LT6604-2.5
10
660425fa
APPLICATIONS INFORMATION
Use Figure 4 to determine the interface between the
LT6604-2.5 and a current output DAC. The gain, or “tran-
simpedance,” is defi ned as A = V
OUT
/I
IN
. To compute the
transimpedance, use the following equation:
A
R
RR
=
+
()
Ω
()
1580 1
12
By setting R1 + R2 = 1580Ω, the gain equation reduces to
A = R1(Ω). The voltage at the pins of the DAC is determined
by R1, R2, the voltage on V
MID
and the DAC output current.
Consider Figure 4 with R1 = 49.9Ω and R2 = 1540Ω. The
voltage at V
MID
, for V
S
= 3.3V, is 1.65V. The voltage at the
DAC pins is given by:
VV
R
RR
I
RR
RR
mV
DAC MID IN
=
++
+
+
=+
••
1
1 2 1580
12
12
26 II
IN
•.48 3Ω
53.6Ω and 392Ω resistors satisfy the two constraints
above. The transformer converts the single-ended source
into a differential stimulus. Similarly, the output of the
LT6604-2.5 will have lower distortion with larger load
resistance yet the analyzer input is typically 50Ω. The 4:1
turns (16:1 impedance) transformer and the two 402Ω
resistors of Figure 5, present the output of the LT6604-2.5
with a 1600Ω differential load, or the equivalent of 800Ω
to ground at each output. The impedance seen by the
network analyzer input is still 50Ω, reducing refl ections in
the cabling between the transformer and analyzer input.
Differential and Common Mode Voltage Ranges
The rail-to-rail output stage of the LT6604-2.5 can process
large differential signal levels. On a 3V supply, the output
signal can be 5.1V
P-P
. Similarly, a 5V supply can support
signals as large as 8.8V
P-P
. To prevent excessive power
dissipation in the internal circuitry, the user must limit
differential signal levels to 9V
P-P
.
The two amplifi ers inside the LT6604-2.5 channel have
independent control of their output common mode voltage
(see the “Block Diagram” section). The following guidelines
will optimize the performance of the fi lter.
V
MID
can be allowed to fl oat, but it must be bypassed to
an AC ground with a 0.01μF capacitor or instability may
be observed. V
MID
can be driven from a low impedance
source, provided it remains at least 1.5V above V
and at
least 1.5V below V
+
. An internal resistor divider sets the
voltage of V
MID
. While the internal 11k resistors are well
matched, their absolute value can vary by ±20%. This
should be taken into consideration when connecting an
external resistor network to alter the voltage of V
MID
.
+
0.1μF
3.3V
+
0.01μF
CURRENT
OUTPUT
DAC
V
OUT
+
V
OUT
660425 F04
R2
R1
I
IN
I
IN
+
R2
R1
=
V
OUT
+
– V
OUT
I
IN
+
– I
IN
1580 • R1
R1 + R2
25
27
4
34
6
2
29
7
1/2
LT6604-2.5
Figure 4
Evaluating the LT6604-2.5
The low impedance levels and high frequency operation
of the LT6604-2.5 require some attention to the imped-
ance matching networks between the LT6604-2.5 and
other devices. The previous examples assume an ideal
(0Ω) source impedance and a large (1k) load resistance.
Among practical examples where impedance must be
considered is the evaluation of the LT6604-2.5 with a
network analyzer.
Figure 5 is a laboratory setup that can be used to char-
acterize the LT6604-2.5 using single-ended instruments
with 50Ω source impedance and 50Ω input impedance.
For a 12dB gain confi guration the LT6604-2.5 requires a
402Ω source resistance yet the network analyzer output is
calibrated for a 50Ω load resistance. The 1:1 transformer,
+
0.1μF
0.1μF
2.5V
–2.5V
+
660425 F05
402Ω
402Ω
NETWORK
ANALYZER
INPUT
50Ω
COILCRAFT
TTWB-16A
4:1
NETWORK
ANALYZER
SOURCE
COILCRAFT
TTWB-1010
1:1
50Ω
53.6Ω
392Ω
392Ω
25
27
4
34
6
2
29
7
1/2
LT6604-2.5
Figure 5
LT6604-2.5
11
660425fa
APPLICATIONS INFORMATION
V
OCM
can be shorted to V
MID
for simplicity. If a different
common mode output voltage is required, connect V
OCM
to a voltage source or resistor network. For 3V and 3.3V
supplies the voltage at V
OCM
must be less than or equal
to the mid supply level. For example, voltage (V
OCM
) ≤
1.65V on a single 3.3V supply. For power supply voltages
higher than 3.3V the voltage at V
OCM
can be set above mid
supply, as shown in Table 1. The voltage on V
OCM
should
not be more than 1V below the voltage on V
MID
. V
OCM
is
a high impedance input.
