Supplementary Information
1. Emittance measurement for sintered Bi0.4Sb1.6Te3
Radiative energy balance for a layer states that the sum of reflectance, transmittance and
absorptance is unity. Based on Kirchhoff’s law, the spectral absortpance is equal to the spectral
emittance for a diffusely reflecting and emitting surface. Thus, assuming sintered Bi0.4Sb1.6Te3 is
spectrally diffuse, we obtained its spectral emittance from 1 - spectral reflectance - spectral
transmittance. We used a Fourier transform infrared (FTIR) spectrometer (Frontier, Perkin Elmer)
to measure the spectral reflectance and transmittance in the wavelength range 7.5 – 13 m. The
corresponding spectral range is identical to the active spectral region for the infrared imaging
camera (SC660, FLIR) used in this study. For a 1 mm thick sintered Bi0.4Sb1.6Te3 sample, we
measured the transmitted and reflected beam intensity with the spectral resolution of 8 cm-1 and
with the scan number of 10. To estimate the spectral transmittance, we took the ratio of the
transmitted intensity through the sample to the transmitted intensity through air. Similarly, to
obtain the spectral reflectance, we took the ratio of the reflected intensity from the sample to the
reflected intensity from a reference sample. The reference was a ~100 nm thick gold film
deposited on top of a 1 mm thick glass substrate employing a ~ 30 nm thick titanium as an
adhesion layer. Figure S1 shows the measured reflectance and absorptance of the Bi0.4Sb1.6Te3
sample. Transmittance data is not included in Fig. S1, since it is trivial. Here, the absorptance (=
emittance) of the sample is nearly uniform and an average value is 0.94. Note that the material
optical property is temperature dependent such that the absorptance measured here is valid only
at a room temperature.
1
Figure S1. Measured absorptance and reflectance of a 1 mm thick sintered Bi0.4Sb1.6Te3 at the
spectral range of 7.5 – 13 m. Absorptance is obtained by 1 – reflectance, since sintered
Bi0.4Sb1.6Te3 is opaque over the corresponding spectral region. An average absorptance over the
corresponding spectral region is 0.94.
2
2. Derivation of the modified Harman relation (Eqn. 3)
A sum of the energy balances for the cold-side surface and the rest part of the sample is
2ks AT
 K w TH  TC   I 2 R   PL T  T0
L

 s   w  I TH  TC  

(S1)
(TH + TC)/2 is approximated as 𝑇̅ and TH-TC is ΔT. Then,
 s   w  IT 

ks AT K w T I 2 R  PL T  T0



L
2
2
2

(S2)
We modify Eqn. S2 to obtain (s-w) ΔT.
 s   w 
2

k A K
I 2 R  PL T  T0
IT   s   w  T  s  w 

2 2T
2T
 L

 
(S3)


We substitute Eqn. S3 into the relation for Vdc.
Vdc  IR   s   w  T  IR 
 s   w 
2
IT

ks A K w I 2 R  PL T  T0



L
2 2T
2T
   w 
 IR  s
2
 s ks
T I  s L ks A
1
A L kA K
I 2 R  PL T  T0
s
w



L
2 2T
2T
   w 
 IR  ZT s
i

2
s


2
 IR
1

 PL2 T  T0
KwL
I 2 RL
1


2k s A 2Tk s A
2Tk s A


(S4)
Since IR is equal to either Vac or sL/A, Eqn. S4 can be modified as


2
 s w 
1

Vdc  Vac 1  ZTi 

 PL2 T  T0
 s 
KwL
I 2  s L2

1



2k s A 2Tk s A2
2Tk s A


3








(S5)
Therefore, ZTi can be expressed as
V
  s 
ZTi   dc  1

 Vac   s   w 
2


  PA T  T0  I 2  s L2 K L 
1 
 w 
2Tks
A2 2ks A 



4
(S6)
3. Boundary conditions for the temperature distribution (Eqn. 5)
For convenience, define Lh = [ksA/(+h)P]0.5. If the temperature is TH at x = 0 and TC at x
= L,
TH  T  0   c1  c2 
I 2
T
   h  PA 0
TC  T  L   c1e L Lh  c2e  L Lh 
I 2
T
   h  PA 0
(S7)
(S8)
An energy balance at x = 0 considers Peltier heating, Joule heating due to electrical contact
resistance, conduction heat flow through the sample and lead wires, and convective heat flow
from the sample to air. Note that dT/dx = (c1ex/Lh – c2e-x/Lh)/Lh.
 s   w  ITH  IRc2  
ks A
 c1  c2   K w TH  T0   hA TH  T0 
Lh
(S9)
Similarly, an energy balance at x = L is expressed as
 s   w  ITC  IRc2 


ks A
c1e L Lh  c2e  L Lh  K w T0  TC   hA T0  TC 
Lh
(S10)
To solve for c1 and c2, we substitute TH in Eqn. S9 by TH in Eqn. S7 and substitute TC in Eqn.
S10 by TC in Eqn. S8. Then, we arrange the equations as
 ks A

 kA

  s   w  I  K w  hA c1   s   s   w  I  K w  hA c2

 Lh

 Lh


  K  hA  I 2 
I 2
   s   w  I 
 T0   w
 IRc2 (S11)
   h  PA
    h  PA

k A

k A

e L Lh  s   s   w  I  K w  hA c1  e L Lh  s   s   w  I  K w  hA c2
 Lh

 Lh

5

  K  hA  I 2 
I 2
  s   w  I 
 T0   w
 IRc2
   h  PA
    h  PA

c1 and c2 are obtainable by solving Eqn. S11 and S12 via a matrix calculation.
6
(S12)