User guide

ACR83 – Reference Manual [e-mail]
Version 1.05
www.acs.com.hk
Page 19 of 49
11.4. Verification Example 2
System unit is bit.
abPINApdu = 00 20 00 01 08 57 A5 30 30 30 30 30 30h
After Lc (08h), the first 11 bits (01010111 101) is control character.
bmFormatString=59h
SpePinPos=11 bits because bmFormatString bit 7 = 0
SpeLeftRight=Left
SpePINTyp=BCD
bmPINBlockString=48h
SpePINSize=4 bits
SpePINLen=8 bytes
bmPINLengthFormat=06h
SpePINLenPos=6 bits
wPINMaxExtraDigit=0108h
SpePinMax=08h
SpePinMin=01h
PIN Input = 1 2 3 4 5
Command Header
SpePINLen
APDU Command
Header
APDU
Length
Offset SpePINPos 7 bits PIN
00 20 00 01 08 Offset 6 bits
SpePINSize
(4bits)
Not used field/may not
exist
PIN
00 20 00 01 08
01010111
101
Input 5 digits
Offset 6 bit relative to
Lc
PIN
00 20 00 01 08
010101
0101 (bits) 1 (bit)= 01010111 101 PIN
00 20 00 01 08 01010101 011 (0101 replace original 01010111 101) PIN
How about the the PIN management?
Because it is Left and BCD arrangement.
PIN (bits)
Original
0 0101 0011 0000 0011 0000 0011 0000 0011 0000 0011
0000 0011 0000
Input 12 34 5 (change to
bit format)
0001 0010 0011 0100 0101
Original
0 0101 0011 0000 0011 0000 0011 0000 0011 0000 0011
0000 0011 0000