CIRCUITED OUTPUT to keep the matching
with the line in sleep mode
DESCRIPTION
The TS615 is a dual operational am plifier featuring a high output current 410mA. These drivers
can be configured differentially for driving signals
in telecommunication systems using multiple carriers. The TS615 is ideally suited for xDSL (High
Speed Asymmetrical Digital Subscriber Line) applications. This circuit is c apable of driving a 10 Ω
or 25Ω load at ±2.5V, 5V, ±6V or +12V power
supply. The TS615 will be able to reach a -3dB
bandwidth of 40MHz on 25Ω load with a 12dB
gain. This device is designed for the high slew
rates to support low harmonic distortion and intermodulation. The TS615 is fitted out with Power
Down function to decrease the consumption. During this sleep state the device displays a short circuit output in order to keep the impedance matching with the line. The TS615 is housed in
TSSOP14 Exposed-Pad plastic package for a
very low thermal resistance.
P
TSSOP14 Exposed-Pad
(Plastic Micro package)
ORDER CODE
Part NumberTemperature RangePackage
TS615IPWT-40, +85°CPW
PW= Thin Shrink Small Outline Package with Exposed-Pad
(TSSOP Exposed-Pad) only available in Tape & Reel (PWT)
PIN CONNECTIONS (top view)
-VCC1
-VCC1
+VCC1
+VCC1
Non Inverting Input1
Non Inverting Input1
Inverting Input1
Inverting Input1
Power Down
Power Down
NC
NC
1
1
2
2
3
3
+ --+
+ -- +
4
4
5
5
6
6
7
7
Top View
Top View
14
14
-VCC2
-VCC2
13
13
Output2Output1
Output2Output1
12
12
+VCC2
+VCC2
Non Inverting Input2
Non Inverting Input2
11
11
10
10
Inverting Input2
Inverting Input2
NC
NC
9
9
NC
NC
8
8
APPLICATION
■ Line driver for xDSL
■ Multiple Video Line Driver
December 2002
Cross Section V iew Showi ng Exposed-Pa d
Cross Section V iew Showi ng Exposed-Pa d
This pad c an be con nected to a (-Vcc) copper ar ea on the PCB
This pad c an be con nected to a (-Vcc) copper ar ea on the PCB
1/27
TS615
ABSOLUTE MAXIMUM RATINGS
SymbolParameterValueUnit
V
T
T
R
R
P
ESD
except
pins 4, 5,
10, 11
ESD
only pins 4,
5, 10, 11
Supply voltage
CC
V
Differential Input Voltage
id
V
inInput Voltage Range
Operating Free Air Temperature Range-40 to + 85°C
oper
Storage Temperature-65 to +150°C
std
T
Maximum Junction Temperature150°C
j
Thermal Resistance Junction to Case4°C/W
thjc
Thermal Resistance Junction to Ambient Area40°C/W
thja
Maximum Power Dissipation (@25°C)3.1W
max.
CDM : Charged Device Model
HBM : Human Body Model
MM : Machine Model
CDM : Charged Device Model
HBM : Human Body Model
MM : Machine Model
Output Short Circuit
1.All voltage values, except differential voltage are with respect to network terminal.
2.Differential voltage are non-inverting input terminal with respect to the inverting input terminal.
3.The magnitude of input and output voltage must never exceed V
4.An output current limitation protects the circuit from transient currents. Short-circuits can cause excessive heating.
Destructive dissipation can result from short circuit on amplifiers.
1)
2)
3)
±7V
±2V
±6V
1.5
2
200
1
1
100
4)
+0.3V.
