Thermocouples
Seebeck coefficient (or thermo power) of a metal A:
PA =
ΔV
ΔT
ΔV: Open circuit Voltage
ΔT: Temperature difference
between thermocouple TA
and reference point TB
PA is material-specific
Seebeck coefficient measured by a thermocouple of metal A
and metal B:
PS = PA − PB
PB is the Seebeck coefficient of metal B.
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Thermoelectric e.m.f. of Various Metals
Thermoelectric e.m.f.
It is possible to obtain
the e.m.f. for any
thermocouple by
subtracting the two Ps
values of the
contacting materials.
ΔVr = ( PA − PB )ΔT
List thermoelectric e.m.f. of metals relative to Platinum at 0ºC
One arm made of high
e.m.f. material and the
other arm with low
e.m.f. material e.g.
Iron v.s. Nickel
Thermocouples have been standardized . Three common
alloys : Chromel, Constantan and Alumel
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Standard Thermocouples
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Thermocouple Performance
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Thermopile
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Semiconductor Thermocouple
Semiconductor materials
show larger thermoelectric
effect than metals.
In practice, the Seebeck
coefficient is related to the
electrical resistivity ρ
Seebeck coefficient of p-type
silicon is 1 mV/K for a sheet
resistance of 200 Ω/sq at 300K.
Thermopiles consists of N
identical p-Si/ Al thermocouples
has been made.
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Figure of Merit for Insulators, Semiconductors and Metals
Figure of merit Z
Z = Ps σ /κ
2
Where σ and κ is the
electrical and
thermal conductivity
respectively.
A high conductivity
ratio is advantageous
because it permits a
high temperature
gradient to be
maintained at a low
power consumption.
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Thermoelectric Effect
• A thermocouple is a closed circuit consisting of tow
junctions , at different temperatures T1 and T2 °C
AB
AB
ETAB
=
E
−
E
T1
T2
1 ,T2
2
2
3
3
= a1 (T1 − T2 ) + a2 (T1 − T2 ) + a3 (T1 − T2 ) + ...
¾ the measured e.m.f. depends on T1, T2
¾ T2 must be known to infer T1
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Five Laws of Thermocouple
•
Five laws of thermocouple behavior
ƒ Law of intermediate metals
ƒ Law of intermediate temperatures
T3
if T2 = 0o C , then ET 1, 0 = ET 1,T 3 + ET 3, 0
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Temperature Measurement Using a Thermocouple
• Thermocouple table
Æ ET,0 for a particular thermocouple
Example of using Type K thermocouple Table:
liquid inside a vessel Å measurement junction.
20 °C (outside) Å reference junction
T3 = 20 °C
Obtained reading : ET1,T3 = ET1,20 =
from table: ET3,0 =
5.3mV
E20,0 = 0.8mV
⇒ ET1,0 = 5.3mV+0.8mV=6.1mV
form table T1=149 °C
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Type J Thermocouple Table
•
Table of type J thermocouple output
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Type J Thermocouple Table (cont.)
•
Table of type J thermocouple output (cont.)
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Type K Thermocouple Table
•
Reference values for Chromel-Alumel thermocouples
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Installation Problem
Type K (Nickel-Chromium v.s. Nickel-Aluminum)
¾ Measure the temperature of high pressure
steam in the pipe , ≈200oC with a chromel
vs. alumel thermocouple
¾ (a) T2 can vary widely , useless
¾ (b) The reference junction is still outside
the control room
¾ (c) The reference junction is now in the
control room , but T2 is still changing
¾ (d) use of law of intermediate temperature
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Automatic Reference Junction Compensation
Circuit of a Thermocouple
•
the e.m.f. source producing ET3,0
¾ ET3,0 =a1T3+a2T32+a3T33+……
T2 is small Æ ET3,0 ≈ a1T3
a circuit giving a millivolt
output signal proportional to T3
metal resistance thermometer , RTD
2
2
R
RT = R0(1+αT+βT +……)
in a deflection bridge
→ ET 3, 0
R2
= a1T3 = Vs αT3
R3
R3
T2
E
ET3,0
RT
,0
R4
Vs
R2 a1
→ Vs
=
R3 α
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AD596/AD597 are
monolithic set point
controllers that include the
amplifier and cold junction
compensation for type J
and K thermocouples.
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Linearity of a Thermocouple
•
Non-linearity
example (from table 8-2) copper vs. constant T.C.
