Thermopile series voltage ($V_{\mathrm{oc}}={\sum }_{i=1}^N{E}_i$)

Use the A-positive terminal order of Seebeck effect for each thermocouple. Its thermocouple measuring junction has temperature $T_{h,i}$ and its reference connections have temperature $T_{c,i}$. Define the difference of Seebeck coefficients as ${\alpha }_i(T)=S_{A,i}(T)-S_{B,i}(T)$. Connect the B reference end of one thermocouple to the A reference end of the next at a common local temperature, so the terminal electric potential differences add:The first A reference terminal is the positive meter input; the final B reference terminal is the negative input. Following this connected path and adding the individual Seebeck effect integrals proves the result. Matched meter leads with common endpoint temperatures cancel as in an isothermal terminal block. The manufacturer's sum formula also allows unequal individual contributions.^[ref]

If all pairs share $T_h,T_c$ and the same nearly constant difference $\alpha$ of Seebeck coefficients throughout that temperature interval, thenThis $\alpha$ is the pair's relative Seebeck coefficient, not either material's individual coefficient or a universal device constant. As an original constructed example, choose $N=20$, $\alpha =40\mu \mathrm{V}{\mathrm{K}}^{-1}$ and $\Delta T=5\mathrm{K}$. Each thermocouple contributes $0.20\mathrm{mV}$ andReversing the meter connections changes the sign. Exchanging $T_h$ and $T_c$ reverses the integrals and hence the electromotive force; this reversible sign response does not make a detector with heat conduction a reversible thermodynamic process. Reversing the wiring of one pair instead gives $(N-2)\alpha \Delta T$, because one contribution is subtracted rather than added. For this example the result is $3.6\mathrm{mV}$, and at equal junction temperatures every contribution vanishes.

The open-circuit condition matters. If the array's total electrical resistance is $R_{\mathrm{pile}}$ and the meter input has electrical resistance $R_m$, the elementary voltage divider model gives $V_m=V_{\mathrm{oc}}{R}_m/(R_m+R_{\mathrm{pile}})$. Thus $R_m\gg R_{\mathrm{pile}}$ is needed to suppress voltmeter loading. Appreciable electric current also permits Joule heating and the thermoelectric heat transfer to disturb the assumed junction temperatures.