CC
kV
kV
V
kV
kV
V
OPERATING CONDITIONS
SymbolParameterValueUnit
V
V
Power Supply Voltage±2.5 to ±6V
CC
+1.5V to +VCC-1.5V
Common Mode Input Voltage
icm
-V
CC
TYPICAL APPLICATION:
Differential Line Driver for xDSL Applications
12
12
11
11
10
10
Vi
Vi
R1
R1
R4
R4
ViVo
ViVo
5
5
4
4
Pw-Dwn
Pw-Dwn
+
+
1/2TS615
1/2TS615
_
_
14
14
R2
R2
GND
GND
R3
R3
3
3
_
_
1/2TS615
1/2TS615
+
+
1
1
6
6
+Vcc
+Vcc
-Vcc
-Vcc
+Vcc
+Vcc
-Vcc
-Vcc
13
13
2
2
12.5Ω
12.5Ω
12.5Ω
12.5Ω
Vo
Vo
1:2
1:2
25Ω100Ω
25Ω100Ω
V
2/27
TS615
ELECTRICAL CHARACTERISTICS
V
= ±6Volts, Rfb=910Ω,T
CC
Note: as described on page 24 (table 71), the TS615 requires a 620Ω feedback res i st or for an optimis ed bandwidth wi t h a gai n of 12 B for
a 12V power supply. Nevertheless, due to production test constraints, the TS615 is tested with the same feedback resistor for 12V and 5V
power su ppl i es (910Ω).
SymbolParameterTest ConditionMin.Typ.Max.Unit
DC PERFORMANCE
V
Input Offset Voltage
io
V
∆
Z
C
CMR
SVR
Differential Input Offset Voltage
io
I
Positive Input Bias Current
ib+
I
Negative Input Bias Current
ib-
Input(+) Impedance82k
IN+
Z
Input(-) Impedance54
IN-
Input(+) Capacitance1pF
IN+
Common Mode Rejection Ratio
20 log (∆V
/∆Vio)
ic
Supply Voltage Rejection Ratio
20 log (∆V
I
Total Supply Current per OperatorNo load1417mA
CC
/∆Vio)
cc
DYNAMIC PERFORMANCE and OUTPUT CHARACTERISTIC
R
Open Loop Transimpedance
OL
-3dB Bandwidth
Full Power Bandwidth
BW
Gain Flatness @ 0.1dB
TrRise Time
TfFall Time
TsSettling Time
SRSlew Rate
V
High Level Output Voltage
OH
V
Low Level Output Voltage
OL
Output Sink Current
I
out
Output Source Current
= 25°C (unless otherwise specified)
amb
T
amb
< T
T
min.
T
amb
T
amb
T
min.
T
amb
T
min.
V
∆
ic
T
min.
V
∆
cc
T
min.
V
out
T
min.
< T
amb
= 25°C
< T
< T
amb
< T
< T
amb
= ±4.5V
< T
< T
amb
=±2.5V to ±6V
< T
< T
amb
= 7Vp-p, RL = 25
< T
amb.
Small Signal V
A
= 12dB, RL = 25
V
Large Signal V
= 12dB, RL = 25
A
V
Small Signal V
= 12dB, RL = 25
A
V
V
= 6Vp-p, AV = 12dB, RL
out
= 25
Ω
= 6Vp-p, AV = 12dB, RL
V
out
= 25
Ω
= 6Vp-p, AV = 12dB, RL
V
out
= 25
Ω
= 6Vp-p, AV = 12dB, RL
V
out
= 25
Ω
R
=25Ω Connected to GND
L
R
=25Ω Connected to GND
L
V
= -4Vp
out
< T
T
min.
V
out
T
min.
amb
= +4Vp
< T
amb
< T
< T
< T
max.
max.
max.
max.
max.
max.
<20mVp
out
Ω
=3Vp
out
Ω
<20mVp
out
Ω
max.
max.
1.253.5
2.1
mV
2.5mV
630
7.8
315
3.2
A
µ
A
µ
Ω
Ω
5863
61
7279
78
Ω
521
8.9
dB
dB
M
Ω
2540
MHz
26
7MHz
10.6ns
12.2ns
50ns
330410V/µs
4.85.1V
-5.5-5.2V
-350-530
-440
330420
mA
365
3/27
TS615
Note: as described on page 24 (table 71), the TS615 requires a 620Ω feedback res i st or for an optimis ed bandwidth wi t h a gai n of 12 B for
a 12V power supply. Nevertheless, due to production test constraints, the TS615 is tested with the same feedback resistor for 12V and 5V
power su ppl i es (910Ω).
