Photonics - Topic 5

Question 1

What are Pyroelectric detectors?

Answer 1

Photon energy converted to heat

uses pyroelectric crystal - Radiation falling on the detector changes its temperature and reduces its polarisation, which changes the charge on its surface.

This can be detected as a transient voltage.

Question 2

Photoconductivity in semiconductors

Answer 2

• An electron-hole pair (EHP) can be generated in a semiconductor by an incoming photon with energy ℎ𝑓 > 𝐸𝑔, where 𝐸𝑔 is the bandgap energy of the semiconductor.
  
  • Increases the conductivity of the semiconductor.
    
    • If there is an electric field across the semiconductor, there will be an increase in current
      
      • (the electric field will separate the EHP, causing them to move in opposite directions).
  • Known as photoconductivity.

Question 3

What bias does a photodiode (pn junction) need to perform as a photodetector?

Answer 3

Reverse bias

The reverse bias increases the potential barrier in the depletion region, resulting in a very small reverse current (the dark current) when there is no light incident on the photodiode.

Question 4

How is current generated in a photodiode (pn junction)?

Answer 4

Thus we can expect the photocurrent due to the light falling on the photodiode to be due to EHPs generated in (or close to) the depletion region.

Question 5

Effect of illumination on the I-V curve of a photodiode

Answer 5

As the optical power incident on the junction is increased more electron hole pairs are created resulting in a larger negative current:

I = -I_s - I₀(P₀)

I_s is dark curent

P₀ is photocurent

Question 6

Explain what happens for increasing optical power on a photodiode in reverse bias

Answer 6

• When the diode is reverse biased, photogenerated EHPs cause a current to flow from n to p, i.e. in the reverse direction.
• Increasing optical power will lead to more EHPs being generated, so the current increases with increasing optical power.
  
  • Since the number of EHPs generated is proportional to the number of incident photons, which is proportional to the power, the photocurrent in the reverse biased region increases linearly with power.

Question 7

What happens if a photodiode is under forward bias?

Answer 7

• If the diode is forward biased, the potential barrier will be lowered and forward current will flow. As the forward voltage is increased, the forward current will eventually dominate over the reverse photocurrent, and the 𝐼 − 𝑉 characteristic will follow that for the unilluminated diode.
• At a certain forward bias, the forward current will balance the reverse photocurrent, so no net current flows. Thus a voltage (𝑉_𝑜𝑐) will appear across an open-circuit photodiode that has light falling on it. For 0 < 𝑉 < 𝑉_𝑜𝑐 the diode operates in photovoltaic mode. It is in this region that pn diode solar cells operate.

Question 8

Eq: Outline Photodiode absorption characteristics

Answer 8

At the interface between the air and photodiode, the power entering the semiconductor is 𝑃₀(1 − 𝑅).

Inside the semiconductor, the power is assumed to decay exponentially, with absorption (or loss) coefficient 𝛼, so that overall the power at a distance 𝑥 into the semiconductor is given by:

𝑃(𝑥) = 𝑃₀ (1 − 𝑅) e^{(−𝛼𝑥)}

Question 9

Eq: Outline Photocurrent in Photdiode

Answer 9

From the fact that power decays exponentially across the semiconductor, we can see that the _power absorbed in the depletion regio_n, which will lead to the generation of photocurrent, is given by:

𝑃(𝑥₁ ) − 𝑃(𝑥₂) = 𝑃₀(1 − 𝑅)[e^{(−𝛼𝑥₁)} − e^{(−𝛼𝑥₂)}]

where 𝑥₁ and 𝑥₂ are the positions of the top and bottom of the depletion region.

If the fraction of absorbed photons that generate an EHP that contributes to the photocurrent is 𝜂_𝑖𝑛𝑡 (the internal quantum efficiency), the photocurrent generated is

𝐼𝑜 = (𝑒𝜂_𝑖𝑛𝑡 / ℎ𝑓) 𝑃₀(1 − 𝑅)[e^{(−𝛼𝑥1)} − e^{(−𝛼𝑥2)}]

Question 10

Eq: What is the responsivity of a photodiode?

Answer 10

The responsivity of a photodiode is the ratio of generated photocurrent to optical power incident on the detector.

mA.mW^-1

Question 11

Eq: What is the external quantum efficiency?

Answer 11

Compares the number of electrons generated to the number of photons incident on the photodiode. The number of photons incident per second is 𝑃₀/ℎ𝑓, and the number of electrons flowing per second is 𝐼₀/𝑒.

