Lubricant Transfer Model at the Head-Disk Interface in Magnetic Storage Considering Lubricant–Lubricant Interaction

Abstract

Interactions among molecularly thin lubricant–solid and lubricant–vapor interfaces become increasingly strong when the slider approaches closer to the disk (or media) surface, which is desirable for higher storage areal densities. Lubricant transfers from the disk to the slider as the head-media spacing decreases below a critical value, leading to loss of protection for the solid thin films. The present work proposes a new criterion for lubricant pickup to occur by improving the lubricant transfer model via considering interactions between the lubricant films, in addition to lubricant–solid interaction. The effect of lubricant roughness is also included in the proposed model, which is then used to predict critical solid distances as a function of lubricant roughness and average lubricant thickness. It is found that the interaction force between the slider solid and the disk lubricant plays a weaker role than that between the lubricant on the slider and the lubricant on the disk in determining the critical solid distance. The physics for the dependence of the critical clearance on the bonding ratio is explained, and a linear model is developed.

Appendices

Appendix 1: Derivation of Lube–Lube Interaction Pressure π _lube in Eq. 6

Considering the lubricant on the disk, the interaction energy with the lubricant with distance of dz on the slider is given by (Fig. 10)

Fig. 10

Schematic of lube–lube interaction of the head-disk interface

$$W_{{\text{lube} - lube}} = - \frac{{A_{\text{LVL}} }}{6\pi }\int_{{D - 2h - 2d_{0} }}^{{D - h - 2d_{0} }} {\frac{{{\text{d}}z}}{{z^{3} }} = } - \frac{{A_{\text{LVL}} }}{12\pi }\left[ {\frac{1}{{\left( {D - h - 2d_{0} } \right)^{2} }} - \frac{1}{{\left( {D - h - h_{\text{p}} - 2d_{0} } \right)^{2} }}} \right]$$

(10)

Assuming the lubricant thickness picked up by the slider is also h, i.e., h _p = h, then the equation reduces to

$$W_{{\text{lube} - lube}} = - \frac{{A_{\text{LVL}} }}{12\pi }\left[ {\frac{1}{{\left( {D - h - 2d_{0} } \right)^{2} }} - \frac{1}{{\left( {D - 2h - 2d_{0} } \right)^{2} }}} \right]$$

(11)

The force per unit area (pressure) for lube–lube interaction is obtained by

$$\pi_{\text{lube}} = - \frac{{\partial W_{{\text{lube} - lube}} }}{\partial D} = \frac{{A_{\text{LVL}} }}{6\pi }\left[ {\frac{1}{{\left( {D - h - 2d_{0} } \right)^{3} }} - \frac{1}{{\left( {D - 2h - 2d_{0} } \right)^{3} }}} \right]$$

(12)

Appendix 2: Effects of Bonding Ratio on the Lubricant Transfer, Eq. 8

When the lubricant bonding ratio equals to zero (BR = 0), A _LVS plays a very weak role in the determination of the critical clearance, as indicated in Fig. 7, and therefore, the $\pi_{\text{slider}}$ term in Eq. 4 can be neglected and reduces to:

$$\frac{{A_{\text{VLS}} }}{{6\pi \left( {h + d_{0} } \right)^{3} }} = \frac{{A_{\text{LVL}} }}{6\pi }\left\{ {\frac{1}{{\left[ {D - 2h - 2d_{0} } \right]^{3} }} - \frac{1}{{\left( {D - h - 2d_{0} } \right)^{3} }}} \right\}$$

(13)

Also neglecting the attraction from the lubricant itself, namely the second term on the RHS of Eq. 13 reduces to:

$$\frac{{A_{\text{VLS}} }}{{6\pi \left( {h + d_{0} } \right)^{3} }} - \frac{{A_{\text{LVL}} }}{6\pi }\frac{1}{{\left[ {D - 2h - 2d_{0} } \right]^{3} }} = 0$$

(14)

Then the critical solid distance can be solved:

$$D_{{{\text{C}}\left( {{\text{BR}} = 0} \right)}} = 2\left( {h + d_{0} } \right) + \sqrt[3]{{\frac{{A_{\text{LVL}} }}{{A_{\text{VLS}} }}}}\left( {h + d_{0} } \right)$$

(15)

To the contrary, when BR is equal to one (BR = 1), there is no mobile lubricant available for transfer, and thus, $\pi_{\text{Lube}} = 0;$ Eq. 4 reduces to:

$$\frac{{A_{\text{VLS}} }}{{6\pi \left( {h + d_{0} } \right)^{3} }} - \frac{{A_{\text{LVS}} }}{6\pi }\frac{1}{{\left[ {D - h - d_{0} } \right]^{3} }} = 0$$

(16)

which is the form widely used in the literature that ignores lube–lube interaction. The solution for the critical solid distance is:

$$D_{{{\text{C}}\left( {{\text{BR}} = 1} \right)}} = h + d_{0} + \sqrt[3]{{\frac{{A_{\text{LVS}} }}{{A_{\text{VLS}} }}}} \left( {h + d_{0} } \right)$$

(17)