March 1992 7
Philips Semiconductors Preliminary specification
Listening-in circuit for line-powered
telephone sets
TEA1085; TEA1085A
Supply; SUP, SREF, VBB, VSSand VA
The line current is divided into I
TR
for the TEA1060 and I
SUP
for the TEA1085/TEA1085A.
The supply arrangement is illustrated in Fig.4.
Fig.4 Supply arrangement.
handbook, full pagewidth
MGR034
TEA1060
V
CC
LN
V
EE SLPE
TEA1085
TEA1085A
V
SS
V
BB
I
line
I
TR
SUP
C20
I
SUP
I
BBO
I
BIAS
I
CC
R1
R9
R38
VA
R20
VOLTAGE
STABILIZER
TR1
TR2
SREF
V
int
LINE
ITR is constant: ITR=V
int
/ R20; I
SUP=Iline
− ICC− I
TR
Where:
A practical value for R20 is 150 Ω. This value of resistance
produces a value for I
TR
= 2 mA and I
SUP
= I
line
− 3 mA.
The TEA1085/TEA1085A stabilizes its own supply voltage
at VBB. Transistor TR1 provides the supplies for the
internal circuits. TR2 is used to minimize the signal
distortion on the line by momentarily diverting the input
current to VSS whenever the instantaneous value of the
voltage V
SUP
drops below the supply voltage VBB. VBB is
fixed to a typical value of 3.6 V but can be increased by
means of an external resistor (R38) connected between
V
int
is an internal temperature compensated
reference voltage with a typical value of
315 mV between SUP and SREF
R20 is a resistor between SUP and SREF
I
CC
is the internal current consumption of the
TEA106X (≈ 1 mA)
VA and VSS or decreased by connecting this resistor
between VA and VBB. The minimum level on VBB is
restricted to 3.0 V; the level of the VBB limiter is also
affected (see application report for further information).
The supply at VBB is decoupled by a 470 µF capacitor.
The DC voltage (V
SUP
− VSS) is determined by the
transmission IC (V
LN−SLPE
); thus:
V
SUP
− VSS = V
LN−SLPE
+ V
int
.
The minimum DC voltage that can be applied to this input
is V
BB(max)
+ 0.4 V.
Where: V
BB(max)
is the worst case supply voltage (this
depends on the setting of R38, which is connected
between VA and VSS).
The internal current consumption of the
TEA1085/TEA1085A (I
SUP0
) is typically 4.2 mA (where
V
SUP
− VSS = 4.5 V, MUTE off). Thus the current available
for powering the loudspeaker is I
SUP
− I
SUP0
.
The current I
SUP0
consists of a bias current of ≈ 0.4 mA for
the circuitry connected to SUP and current I
BB0
of≈ 3.8 mA
which is used for the circuitry connected to VBB(see Fig.4).