MT91600 Mitel Networks Corporation, MT91600 Datasheet - Page 6

• Manufacturer Part Number: MT91600
• Description: Programmable Slic
• Manufacturer: Mitel Networks Corporation

MT91600
Step 1: Gain Setting (R12, R13, R14, R15)
Gain 2 - 4 = 20 Log [ R13 / R12]
Gain 4 - 2 = 20 Log [0.891 * [R14 / R15)]
Step 2: Impedance Matching (R11, R18, R19, C8)
a) Z
600/(220+220) = (k*600)/R11
let k = 125
b) In general,
kZ
where:
Z
Z
Since we are dealing with a resistive load in this
example Z
kZ
(125 * 600)= R18
Step 3: Network Balance Impedance (R16, R17)
R17 + R16
R17 + R16
set R17 = 100k , R16 becomes 138k .
Complex Line Impedance, Z
In situations where the characteristic impedance of
the line Z
component values for impedance matching (R11,
R18, R19, C8) is as follows:
6
LZ
LZ
o
o
R12 = 112.2k , R13 = 100k .
R14 = 100k , R15 = 100k .
R16 = 138k , R17 = 100k .
-1 dB = 20 Log [R13 / R12]
-1 dB = 20 Log [0.891 * [R14 / R15)]
R11 = 55k
R18 = 75k
= R18, for a resistive load.
= [R18 + (R19 // C8)], for a complex load.
= Z
= R18
R17
R17
o
Z
/ ( R1+R2) = kZ
o
LZ
Gain 2 - 4 = -1dB, Gain 4 - 2 = -1dB
= 600 , Z
LZ
o
= [1.782 * Z
= 0.4199
= R18, and therefore:
is a complex value, determining the
NB
o
= 600 , R1=R2= 220
[1 + R13 / R12)]
/ R11
o
/ ( Z
o
+Z
o
NB
) * ( R13 / R12 )]
Given Z
Z
where, kZ
Choose a standard value for C8 to find a suitable
value for k.
Since 1nF exists, let C8 = 1nF then,
k = 120nF / C8
k = 120nF / 1nF
R18 = k * 220
R18 = 120 * 220
R18 = 26400
R19 = k * 820
R19 = 120 * 820
R19 = 98400
From (Equation 1)
R11 = k * (R1 + R2)
R11 = 120 * (220
Power Sharing Resistor (R23)
To determine the value of R23, use the following
equations:
R23(max)= |Vbat(min)| - 100 - (2*R2 + Lr + DCRP)
R23(min)= |Vbat(max)| - Pd(max) - R2
where,
Vbat(min/max) = the expected variation of Vbat.
R2 = the feed resistor.
Lr = maximum DC loop resistance.
DCRP = DC resistance of the phone set.
Pd(max) = the maximum power dissipation of the
ring drive transistor Q3.
If R23(max) > R23(min), then set R23 to be the
geometric center:
R23 = Square Root (R23(max) * R23(min))
o
/ ( R1+R2) = kZ
k =120
R18 = 26k4 , R19 = 98k4
R11 = 52k8
o
= 220
o
= [R18 + (R19 // C8)]
30mA
40mA
+ (820
o
+ 220 )
/ R11
Preliminary Information
1.6mA
// 120nF)
(Equation 1)