AC CalorimetryPrinciple, Schematic, Examples
Chaochao Pan
PHYS 590B 18F
1
𝐼 = 𝐼0cos1
2ω𝑡
𝑃 = 𝐼02𝑅𝑐𝑜𝑠2
1
2ω𝑡
𝑇 = 𝑇𝑑𝑐 + 𝑇𝑎𝑐cos(ω𝑡 + φ)
𝐶𝑠 𝑇𝑠 = 𝑄𝑠 = 𝑃 − 𝐾𝑏(𝑇𝑠 − 𝑇𝑏)
𝑃
𝐾𝑏
Simplified model
𝑇𝑎𝑐 =𝑃02𝐾𝑏
1 + 𝜔2𝜏2 −1/2
Simple but requires knowledge of R(P, T)!
𝜏 =𝐶𝑠𝐾𝑏
Physics 590 B Elena Gati2
𝑇θ = 𝑇𝑑𝑐 + 𝑇𝑎𝑐cos(ω𝑡 + φ)
If ω2 τθ2 + τℎ
2 ≪ 1
𝑎𝑛𝑑 ωτ𝑠 ≫ 1
𝑇𝑎𝑐 =𝑃02ω𝐶
1 +1
ω2τ𝑠2 + ω2(τθ
2+τℎ2 )
−1/2
geometric correction term
𝐾𝑠 = ∞
𝐾𝑠 ≠ ∞
τ1 = τ𝑠
τ22 = τθ
2 + τℎ2 + τ𝑖𝑛𝑡
2
Time to reach thermal equilibrium
Sullivan and Seidel, Phys. Rev. 173, 679 (1968) 3
τ1 = 3𝑠𝑒𝑐, τ2 = 7 × 10−4𝑠𝑒𝑐
0.8
1 10 100 1000
T ac*
ω/(
P0/2
C)
ω
𝑇𝑎𝑐 =𝑃02ω𝐶
1 +1
ω2τ12 + ω2τ2
2
−1/2
ω2τ22 ≪ 1
Slow enough for heater and thermometer to catch up sample temperature
ωτ1 ≫ 1
Quick enough for sample to decouple from thermal bath
𝑇𝑎𝑐𝑃02ω𝐶
−1
~1
𝑓 =ω
2𝜋2Hz – 20Hz for this case
4
Sullivan and Seidel, Phys. Rev. 173, 679 (1968)
5
Sample: 20 µmThermocouple: 12 µmHeating wire: 3 µm
Drawbacks(i) Thermocouple
calibration(ii) Absolute value
Advances in Solid State Physics. Springer, Berlin,
Heidelberg, 2003. 889-913.Eichlerand Gey, Rev. Sci. Instrum. 50, 1445 (1979)
6
Advances in Solid State Physics. Springer, Berlin,Heidelberg, 2003. 889-913.
7
Current directly applied to sample
PHYSICAL REVIEW B 66, 064428 (2002)H. Wilhelm and D. Jaccard
• C(T) vs. ρ(T)• 𝛻𝑇• 𝛻𝑃
CePd2+xGe2-x
For x = 0.02, open circle by ac calorimetry, filled circle by resistivityFor x = 0, open and filled square are two samples measured by resistivity
Antiferromagnetic transition TN vs P
Unpressurized thermometer
Kazunori Umeo, REVIEW OF SCIENTIFIC INSTRUMENTS 87, 063901 (2016)
Grain size 0.25 μm.
0.3K ≤ 𝑇 ≤ 30𝐾
8
Cs = Ctot-Cbackground
Conclusion
• Sensitive to heat capacity change
• Absolute value not as accurate as adiabatic calorimetry
• Suitable in pressure cell
9
Thank you for attentions
Questions?
Top Related