SymbolParameterTest ConditionMin.Typ.Max.Unit
NOISE AND DISTORTION
eNEquivalent Input Noise VoltageF = 100kHz2.5nV/√Hz
iNpEquivalent Input Noise Current (+)F = 100kHz15pA/√Hz
iNnEquivalent Input Noise Current (-)F = 100kHz21pA/√Hz
Vcc=±6V, Rfb=620Ω, Load = 25Ω
Vcc=±2.5V, R f b=910Ω, Load =10Ω
IF Bw = 10Hz
Smoothing=19.247MHz
on 10ns/div scale
50M
14/27
TS615
C
t
INTERMODULATION DISTORTION PRODUCT
A non-ideal output of the amplifier can be described by the following development :
VoutC0C1VinC2V
++ +=
2
in
…
CnV
n
in
due to a non-linearity in the input-output amplitude
transfer. In the case of the input is Vin=Asinωt, C
is the DC component, C1(Vin) is the fundamental,
C
is the amplitude of the harmonics of the output
n
signal V
out
.
A one-frequency (one-tone) input signal contributes to a harmonic distortion. A two-tones input
signal contributes to a harmonic disto rtion and intermodulation product.
This intermodulation product or intermodulation
distortion study of a two-tones input signal is the
first step of the amplifier characterization of driving
capability in the case of a multi-tone signal.
In this case :
C
+A
()
2
C
+
…
()
n
VinA
tsinB
ω
1
A
tsinB
ω
1
tsinB
ω
1
2
tsin+
ω
2
n
tsin+
ω
2
tsin+=
ω
2
In this expression, we recognize the second order
intermodulation IM2 by the frequencies (ω
and (ω
IM3 by the frequencies (2ω
+2ω2) and (ω1+2ω2).
(−ω
1
) and the third order intermodulation
1+ω2
), (2ω1+ω2),
1-ω2
The measurement of the intermodulation product
of the driver is achieved by using the driver as a
0
mixer by a summing amplifier configuration. By
this way, the non-linearity problem of an external
mixing device is avoided.
Figure 45 : Non-inve r t ing S umming Amp lif ier
1kΩ
1kΩ
1kΩ
1kΩ
49.9Ω
Vin1
Vin1
50Ω
50Ω
1:√2
1:√2
North Hills
North Hills
0315PB
0315PB
400Ω
400Ω
Vin2
Vin2
50Ω
50Ω
49.9Ω
1:√2
1:√2
North Hills
North Hills
0315PB
0315PB
49.9Ω
49.9Ω
400Ω
400Ω
49.9Ω
49.9Ω
49.9Ω
49.9Ω
1kΩ
1kΩ
11
11
+Vcc
+Vcc
+
+
1/2TS615
1/2TS615
_
_
13
13
10
10
Rfb1
Rfb1
Rg1
Rg1
Vout diff.
Vout diff.
Rg2
Rg2
Rfb2
Rfb2
_
_
1/2TS615
1/2TS615
1kΩ
1kΩ
+
+
-Vcc
-Vcc
33
33
33Ω
33Ω
49.9Ω
49.9Ω
49.9Ω
49.9Ω
Ω
Ω
1-ω2
√2:1
√2:1
100Ω50Ω
100Ω50Ω
North Hills
North Hills
0315PB
0315PB
)
V
outC0C1
+=
and :
C+
()
1
C
2
2
------- A
–
2+C2AB
3C3A2B
----------------------- -+2
2
3C3A2B
----------------------- -+
2
3
A
tsi nB
ω
1
cost B
2
()
C3A33
+
()
–
()
…
V
outC0C2
3
ω
A
()
A
ω
2
ω
1
–
ω1ω
()
3
ω
–
ω
1
ω
1
C
+
+=
t2A2B
2
tsinB
ω
1
tsinB
1
ω
2
t
cos–tcos
2
3
-------
¥+
4
tsinB33
1
1
tsin
-- -–2
ω
2
2
1
–
2ω+
tsin
-- 2
2
n
V
()
n
in
A2B2+
---------------------
2
ω
tsin+
ω
2
tsin+
2
2
tcos+
ω
2
–
ω
ω
()
1
2
tsin+
ω
2
+
ω1ω
()
ω
()
1
tsin2AB
1
tsin
2
2ω+
tsin
2
2
ω
sin++sin+
2
The following graphs show the IM2 and the IM3 of
the amplifier in different configuration. The
two-tones input signal is achieved by the multisource generator Marconi 2026. Each tone has
the same amplitude. The measurement is
achieved by the spectrum analyzer HP3585A .
