IN221 – Semiconductor Based
Sensors and Actuators – Part 2
Charge Transport
Continuity Equation
Sensors Based on Modulating
Currents in Semiconductors
Thermistors (Thermoresistors)
In metals, as the temperature increases, the resistance increases since the mean collision time decreases.
Thus, the resistivity increases. Thus the coefficient of thermal resistance is positive – positive temperature
coefficient (PTC). These temperature sensors are called Resistance Temperature Detectors (RTDs).
In semiconductors:
If the semiconductor is intrinsic or lightly doped – increasing the temperature increases ni as well as reduces
the mean collision time. However, the increase in ni dominates and therefore the resistivity goes down. Thus
we have a negative temperature coefficient (NTC)
𝜇=
𝑞𝑡𝑐𝑜𝑙
𝑚𝑒𝑓𝑓
𝜌𝑖 =
1
~
𝑞𝑛𝑖 (𝜇𝑛 +𝜇𝑝 )
𝑚
𝑞 2 𝑡𝑐𝑜𝑙
𝑛𝑖
If we dope the semiconductor heavily (and the semiconductor is in extrinsic operation), the number of free
carriers is now fixed (equal to the dopant concentration). Therefore the reduced mean collision time with
increasing temperature dominates and the resistance increases with temperature (PTC).
This is true only in extrinsic operation (for all practical purposes around room temp). In the journey from
freezeout-extrinsic and extrinsic to intrinsic, the physics is different.
𝜌𝑛 =
1
𝑞𝜇𝑛 𝑛
=
𝑚𝑒𝑓𝑓
2
𝑞 𝑡𝑐𝑜𝑙 𝑁𝐷
These temperature sensors based on semiconductors are called thermistors.
Conventional model for NTC thermistors:
1 1
𝑅 𝑇 = 𝑅0 exp[𝐵
−
]
𝑇 𝑇0
T0 = reference temperature (typically 25C)
R0 = Resistance at T0
B = characteristic temperature of the material
Relative sensitivity = (dR/dT)(1/T)
Photosensors (Photoresistors)
When light falls on a semiconductor, electron-hole pairs are generated.
The excess carriers result in a reduction in resistivity and this can be measured.
I will assume that all of you can now calculate resistivity when you have the mobility and carrier concentrations
Question 1: When a n-type semiconductor sample is bathed in light, what is the number of excess holes created in steady state ?
No electric field, the whole sample is bathed in light and therefore no concentration gradients
If we assume low level injection
𝛿𝑝𝑛
𝑝𝑛 − 𝑝𝑛0
𝑅=
=
𝜏𝑝
𝜏𝑝
Generation rate = G (this is number of photons/sec = no of electron hole pair created per sec after accounting for efficiency of
absorption
In steady state, this becomes,
0=𝐺−𝑅
𝑝𝑛 = 𝑝𝑛0 + 𝜏𝑝 G
Photosensors (Photoresistors)
Question 2: From this previous example, if the light is turned off at t=0, after how much time will the sample return to equilibrium
levels of carrier concentration?
Generation rate = G =0
𝑑𝑝𝑛
𝛿𝑝𝑛
= −𝑅 = −
𝑑𝑡
𝜏𝑝
Note that
𝑑𝑝𝑛 𝑑 𝑝𝑛0 + 𝛿𝑝𝑛
𝑑𝛿𝑝𝑛
=
=
𝑑𝑡
𝑑𝑡
𝑑𝑡
Therefore,
𝑑𝛿𝑝𝑛
𝛿𝑝𝑛
= −𝑅 = −
𝑑𝑡
𝜏𝑝
𝛿𝑝𝑛 (t)= 𝛿𝑝𝑛 𝑡 = 0 exp(−𝑡/𝜏𝑝 )
𝛿𝑝𝑛 (t)= 𝐺𝜏𝑝 exp(−𝑡/𝜏𝑝 )
𝑝𝑛 (𝑡) = 𝑝𝑛0 + 𝛿𝑝𝑛 (t)= pn0 + 𝐺𝜏𝑝 exp(−𝑡/𝜏𝑝 )
Photosensors (Photoresistors)
Question 3: When light falls on one side of a long semiconductor, the light is rapidly
absorbed (Beer-Lamberts Law).
