参数资料
型号: LOG101AID
英文描述: Precision LOGARITHMIC AND LOG RATIO AMPLIFIER
中文描述: 精密对数和对数比率放大器
文件页数: 12/13页
文件大小: 281K
代理商: LOG101AID
www.ti.com
LOG101
8
SBOS242B
A
2
A
1
I
1
Q
1
Q
2
V
OUT = (1V) LOG
I
1
I
2
I
2
I
1
I
2
++
––
R
2
V
OUT
V
L
R
1
V
BE1
V
BE2
also
VV
RR
R
V
RR
R
nV
I
OUT
L
OUT
T
=
+
=
+
12
1
12
1
2
log
VV
I
OUT = () log
1
2
Using the base-emitter voltage relationship of matched
bipolar transistors, the LOG101 establishes a logarith-
mic function of input current ratios. Beginning with the
base-emitter voltage defined as:
VV
I
where V
kT
q
BE
T
C
S
T
==
ln
:
k = Boltzman’s constant = 1.381 10–23
T = Absolute temperature in degrees Kelvin
q = Electron charge = 1.602 10–19 Coulombs
IC = Collector current
IS = Reverse saturation current
From the circuit in Figure 11, we see that:
VV
V
LBE
BE
=
12
Substituting (1) into (2) yields:
VV
I
V
I
LT
S
T
S
=
1
2
ln
ln
If the transistors are matched and isothermal and
VTI = VT2, then (3) becomes:
VV
I
VV
I
and
ce
xx
Vn V
I
LT
SS
LT
=
=
1
12
1
2
10
1
2
23
ln
– ln
ln
sin
ln
. log
log
where n = 2.3
INSIDE THE LOG101
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(9)
(10)
(11)
FIGURE 11. Simplified Model of a Log Amplifier.
(8)
or
It should be noted that the temperature dependance
associated with VT = kT/q is internally compensated on
the LOG101 by making R1 a temperature sensitive resis-
tor with the required positive temperature coefficient.
DEFINITION OF TERMS
TRANSFER FUNCTION
The ideal transfer function is:
VOUT = 1V log (I1/I2)
Figure 12 shows the graphical representation of the transfer
over valid operating range for the LOG101.
ACCURACY
Accuracy considerations for a log ratio amplifier are some-
what more complicated than for other amplifiers. This is
because the transfer function is nonlinear and has two
inputs, each of which can vary over a wide dynamic range.
The accuracy for any combination of inputs is determined
from the total error specification.
FIGURE 12. Transfer Function with Varying I2 and I1.
(5)
3.0
3.5
2.0
2.5
1.0
1.5
0.5
0.0
–3.0
–3.5
–2.0
–2.5
–1.0
–0.5
–1.5
1nA
10nA
100nA 1
A
10
A
100
A
1mA
10mA
100pA
V
OUT
(V)
I 2 =
100pA
I 2 =
1nA
I 2 =
10
nA
I 2 =
100nA
I 2 =
1
A
I 2 =
10
A
I 2 =
100
A
I 2 =
1mA
I
1
V
OUT = (1V) LOG (I1/I2)
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