Partial Transient Liquid-Phase Bonding, Part II: A Filtering Routine for Determining All Possible Interlayer Combinations

Abstract

Partial transient liquid-phase (PTLP) bonding is currently an esoteric joining process with limited applications. However, it has preferable advantages compared with typical joining techniques and is the best joining technique for certain applications. Specifically, it can bond hard-to-join materials as well as dissimilar material types, and bonding is performed at comparatively low temperatures. Part of the difficulty in applying PTLP bonding is finding suitable interlayer combinations (ICs). A novel interlayer selection procedure has been developed to facilitate the identification of ICs that will create successful PTLP bonds and is explained in a companion article. An integral part of the selection procedure is a filtering routine that identifies all possible ICs for a given application. This routine utilizes a set of customizable parameters that are based on key characteristics of PTLP bonding. These parameters include important design considerations such as bonding temperature, target remelting temperature, bond solid type, and interlayer thicknesses. The output from this routine provides a detailed view of each candidate IC along with a broad view of the entire candidate set, greatly facilitating the selection of ideal ICs. This routine provides a new perspective on the PTLP bonding process. In addition, the use of this routine, by way of the accompanying selection procedure, will expand PTLP bonding as a viable joining process.

Appendices

Appendix A

For half of the IC, the RC and diffusant element volumes are given by

$$ V_{\text{RC,bin}} = \frac{{{\text{th}}_{\text{RC}} }}{2}A $$

(A1)

$$ V_{{{\text{D}}_{\text{x}} }} = {\text{th}}_{\text{D}} \cdot A $$

(A2)

Because the cross-sectional area (A) is constant across the IC, volumes can be simplified to the interlayer thicknesses. These values are then multiplied by room-temperature densities and divided by atomic weights (values taken from Reference 10) to produce molar amounts:

$$ n_{\text{RC,bin}} = \frac{{{\text{th}}_{\text{RC}} }}{2}\frac{{\rho_{\text{RC}} }}{{{\text{at}} . {\text{wt}}_{\text{RC}} }} $$

(A3)

$$ n_{{{\text{D}}_{\text{X}} }} = {\text{th}}_{\text{D}} \frac{{\rho_{{{\text{D}}_{\text{X}} }} }}{{{\text{at}} . {\text{wt}}_{{{\text{D}}_{\text{X}} }} }} $$

(A4)

The number of diffusant element moles (Eq. [A4]) is the same for the binary and ternary cases.

The at. pct of the RC element and a diffusant element are then given by

$$ {\text{at}} .\,{\text{pct}}_{\text{RC,bin}} = \frac{{100 \cdot n_{\text{RC,bin}} }}{{n_{\text{RC,bin}} + n_{{{\text{D}}_{\text{X}} }} }} $$

(A5)

$$ {\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{X}} , {\text{bin}}}} = \frac{{100 \cdot n_{{{\text{D}}_{\text{X}} }} }}{{n_{\text{RC,bin}} + n_{{{\text{D}}_{\text{X}} }} }}. $$

(A6)

To calculate the ternary composition, the complete volume of the RC,

$$ V_{\text{RC,ter}} = {\text{th}}_{\text{RC}} \cdot A , $$

(A7)

is used to calculate the number of moles present in the complete RC layer:

$$ n_{\text{RC,ter}} = {\text{th}}_{\text{RC}} \frac{{\rho_{\text{RC}} }}{{{\text{at}} . {\text{wt}}_{\text{RC}} }}. $$

(A8)

The at. pct of the RC element and one of the diffusant elements are given by Eqs. [A9] and [A10]:

$$ {\text{at}} .\,{\text{pct}}_{\text{RC,ter}} = \frac{{10 \cdot n_{\text{RC,ter}} }}{{n_{\text{RC,ter}} + n_{{{\text{D}}_{\text{A}} }} + n_{{{\text{D}}_{\text{B}} }} }} $$

(A9)

$$ {\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{X}} , {\text{ter}}}} = \frac{{100 \cdot n_{{{\text{D}}_{\text{X}} }} }}{{n_{\text{RC,ter}} + n_{{{\text{D}}_{\text{A}} }} + n_{{{\text{D}}_{\text{B}} }} }}. $$

(A10)

If the ternary composition resides on the line connecting the two binary compositions, then it will satisfy the basic equation for a line, shown in Eq. [A11] with the appropriate values inserted:

$$ {\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{B}} , {\text{ter}}}} = {\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{B}} , {\text{bin}}}} -\frac{{{\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{B}} , {\text{bin}}}} }}{{{\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{A}} , {\text{bin}}}} }}{\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{A}} , {\text{ter}}}} . $$

(A11)