𝐼 = 𝐼0 exp(−𝛼𝑥)
We should look at the absorption versus wavelength characteristics to identify the
characteristic absorption depth. For blue-green light this is small (< 1 um) in silicon.
Just after is thin region, we don’t have any generation rate => G =0
What is the situation in steady state?
𝐷𝑝 𝑑2 𝛿𝑝𝑛 𝛿𝑝𝑛
0=
−
𝑑𝑥 2
𝜏𝑝
Boundary condition, at x → ∞, 𝛿𝑝𝑛 → 0
𝑝𝑛 𝑥 = 𝑝𝑛0 + (𝑝𝑛 0 − 𝑝𝑛0 ) exp(−𝑥/𝐿𝑝 )
Because we have a concentration gradient, we will now have a diffusion current. What will
this current be?
Question 4: As a homework, try what happens if the semiconductor is of finite width
W
Boundary condition, at x = 𝑊, 𝛿𝑝 = 0
𝑛
Most Generation happens here
Photosensors (Photoresistors)
Question 5: Let us say a light pulse is applied to the semiconductor as shown, what is
the dynamics in the absence of an electric field?
After the light is switched off, we don’t have any generation rate => G =0
𝑑𝛿𝑝𝑛 𝐷𝑝 𝑑2 𝛿𝑝𝑛 𝛿𝑝𝑛
=
−
𝑑𝑡
𝑑𝑥 2
𝜏𝑝
Can you imagine the solution?
𝑝𝑛 𝑥, 𝑡 = 𝑝𝑛0 +
𝑃𝑎𝑟𝑒𝑎
4𝜋𝐷𝑝 𝑡
𝑥2
𝑡
exp
−
−
1
4𝐷𝑝 𝑡 𝜏𝑝
2
Question 6: What happens, if in Question 5, we have a constant electric field In the semiconductor?
𝑑𝛿𝑝𝑛
𝑑𝑡
𝜇𝑝 𝜉𝑑𝛿𝑝𝑛 𝐷𝑝 𝑑2 𝛿𝑝𝑛 𝛿𝑝𝑛
=−
+
−
𝑑𝑥
𝑑𝑥 2
𝜏𝑝
Can you imagine the solution?
𝑑𝜉
= 0 ,𝜉 ≠ 0
𝑑𝑥
Thermoelectricity – Seebeck Effect
VA
VB
TA>TB
TA
TB
If you take a metal and heat one end (A) and cool the other end (B), we have a voltage drop across the two ends with the hot
end being at a higher potential than the cold end, VAB>0
If you take a n-doped semiconductor and heat one end (A) and cool the other end (B), we have a voltage drop across the two
ends with the hot end being at a higher potential that the cold end, VAB>0
If you take a p-doped semiconductor and heat one end (A) and cool the other end (B), we have a voltage drop across the two
ends with the hot end being at a lower potential that the cold end, VAB<0
This phenomena of a temperature gradient in a material causing a voltage gradient is called the Seebeck Effect.
It is quantified by the Seebeck coefficient, S where for metals and
Δ𝑉 𝑉𝐵 − 𝑉𝐴
𝑆=
=
Δ𝑇 𝑇𝐴 − 𝑇𝐵
Note that the sign of S is defined by the voltage at the cold end with respect to the hot end. Therefore for metals and n-dope
semiconductors, S<0 and for p-type semiconductors S>0.
