Datasheet

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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
BE
1
V
BE
2
also
VV
RR
R
V
RR
R
nV
I
I
OUT
L
OUT
T
=
+
=
+
12
1
12
1
1
2
log
VV
I
I
OUT
= ()• log1
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
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
I
C
= Collector current
I
S
= Reverse saturation current
From the circuit in Figure 11, we see that:
VV V
LBE BE
=
12
–
Substituting (1) into (2) yields:
VV
I
I
V
I
I
LT
S
T
S
=
1
1
1
2
2
2
ln – ln
If the transistors are matched and isothermal and
V
TI
= V
T2
, then (3) becomes:
VV
I
I
I
I
VV
I
I
and ce
xx
VnV
I
I
LT
SS
LT
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 V
T
= kT/q is internally compensated on
the LOG101 by making R
1
a temperature sensitive resis-
tor with the required positive temperature coefficient.
DEFINITION OF TERMS
TRANSFER FUNCTION
The ideal transfer function is:
V
OUT
= 1V • log (I
1
/I
2
)
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 I
2
and I
1
.
(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
1µA
10µA
100µA
1mA
10mA
100pA
V
OUT
(V)
I
2
= 100pA
I
2
= 1nA
I
2
=
1
0
n
A
I
2
= 100nA
I
2
= 1µA
I
2
= 10µA
I
2
= 100µA
I
2
= 1mA
I
1
V
OUT
= (1V) • LOG (I
1
/I
2
)