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參數(shù)資料
| 型號： | TEA7063FP |
| 廠商： | 意法半導(dǎo)體 |
| 英文描述： | SPEECH CIRCUIT WITH POWER MANAGEMENT |
| 中文描述： | 語音電路電源管理 |
| 文件頁數(shù)： | 11/15頁 |
| 文件大小： | 228K |
| 代理商： | TEA7063FP |
7. INTERNAL DESCRIPTION OF CURRENT
MANAGEMENT
7.1 Internal Power Supply Management
R9 fixesthe line power supply management.
R9 fixesthe values of: I
ear
, I
up
, I
ref
and ILS.
A current line information is used to modifie the
values of I
ear
, I
up
, I
ref
and ILS between a minimum
and a maximum values.
On Fig 10:
The transmit output stage is representedby a cur-
rent source (I
tr
). The I
tr
value depends of the DC
voltageon V
LINE
(pin 16) and R
ZAC
value.
The other internal stages connected to V
LINE
(pin
16) are represented by a constant 1.3mA current
source.
7.2 DCCharacteristics
(internal)
The DC characteristic is equals to:
V
LINE
(pin 16) = V
C
(pin17) + R6
I
int
Iint is the sum of all the current sources con-
nected on VC (pin 17):
[I
p
+ I
ref
+ Vpin17 / (r7 + r8)]
- I
p
is the bias internal operational amplifiers
power supply.
- I
ref
= 1 / 3
(V
refi
/ R9); with V
refi
= 470mV
- I
ref
= 156 / R9mA
The current line information changesIint value;
at low linecurrent (6mA): I
int
= 4V / R9 + 60
μ
A
at low linecurrent (IL = ILb): I
int
=8V / R9 + 60
μ
A
7.3 Microcontroller Supply
(internal)
I
up
= [(p2 / r2)
I
ref
+ 0.3] mA = [(p2 / r2)
156 /
R9) + 0.3] mA
The current line information changesp2/r2 value;
at low linecurrent (6mA): p2 / r2 = 70
at a linecurrent (IL = ILb): p2/ r2 =820
7.4 Earphone Current Supply
(internal)
I
ear
= (p1 / r1)
I
ref
mA= (p1 / r1)
(156 / R9) mA
The current line information changesp1/r1 value;
at low linecurrent (6mA): (p1 / r1) = 200
at a linecurrent (IL = ILb): p1/ r1 =2700
The maximum peak dynamic on the earphone is:
V
pear
= Z
ear
I
ear
7.5 Transmit Output Stage
(internal)
The output stage bias current depends of the DC
voltageon pin 16 and on R
ZAC
impedance.
I
tr
=0.1425
V
LINE
0.517
connectedbetweepin3 and the ground)
R
(R
Z
is the resistor
7.6 LoudspeakerCurrent Source
(internal)
The current source for external peripherals has
two slopes:
- First slope; before I
ear
, I
up
, I
tr
and I
int
are stabi-
lized at theirmaximum values: (IL = ILb)
ILS = 0.285
IL
- Second slope; after I
ear
, I
up
, I
tr
and I
int
are sta-
bilizedat theirmaximum values: (for IL > ILb)
(ILS) = 0.91
(I
LINE
I
ear
, I
up
, I
tr
and I
int
are stabilized at their maximum
values between 16 and 26mA, the absolute IL
value depends of R9 value. The line current (ILb)
where I
ear
, I
up
,
Itr
, I
int
are stabilized at their maxi-
mum values and where the slope of ILS change
is:
ILb =I
ear
+
I
up
+
I
tr
+
I
int
+
1.3
0.715
7.7 Numerical Example
1)R9 =100K
; R6 = 25K
DC characteristic = 6V for I
int
max:
= 5V for I
int
min:
I
int
min (IL = 6mA) = 4 /100K + 60 = 100
μ
A
I
int
min (IL = ILb) = 8 /100K+ 60 = 140
μ
A
Vpin17 = 2.5V
R6 = 25K
Vpin16min (IL = 6mA) = 2.5 + 25
100E- 3 = 5V
Vpin16max (IL = ILb) = 2.5 + 25
140E- 3 = 6V
CurrentSources
I
up
min (IL= 6mA) = 0.4mA
I
up
max (IL = ILb) = 1.6mA
I
ear
min (IL = 6mA) = 0.3mA
I
ear
max (IL = ILb) = 4.2mA
with R
Z
= 75
I
tr
min (IL = 6mA) = 2.6mA
I
tr
max (IL = ILb) = 4.5mA
ILS min (IL = 6mA) = 1.3mA
IL ILb
+
4.2
+
4.5
+
0.14
+
1.3
0.715
mA
ILb = 16.5mA
ILS (for IL = ILb) = 0.285
ILb = 4.7mA
at IL = 100mA:
(ILS) = 0.91
(IL) = 0.91
(100 - 16.5) = 76mA
ILS = 4.7 + 76 = 80.7mA
2)
R9 =56K
; R6 = 18K
DC characteristic = 6.1V for I
int
max:
Χ χηαραχτε
= 4.8Vfor I
int
min:
TEA7063
11/15
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