5 10 6 t = = 9 = 10 s = 1ns 4
Speed of Response The detector s rise time is on the order of its transit time. The bandwidth is limited to: f = 3 db 0.35 In addition to the transit time, the detector s capacitance and load resistance also limit the response speed. t r
Speed of Response Detector Equivalent Circuit i d (ρp) C d R L v C d : Diode Capacitance The rise time of this circuit is: t = 2.19RC r L D
Speed of Response The 3-dB bandwidth is f 0.35 0.35 3 db = = = tr 2.19RLCd 1 2π R C L d This establishes the relationship between the load resistance and the bandwidth. The larger the load resistance, the smaller the bandwidth.
Metal Semiconductor Metal PD V d1 V d2 g The energy band diagram for an MSM PD in thermal equilibrium is shown above, where the electron Schottky barrier heights and built in voltages are labeled Φ n1 and Φ n2, V d1 and V d2, respectively. Φ p2 is the hole Schottky barrier height from the second contact, and g is the gap length between the contacts.
43 Conventional Versus Inverted MSM Photodetectors MSM Advantage: Low capacitance per unit area compared to typical PIN PDs MSM Disadvantage: Shadowing by interdigitated electrodes reduces responsivity Inverted (I-) MSM: By inverting, low capacitance per unit area is maintained and responsivity is dramatically improved because the fingers are on the bottom
Thin Film I-MSM Photodetector Silicon host substrate GaAs thin film (1 µm thick) I-MSM PD 44
Metal-Semiconductor-Metal Photodetector (MSM PD) and LED Device Structure MSM PD material Undoped GaAs (1000 nm) AlAs (200 nm) GaAs (Substrate) LED Material GaAs (20 nm) GaAlAs (1000 nm) GaAs (1000 nm) GaAlAs (1000 nm) GaAs (20 nm) AlAs (200 nm) GaAs (50 nm) GaAs (Substrate) Each device is grown separately An AlAs sacrificial etch layer is included in the growth Different structures can be used to optimize the performance of each individual device 45
How the I-MSM Works Incident photons Electric field line EHP #3 Lines of equipotential EHP #2 0 V EHP #1 +5 V 46
Frequency Response
Current-Voltage Characteristic I D is called the dark current. This is the current that flows when no photons are incident (this is just the diode reverse leakage current). It is due to thermal generation of minority charge carriers. Small optical signals are masked by the dark current. For good signal reception, the received power must greatly exceed levels of power which generate currents on the order of the dark current. 48
Current-Voltage Characteristic Typical Dark Current Values Silicon InGaAs Germanium 2 na 50 na 500 na Example: Compute the responsivity from the v d i d curve given several slides back. Solution: i d = -10 µa when P = 20 µw, so that i d 10 ρ = = = 0.5 µ A/ µ W P 20
Current-Voltage Characteristic Example: Let I D = 2 na in the previous example. What is the minimum detectable optical power if we can detect a signal current equal to (or greater than) the dark current? Solution: We want i = 2 na as a minimum. i = ρ P i 2nA P = = = 4nW ρ 0.5nA / nw We can detect powers as low as 4 nw.
Noise and Detection
Noise degrades signals. Without noise, it would not matter how little optical power arrived at the receiver. Signal quality is measured in several ways. Analog systems: The signal-to-noise ratio (SNR) is the measure. Digital systems: The bit-error-rate (BER) is the measure.
Thermal and Shot Noise
Thermal Noise Recall the simple receiver circuit : P V b - + i R L v
Even if P = 0 (and the photodiode dark current is zero), a current i = i NT will exist in resistor R L. It has zero average value, but it is random, like: i NT t
The current arises from the random thermal motion of the electrons. The instantaneous noise power is R L i 2 NT The average thermal noise power is R L i 2 NT R L i 2 NT R L i 2 NT t
i 2 NT = mean square thermal noise current. It is given by : i 2 NT = 4kT f /R L k = 1.38 x 10-23 J /K, Boltzmann constant T = temperature, K f = receiver s bandwidth. Usually f is a bit larger than the information bandwidth.
The load resistor s equivalent circuit looks like: i 2 NT = 4kT f /R L R L v where R L is an ideal (noiseless) resistor.
Shot Noise It is caused by the discrete nature of charge carriers (electrons and holes). Consider the single emitted photoelectron, shown for vacuum phototube. Cathode hf - - + electron Anode V b i R L
A current exists in the circuit during the transit time (τ) of the emitted electron. τ = time for travel from cathode to anode. The electron recombines at the anode with a positive ion. The current caused by a single electron looks something like : h(t) t τ
Every electron produces the same current pulse shape. Consider constant optical power P incident on the detector. The expected current is : i i = (ηe / hf)p t This current is made up of numerous pulses of the type shown by h(t).
