Electro-chemical device and method for manufacturing the same
一种制造方法、电化学的技术,应用在电解质蓄电池制造、最终产品制造、电容器外壳/封装等方向,能够解决泄漏导通、引线短路、损伤树脂层等问题,达到高品质电化学装置、抑制不良情况、不良情况改善的效果
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Embodiment 1
[0112] In Example 1, under the same conditions as in Comparative Example 1, the portion to be thermocompression-bonded was subjected to press working as described above. In this stamping process, the thermocompression bonding of the Al package (3.0kgf / cm 2 )(2.94×10 5 Pa) Higher pressure (4.0kgf / cm 2 )(3.92×10 5 Pa), carry out stamping treatment on the range of predetermined thermocompression bonding. As a result, the thickness of the Al lead after the planarization process was reduced to 80 μm. The ratio α=Z1 / Z2=80 μm / 166 μm=48%. In addition, curvature R=0.14% (=5μm / 3.5mm), area S1=0.009mm 2 .
[0113] Next, thermocompression bonding of the lead sealing portion was performed in the same manner as in Comparative Example 1. When measuring yield β1, no electrolyte solution was injected into the package. When measuring the pass rate β2, inject electrolyte. In Example 1, all the conditions were the same as those of Comparative Example 1 except for the lead wire punching. ...
Embodiment 2
[0159] (Example 2) Case of ratio α=50%
[0160] All the conditions were the same as in Example 1 except that the thickness Z1 of the Al lead was 90 μm (punching completed) and Z2=180 μm (90 μm×2). In this case, α=50%. Curvature R=0.14% (=5μm / 3.5mm)% (area S1=0.009mm 2 ). The depth of the bent portion is 5 μm. As a result, the pass rate β1 of the empty cell before the electrolyte solution was inserted was 80%, and the effect of lead punching (deburring) appeared, but it was lower than the pass rate (90%) of Example 1. The reason for this is considered to be that since the resin layer is thinner than the thickness of the lead wires, the lead wires cannot be sufficiently covered and a short circuit occurs. Also, the pass rate β2 related to liquid leakage after electrolyte solution injection was 90%.
[0161] That is, the following results are obtained.
[0162] Z1=90μm
[0163] Z2=180μm
[0164] α=50%
[0165]R=0.14% (=5μm / 3.5mm)
[0166] S=0.009mm 2
[0167] Pass rat...
Embodiment 3
[0241] (Example 3) Case of ratio α=56%
[0242] All the conditions were the same as in Example 1 except that the thickness Z1 of the Al lead was 90 μm (punching completed) and Z2=160 μm (80 μm×2). In this case, α=56%, and the following results are obtained.
[0243] Z1=90μm
[0244] Z2=160μm
[0245] α = 56%
[0246] R=0.14% (=5μm / 3.5mm)
[0247] S=0.009mm 2
[0248] Pass rate β1=82%
[0249] Pass rate β2 = 75%
[0250] Final pass rate β=61.5%
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Abstract
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