As follows from appendix A of Nowick and Berry $^{2}$, the effective spring constant of a resonance mode with a complex position dependent Young’s modulus is given by: Supplementary Figure A: Schematic of the thermodynamic coupling mechanism via the piezoresistive, resistive heating and thermal expansion effects. Relevant variables in the mechanical, electrical and thermal domain are shown. Coupling mechanisms are represented by arrows. 10 Where the integrals run over the volume V of the resonator. Combining this with equation (A6), it is found that the feedback coefficient β can be expressed as: Where A is the cross-sectional area of the piezoresistive conductor, and γ$_{k}$ is a factor which accounts for the non- uniformity of Y ∗ $_{eff}$. If Y ∗ eff is constant over the whole volume, γ$_{k}$ = 1. For the resonator under consideration, the thermodynamic feedback effect is large inside the narrow engine beam where the strain and current density are concentrated. It is therefore a good approximation to assume that Y ∗ eff is given by equation (A6) inside the volume V$_{engine}$ of the engine beam and is equal to Y everywhere else in the resonator. In this case the factor γ$_{k}$ is given by: k ∗ eff = k ( ∫ V Y ∗ $_{eff}$( z ) ε 2 $_{ac}$( z ) d z ∫ V Y ε 2 $_{ac}$( z ) d z ) = k (1 − βI 2 $_{dc}$) (A7) β = γ$_{k}$γ$_{Z}$ ρ$_{dc}$π$_{l}$α$_{te}$Y A $^{2}$( γ$_{h}$ + iωρ$_{d}$c$_{p}$ ) = γ$_{Z}$β$_{0}$ (A8) γ$_{k}$ = ∫ V$_{engine}$ ε 2 $_{ac}$( z ) d z ∫ V ε 2 $_{ac}$( z ) d z (A9) k$_{h}$ ∇ $^{2}$T$_{ac}$ + p$_{pr}$ = ic$_{p}$ρ$_{d}$ωT$_{ac}$ (A10) For the lowest frequency in-plane bending resonance mode this fraction is found to be γ$_{k}$ = 0 . 11 using a finite element simulation. To evaluate γ$_{h}$ the AC heat equation is to be solved: Since the dimensions of the engine beam are much smaller than the thermal wavelength λ$_{h}$ = √ 8 π $^{2}$k$_{h}$ / ( c$_{p}$ρ$_{d}$ω$_{0}$ ) ≈ 26 µ m, the heat equation can be simplified by assuming that the AC resistive heating density p$_{pr}$ is zero outside the engine beam and that p$_{pr}$ and T$_{ac}$ are independent of position inside the engine beam. The beam can be treated as a point source at y = 0 with temperature T$_{ac,}$$_{0}$e $^{iωt}$. The corresponding solution of the 1-dimensional heat equation outside the beam is: Since the heat conduction from the 2 ends of the beam needs to be equal to the difference between the generated and stored energy in the beam, it follows from equation (A3) that: Where L =800 nm is the beam length. Substitution of equation (A11) yields: T ( x, t ) = T$_{ac,}$$_{0}$$_{e}$− (1+ i )2 π | y | /λ$_{h}$ + iωt (A11) − 2 k$_{h}$ ∇ T$_{ac}$ |$_{y}$$_{=0}$ = L ( p$_{pr}$ − ic$_{p}$ρ$_{d}$ωT$_{ac}$ ) = Lγ$_{h}$T$_{ac}$ (A12) γ$_{h}$ = 4 π (1 + i ) k$_{h}$/ ( Lλ$_{h}$ ) (A13) The material constants along the [100] direction of n -type silicon are Y = 130 GPa, π$_{l}$ = − 102 × 10 − 11 Pa, α$_{te}$ = 2 . 6 × 10 − 6 K − $^{1}$, ρ$_{d}$ =2329 kg/m $^{3}$, c$_{p}$ = 702 J/kg · K and k$_{h}$ =113 W/(K · m). The cross-sectional area is A =280 × 1500 nm $^{2}$.