In this range of frequency, printed circuit board
parasites can affect the closed-loop performance.
The implementation of a proper ground plane in
both sides of the PCB is mandatory to provide low
inductance and low resistance common return.
Most important for controlling the gain flatness
and the bandwidth are stray capac itances at the
output and inverting input. For minimizing the coupling, the space between signal lines and ground
plane will be increased. Connec tions of the feedback component s mu st be as short as possible in
order to decrease the associated inductance
which affect high frequen cy gain errors. It is very
important to choose external components as small
as possible such as surface mounted devices,
SMD, in orde r to minimize the size of all the DC
and AC connections.
THERMAL INFORMATION
The TS615 is housed in an Exposed-Pad plastic
package. As described on the figure 56, this package uses a lead frame upon which the dice is
mounted. This lead frame is exposed as a thermal
pad on the underside of the package. The thermal
contact is direct with the di ce. This thermal path
provide an excellent thermal performance.
Figure 55 : Exposed-Pad Package
DICE
DICE
Bottom Vie w
Side View
Side View
Bottom Vie w
DICE
DICE
Cross Section Vie w
Cross Section Vie w
Figure 56 : Evaluation Board
1
1
The thermal pad is electrically isolated from all
pins in the package. It should be soldered to a
copper area o f th e PC B underneath the pac kage.
Through these thermal paths within this copper area, heat can b e conducted away from the pack age. In this case, the copp er area should be connected to (-
V
).
CC
18/27
Figure 57 : Schematic Diagram
J105
J105
J106
J106
J107
J107
J108
J108
J109
J109
Differential Amplifier
Differential Amplifier
J106
J106
J109
J109
R106
R106
4
R107
R107
R102R101
R102R1 01
R103
R103
R108
R108
R109
R109
R104
R104
R110
R110
R105
R105
R107
R107
R102
R102
R110
R110
R105
R105
4
+
+
1/2TS615
1/2TS615
_
_
2
2
5
5
R114
R114
R111
R111
R112
R112
R115
R115
10
10
_
_
1/2TS615
1/2TS615
13
13
+
+
11
11
R113
R113
4
4
+
+
1/2TS615
1/2TS615
_
_
2
2
5
5
R114
R114
R111
R111
R112
R112
R115
R115
10
10
_
_
1/2TS615
1/2TS615
13
13
+
+
11
11
R113
R113
TS615
Non-Inverting Amplifier
Non-Inverting Amplifier
4
R107
J106
J106
R118
R118
R116
R116
R119
R119
R117
R117
R118
R118
R116
R116
R119
R119
R117
R117
J110
J110
R120
R120
Inverting Amplifier
Inverting Amplifier
J111
J111
R121
R121
J110
J110
R120
R120
J111
J111
R121
R121
J108
J108
Non-In ve r tin g S um m ing Amplif i er
Non-In ve r tin g S um m ing Amplif i er
J105
J105
J106
J106
R107
R102
R102
R109
R109
R104
R104
R101
R101
R106
R106
R107
R107
R102
R102
4
+
+
1/2TS615
1/2TS615
_
_
2
2
5
5
R114
R114
R111
R111
R115
R115
_
_
10
10
1/2TS615
1/2TS615
13
13
+
+
11
11
R113
R113
4
4
+
+
1/2TS615
1/2TS615
_
_
2
2
5
5
R114
R114
R111
R111
R118
R118
R116
R116
R119
R119
R117
R117
R118
R118
R116
R116
J110
J110
R120
R120
J111
J111
R121
R121
J110
J110
R120
R120
PowerSupply
PowerSupply
C102
C102
100nF
100nF
J101
J101
+Vcc