use: 0°C ~ 400 °C
at 200 °C Æ 9286μV
ideal straight line value: 10435 μV Æ -5.5%
non-linearity at 200 °C
• Current transmission is used in industry ,
ET1,0 : millivolts
Æ current : 4 ~ 20 mA
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Thermodiodes
eV
− 1)
i = is (exp
kT
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A PTAT device
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Sensitivity of a Diode Temperature Sesnsor
The forward
junction voltage
can be used to
measure
temperature
between 50 and
300 K.
Sensitivity:
-2.27(mV/K)
Example: 1N4148 low
cost electronic
thermometer, 0.1º
accuracy
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壓電晶體原理
壓電效應（Piezoelectric effect）是許多種單一晶體具有的
一種現象。當單一晶體，如石英晶體（Quartz crystal ），
承受彈性形變時，在某些晶面上會產生電荷，石英體具有
六角形結構，矽（Si）及氧（O）原子佔在角上的位置（
圖一）。
（圖一）
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壓電晶體原理(續)
今考慮這種晶體在 z 方向的一斷面：
（圖二 a）
（圖二 b）
（圖二 c）
沿 x-軸方向的壓力會在與 x-軸垂直的平面上產出電荷（圖二 b）；
沿 y-軸方向的壓力也同樣會在與 x-軸垂直的平面上產生出電荷（圖
二 c）。圖二 b 的情形，稱做縱向壓電效應（Longitudinal
piezoelectric effect），因其壓力（或負荷）是在電荷平面；圖二 c
的情形，稱做橫向壓電效應（Transverse piezoelectric effect），因其
負荷是在與電荷平面垂直的方向。
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壓電晶體感測元件（Piezoelectric Sensing Element）
•Crystal:
x=
displacement
1
F
k
applied force
k: stiffness～2 ×109 Nm-1
Δx
1/k
(s ) =
1 2 2ξ
ΔF
s + ωn s + 1
2
ωn
fn = ωn /2π , 10 ~ 100 kHz
ξ ≈ 0.01
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壓電晶體感測元件（Piezoelectric Sensing Element）
•Piezoelectric crystal:
Direct piezoelectric effect
q = Kx
deformation of crystal
charge
q=
K
F = dF
k
d = K/k
coulombs newton-1
= Charge sensitivity to force
Inverse piezoelectric effect
x = dV
applied voltage
Mechanical displacement
d = (meter volt-1) = (coulombs newton-1)
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Charge Sensitivity to Force
•
Piezoelectric materials
d(PCN-1)
Natural
Ceramic
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Quartz
2.3
Tourmaline
1.9, 2.4
Lead
265
Zirconate-titanate鈷酸鹽
265
Lead mataniobate
80
PVDF film
23
ZnO (bulk)
11.7
ZnO (film)
12.4
BatiO3
190
PZT
370
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Piezoelectric Force Sensor
crystal →charge generator // CN
dq
dx
=K
iN =
dt
dt
ΔiN
(s) = Ks
Δx
1
1
RL
= C N s + C Cs +
,Z=
Z
RL
1 + R L (C N + C C )s
Δ VL
RL
(s) =
Δ iN
1 + R L (C N + C C )s
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Overall Transfer Function
Δ VL
Δ VL Δ i N Δ x
( s) =
ΔF
Δ iN Δ x Δ F
1/k
RL
Ks
=
1 2 2ξ
1 + R L (C N + Cc )s
s +1
s +
2
ωn
=
ωn
K
1
R L (C N + Cc )s
1
k (C N + Cc ) 1 + R L (C N + Cc )s 1 s2 + 2ξ s + 1
2
ωn
=
ωn
d
1
τs
(C N + Cc ) ( 1 + τs) ( 1 s2 + 2ξ s + 1)
2
ωn
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ωn
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Discussion on Force Measurement
•Steady-state sensitivity = d/(CN+CC)
CC = f (length of cable)
1
τs
•
G(s) =
, τ = R L (C N + CC )
τs + 1 1 s2 + 2ξ s + 1
ω n2
ωn
the system cannot be used for measuring d.c. and slowly varying forces
If CN = 1600pF, CC = 600pF, RL=1MΩ,
τ=2.2 ×10-3 sec,
1/τ= 455 rad sec-1 or 72 Hz
useful range: 3/τ~ 0.2 Wn
=216 Hz ~ 5.4 kHz where 0.95≦∣G(jw) ∣ ≦0.5, ψ～0°
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Frequency Responses
Use of charge
amplifier:
output ∫iN dt,
output charge
q, nonzero
output for a
steady force
input.
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Use of Charge Amplifier
•In the figure:
i1 = iF + iqF = CF(V--Vout)
i+ = i- = 0
V- = V+ =0
i1 = i F =
dq F
dV
= −C F out
dt
dt
•No potential drop across CN and CC
i1 = i N =
• dVout
dt
=−
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dq
dt
1 dq
q
, Vout = −
C F dt
CF
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Ideal Charge Amplifier
•Overall transfer function with ideal charge amplifier:
ΔVout
ΔVout ΔiN Δx
(s) =
(s)
ΔF
Δ i N Δx Δ F
=-
1/k
1
Ks
1
2ξ
C Fs
( 2 s2 +
s + 1)
ωn
ωn
-1
1
=
d
C F ( 1 s2 + 2ξ s + 1)
2
ωn
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ωn
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Discussion
1.
2.
Steady state sensitivity is d/CF, depends only on the capacitance
CF of the charge amplifier and independent of transducer and
cable capacitance.
Ideal case: ∣G(jω) ∣ = 1, at ω = 0.
in practice RF, CF introduce
τ Fs
, τ F = R FC F
1 + τ Fs
making RF and CF large
RF = 108Ω, CF = 104pF →τF = 1.0 sec
∣G(j ω) ∣= 0.95 at ω = 3 rad/sec → f≒0.5 Hz
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