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CA3282 Folha de dados(PDF) 7 Page - Intersil Corporation |
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CA3282 Folha de dados(HTML) 7 Page - Intersil Corporation |
7 / 10 page 7 For the CA3282, the maximum positive output current rating is 1A when one output is ON. When ALL outputs are ON, the rating is reduced to 0.625A because the total maximum cur- rent is limited to 5A. For any given application, all output driv- ers on a chip may or may not have a different level of loading. The discussion here is intended to provide relatively simple methods to determine the maximum dissipation and current ratings as a general solution and, as a special solu- tion, when all switched ON outputs have the same current loading. General Solution A general equation for dissipation should specify that the total power dissipation in a package is the sum of all signifi- cant elements of dissipation on the chip. However, in Power BiMOS Circuits very little dissipation is needed to control the logic and predriver circuits on the chip. The over-all chip dis- sipation is primarily the sum of the I2R dissipation losses in each channel where the current, I is the output current and the resistance, R is the NMOS channel resistance, rDS(ON) of each output driver. As such, the total dissipation, PD for n output drivers is: This expression sums the dissipation, PK of each output driver without regard to uniformity of dissipation in each MOS channel. The dissipation loss in an NMOS channel is: where the current, I is determined by the output load when the channel is turned ON. The channel resistance, rDS(ON) is a function of the circuit design, level of gate voltage and the chip temperature. Refer to the Electrical Specifications values for worse case channel resistance. The temperature rise in the package due to the dissipation is the product of the on-chip dissipation, PD and the package Junction-to-Case thermal resistance, θJC. To determine the junction temperature, TJ, given the case (heat sink tab) temperature, TC, the linear heat flow solution is: or Since this solution relates only to the package, further consideration must be given to a practical heat sink. The equation of linear heat flow assumes that the Junction-to- Ambient thermal resistance, θJA, is the sum of the thermal resistance from Junction-to-Case and the thermal resistance from Case (heat sink)-to-Ambient, θCA. The Junction-to- Ambient thermal resistance, θJA is the sum of all thermal paths from the chip junction to the ambient temperature (TA) environment and can be expressed as: Equation 3 and Equation 3A may be expressed as: or Not all Integrated Circuit packages have a directly definable case temperature because the heat is spread thru the lead frame to a PC Board which is the effective heat sink. Calculation Example 1 For the CA3282, θJC =3oC/W and the worst case junction temperature, as an application design solution, should not exceed 150oC. For any given application, Equation 1 deter- mines the dissipation, PD. Assume the package is mounted to a heat sink having a thermal resistance of 6oC/W and, for a given application, the dissipation, PD = 3W. Assume the operating ambient tem- perature, TA = 100 oC. The calculated Junction-to-Ambient thermal resistance is: θJA = θJC + θCA = 9oC/W The solution for junction temperature by Equation 5 is : TJ = 100 oC + 3W x 9oC/W = 127oC Calculation Example 2 Using the CA3282 maximum Junction-to-Ambient Thermal Resistance, θJA value of 45oC/W (no external heat sink) and the worst case Junction Temperature, TC of 150 oCwehave an application design solution for the maximum ambient temperature or dissipation. For example; Using Equation 1 and assuming a device dissipation, PD of 1W, the maximum allowable Ambient Temperature, TA from Equation 5A is calculated as follows: TA = 150 oC - 1.0W x 45oC/W = 105oC Equal Current Loading Solution Where a given application has equal current loading in the output drivers, equal rDS(ON) and temperature conditions may be assumed. As such, a convenient method to show rating boundaries is to substitute the dissipation Equation 2 into the junction temperature Equation 3. For m outputs that are ON with equal currents, where I = I1 =I2..... = Im,we have the following solution for dissipation: P D P k k1 = n ∑ = (EQ. 1) P k I 2 r DS ON () × = (EQ. 2) T J T C P D θ JC × + = (EQ. 3) T C T J P D θ JC × – = (EQ. 3A) θ JA θ JC θ CA + = (EQ. 4) T J T A P D θ JA × + = (EQ. 5) T A T J P D θ JA × – = (EQ. 5A) P D mP k × mI × 2 r DS ON () × == (EQ. 6) I T J T C – m θ JC r DS ON () × × ----------------------------------------------------- = (EQ. 7) CA3282 |
Nº de peça semelhante - CA3282 |
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Descrição semelhante - CA3282 |
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