+Vcc
J102
J102
GND
GND
J103
J103
-Vcc
-Vcc
C103
C103
C103
100nF
100nF
100nF
J10412
J10412
3
3
Power down
Power down
J112
J112
+Vcc
+Vcc
C105
C105
+Vcc
+Vcc
C101
C101
100uF
100uF
C104
C104
100uF
100uF
+Vcc
+Vcc
-Vcc
-Vcc
100nF
100nF
3
3
4
4
+
+
1/2TS615
1/2TS615
_
_
5
5
C106
C106
100nF
100nF
-Vcc
-Vcc
C107
C107
100nF
100nF
100nF
100nF
+Vcc
+Vcc
_
_
10
10
1/2TS615
1/2TS615
+
+
11
11
C108
C108
R122
R122
-Vcc
-Vcc
6
6
2
2
1
1
-Vcc
-Vcc
12
12
14
14
-Vcc
-Vcc
-Vcc
-Vcc-Vcc
Expo sed-P ad
Expo sed-P ad
13
13
Differentia l Amplifier
Differentia l Amplifier
R107
J106
J106
J109
J109
R107
R102
R102
R110
R110
R105
R105
4
4
+
+
1/2TS615
1/2TS615
_
_
2
2
5
5
R114
R114
R111
R111
R112
R112
R115
R115
10
10
_
_
1/2TS615
1/2TS615
13
13
+
+
11
11
R113
R113
R118
R118
R116
R116
R119
R119
R117
R117
J110
J110
R120
R120
J111
J111
R121
R121
19/27
TS615
Figure 58 : Component Locations - Top Side
Figure 59 : Component Locations - Bottom Side
Figure 60 : Top Side Board Layout
Figure 61 : Bottom Side Board Layout
20/27
TS615
g
)
)
NOISE MEASUREMENT
Figure 62 : Noise Model
+
eN
eN
+
TS615
TS615
_
_
R2
R2
iN+
iN+
R3
R3
N3
N3
iN-
iN-
N2
N2
R1
R1
N1
N1
output
output
HP3577
HP3577
Input noi s e:
Input noi s e:
8nV/√Hz
8nV/√Hz
eN : input voltage noise of the amplifier
iNn : negative input current noise of the amplifier
iNp : positive input current noise of the amplifier
The closed loop gain is :
R
A
g1
V
The six noise sources are :
2
=
V
iNn R2×
54
V
=
V1eN
=
V3iNp R
4
V
61
V
×
2
R
------ -
–=
1
R
R
------ -
+
R
fb
--------- -+==
R
2=
kTR
2
R
------ -
1
+
1
R
2
R
31
×
2
1
-------
+
××
1
R
4
1
kTR
4
3=
kTR
Assuming the thermal noise of a resistance R as:
4kTR F∆
with
∆F the specified bandwidth.
On 1Hz bandwidth the thermal noise is reduced to
4kTR
k is the Boltzmann’s constant equals to
1,374.10-23J/°K. T is the temperature (°K).
The output noise eNo is c alculated using the Su-
perposition Theorem. But it is not the sum of all
noise sources. The output noise is the square root
of the sum of the square of each noise source.
NoV12V22V32V42V52V6
+++++ eq1
2
,=
(
2
2
=
eNo2eN2g2iNn2R
2
2
R
------ -
+
4
×eq2(
1
R
+
kTR
×
21
×R
++
14
kTR
2
+
iNp
2
R
-------
+
1
R
3
×g2×
2
4
kTR
×
2
3
,
The input noise of the instrumentation must be extracted from the measured noise value. Th e real
output noise value of the driver is:
eNoMeasured
()
2
instrumentation
–eq3
()
2
,=
()
The input noise is called the Equivalent Input
Noise as it is not directly measured but it is evaluated from the meas urement of the output divided
by the closed loop gain (eNo/g).
After simplification of the fourth and the fifth term
of (eq2) we obtain:
2
2
=
eNo2eN2g2iNn2R
+
… g4kTR
×eq4(),
+
×R
21
×
2
R
------ -
+
+
1
R
2
+
iNp
2
4
kTR
×
3
×g2×
3
2
Measurement of eN:
We assume a short-circuit on the non-inverting input (R3=0). (eq4) comes:
=
eNoeN2g2iNn2R
+
×eq5(),
×
2
+
g4kTR
2
×
2
In order to easily extract the value of eN, the resistance R2 will be chosen as low as possible. In the
other hand, the gain must be large enough.