Example : t The pulses start at random times, t N. The total current is the sum of these pulses. i = Σ N h(t - t N )
Total current: i (ηe /hf)p t The average current is still: i = (ηe /hf)p but noise is superimposed onto this current. This is shot noise.
The shot noise current is: i NS = Σ N h(t - t N ) - (ηe /hf)p i 2 NS = 2eI f f = receiver s bandwidth I = average current. I = i S + I D where i S I D = average of the signal current = average dark current
The equivalent circuit for shot noise is just a noise current generator, as shown below. i 2 NS = 2eI f
Signal-to-Noise Ratio
Consider the equivalent circuit of a photodiode receiver. R s i S C d R d v C d = diode s junction capacitance (small) R d = diode s junction resistance (large) R s = diode s bulk series (n and p) resistance (small)
i S is the photocurrent. As before, it is given by i s = (ηep/hf) = ρp For simplicity, assume R s = 0 and R d = infinite. Also neglect C d for purposes of noise calculations, since it does not affect the noise in the circuit. The simplified receiving circuit, including all sources of thermal and shot noise is now:
i S i 2 NT i 2 NS R L We will use this circuit to compute SNR.
Constant Power SNR Let the incident optical power P be a constant. This corresponds to a binary 1 in a digital system. Compute the SNR. SNR = average signal power / average noise power These are the electrical powers. From the equivalent circuit, we see that SNR = (R L i 2 S) / (R L i 2 NS + R L i 2 NT) SNR = P ES / (P NS + P NT ) SNR = i 2 S / ( i 2 NS + i 2 NT)
These equations are general. For the special case where P = a constant: i S = i S = (ηe / hf)p = ρp P ES = R L i 2 S = (ηe P/ hf) 2 R L P NT = R L i 2 NT = (4kT f / R L ) R L = 4kT f P NS = R L i 2 NS = 2e[ I D + (ηe P/ hf)] f R L Then SNR = [(ηe P/ hf) 2 R L ] {2e[ I D + (ηe P/ hf)] f R L }+ 4kT f
Example: Light source is an LED, 10 mw output power, λ = 0.85 µm. The system losses are: coupling loss = 14 db fiber loss = 20 db connector losses = 10 db Total loss = 44 db
Compute the received power. db = log (P R /P T ) - 44 = log (P R /P T ) (P R / P T ) = 10-4.4 P R = 10 x 10-4.4 = 10(3.98 x 10-5 ) P R = 4 x 10-4 mw Alternative calculation for the received power: P T = 10 mw = 10 dbm Loss = - 44 db P R = - 34 dbm
Check : dbm = 10 log P R -34 = 10 log P R P R = 10-3.4 = 4 x 10-4 mw This result checks. The receiver has the following characteristics: ρ = 0.5 A/W (responsivity) I D = 2 na (dark current) R L = 50 Ω (load resistance)
f = 10 MHz (receiver bandwidth) T = 300 K (27 0 C) receiver temperature Find : Signal current and power Shot noise power Thermal noise power SNR
Solution : Signal Current i S = ρp R = 0.5(4 x 10-4 ) = 2 x 10-4 ma i S = 0.2 µa = 200 na i S = 200 na >> I D = 2 na Signal Power P ES = R L i 2 S = 50(0.2 x 10-6 ) 2 = 2 x 10-12 W Shot Noise Power P NS = 2e i S R L f = 2(1.6 x 10-19 ) (0.2 x 10-6 ) (50)10 7 P NS = 3.2 x 10-17 W
Thermal Noise Power P NT = 4kT f = 4(1.38 x 10-23 ) 300 x 10 7 P NT = 1.66 x 10-13 W Note P NT >> P NS Thus, we have a thermal-noise limited system, and SNR = P ES / P NT = 2 x 10-12 /1.66 x 10-13 = 12 SNR db = 10 log SNR SNR db = 10 log 12 = 10.8 db
If the system were shot-noise limited, then SNR = ηp /2hf f = i s /2e f SNR = 0.2 x 10-6 / 2(1.6 x 10-19 ) 10 7 = 62,500 SNR db = 10 log 62,500 = 48 db Note the improvement if the system were shot-noise limited. We can approach this higher SNR if we use a photodetector with internal gain, or use heterodyne detection.
References 1. Joseph Palais. Fiber Optics Communications, 5 th Edition. New Jersey: Prentice Hall, 2004. 2. Frank L. Pedrotti and Leno Pedrotti. Introduction to Optics. New Jersey: Prentice Hall, 1993 3. Pallab Bhattacharya. Semiconductor Optoelectronic Devices, 2 nd Edition. New Jersey: Prentice Hall, 1997.