The life of a sputter target is typically quantified in terms of units of power and time, like kilowatt/hours. For a target being sputtered at 500 watts for a total duty cycle of 100 hours thats 50 kilowatt/hours. Target life is also a function of the sputter yield of the material or how many target atoms are ejected for each gas ion (typically argon) that strikes the surface. For example when an argon ion with a mass of 39.948, hits a light material, like carbon, with an atomic mass of 12.01, about 3 carbon atoms are ejected. For materials like platinum with an atomic mass of 195.09, nearly 5 argon ions have to strike the target to get one platinum atom out (Ref 1).
Other factors that affect sputter yield include the bias voltage used to accelerate the argon ion to the target surface and the incident angle of the collision. In addition, there are big differences in sputter yield for metals verses oxides. Typically oxides will last many more kilowatt/hours than metals like aluminum.
One of our resident thin film deposition experts, Rob Belan, recommends that a target be replaced when the trench depth of the race track is ¾ of the total target thickness. For a ¼ thick target there will be 0.062 of material remaining at the bottom of the trench. He adds that if you are particularly careful you may be able to sputter the trench to a thickness of 0.031, but beyond that you are risking a complete burn through.
A handy online calculator for sputter yield can be found at TU Weins Institute fur Angewandte Physik. Im not sure how accurate it is, but it does list out several single element metals and their yields:
https://www.iap.tuwien.ac.at/www/surface/sputteryield
For a list of sputter yields using other ions, such as xenon and neon, there is an expansive data base on the web site of the National Physical Laboratory. Their data base also includes sputter yields at various powers (Ref 2).
So to determine when it is time to change out a sputter target you will need to have a depth gauge, either digital or dial. Check the depth of the trench in the race track after every deposition until you get a feel for the number of kilowatt/hours it takes to thin the target out to 25% of its original thickness (Ref 3). For those looking for an in-situ, real time method for measuring target thickness during sputtering, check out the publication from Alex Leybovich of TOSOH SMD who used ultrasonic time of flight measurements to monitor the health of sputter targets and target bonding during thin film depositions (Ref 4).
References:
1. Argonne National Laboratory, Noble Gas Sputtering Calculations using TRIM, https://www.osti.gov/biblio/
2. National Physical Laboratory of the UK, http://www.npl.co.uk/science-technology/surface-and-nanoanalysis/services/sputter-yield-values
3. MSC Direct, https://www.mscdirect.com/browse/Measuring-Inspecting/Dimensional-Measuring-Tools/Depth-Gages/?navid=&cid=ppc-bing-New%20-%20Measuring%20%26%20Inspecting%20-%20Product%20-%20PPC%20-%20Exact_I4PUCfCI_depth%20gauge_be__c_&mkwid=I4PUCfCI|dc&pcrid=&utm_source=bing&utm_medium=cpc&utm_campaign=New%20-%20Measuring%20%26%20Inspecting%20-%20Product%20-%20PPC%20-%20Exact&utm_term=depth%20gauge&utm_content=Depth%20Gages
4. In-situ real time sputtering source health monitoring using ultrasonics, Alex Leybovich, TOSOH SMD, Grove City, OH, , https://www.sciencedirect.com/science/article/pii/S?via%3Dihub
Category: Deposition Equipment
Sub-Category: Sputtering Targets
Link to Acetron
Related Topics: Sputtering, Process
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Roughly three-quarters of the power applied to a sputter gun ends up heating the coolant water. Understanding the heat transfer processes and heat dissipation is critical to successful sputtering. For a given sputter gun, four factors determine the maximum power density the target will accept: (1) the type of target material; (2) its thermal conductivity; (2) heat transfer from target to guns cooling water; and (4) the waters flow rate. The type of material determines its brittleness, thermal expansion coefficient, fragility, etc which determines the targets response to thermal stress. Applying a high power density to a material with a poor thermal conductivity (compared to, say Cu or Au) creates a large temperature difference between the targets top and bottom surfaces causing thermal stress. For fragile materials (such as some ceramics and semiconductors) stress may crack the target. For stronger, ductile materials (many metals) the target may melt. The efficiency of target cooling depends on the thermal resistance between the targets back surface (the surface not being sputtered) to the guns cooling water. Directly-cooled targets, where the water flow is in contact with the targets back surface, have low thermal resistance. By comparison, indirectly-cooled targets (which are clamped to a copper plate in the sun and cooled by water flowing across the plates back surface, not the targets) have high thermal resistance. At the atomic level, the interface between targets back surface and copper plates front surface has a small contact area for thermal conduction. Over the remaining area, heat transfer is by radiation. (Interestingly, the thermal resistance of such an interface under vacuum may be 4-10 higher than its thermal resistance in air.) Water flow through the sputtering gun should be sufficient to remove the total power. As a guideline 1 gal/min (~4 L/min) dissipating 4 kW power causes a rise in the water temperature of ~15C. For coolant liquids other than water, remember to factor in the liquids specific heat. Suggested guidelines for calculating maximum power are: for a high thermal conductivity, directly-cooled, metal target such as Al, the maximum power density is ~250 W/in^2 for the targets (front-side) surface area. For other types of (directly-cooled) targets, de-rate the maximum power density based on the target materials thermal conductivity compared to aluminum. As an example, for a target with a thermal conductivity 1/10th that of Al, the maximum power density should be 25 W/in^2. For a high thermal conductivity, indirectly-cooled, metal target such as Al, the power density limit is ~100 W/in^2 for the targets (front-side) surface area. Again, de-rate this power density for materials with lower thermal conductivities. Two important final points: (1) To prevent target cracking, fragile targets may need to have maximum power density de-rated even further than indicated by the relative thermal conductivities; and (2) some target materials crack no matter what power density is used.
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