R1=10Ω, R2=910Ω, R3=0, Gai n=92
Equivalent Input Noise: 2.57nV/√Hz
Input Voltage Noise: eN=2.5nV/√Hz
Measurement of iNn:
R3=0 and the output noise equation is still the
(eq5). This time the gain must be decreased to decrease the thermal noise contribution.
R1=100Ω, R2=910Ω, R3=0, Gain=10.1
Equivalent Input Noise: 3.40nV/√Hz
Negative Input Current Noise: iNn =21pA/√Hz
Measurement of iNp:
To extract iNp from (eq3), a resist ance R3 is connected to the non-inverting input. The val ue of R3
must be chose n in o rder to keep i ts therm al noise
contribution as low as possible against the iNp
contribution.
R1=100Ω, R2=910Ω, R3=100Ω, Gain=10.1
Equivalent Input Noise: 3.93nV/√Hz
Positive Input Current Noise: iNp=15pA/√Hz
Condit i ons: frequenc y=100kHz, V
Instrumentation: Spectrum Analyzer HP3585A
(input noi se of the HP358 5A: 8nV/√Hz)
CC
=±2.5V
21/27
TS615
+
-VCC
+VCC
10µF
+
10nF
TS615
10µF
+
10nF
-
+
-VCC
+VCC
10µF
+
10nF
TS615
10µF
+
10nF
-
POWER SUPPLY BYPASSIN G
A proper power suppl y bypass ing com es very im portant for optimizing the pe rformanc e in high frequency range. Bypass capacitors should be
placed as close as possible to the IC pins to improve high frequency bypassing. A capacitor
greater than 1µF is necessary to minimize the distortion. For a better quality bypassing a capacitor
of 10nF is added following the same condition of
implementation. These bypass capacitors must be
incorporated for the negative and the positive supply.
Figure 63 : Circuit for Power Supply Bypassing
SINGLE POWER SUPPLY
The following figure show the case of a 5V single
power suppl y configuration
Figure 64 : Circuit for +5V single supply
+5V
+5V
10µF
10µF
+
IN
IN
+5V
+5V
R1
R1
820Ω
820Ω
R2
R2
820Ω
820Ω
Rin
Rin
1kΩ
1kΩ
+ 1µF
+ 1µF
10nF
10nF
+
½ TS615
½ TS615
_
_
G
G
R
R
+
+
G
G
C
C
R
R
910Ω
910Ω
100µF
100µF
fb
fb
OUT
OUT
Rs
Rs
Rload
Rload
necessary to assume a positive output dynamic
range between 0V and + V
ering the values of V
OH and VOL, the am pl ifier will
supply rails. Consid-
CC
provide an output dynamic from +0.5V to 10.6V on
25Ω load for a 12V supplying, from 0.45V to 3.8V
on 10 Ω load for a 5V supplying.
The amplifier must be biased with a mid supply
(nominally +V
/2), in order to maintain the DC
CC
component of the signa l at this val ue. Several options are possible to provide this bias supply (such
as a virtual ground using an operational amplifier),
or a two-resistance di vider which is the cheapest
solution. A high resistance value is required to limit the current consumption. On the other hand, the
current must be high enough to bias the non-inverting input of t he amplifier. If we consider this
bias current (30µA max.) as the 1% o f the current
through the resistance divider to keep a stable mid
supply, two resistances of 2.2kΩ can be used i n
the case of a 12V power supply and two resistances of 820Ω can be used in the case of a 5V power
supply.
The input provides a hig h pass filter with a break
frequency below 10Hz which is necessary to remove the original 0 volt DC component of the input
signal, and to fix it at +V
CC
/2.
CHANNEL SEPARATION - CROSSTALK
The following figure show the crosstalk from an
amplifier to a second amplifier. This phenomenon,
accented in high frequ encies, is unavoidable and
intrinsic of the circuit.
Nevertheless, the PCB layout has also an effect
on the crosstalk level. Capacitive coupling between signal wires, distance between critical signal nodes, power supply bypassing, are the most
significant points.
