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?