PI2127 Datasheet, PDF (15/19 Pages) Vicor Corporation – 60 Volt, 12 Amp Full-Function Active ORing Solution


At maximum response time:
I _PEAK
=
V_S * t _RVS
L_PARASITIC
=
45V *150ns
60nH
= 112.5A
Avalanche Energy:
E _AS
=
1 _* 1.3 * BV_DSS
2 1.3 * BV_DSS − V_S
* L_PARASITIC
* I _PEAK ²
E _AS
=
1
2
*
1.3*
1.3 * 60V
60
− 45V
* 60nH
*112.5 A ²
= 897μJ
The avalanche energy is well below the total MOSFET
specified peak current of 150A for 300ns and below
the rated avalanche energy. The specified energy
can be calculated from Single Pulse Avalanche
Current as specified in the Absolute Maximum Ratings
table:
1
2
⋅1.3
*
BV
DSS
⋅ I _AS ⋅ t _AV
=
1
2
1.3
⋅
60V
⋅
33
A
⋅11μs
=
14mJ
Figure 23 : Two PI2127 in High Side ORing
configuration
VC bias through Constant current circuit
Select an NPN transistor with V_CEO equal or higher
than the input voltage (Vin) plus any expected
transient voltage and capable of handling the
expected maximum power dissipation. Any NPN
transistor with V_CEO ≥ 60V in a small footprint is
suitable. An exemplary NPN is the BC846 from NXP
Semiconductors:
From the BC846 datasheet:
NPN general-purpose transistor
V_CEO = 65V Collector-Emitter maximum voltage
I_C = 100mA maximum collector current
h_FE = 110 minimum at I_C=2mA
V_BE = 0.580V to 0.70V Base-Emitter voltage
at I_C = 2mA and 25°C
R_θJ-A = 500°C/W Junction to ambient thermal
resistance.
Select Zener Diode: A Zener diode with low bias
current and V_Z=10 in small foot print is suitable for this
application. An exemplary Zener diode is the
MM3Z10VST1 the from ON Semiconductor
From the MM3Z10VST1 datasheet:
10V, 200mW Zener Diode
V_Z = 9.80V to 10.2V Zener voltage range
I_R = 10μA will hold the Zener breakdown voltage
at 9.8V
R _LIMIT
= ^{VZ _ MIN −VBE (on)}
I _{VC _ MAX}
=
9.8V − 0.7V
2.1mA
= 4.33kΩ
Or 4.32kΩ 1%
I _{B _ MAX}
= ^{I C _ MAX}
h_{FE _ MIN}
=
3mA
110
=
27.27μA
R_Z Calculation:
Use 120μA as minimum for the Zener diode reverse
leakage current and Q2 base current combined.
R_Z
= ^{Vin _ MIN − VZ _ MAX}
I _Z + I _{B _ MAX}
=
40V − 10.2V
120μA
= 248kΩ
Select R_Z= 249kΩ 1%
Maximum Q1 collector current:
I _{C _ MAX}
= ^{VZ _ MAX − VBE _ MIN}
R_{LIMIT _ MIN}
=
10.2V −
4.32kΩ
0.50V
* 0.98
= 2.29mA
Maximum Q2 power dissipation
Pd _Q1 = I _{C _ MAX} * [Vin_MAX − V_VC−CLM − (V_{Z _ MIN} − V_{EB _ MAX} )]
Pd_Q1 = 2.29mA*[45V −11V − (9.8V − 0.7V )] = 57mW
Transistor temperature rise
T_RISEQ1
=
Pd_Q1 * R_{θJ −A}
= 57mW *500 ^°C
W
=
28.50°C
Picor Corporation • picorpower.com
PI2127
Rev 1.3
Page 15 of 19