Figure 65 : Crosstalk vs. Frequency
AV=+4, Rfb=620Ω, VCC=±6V, Vout=2Vp
-50
-60
-70
-80
-90
The TS615 operates from 12V down to 5V power
supplies. This is achieved with a d ual power s up-
ply of ±6V an d ±2.5V or a s ingle power s upply of
12V and 5V referenced to the grou nd. In t he cas e
of this asymmet rical supp lying, a new biasing is
22/27
-100
CrossTalk (dB)
-110
-120
-130
10k100k1M10M
Frequency (Hz)
TS615
0 1020304050
-6
-5
-4
-3
-2
-1
0
Vout
Vpdw
Disabled Output
Enabled Output
(Volts)
Time (µs)
POWER DOWN MODE BEHAVIOUR
Figure 66 : Equivalent Schematic
V
cc +
3
.
+
Vcc -
V
+
V
.
A1
_
cc -
A2
_
cc
+
..
.
pdw
R
1
14
.
.
12
-Vcc
-Vcc
Rpdw
.
..
13
4
5
11
10
2
Ouput 1
Ouput 2
POWER
DOWN
pin6
POWER
DOWN
pin6
Please note that the short circuited output in power down mode is referenced to (-
V
). No problem
CC
appears when used in differential mode. Nevertheless, when used in single ended on a load referenced to GND, the (-
V
) level contributes to a
CC
current consumption through the load. As described on t he Figure 68, t he interest of featuring
an output short circuit in power down mode is to
keep the best impedance matchi ng between the
system and the twisted pair telephone line when
the modem is in sleep mode. By this way, the modem can be waked-up with a sign al from the line
without any damage of this signal. This concept is
particularly intended for the ADSL over voice modems, where the modem in sleep mode, must be
waked-up by the phone call.
Figure 67 : Matching in Sleep Mode
Figure 68 : Standby Mode. Time On>Off
Figure 69 : Standby Mode. Time Off>On
1
0
Disabled Output
−1
−2
(Volts)
−3
−4
−5
−6
012345
Time (µs)
Enabled Output
Vout
Vpdw
Figure 70 : Standby Mode. Input/Output Isolation
vs. Frequency
AV=+4, Rfb=620Ω, VCC=±6V, Vout=3Vp
Consumption=80µA
Consumption=80µA
12.5Ω
12.5Ω
TS615
TS615
POWER DOWN
POWER DOWN
5Ωmax.
5Ω max.
12.5Ω
12.5Ω
The system can be waked-up
The system can be waked-up
from the line
from the line
Matching
Matching
25Ω
25Ω
Transformer
Transformer
1:2
1:2
Line (100Ω)
Line (100Ω)
0
-10
-20
-30
-40
-50
-60
-70
-80
Isolation (dB)
-90
-100
-110
-120
-130
10k100k1M10M
Frequency (Hz)
23/27
TS615
CHOICE OF THE FEEDBACK CIRCUIT
Table 71 : Closed-Loop Gain - Feedback Compo-
nents
V
CC
±6
±2.5
(V)
Gain
+1750
+2680
+4620
+8510
-1680
-2680
-4620
-8510
+11.1k
+21k
+4910
+8680
-11k
-21k
-4910
-8680
Rfb (Ω)
INVERTING AMPLIFIER BIASING
In this case a resistance is necessary to achieve a
good input biasing, as R on (fig.30).
This resistance is calculated by assuming the negative and positive input bias current. The aim is to
make the compensation of the offset bias current
which could affect the input offset voltage and t he
output DC component . A ssumin g I b-, Ib+ , R
in, Rfb
and a zero volt output, the resistance R comes: R
= R
in // Rfb .
Figure 72 : Compensation of the Input Bias
Current
ACTIVE FI LT ER I N G
Figure 73 : Low-Pass Active Filtering. Sallen-Key
C1
C1
1R2
1R2
R
R
IN
IN
The resistors R
fb and RG give d irectly the gain of
+
+
C2
C2
RG
RG
_
_
TS615
TS615
910Ω
910Ω
R
R
fb
fb
OUT
OUT
25Ω
25Ω
the filter as a classical non-inverting amplification
configuration :
R
A
V
g1
fb
--------- -+==
R
g
Assuming the following expression as the response of the system:
the cutoff frequency is not gain depe ndent and it
Vcc+
TS615
TS615
Vcc+
Vcc-
Vcc-
Output
Output
Load
Load
_
_
+
+
R
R
comes:
------------------------------------- -=
ω
c
R1R2C1C2
the damping factor comes:
1
-- -
=
ζ
ω
()
cC1R1C1R2C2R1C1R1
2
1
g–++
The higher the gain the more sensitive the damping factor is. When the gain is hig her than 1 it is
preferable to use som e very stable resistors and
capacitors values.
