Datasheet

LT6604-10
10
660410fb
APPLICATIONS INFORMATION
In Figure 3 the LT6604-10 is providing 12dB of gain. The
gain resistor has an optional 62pF in parallel to improve
the passband fl atness near 10MHz. The common mode
output voltage is set to 2V.
Use Figure 4 to determine the interface between the
LT6604-10 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 =
402 •R1
(R1+ R2)
(Ω)
By setting R1 + R2 = 402, the gain equation reduces to
A = R1 ().
The voltage at the pins of the DAC is determined by R1, R2,
the voltage on V
MID
and the DAC output current (I
IN
+
or
I
IN
–
). Consider Figure 4 with R1 = 49.9 and R2 = 348.
The voltage at V
MID
, for V
S
= 3.3V, is 1.65V. The voltage at
the DAC pins is given by:
V
DAC
= V
MID
•
R1
R1+R2+ 402
+I
IN
•
R1• R2
R1+R2
= 103mV+I
IN
• 43.6Ω
I
IN
is I
IN
+
or I
IN
–
. The transimpedance in this example is
50.4.
Evaluating the LT6604-10
The low impedance levels and high frequency operation
of the LT6604-10 require some attention to the matching
networks between the LT6604-10 and other devices. The
previous examples assume an ideal (0) source impedance
and a large (1k) load resistance. Among practical examples
where impedance must be considered is the evaluation of
the LT6604-10 with a network analyzer.
–
+
100
100
0.1µF
0.01µF
5V
–
+
V
IN
–
V
IN
+
25
27
4
34
6
2
29
7
660410 F03
V
OUT
+
V
OUT
–
62pF
62pF
+
–
2V
V
t
3
2
1
0
V
IN
+
V
IN
–
V
t
3
2
1
0
V
OUT
+
V
OUT
–
500mV
P-P
(DIFF)
1/2
LT6604-10
–
+
0.1µF
0.01µF
3.3V
–
+
25
27
V
OUT
+
I
IN
+
I
IN
–
V
OUT
–
4
34
6
2
29
7
660410 F04
CURRENT
OUTPUT
DAC
R1
R1
R2
R2
=
V
OUT
+
– V
OUT
–
I
IN
+
– I
IN
–
402 • R1
R1 + R2
1/2
LT6604-10
Figure 3
Figure 4