LOG112AIDRE4 Datasheet, PDF (12/20 Pages) Texas Instruments – Precision LOGARITHMIC AND LOG RATIO AMPLIFIERS

ERRORS RTO AND RTI
As with any transfer function, errors generated by the func-
tion may be Referred-to-Output (RTO) or Referred-to-Input
(RTI). In this respect, log amps have a unique property: given
some error voltage at the log amp’s output, that error corre-
sponds to a constant percent of the input regardless of the
actual input level.
LOG CONFORMITY
For the LOG112 and LOG2112, log conformity is calculated
the same as linearity and is plotted I₁/I₂ on a semi-log scale.
In many applications, log conformity is the most important
specification. This is true because bias current errors are
negligible (5pA compared to input currents of 100pA and
above) and the scale factor and offset errors may be trimmed
to zero or removed by system calibration. This leaves log
conformity as the major source of error.
Log conformity is defined as the peak deviation from the best
fit straight line of the V_LOGOUT versus log (I₁/I₂) curve. This is
expressed as a percent of ideal full-scale output. Thus, the
nonlinearity error expressed in volts over m decades is:
V_{LOGOUT (NONLIN)} = 0.5V/dec • 2NmV
(7)
where N is the log conformity error, in percent.
INDIVIDUAL ERROR COMPONENTS
The ideal transfer function with current input is:
V_LOGOUT
=
(0.5V)LOG


I₁
I₂


(8)
The actual transfer function with the major components of
error is:
V_LOGOUT
=
(0.5V)
(1±
∆K)

log_
I₁
I₂
–
–
I_B1
I_B2


±
Nm ±
V_OSO
(9)
The individual component of error is:
∆K = gain error (0.10%, typ), as specified in the specifica-
tion table.
I_B1 = bias current of A₁ (5pA, typ)
I_B2 = bias current of A₂ (5pA, typ)
N = log conformity error (0.01%, 0.13%, typ)
0.01% for m = 5, 0.13% for m = 7.5
V_OSO = output offset voltage (3mV, typ)
m = number of decades over which N is specified
For example, what is the error when:
I₁ = 1µA and I₂ = 100nA
(10)
(11)
V_LOGOUT
=
(0.5
±
0.001)

log_
10^−6
10^−7
−
−
5
5
• 10^−12
• 10^−12


±
(2)(0.0001)5
±
3.0mV
= 0.505V
Since the ideal output is 0.5V, the error as a percent of the
reading is:
% error = ^0.505V • 100% = 1.01%
0.5
(12)
For the case of voltage inputs, the actual transfer function is:
(13)
V_LOGOUT
=
(0.5V) (1±
∆K)log




V₁
R₁
V₂
R₂
– I_B1
– I_B2
± ^EOS1
R₁
± ^EOS2
R₂




± Nm ±
V_OSO
Where
E_{OS1 and} E_OS2
R₁
R₂
(offset error) are considered to be
zero for large values of resistance from external input current
sources.
12
LOG112, 2112
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