𝜂_𝑒𝑥𝑡 = 𝑅_𝑠 (ℎ𝑐 / 𝑒λ)

𝜂_𝑒𝑥𝑡 can be expressed as:

𝜂_𝑒𝑥𝑡 = 𝜂_𝑖𝑛𝑡(1 − 𝑅)[e^{(−𝛼𝑥₁ )} − e^{(−𝛼𝑥₂)}]

Question 12

What is the Spectral Response of a photodiode?

Answer 12

• Rearranging 𝜂_𝑒𝑥𝑡 = 𝑅_𝑠 (ℎ𝑐 / 𝑒λ) we can see that responsivity is proportional to the wavelength.
  
  • only true for photons with energy greater than the bandgap energy
• Real case
  
  • The real responsivity deviates from the ideal at short wavelengths due to the wavelength dependence of the absorption coefficient in the external quantum efficiency
  • At long wavelengths below the bandgap energy the responsivity tails off in a smooth way due to the continuous nature of band edge _(_i.e. cut off of the band gap energy is not abrupt in real materials).

Question 13

p-i-n junction diagam, energy plot, fixed charge density plot and electric field plot

Answer 13

Question 14

Eq: Poisson’s equation for pin junction

Answer 14

Where N_D is the charge density

and the denominator consists of the permittivity constants.

Question 15

p-i-n junction design considerations: increasing the length of the depletion region

Answer 15

• The external quantum efficiency can be increased to almost 100% by increasing the length of the depletion region.
  
  • In practice: l_depletion >> 1 / α
    
    • where 1 / α is the attenuation length scale
• However, increasing he length will also affect the bandwidth of the photodiode:
  
  • It increases the response time as the photo-generated carriers have further to drift
  • It reduces the capacitance of the photodiode, which will increase the bandwidth

Question 16

What is carrier transit time and what does it depend on?

Answer 16

• It is the time taken for the photogenerated carriers to cross the depletion region and enter the doped region where they are majority carriers.
  
  • The transit time depend on → drift velocity, which depends on → electric field in the intrinsic region.
  • drift velocity → increased
    
    • high reverse bias across the pin photodiode.

Question 17

Eq: Carrier velocity and drift time in phototiode

Answer 17

Carrier velocity:

𝑣 = μ𝐸 = μ𝑉/x

μ is the Carrier mobility (m² /V/s)

Drift time:

𝑡_𝑐 = 𝑥 / 𝑣 = 𝑥 ² / μ𝑉

x is the Junction width (m)

Question 18

What happens to the drift velocity at high electric fields?

Answer 18

𝑡𝑐 = 𝑥 / 𝑣_{𝑠𝑎<em>t</em>}

Drift velocity saturates at large electric field values and electron and hole velocities become similar.

Saturated velocity ⇒ shortest drift time (and hence highest bandwidth).

Question 19

What is the Photodiode equivalent circuit

Answer 19

• The space charge regions of a pin photodiode are like charges on the plates of a capacitor, so the photodiode can be modelled as a current source in parallel with a capacitor with capacitance:
  
  • 𝐶_𝑗 = 𝜀_𝑟𝜀₀𝐴 / 𝑥
• Photodiode will be connected to a load with resistance 𝑅_L
• Forms a low-pass filter, with time constant 𝑅_L𝐶_𝑗

Question 20

Photodiode junction discharge time

Answer 20

Question 21

Eq: Photodiode junction diffusion time

Answer 21

𝑡_𝑑 = 𝑙² / 2 D_e

Question 22

How is diffusion time minimised in photodiodes?

Answer 22

In well designed photodiodes the diffusion time is minimised by using a thin top p-region such that

t_d << t_c

Question 23

How is the overall bandwidth of the phototdiode determined?

Answer 23

• The overall bandwidth of a photodiode is determined by the combination of the responses due to the drift, diffusion and RC times
  
  • not simple to calculate the combined effect
  • by comparing the drift, diffusion and RC times it may be possible to identify a dominant factor

Question 24

Design guidelines for photodiode

Answer 24

• Design guidelines (for overall bandwidth ):
  
  • Use thin p-layer to so that diffusion time 𝑡_𝑑 << 1/B
  • Chose intrinsic layer width such that drift time 𝑡_𝑐 < 1 / (2B)
  • Chose junction capacitance (PD area) so that 𝑡_RC < 1 / (2πB)
  • Note that there is a trade-off between junction capacitance (𝑡_RC) and drift time

Question 25

pin photodiode structure (top entry)

Answer 25

• Thin p-region
• Anti-reflection (AR) coating to reduce reflections at air semiconductor interface
• Thick intrinsic region (depletion region) to maximise absorption
• Small area to reduce junction capacitance