PBL3853N データシートの表示（PDF） - Ericsson
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PBL3853N Datasheet PDF : 16 Pages
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PBL 3853
What can be done, is to use more of the
transmitter ouput signal for charging the
VDC via the T1,. The transmitter output
signal passes both R12 and R13. The
transmit current that passes R12 is lost
regarding the charging of VDC but it
can not be zero because it is exactly
that signal that via the amplifiers and
followers T1 and T2 drives the current
through R12. Caution has to be taken
so that spread figures do not eat up all
the won current. If the specification of
DC-characteristic towards the line has
to be maintained when the quota of
R12/R13 is made smaller, the R5 must
most probably be increased. What has
to be understood is that the current
consumption of the IC circuit it self (like
in the given figure. 4.) can not be
lowered and that it at 16 mA line current
not only consists of approx. 540 mA into
pin 4, but also that of the 3.2 mA into
pin 1, some 600 mA is used to power
up the IC. This supply current to the
circuit will be multiplied by the R13/R12
ratio but it will not participate in giving
any signal out. What is left of the
current that also passes pin 2, to
produce output signal, is 2.6 mA. Of this
current 1.9 mAp is modulated. This
current in its turn will be multiplied by
R12 and R13 (the function being 1+R12/
R13) and the result
4.2 mAp will give in the load of 490 Ω
(600 Ω//2.7 k) a signal swing of approx.
2 Vp. This value will fulfil a typical
requirement of 1.4 Vp swing with margin.
Even in a case, that the differencies
in the in- and out-current respective
voltage are satisfactory, there might be
a need for redimensioning. In case that
a lower DC mask is desirable and there
is a possibility to accept a lower VDC it
can be made with RA for the DC mask
(RC and RD must be high ohmic) and
sinking the voltage for the shunt
regulator VDC (simply drawn as a diode
in the fig. 4). What has to be observed
is that the supply for the receiver
8
T1
Shunt
regulator
18
D1
-Line
Figure 20.
VDC
C10
amplifier is also lowered this way and
hence the output swing of it. In contrary
if a higher DC mask is desired and a
higher line voltage is acceptable, both
can thus be increased (the line voltage
is increased with RB).
A higher current out of VDC can be
acchieved by altering R12, R13 and R5
but at cost of the lowest line current the
circuit will work with, which will increase
accordingly. The opposite is valid if less
current is required out of VDC it will say
that the circuit will work to lower line
current. The lowest line current the
circuit will work to full specification is
(with the increased DC mask) 5.5 mA
but in this case only 1 mA can be taken
out of VDC (with no current out of
receiver output).
The circuitry can be made simpler in
case the requirement for the voltage
and/or current difference between the
line - and VDC is smaller. If the
requirement for the smallest possible
voltage between the line voltage and
the voltage at VDC gets easier, which
will say that the difference is that large
(1-3 V dependent on spec.) so that the
AC signal on the line never goes that
much negative that it reaches the VDC
level, the transistor T2, which is used to
shunt current past the VDC in order not
to disturbe the line impedance, can be
omitted and the pins 7 and 8 connected
together. To ensure that high enough
signal can be taken out from the
receiver when a low VDC is used a
shunt element should be connected
between VDC and pin 18 (see fig. 20).
If the requirement on the difference
between the line current and the current
out from VDC is not that critical, it
means that the circuit can dispose more
current for its own function. The first
thing in this case is to see if the active
transmitter output impedance towards
the line is necessary. (in case it is used
in the first place). The active impedance
towards the line is used to save current
and it functions as follows. The transmit-
ter generates at the same time as it
transmits a required impedance towards
the line by taking a part of the signal on
the line and feeding it back to the
transmitter amplifier, thus saving the
current that would have been necessary
to drive the signal out on the line.
Because the transmitter amplifier is not
current loaded by the impedance that it
generates itself, the current need with
active impedance will be: 600 Ω line
impedance parallel with the 2.7 k (R1)
which with transmit swing of 2 Vp needs
4.1 mAp transmitter current instead if
the circuit would have a passive
impedance of 600 Ω the current needed
would be 6.7 mAp. The current need
without active impedance increases with
2.6 mAp. Half of this current flows
through R12 and generates no charge
current for VDC, hence if the difference
between the line current and VDC
current increases by 1.3 mA it would
result that the active impedance could
be changed against a passive. This is
done by omitting C1, short R2, decrease
R1 to a suitable value for the impedance
towards line (R1 can be a complex
network) and adjust back the DC mask
to compensate for the lower voltage
drop across R1. The pin 5 can be left
open or connected with a small
R13
+Line
R12
PBL 3853 1
5
4
+V
10
12
M
Rx
13
9
-
7
8
-
-
18
Tx
16
T2
T1
Shunt
regulator
( ) C10
VDC
D1
6 14
11 3 2
17
15
( ) Maybe a
current
generator
-Line
Figure 21.
13
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