In the case of R1=R2:
R
2C2C
------------------------------------=
ζ
2C1C
fb
--------- -–
1
R
g
2
TS615
INCREASING THE LINE LEVEL BY USING AN
ACTIVE IMPEDANCE MATCHING
With a passive mat ching, the output signal amplitude of the driver must be twice the amplitude on
the load. To go beyond this limitation an active
matching impedance can be used. With this technique, it is possible to keep a good impedance
matching with an amplitude on the load higher
than the half of the output driver amplitude. This
concept is shown in figure 74 for a differential line.
Figure 74 : T S615 as a differential line driver with
an active impedance matching
µ
100n
Vcc+
1k
Vi
100n
100n
10µ
1k
GND
ViVo
1/2 R1
1/2 R1
+
_
+
_
R2
R3
Vcc/2
R5
Vcc+
Rs1
GND
Vo°
Vcc+
GND
Vo°
Rs2
R4
1
10n
Vo
1:n
Hybrid
&
RL
Transformer
100Ω
2Vi
Vi Vo°–
()
---------
---------------------------and
R1
As Vo
° equals Vo without load, the gain in this
R2
Vi Vo+
()
------------------------,
R3
case becomes :
2R2
----------R1
R2
1
–
------- R3
R2
------- -++
R3
G
Vo noload
()
--------------------------------Vi
1
------------------------------------==
The gain, for the loaded system will be (eq1):
2R2
----------R1
R2
1
------- -–
R3
R2
------- -++
R3
,==
()
GL
Vo withload
()
-------------------------------------Vi
1
1
-- -
------------------------------------ e q1
2
As shown in figure76, this system is an ideal generator with a synthesized impedance as the internal impedance of the system. From this, the output voltage becomes:
VoViG
()
RoIout
–=eq2
()
,
()
with Ro the synthesized impedance and Iout the
output current. On the other hand Vo can be expressed as:
Component Calculation
Let us consider the equivalent c ircuit for a single
ended configuration, Figure75.
Figure 75 : Single ended equivalent circuit
+
1/2
1/2
+
Vi
Vi
R1
R1
R2
R2
_
_
R3
R3
Vo°
Vo°
Rs1
Rs1
-1
-1
Vo
Vo
1/2
1/2
RL
RL
Let us consider the unloaded system . Assumi ng
the currents through R1, R2 and R3
as respectively:
2R2
----------R1
R2
------- -–
R3
R2
------- -++
R3
Rs1Iout
----------------------- eq 3
1
R2
------- -–
R3
,–=
()
Vi 1
Vo
------------------------------------------------
1
By identification of both equations (eq2) and
(eq3), the synthesized impedance is, with
Rs1=Rs2=Rs:
Ro
Rs
----------------- e q 4
,=
()
R2
1
------- -–
R3
Figure 76 : Equivalent schematic. Ro is the
synthesized impedance
Ro
Vi.Gi
Iout
1/2RL
25/27
TS615
Unlike the level Vo° required for a passive impedance, Vo
us write Vo
° will be smaller than 2Vo in our case. Let
°=kVo with k the m at ching f actor vary -
ing between 1 and 2. Assum ing that the current
through R3 is negligible, it comes the following resistance divider:
Ro
kVoRL
----------------------------- -=
RL 2Rs1+
After choosing the k factor, Rs will equal to
1/2RL(k-1).
A good impedance matching assume s:
1
Ro
-- -RL eq5
,=
()
2
From (eq4) and (eq5) it becomes:
R2
------- -1
R3
2Rs
----------- eq 6
,–=
()
RL
By fixing an arbitrary value of R2, (eq6) gives:
R3
R2
-------------------- -=
2Rs
1
-----------–
RL
Finally, the values of R2 and R3 allow us to extract
R1 from (eq1), and it comes:
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