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About Jessicale

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  1. Hi, I want to ask for your help on my L5973D. I used to use L5973D to produce voltage of 3.3V/4.2V according to the typical circuit in the datasheet. The figures are same while the feedback resistance are different. The input is 24V DC. They worked well without anything wrong. (Test application circuit in the datasheet) However, I met some problems when I want to produce 5V voltage according to the circuit above. I tried to turn the R1 into 10K. Theoretically,it will produce 5V, but when I connected it with 24V power,L5973D burned up. In the same circuit, the voltage is 3.3V when R1 is 5.6K. It worked well. I am wondering why...So I’m here to ask for your help.Sorry for my poor English,but can anyone give me some suggestions? Here are some details of L5973D: Series:L5973 Operating-Temperature: -40°C ~ 150°C (TJ) Mounting-Type: Surface Mount Output-Type: Adjustable Function: Step-Down Supplier-Device-Package: 8-HSOP Number-of-Outputs: 1 Topology: Buck Frequency-Switching: 250kHz Current-Output: 2.5A Output-Configuration: Positive Voltage-Output-Min-Fixed: 1.235V Voltage-Output-Max: 35V Synchronous-Rectifier: No Voltage-Input-Min: 4V Voltage-Input-Max: 36V Output-Voltage: 35 V Input-Voltage-MAX: 36 V Output-Current: 2.5 A Switching-Frequency: 212 kHz to 280 kHz
  2. To replace my overheating linear regulator DC-DC converter, I've designed this new board. This time I used a couple of switching 12V regulators .Here's the data sheet for reference: http://www.kynix.com/uploadfiles/pdf9675/LM2575-12BN.pdf I'm using two 1 amp regulators because my local supplier doesn't have a 2 amp version. Here are the schematics. Here's the board: There are no pictures of the board because it doesn't exist yet. This time, I did the math first and didn't blindly built the thing. The red arrows highlight the single-point grounding I tried to make, as recommended in the datasheet. Is that design correct?. Here is my math for calculating the temperature rise from ambient temperature. According to the LM2575 datasheet, the power dissipation is calculated as follows: PowerDissipated=Vin.Iq+(Vo/Vin).Iload.Vsat Vo=12V,Vin=18V,Vsat=1.4V,Iq=10mA,Iload=1A PowerDissipated=0.18W+0.93W=1.11W Again, from the LM2575 datasheet, its termal resistance (junction to ambient, in worst case) is 65°C/W, which would keep the regulators just below 100°C considering a ambient temperature of 25°C. If I use my heatsinks with 20°C/W and the regulator thermal resistance (junction to case) of 2°, that would keep the regulators under 50°C My questions are: 1. Is my heat dissipation calculations correct? 2. Is my board designed correctly, especially regarding the required single-point grounding for the regulator pins? These points are marked by the red arrows on the board image.
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