Table 1. Output Common Mode Range for Various Supplies
SUPPLY
VOLTAGE
DIFFERENTIAL OUT
VOLTAGE SWING
OUTPUT COMMON MODE RANGE
FOR LOW DISTORTION
3V 4V
P-P
2V
P-P
1V
P-P
1.4V ≤ V
OCM
≤ 1.6V
1V ≤ V
OCM
≤ 1.6V
0.75V ≤ V
OCM
≤ 1.6V
5V 8V
P-P
4V
P-P
2V
P-P
1V
P-P
2.4V ≤ V
OCM
≤ 2.6V
1.5V ≤ V
OCM
≤ 3.5V
1V ≤ V
OCM
≤ 3.75V
0.75V ≤ V
OCM
≤ 3.75V
±5V 9V
P-P
4V
P-P
2V
P-P
1V
P-P
–2V ≤ V
OCM
≤ 2V
–3.5V ≤ V
OCM
≤ 3.5V
–3.75V ≤ V
OCM
≤ 3.75V
–4.25V ≤ V
OCM
≤ 3.75V
NOTE: The voltage at V
OCM
should not be more than 1V below the voltage
at V
MID
. To achieve some of the output common mode ranges shown in the
table, the voltage at V
MID
must be set externally to a value below mid supply.
The LT6604-2.5 was designed to process a variety of
input signals including signals centered on the mid-sup-
ply voltage and signals that swing between ground and
a positive voltage in a single supply system (Figure 1).
The allowable range of the input common mode voltage
(the average of V
IN
+
and V
IN
in Figure 1) is determined
by the power supply level and gain setting (see “Electrical
Characteristics”).
Common Mode DC Currents
In applications like Figure 1 and Figure 3 where the LT6604-
2.5 not only provides lowpass fi ltering but also level shifts
the common mode voltage of the input signal, DC currents
will be generated through the DC path between input and
output terminals. Minimize these currents to decrease
power dissipation and distortion.
Consider the application in Figure 3. V
MID
sets the output
common mode voltage of the 1st differential amplifi er inside
the LT6604-2.5 channel (see the “Block Diagram” section)
at 2.5V. Since the input common mode voltage is near 0V,
there will be approximately a total of 2.5V drop across
the series combination of the internal 1580Ω feedback
resistor and the external 402Ω input resistor. The result-
ing 1.25mA common mode DC current in each input path,
must be absorbed by the sources V
IN
+
and V
IN
. V
OCM
sets
the common mode output voltage of the 2nd differential
amplifi er inside the LT6604-2.5 channel, and therefore sets
the common mode output voltage of the fi lter. Since, in
the example of Figure 3, V
OCM
differs from V
MID
by 0.5V,
an additional 625μA (312μA per side) of DC current will
ow in the resistors coupling the 1st differential amplifi er
output stage to the fi lter output. Thus, a total of 3.125mA
is used to translate the common mode voltages.
A simple modifi cation to Figure 3 will reduce the DC com-
mon mode currents by 36%. If V
MID
is shorted to V
OCM
the common mode output voltage of both op amp stages
will be 2V and the resulting DC current will be 2mA. Of
course, by AC-coupling the inputs of Figure 3, the common
mode DC current can be reduced to 625μA.
Noise
The noise performance of the LT6604-2.5 channel can be
evaluated with the circuit of Figure 6. Given the low noise
output of the LT6604-2.5 and the 6dB attenuation of the
transformer coupling network, it is necessary to measure
the noise fl oor of the spectrum analyzer and subtract the
instrument noise from the fi lter noise measurement.
