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PBL3853SO-T 参数 Datasheet PDF下载

PBL3853SO-T图片预览
型号: PBL3853SO-T
PDF下载: 下载PDF文件 查看货源
内容描述: 环球巡回演讲 [Universal Speech Circuit]
分类和应用: 电信集成电路电信电路电话电路光电二极管
文件页数/大小: 16 页 / 259 K
品牌: ERICSSON [ ERICSSON ]
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PBL 3853  
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).  
out from VDC is not that critical, it  
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  
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 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 the current needed  
would be 6.7 mAp. The current need  
without active impedance increases with  
2.6 mAp. Half of this current flows  
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  
current that also passes pin 2, to  
through R12 and generates no charge  
current for VDC, hence if the difference  
between the line current and VDC  
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 //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  
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  
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  
+Line  
R13  
R12  
regulator VDC (simply drawn as a diode  
in the fig. 4). What has to be observed  
is that the supply for the receiver  
9
1
PBL 3853  
-
-
5
4
T2  
7
8
+V  
10  
T1  
V
DC  
-
Shunt  
regulator  
C10  
D1  
12  
13  
18  
16  
(
)
T1  
8
M
Rx  
Tx  
VDC  
C10  
Shunt  
regulator  
6
14  
17  
11  
3
2
15  
Maybe a  
current  
generator  
D1  
-Line  
18  
( )  
-Line  
Figure 21.  
Figure 20.  
13  
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