Thermoelectricity – Physics
of
Seebeck
Effect
E
E
VB
VA
TA
TA>TB
TB
VA
f(E)
f(E)
1. Since the fermi-functions smoothens out at higher temperatures, there are more carriers available
for conduction at the hotter end as compared to the cold end.
i.e. there is a concentration gradient
2. In the hotter side, the carriers have more velocity as compared to the colder side. i.e. the Diffusion
coefficient is greater at the hotter side as compared to the colder side.
VA>VB, S<0
+ ions
TA
VB
electrons
TB
TA>TB
N-type or metals
Due to 1 and 2, we have diffusion of carriers from the hotter side to the colder side i.e. a
current is established.
VA<VB, S>0
VA
In metals and n-doped semiconductors – the electrons diffuse from hotter to colder side. Thus they
leave behind + vely charged metal ions or donor ions on the hotter side and rush to the colder side.
- ions
holes
Therefore the potential at the hotter side is more than the potential at the colder side. S<0.
In p-doped semiconductors, the holes diffuse from the hotter side to the colder side and leave
TA
TA>TB
behind – vely charged acceptor ions at the hotter side. Thus, the potential at the hotter side is less
than the potential at the colder side. Therefore S>0.
P-type
(Generally, you can imagine S to have the same sign as the sign of the charge on the carriers)
Once this potential develops, it will oppose the diffusion of carriers and a balance between
drift and diffusion is reached
VB
TB
Thermoelectricity – Seebeck Effect
Quantifying the Seebeck Coefficient
In metals or n-doped semiconductors, the average energy of the electron at a temperature T is
𝐸𝑎𝑣
3
5𝜋 2 𝑘𝑇
𝑇 = 𝐸𝐹0 1 +
5
12 𝐸𝐹0
2
(Don’t worry about deriving or memorizing this)
If we imagine a small temperature gradient dT, the electrons from the hotter side would diffuse to the colder side and result in
a voltage drop of dV. If we now consider another electron crossing this section, it would have to give up energy to overcome
this potential drop.
𝐸𝑎𝑣 𝑇 + 𝑑𝑇 − 𝐸𝑎𝑣 𝑇 = −𝑞 𝑑𝑉
𝜋 2 𝑘 2 𝑇𝛿𝑇
−𝑞 𝑑𝑉 =
2𝐸𝐹0
Substituting for the average energy and ignoring the (dT)2 term,
𝑑𝑉
𝜋 2𝑘2𝑇
𝑆=
=−
𝑑𝑇
2𝑞𝐸𝐹0
The Seebeck coefficient is dependent on the temperature as well as material (EF0).
Thermoelectricity – Seebeck Effect
Odd Behavior of the Sign of the Seebeck Coefficient in some materials
In some materials, the Seebeck coefficient has the opposite coefficient as expected.
i.e. in some metals, the Seebeck coefficient is positive. Why?
This is because, with an increase in temperature the lattice vibrations also increase. Therefore even though the electron
concentration is higher at the hotter end, the effective mobility and diffusion coefficient of the electrons becomes smaller than
those at the colder end – and in such materials this leads to the effect of electrons from the colder end diffusing to the hotter
side due to their larger diffusion coefficient (diffusion coeffient and concentration gradient product is larger)
Thermocouples
In thermocouples, we have two different metals (could also be semiconductors) with
different Seebeck coeff – say SA and SB.
The metals are joined at one end to create a junction.
The free ends are typically at some ambient (temperature Tamb) and the junction is
placed at a temperature that is to be measured (temperature Tm).
The potential of the junction is say V0
The potential of the free end of metal A is 𝑉0 + 𝑆𝐴 (𝑇𝑚 − 𝑇𝑎𝑚𝑏 )
The potential of the free end of metal B is 𝑉0 + 𝑆𝐵 (𝑇𝑚 − 𝑇𝑎𝑚𝑏 )
The potential difference between the two free ends is (𝑇𝑚 − 𝑇𝑎𝑚𝑏 )(SA − SB )
Thus measuring this potential is an indirect measurement of Tm
Junctions