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<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'>True, You might get lucky and find some. But the app notes tell
you how it is done. <o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'>http://www.vishay.com/docs/53044/epicappn.pdf<o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'>From what I can see you want a small resistor – they use 0603<o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'>You need a element that can get to 400oC – they use Ta2N, they
also list other possibilities. A bit of trial and error with different metal
films maybe in order<o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'>Take the protective coating off<o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'>Calculated the current and time needed – equations below<o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'> <o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><b><span style='font-size:9.0pt;
font-family:"Helvetica-Bold","sans-serif"'>The Heating Resistor<o:p></o:p></span></b></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>is a very precisely patterned Ta</span><span
style='font-size:6.5pt;font-family:"Helvetica","sans-serif"'>2</span><span
style='font-size:9.0pt;font-family:"Helvetica","sans-serif"'>N block which
dimensions<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>are:<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>Length L, width W and thickness T.<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>The ohmic value is controlled by the
resistivity of Ta</span><span style='font-size:6.5pt;font-family:"Helvetica","sans-serif"'>2</span><span
style='font-size:9.0pt;font-family:"Helvetica","sans-serif"'>N and<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>by L, W and T.<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>R = (r x L)/(W x T) (1)<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>The electrical energy is also controlled
through R<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>E = R x I x I x t (2)<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>The electrical energy is transformed into
heat and the<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>adiabatic heating of the resistor comes
from the following<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>equation:<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>E = m x c x </span><span
style='font-size:9.0pt;font-family:SymbolMT'>Δ</span><span style='font-size:
9.0pt;font-family:"Helvetica","sans-serif"'>T = L x W x T x d x C x </span><span
style='font-size:9.0pt;font-family:SymbolMT'>Δ</span><span style='font-size:
9.0pt;font-family:"Helvetica","sans-serif"'>T (3)<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>density d and specific heat C are Ta</span><span
style='font-size:6.5pt;font-family:"Helvetica","sans-serif"'>2</span><span
style='font-size:9.0pt;font-family:"Helvetica","sans-serif"'>N physical
constants,<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>providing Ta</span><span
style='font-size:6.5pt;font-family:"Helvetica","sans-serif"'>2</span><span
style='font-size:9.0pt;font-family:"Helvetica","sans-serif"'>N deposition is
perfectly under control.<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>Let us consider standard dimensions L = W
= 50 μm,<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>T = 2 μm, then per equation (1) R = 2 </span><span
style='font-size:9.0pt;font-family:SymbolMT'>Ω<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>Assuming that we want a 400 °C adiabatic
heating of the<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>resistor the electrical energy E can be
obtained from<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>equation (3).<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>E = 0.005 x 0.005 x 0.0002 x 16 g x 0.14
J/(g x °C) x 400 =<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>4.5 μJ<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>As a matter of fact and from some figures
we have got back<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>from customers the actual electrical
energy which is<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>necessary to fire up a critical mass of
pyrotechnical material<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>is about 2 to 10 times the level of
energy which is necessary<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>to heat up the resistor alone by 400 °C.<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>In our case, and in the worst case, a 50 μJ
energy firing pulse<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>will be convenient. <o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>From equation (3) one can calculate the
combinations R, I, t.<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>With R = 2 </span><span style='font-size:
9.0pt;font-family:SymbolMT'>Ω<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>I = 1 A</span><span style='font-size:
9.0pt;font-family:SymbolMT'>, </span><span style='font-size:9.0pt;font-family:
"Helvetica","sans-serif"'>t = 25 μs<o:p></o:p></span></p>
<p class=MsoNormal style='text-autospace:none'><span style='font-size:9.0pt;
font-family:"Helvetica","sans-serif"'>I = 0.7 A, t = 50 μs<o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:9.0pt;font-family:"Helvetica","sans-serif"'>I
= 0.5 A, t = 100 μs</span><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'><o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'><o:p> </o:p></span></p>
<blockquote style='margin-top:5.0pt;margin-bottom:5.0pt'><pre> <o:p></o:p></pre></blockquote>
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