Substituting Eqs. [A6] and [A10] into the right-hand side of Eq. [A11] yields

$$ {\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{B}} , {\text{ter}}}} \,{ \mathop {= } ^ { ?}} \,\frac{{100 \cdot n_{{{\text{D}}_{\text{B}} }} }}{{n_{\text{RC,bin}} + n_{{{\text{D}}_{\text{B}} }} }} - \frac{{\frac{{100 \cdot n_{{{\text{D}}_{\text{B}} }} }}{{n_{\text{RC,bin}} + n_{{{\text{D}}_{\text{B}} }} }}}}{{\frac{{100 \cdot n_{{{\text{D}}_{\text{A}} }} }}{{n_{\text{RC,bin}} + n_{{{\text{D}}_{\text{A}} }} }}}} \cdot \frac{{100 \cdot n_{{{\text{D}}_{\text{A}} }} }}{{n_{\text{RC,ter}} + n_{{{\text{D}}_{\text{A}} }} + n_{{{\text{D}}_{\text{B}} }} }}. $$

(A12)

Since n _RC,bin = n _RC,ter/2, simplifying Eq. [A12] yields

$$ {\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{B}} , {\text{ter}}}} \,{\mathop {=} ^ {?}}\, \frac{100 \cdot n_{ {\text{D}}_{\text{B}} } } {{{{{n_{\text{RC,ter}} }}} /2} + n_{{\text{D}}_{\text{B}} } } - \frac{{100 \cdot n_{{D_{B} }} }}{{{{{n_{\text{RC,ter}} }}} /2} + n_{{\text{D}}_{\text{B}} } } \cdot \frac{{{{{{n_{\text{RC,ter}} }}} /2} + n_{{{\text{D}}_{\text{A}} }} }}{{\left( {n_{\text{RC,ter}} + n_{{{\text{D}}_{\text{A}} }} + n_{{\text{D}}_{\text{B}} } } \right)}} $$

(A13)

Creating a common denominator produces

$$ {\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{B}} , {\text{ter}}}}\, {\mathop {=} ^ {?}}\, \frac{{100n_{{{\text{D}}_{\text{B}} }} \cdot \left( {n_{\text{RC,ter}} + n_{{{\text{D}}_{\text{A}} }} + n_{{{\text{D}}_{\text{B}} }} } \right)-100n_{{{\text{D}}_{\text{B}} }} \cdot \left( {{{{{n_{\text{RC,ter}} }}} /2} + n_{{{\text{D}}_{\text{A}} }} } \right)}}{{\left( {n_{\text{RC,ter}} + n_{{{\text{D}}_{\text{A}} }} + n_{{{\text{D}}_{\text{B}} }} } \right) \cdot \left( {{{{{n_{\text{RC,ter}} }}} /2} + n_{{{\text{D}}_{\text{B}} }} } \right)}} $$

(A14)

And simplifying the numerator yields

$$ {\text{at}} .\,{\text{pct}}_{{{\text{D}}_{\text{B}} , {\text{ter}}}} \,{\mathop {=} ^ {?}}\, \frac{{100n_{{{\text{D}}_{\text{B}} }} \cdot \left( {{{{{n_{\text{RC,ter}} }}} /2} + n_{{{\text{D}}_{\text{B}} }} } \right)}}{{\left( {n_{\text{RC,ter}} + n_{{{\text{D}}_{\text{A}} }} + n_{{{\text{D}}_{\text{B}} }} } \right) \cdot \left( {{{{{n_{\text{RC,ter}} }}} /2} + n_{{{\text{D}}_{\text{B}} }} } \right)}} $$

(A15)

which simplifies to Eq. [A10] and proves that the ternary composition resides on the line connecting the binary compositions.

Appendix B

For a binary system, the conversion from atomic (c) to weight (w) percent is accomplished by calculating the weighting factors shown in Eqs. [B1] and [B2] and substituting them into the ratio shown in Eq. [B3]. The result of this substitution is shown in Eq. [1] (Section II–D).

$$ p_{\text{A}} = c_{\text{A}} \cdot {\text{at}} . {\text{wt}} ._{\text{A}} $$

(B1)

$$ p_{\text{B}} = (100--c_{\text{A}} )\cdot {\text{at}} . {\text{wt}} ._{\text{B}} $$

(B2)

$$ w_{\text{A}} = \frac{{100 \cdot p_{\text{A}} }}{{p_{\text{A}} + p_{\text{B}} }}. $$

(B3)

In similar fashion, the conversion from weight to volume (v) percent is accomplished by calculating the weighting factors shown in Eqs. [B4] and [B5] and substituting them into Eq. [B6], resulting in Eq. [2] (Section II–D).

$$ q_{\text{A}} = \frac{{w_{\text{A}} }}{{\rho_{\text{A}} }} $$

(B4)

$$ q_{\text{B}} = \frac{{100 - w_{\text{A}} }}{{\rho_{\text{B}} }} $$

(B5)

$$ v_{\text{A}} = \frac{{100 \cdot q_{\text{A}} }}{{q_{\text{A}} + q_{\text{B}} }} $$

(B6)