lt6109-1 Linear Technology Corporation, lt6109-1 Datasheet - Page 19
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lt6109-1
Manufacturer Part Number
lt6109-1
Description
High Side Current Sense Amplifer With Reference And Comparators
Manufacturer
Linear Technology Corporation
Datasheet
1.LT6109-1.pdf
(28 pages)
APPLICATIONS INFORMATION
R3 should be chosen to allow sufficient V
tor output rise time due to capacitive loading.
R2 can be calculated:
For very large values of R2 PCB related leakage may
become an issue. A tee network can be implemented to
reduce the required resistor values.
The approximate total hysteresis will be:
For example, to achieve I
total hysteresis, R6 = 3.57k. Choosing R1 = 35.7k, R3 =
10k and V
The analog output voltage will also be affected when the
comparator trips due to the current injected into R6 by
the positive feedback. Because of this, it is desirable to
have (R1 + R2 + R3) >> R6. The maximum V
caused by this can be calculated as:
R2 =
V
∆V
HYS
OUTA
R1• V
= 10mV +R1•
+
= V
= 5V results in R2 = 4.12M.
(
+
+
•
– 400mV
R1+R2 +R3 +R6
V
HYS(EXTRA)
V
V
V
UNDER
+
+
+
)
R2 +R3
R6
R3
– V
– 400mV
(
R
SENSE
I
LOAD
HYS(EXTRA)
R
= 100µA with 50mV of
IN
Figure 14. Inverting Comparator with Added Hysteresis
10
1
4
SENSEHI
SENSELO
OUTC1
LT6109-1
OL
and compara-
•R3
OUTA
)
–
+
error
V
V
+
–
V
V
+
9
+
In the previous example, this is an error of 4.3mV at the
output of the amplifier or 43µV at the input of the amplifier
assuming a gain of 100.
When using the comparators with their inputs decoupled
from the output of the amplifier, they may be driven directly
by a voltage source. It is useful to know the threshold
voltage equations with the additional hysteresis. The input
falling edge threshold which causes the output to transition
from high to low is:
The input rising edge threshold which causes the output
to transition from low to high is:
Figure 14 shows how to add additional hysteresis to an
inverting comparator.
R7 can be calculated from the amplifier output current which
is required to cause the comparator output to trip, I
–
+
R7 =
R2
V
V
TH(F)
TH(R)
400mV
400mV
I
= 400mV •R1•
= 410mV •R1•
OVER
V
5
–
OUTA 8
INC1
, Assuming R1+R2
6
610912 F14
LT6109-1/LT6109-2
R1 VTH
R1
R1
1
1
R6
R7
+
(
+
R2 +R3
R2
1
1
)
>> R7
–
R2 +R3
V
+
•R1
19
OVER
610912f
.