LMH6555EVAL National Semiconductor, LMH6555EVAL Datasheet - Page 16

• Manufacturer Part Number: LMH6555EVAL
• Manufacturer: National Semiconductor

Specifications of LMH6555EVAL

• Lead Free Status / Rohs Status: Not Compliant

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For applications similar to the one shown in Figure 12, the
following conditions should be maintained:
1.
2.
3.
Consider a numerical example (R
refers to R
Assume:
V
(I
erational U1 collector voltage = 6V.
Here are the series of steps to take in order to carry out this
design:
FIGURE 13. Norton Equivalent of the Input Circuitry Tied
a. Select the R
b. Find the value of R
PP
CC
) = 15 mA
The LMH6555 differential output voltage has to comply
with the ADC full scale voltage (800 mV
The LMH6555 input Common Mode Voltage Range is
observed. “CMVR”, as specified in the Electrical
Characteristics table, is to be between −0.3V and 2.0V
for the specified CMRR.
U1 collector voltage swing must to be observed so that
the U1 output transistors do not saturate. The expected
operating range of these output transistors is defined by
the specifications and operating conditions of U1.
collector voltage (6V in this case) with 1V extra as
margin because of LMH6555 loading.
R
Choose 169Ω, 1% resistors for R
(800 mV
Norton equivalent as shown in Figure 13.
I
I
I
The entirety of the Norton source differential component
will flow through the feedback resistors within the
LMH6555 and generate an output. Therefore:
N
N
N
L
=
= I
(common mode) = (V
(differential) = I
= [10 - (6+1)] V / (10+ 7.5) mA = 171Ω
to Q1 within the LMH6555 in Figure 12
N
S1
10V,
(common mode) + I
& R
PP
PP
). To do so, convert the input stage into its
with 10 mA quiescent (I
L
S2
value which allows compliance with the U1
).
U1
PP
S
to get the proper swing at the output
* R
peak-to-peak
L
CC
/ (R
N
– I
(differential)
L
cQ
L
+ R
refers to R
* R
L
S
cQ
L
+ R
) / (R
), and minimum op-
collector
G
PP
)
L
in this case).
+ R
L1
& R
S
+ R
current
L2
20127748
, R
G
)
S
16
Figure 14 shows the complete solution using the values de-
rived above, with the node voltages marked on the schematic
for reference.
c.
I
→
R
resistances).
So, in this case:
R
Choose 1.15 kΩ, 1% resistors for R
With R
voltage(s) minimum is not violated due to the loading
effect of the LMH6555 through R
to ensure that the LMH6555's CMVR is also not violated.
The “V
13) would need to be calculated. Use the Common
Mode component of the Norton equivalent source from
above, and write the KCL at the V
V
R
V
+ R
→
With V
and low) and the low end of the U1 collector voltage
(V
If necessary, steps “a” through “c” would have to be
repeated to readjust these values.
V
I
values derived)
→
= 8.44V
→
= 6.22V
V
→
= 0.721V
→
= 0.649V
N
N
x
x
C
IN
G
S
E
_High = 7.05 mA, I
V
V
V
V
V
C
(differential) * R
R
/ R
/ R
= (169 * 15 mA
= 25Ω.
x
) can be derived to be within the acceptable range.
= V
C
C
IN
IN
= 39Ω (R
G
= V
S
_High = 0.4595 * 169 / 1358 + 7.05 mA (1150 + 39)
_Low = 0.4595 * 169 / 1358 + 5.19 mA (1150 + 39)
)
= 0.4595V
_High = 0.4595 * (1358- 39) / 1358 + 7.05 mA * 39
_Low = 0.4595 * (1358- 39) / 1358 + 5.19 mA * 39
= (R
E
E
x
L
x
X
X
” node voltage within the LMH6555 (see Figure
+ V
+ V
calculated, both the input voltage range (high
and R
R
(R
L
L
x
x
* I
N
/ R
/ R
/ R
F
– R
PP
and R
S
N
N
N
* R
defined, ensure that the U1 collector
+ I
G
= 12.6 mA + I
= 12.6 mA + (V
) / R
F
PP
N
F
= 800 mV
/ 0.8) – R
G
N
(R
* 430/ 0.8) – 39 – 169 = 1154Ω
_Low = 5.19 mA (based on the
are internal LMH6555
N
S
+ I
+ R
N
G
R
PP
G
)
G
N
– R
CC
(common mode); with
S
x
. Also, it is important
L
node as follows:
– I
S
where R
.
cQ
R
L
)/ (R
F
= 430Ω,
L
+ R
S