4.1. Input Data for Calculus with MIL-HDBK-217standard
This paper continues the reliability calculus within [
4] which was focused on output capacitor bank’s failure rate calculation according to MIL-HDBK-217 rev.2 cap.10.1 – Capacitors, thus we use the same schematic for the converter from [
1] where SPICE simulation was used for multiple-constraint choice of capacitor bank (
Figure 3) with the same parameters (
Table 1). Part Stress Analysls Prediction will be used further. This method is applicable when most of the design is completed and a detailed part list including part stresses is available. It can also be used during later design phases for reliability trade-offs vs. part selection and stresses.
-
A.
Reliability calculation for the MOSFETs.
According to subchapter 6.9 (Specification MIL-S-19500) from MIL217 standard we can calculate the failure rate for FET transistors using the following formula
expressed in [F/10
6 h] or [FIT].
For MOSFETs λ
base = 0.06, thermal stres factor
with Tj = junction temperature (about 40 °C for
High-MOSFET and about 49 °C for λ
Low-MOSFET); Quality factor π
Q = 1
, for JAN category (Joint Army Navy – i.e., parts are the highest quality level of manufacturing), Environment factor π
E = 1
, for GB (Ground benign environment) – so, we have
The environment that the device will see is a factor along with the type of packaging technology (ceramic vs plastic or metal packaging).
-
B.
Reliability calculation for capacitors.
We show here the equation for failure rate of either electrolytic or ceramic capacitors stated by above mentioned standard:
Because our investigated DC-DC converter is a synchronous buck converter who has a series structure from reliability calculation point of view, we apply the parts count approach. Thus, the overall system failure rate can be written as being the sum of all components’ failure rate.
Equation (14) illustrates the total failure rate [
17,
18] with N = total number and λ
i = the failure rate for ith component.
The calculus in [
1] had has the following results for the three types of capacitors used in converter (according with
Table 1):
and for the entire bank:
4.2. Input Data for Calculus with SN 29500 Standard
To predict reliability using the SN 29500 standard, follow these steps:
1. Define the System - start by outlining your system, sub-systems, and components. This can be done using a reliability prediction tool that supports SN 29500.
2. Gather data - collecting accurate data for each component. This includes environmental conditions, operational profiles, and stress factors.
3. Apply Failure Rates - using the failure rate data provided in the SN 29500 standard. This data is specific to different types of electronic and electromechanical components.
4. Calculate Failure Rates - input the data into the prediction tool. The tool will use the SN 29500 models to calculate the failure rates for each component and the overall system.
5. Analyze Results - review the calculation of failure rates to identify potential reliability issues and areas for improvement.
The SN29500 standard, developed by Siemens, is used to estimate the failure rates of electronic components for reliability analysis. Here are the key components of SN29500:
1. Failure Rate data - it provides expected failure rates for various types of electronic components, such as resistors, capacitors, and integrated circuits³.
2. Temperature factors - adjustments based on the operating temperature of the components¹.
3. Voltage factors - adjustments based on the operating voltage.
4. Current factors - adjustments based on the operating current.
5. Stress factors - adjustments based on the percentage of time the component is under stress.
These factors help in calculating the Failure In Time (FIT) rate, which is crucial for assessing the reliability of components in safety-critical applications. SN 29500 failure rates calculation for MOSFETs are valid for:
-
A.
Reference conditions – stated as:
Failure criterion, i.e., complete failures and changes of major parameters leading to failure in the majority of applications.
Time interval, i.e., the operating interval of time need to be greater than 1000 h.
Operating voltage, i.e., about 50% of the maximum permissible voltage.
Junction temperature θj, stated in Tables 1,2 and 3 (page 4 from SN 29500-3: expected values for discrete semiconductors - Siemens Norm, Edition 2009-06)
Description of environment, i.e., the same statement like in IEC 60721 parts 3-1,3-2, and 3-3 are valid.
Operating mode, i.e., continuous duty under constant stress.
Table 3.
Elected parameters for the converter under study.
Table 3.
Elected parameters for the converter under study.
