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HRW0202A

Hitachi Semiconductor
Part Number HRW0202A
Manufacturer Hitachi Semiconductor
Description Silicon Schottky Barrier Diode for Rectifying
Published Mar 26, 2005
Detailed Description HRW0202A Silicon Schottky Barrier Diode for Rectifying ADE-208-209E (Z) Rev 5 Jul. 1998 Features • Low forward voltage ...
Datasheet PDF File HRW0202A PDF File

HRW0202A
HRW0202A

Overview
HRW0202A Silicon Schottky Barrier Diode for Rectifying ADE-208-209E (Z) Rev 5 Jul.
1998 Features • Low forward voltage drop and suitable for high effifiency rectifying.
• MPAK Package is suitable for high density surface mounting and high speed assembly.
Ordering Information Type No.
HRW0202A Laser Mark S17 Package Code MPAK Outline 3 2 1 (Top View) 1 Anode 2 Anode 3 Cathode HRW0202A Absolute Maximum Ratings (Ta = 25°C) *1 Item Repetitive peak reverse voltage Average rectified current Non-Repetitive peak forward surge current Junction temperature Storage temperature Symbol V *2 RRM Value 20 200 Unit V mA A °C °C I O*2 I FSM Tj Tstg *3 3 125 –55 to +125 Notes 1.
Two device total Notes 2.
See from Fig.
4 to Fig.
7 Notes 3.
10msec sine wave 1 pulse Electrical Characteristics (Ta = 25°C) *1 Item Forward voltage Reverse current Thermal resistance Symbol VF IR Rth(j-a) Min — — — Typ — — 350 Max 0.
40 50 — Unit V µA Test Condition I F = 100 mA VR = 20V ° C/W Polyimide board *2 Notes 1.
Per one device Notes 2.
Polyimide board 20hx15wx0.
8t 1.
5 3.
0 1.
5 1.
5 0.
8 Unit: mm 2 HRW0202A Main Characteristic 1.
0 Pulse test 10 –1 10 –3 Pulse test –4 Forward current I F (A) 10 –2 Ta = 75°C Reverse current I R (A) 10 Ta = 75°C 10 –5 10 –3 Ta = 25°C 10 –6 10 –4 Ta = 25°C 10 –5 10 –7 10 –6 0 0.
1 0.
2 0.
3 0.
4 0.
5 0.
6 10 –8 0 Forward voltage VF (V) Fig.
1 Forward current Vs.
Forward voltage 15 20 5 10 Reverse voltage VR (V) 25 Fig.
2 Reverse current Vs.
Reverse voltage 10 2 f = 1MHz Pulse test Capacitance C (pF) 10 1.
0 1.
0 10 Reverse voltage VR (V) 40 Fig.
3 Capacitance Vs.
Reverse voltage 3 HRW0202A Main Characteristic 0.
08 Forward power dissipation Pd (W) 0A t 0.
25 D=1/6 Reverse power dissipation Pd (W) 0V t D= \ T 0.
06 T t Tj =25°C D= T\ 0.
20 T t Sin( ˘=180°) D=1/3 D=1/2 DC D=5/6 D=2/3 D=1/2 Sin( ˘=180°) Tj =125°C 0.
15 0.
04 0.
10 0.
02 0.
05 0 0 0 0.
02 0.
04 0.
06 0.
08 @ @IF @(A) 0.
1 0 5 10 15 20 25 Forward...


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