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AN319 датащи(PDF) 4 Page - STMicroelectronics |
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AN319 датащи(HTML) 4 Page - STMicroelectronics |
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4 / 7 page ![]() AN319 APPLICATION NOTE 4/7 CONCLUSION The Transil is a must in relay drive circuits. It guarantees a reliable and efficient protection while reducing the delay between the coil drive turn-off and the contact release. EXAMPLE OF APPLICATION: CALCULATION AND CHOICE OF A TRANSIL We wish to protect the transistor in Figure 5 by a Transil whose clamping voltage must not under any cir- cumstances exceed VCL = 85V The clamping voltage is given by: VCL = VBR + Rd IC In the case of repetitive overload, the current is small enough such that the Rd Ic term can be neglected. Thus VCL approximates to: VCL = VBR The calculation is therefore simplified, and the Transil can be selected according to its thermal resistance. Mean Power Determination: PAV A rough value can be obtained by assuming that all the energy contained in the inductance is absorbed by the Transil. This is true when VBR >> VCC. Stand Off Voltage Selection The supply voltage varies between 9.6V and 14.4V. The stand off voltage of the diode VRM will be therefore be greater than or equal to 14.4V. VCL Determination From the data sheets we can see that for a low current, VCL = VBR. Tj Calculation Tj = Tamb + PAV x Rth = 50 + 90 = 140°C < Tj max (150°C) This value is consistent with the characteristics of the BZW04- 61B Transil (F 126 case) but the safety margin is somewhat low, so a device from the 1500 W series is preferable as a first choice. { 1.5 KE 75CP (CB– 429 case): VBR max = 82.5V, Rth = 75°C/W, VRM = 64.1V > 14.4V Tj = 50 + 68 = 118°C Tj max = 175°C Temperature Correction Voltage at 118°C is: VCL (118°C) = VCL (25°C) [1 + αT (118 - 25)] = [1 + 10.5 x 10-4 (93)] 82.5 = 90.5V This value is too high. Definitive Choice 1.5 KE 68CP: VBR max = 74.8V, VRM = 58.1V > 14.4V VCL (118°C) = 1.098 x 74.8 = 82.5V ⇒ 1.5 KE 68CP is suitable. PAV 1 2 --- LI 2 f ⋅⋅ 1 2 --- 035 12 24 + 45 ------------------ 2 50 ⋅⋅ ⋅ 09W == = |
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