Capacitor type |
Temperature |
PCF0J471MCL6GS – Polymer electrolytic SMD Can-type, ESR = 18 mΩ – from Nichicon |
59 °C |
GRM32ER60J107ME20 – MLCC, SMD -Class II (X7R) ESR = 8 mΩ - from Murata |
32 °C |
The expected values under reference conditions λref,i that are found in above mentioned
Tables from standard should be understood for operation under above stated reference conditions. From
Table 1-Failure rate for transistors from standard we can found that λref (power-MOSFET) = 200 FIT for θj = 125 °C. The conversion from reference condition to operating conditions is as follow:
where π
T, is temperature dependence factor:
with: T
Uref = θ
Uref + 273 [°K], T
1 = θj,1 + 273 [°K], T
2 = θj,2 +273[°K], and A, E
a1, E
a2, θUref - are constant, is found in specific table within standard, θj
,1 - is reference virtual (equivalent) junction temperature in [°C], θj
,2 - is the actual junction temperature in [°C] obtained from thermal IR scanning from [y]. For power transistors we have:
A = 1, Ea1 = 0.4 eV, Ea2 = 0.7 eV, θUref = 40 °C, and
θj,2 = θU + Δ θ with θU = mean ambient temperature of the component in °C, and
Δ θ = P ×Rth with P = operating power dissipation, Rth = thermal resistance (junction to ambient).
For Q1 – High MOSFET we have θj,2 - HighMOSFET = 40 °C and from transistor’s datasheet:
R(DS)on = 6.2 mΩ, Coss = 430 pF (output MOSFET capacitance), Rth(J-C) = 20 °C/W (thermal resistance - junction to case)
Hence, in order to calculate the failure rate value :
For Q2 – Low MOSFET we have θj,2 - LowMOSFET = 49 °C
from transistor’s datasheet:
R(DS)on = 2.5 mΩ, Coss = 2070 pF (output MOSFET capacitance), Rth(J-C) = 1.4 °C/W (thermal resistance - junction to case)
Hence, in order to calculate the failure rate value :
-
B.
Operating stress conditions
Operating stress conditions are typical in industrial environment pretty similar to IEC 60654-1 class C – i.e., sheltered location who have an average temperature of 40°C during long period of time. θ1 = 40°C from standard table means 25°C from ambient temperature plus the internal self-heating due to ripple current through capacitor. θ2 is the actual capacitor’s temperature that was obtained from SPICE simulation [
14,
15] and is shown in
Table 2.
Taking account of
Section 3 and after doing the extracting work of data from SN29500 standard’s tables we found that [
1] :
-
For polymer capacitor we have: C2 = 1.9; C3 =3; U
max = 6.3 V; U
ref /U
max = 0.8; U = 1.2V ;U /U
max =0.19048; A = 0.4; E
a1= 0.14; E
a2 = 0; θ
Uref = 40°C; θ1 = 40°C; θ2 = 28°C (from
Table 2 within standard); T
Uref = 313[°K], T1 =313[°K], T2 = 301 [°K]; zref = -3632327.923 [1/eV], z = -1.478139959 [1/eV] results:
λref-polymer electrolytic = 3 ; πU = 2.735932892; πT = 0.542388224; πQ =2;
-
For MLCC capacitor we have: C2 = 1; C3 =4; U
max = 6.3 V; U
ref /U
max = 0.5; U = 1.2V; U /U
max =0.19048; A = 1; E
a1= 0.35; E
a2 = 0; θ
Uref = 40°C; θ1 = 40°C; θ2 = 32°C (from
Table 2 within standard); T
Uref = 298[°K], T1 =313[°K], T2 = 305 [°K]; z
ref = -362327.923 [1/eV], z = -1.478139959 [1/eV] results:
λref-MLCC = 2; πU = 3.71; πT = 0.712242068; πQ =2;
-
For ceramic through hole HF capacitor we have: C2 = 1; C3 =4; Umax = 25 V; U = 1.2V ;Uref /Umax = 0.5; U /Umax =0.048; A = 0.4; Ea1= 0.14; Ea2 = 0; θUref = 40°C; θ1 = 40°C; θ2 = 32°C (see Table II.); TUref = 313°K, T1 =313°K, T2 = 301 [°K]; zref = -3632327.923 [1/eV], z = -1.478139959 [1/eV] results:
λref-ceramic cap HF = 3; πU = 2.735932892; πT = 0.542388224; πQ =2;