TIDUDG1 July   2022

 

  1.   Description
  2.   Resources
  3.   Features
  4.   Applications
  5.   5
  6. 1System Description
  7. 2System Overview
    1. 2.1 Block Diagram
    2. 2.2 Design Considerations
      1. 2.2.1 MIL-STD-1275E versus MIL-STD-1275D
      2. 2.2.2 Reverse Polarity Event
      3. 2.2.3 Voltage Spike Event
      4. 2.2.4 Voltage Spike Event: Component Selection
      5. 2.2.5 Voltage Surge Event
      6. 2.2.6 Voltage Surge Event: Component Selection
    3. 2.3 Highlighted Products
      1. 2.3.1 LM7480-Q1
      2. 2.3.2 LM5069
  8. 3Hardware, Testing Requirements and Test Results
    1. 3.1 Hardware Requirements
    2. 3.2 Test Setup
    3. 3.3 Test Results
  9. 4Design and Documentation Support
    1. 4.1 Design Files
      1. 4.1.1 Schematics
      2. 4.1.2 BOM
    2. 4.2 Documentation Support
    3. 4.3 Support Resources
    4. 4.4 Trademarks

2.2.6 Voltage Surge Event: Component Selection

To protect against the voltage surge event defined in MIL-STD-1275E, a two-stage clamping solution is used to deal with the large amount of energy, while helping to reduce overall solution size, and to allow for power to be provided throughout the entire surge event.

The advantage of using a two-stage clamping topology is that it reduces the amount of thermal and electrical stress on the MOSFET. For example, in the reference design, the first stage clamps the voltage from 100 V to 50 V, and the second stage goes from 50 V to 33 V. This means that Q1 must be able to handle a maximum of 4.28 A with a drain-to-source voltage of 50 V during the surge event.

To determine the MOSFET, the safe operating stress of the MOSFET must be calculated. There is a MIL-STD-1275E SOA calculator that can be downloaded to help determine if a MOSFET is capable of handling the surge event. The first step is to approximate the surge envelope as a square pulse. To do that, find the total area under the curve (which equates to the energy) and then translate that to a square pulse for easier calculations.

  1. The first portion of the surge event is the maximum 100 V for 50 ms. This means Q1 detects a 50-V drop across it, and the system pulls about 4.28 A at its maximum load. This results in a power dissipation on the MOSFET at 50 V × 4.28 A = 214 W for 50 ms, or 10.7 J of energy dissipation.
  2. The second portion of the surge event is the linear decrease from 100 V at 50 ms to 33 V at 500 ms, which means the FET detects a voltage drop of 50 V to 0 V over that span. This results in about 107 W for 336 ms assuming a linear line between 100 V and 33 V or 36 J of energy dissipation
  3. After calculating each segment of the surge event, the total power dissipation is 10.7 J + 36 J resulting in a total of 46.7 J.
  4. Next, to approximate the equivalent time of the square wave, take the maximum power (214 W) and the total energy that was just calculated (46.7 J) and solve for the time which is 46.7 J / 214 W = 218 ms. Now, approximate that the MOSFET must withstand 214 W for 218 ms.

Calculate the amount of power the MOSFET can withstand for 218 ms at 25°C. The following equations are used to determine the safe operating area (SOA) of the MOSFET.

Equation 1. S O A t = a   ×   t m
Equation 2. m =   l n ⁡ ( S O A ( t 1 ) / S O A ( t 2 ) ) l n ⁡ ( t 1 / t 2 ) , a =   S O A ( t 1 ) t 1 m

To solve for m and a, the information comes from the SOA graph found in the MOSFET data sheet. From the SOA graph, the important information is the time and the current capabilities at 50 V. So in this design, the IXTT88N30P was used, and using it as an example, the SOA(t1) = 23 A for t1 = 0.01 s. While SOA(t2) = 13 A for t2 = 1 s. Plugging in the numbers for the IXTT88N30P results in an SOA of 15.7 A, meaning that the IXTT88N30P can handle 439.6 W for 252 ms. Which indicates this is more than enough for a 120-W system at 25°C.

The second part of the SOA calculation is to calculate for thermal derating. To determine the thermal derating, the following equations are used:

Equation 3. S O A T C = S O A 25 º C   ×   T J , A B S M A X -   T C T J , A B S M A X -   25 º C
Equation 4. T C = T A + R θ C A × I L O A D , M A X 2 × R D S O N ( T J )

TA is the ambient temperature of the system. For example, calculating the SOA of the IXTT88N30P at a TA = 100°C results in an adjusted SOA of 6.15 A. Which means that the MOSFET can handle 172.2 W for 252 ms at 100°C.