Example: With the IC removed and the 25Ω resistors
grounded, Figure 6, measure the total integrated noise (e
S
)
of the spectrum analyzer from 10kHz to 2.5MHz. With the
IC inserted, the signal source (V
IN
) disconnected, and the
Figure 6
+
0.1μF
0.1μF
2.5V
–2.5V
+
1/2
LT6604-2.5
R
IN
R
IN
25Ω
25Ω
660425 F06
SPECTRUM
ANALYZER
INPUT
50Ω
V
IN
COILCRAFT
TTWB-1010
1:1
25
27
4
34
6
2
29
7
LT6604-2.5
12
660425fa
APPLICATIONS INFORMATION
input resistors grounded, measure the total integrated noise
out of the fi lter (e
O
). With the signal source connected, set
the frequency to 100kHz and adjust the amplitude until
V
IN
measures 100mV
P-P
. Measure the output amplitude,
V
OUT
, and compute the passband gain A = V
OUT
/V
IN
. Now
compute the input referred integrated noise (e
IN
) as:
e
ee
A
IN
OS
=
()()
22
Table 2 lists the typical input referred integrated noise for
various values of R
IN
.
Table 2. Noise Performance
PASSBAND
GAIN R
IN
INPUT REFERRED
INTEGRATED NOISE
10kHz TO 2.5MHz
INPUT REFERRED
INTEGRATED NOISE
10kHz TO 5MHz
4 402Ω 18μV
RMS
23μV
RMS
2 806Ω 29μV
RMS
39μV
RMS
1 1580Ω 51μV
RMS
73μV
RMS
Figure 7 is plot of the noise spectral density as a function
of frequency for an LT6604-2.5 channel with R
IN
= 1580Ω
using the fi xture of Figure 6 (the instrument noise has
been subtracted from the results).
The noise at each output is comprised of a differential
component and a common mode component. Using a
transformer or combiner to convert the differential outputs
to single-ended signal rejects the common mode noise and
gives a true measure of the S/N achievable in the system.
Conversely, if each output is measured individually and the
noise power added together, the resulting calculated noise
level will be higher than the true differential noise.
Power Dissipation
The LT6604-2.5 amplifi ers combine high speed with large
signal currents in a small package. There is a need to en-
sure that the die’s junction temperature does not exceed
150°C. The LT6604-2.5 has an exposed pad (pin 35) which
is connected to the negative supply (V
). Connecting the
pad to a ground plane helps to dissipate the heat generated
by the chip. Metal trace and plated through-holes can be
used to spread the heat generated by the device to the
backside of the PC board.
Junction temperature, T
J
, is calculated from the ambient
temperature, T
A
, and power dissipation, P
D
. The power
dissipation is the product of supply voltage, V
S
, and total
supply current, I
S
. Therefore, the junction temperature is
given by:
T
J
= T
A
+ (P
D
θ
JA
) = T
A
+ (V
S
• I
S
θ
JA
)
where the supply current, I
S
, is a function of signal level,
load impedance, temperature and common mode voltages.
For a given supply voltage, the worst-case power dissipation
occurs when the differential input signal is maximum, the
common mode currents are maximum (see Applications
Information regarding Common Mode DC Currents), the
load impedance is small and the ambient temperature is
maximum. To compute the junction temperature, measure
the supply current under these worst-case conditions, use
43°C/W as the package thermal resistance, then apply the
equation for T
J
. For example, using the circuit in Figure 3
with DC differential input voltage of 1V, a differential
output voltage of 4V, no load resistance and an ambient
temperature of 85°C, the supply current (current into V
+
)
measures 37.6mA per channel. The resulting junction
temperature is: T
J
= T
A
+ (P
D
θ
JA
) = 85 + (5 • 2 • 0.0376
• 43) = 101°C. The thermal resistance can be affected by
the amount of copper on the PCB that is connected to V
.
The thermal resistance of the circuit can increase if the
Exposed Pad is not connected to a large ground plane
with a number of vias.
FREQUENCY (MHz)
0.01
0
30
40
50
0.1 1 10
660425 F07
20
10
0
60
80
100
40
20
NOISE SPECTRAL DENSITY (nV
RMS
/√Hz)
INTEGRATED NOISE (μV
RMS
)
SPECTRAL DENSITY
INTEGRATED
Figure 7. Input Referred Noise, Gain = 1

LT6604IUFF-2.5#PBF

Mfr. #:
Manufacturer:
Analog Devices Inc.
Description:
Differential Amplifiers Dual Differential Amplifier and 2.5MHz Lowpass Filter
Lifecycle:
New from this manufacturer.
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