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参数资料
| 型号： | 4127 |
| 厂商： | Texas Instruments, Inc. |
| 元件分类： | 对数放大器 |
| 英文描述： | Logarithmic Amplifier(对数放大器) |
| 中文描述： | 对数放大器（对数放大器） |
| 文件页数： | 5/8页 |
| 文件大小： | 107K |
| 代理商： | 4127 |
4127
5
I
R
I
MAX
E
O = –A log10
,
where A
≈
(26mV)
, R
T ≈ 520
The external resistor R
1 sets the reference current IR and
resistor R
2 sets the scale-factor “A”. R1 and R2 must be
trimmed to the desired values, but the approximate relation-
ships are shown in Typical Performance Curves.
The relationship between the input current, I
S, and the output
voltage, E
O, in terms of the externally adjusted parameters,
I
R and “A”, is illustrated in Typical Performance Curves.
This relationship is, of course, restricted to values of I
S
between 1nA and 1mA and output voltages of less than
±10V.
CHOOSING THE OPTIMUM
SCALE FACTOR AND REFERENCE CURRENT
To minimize the effects of output offset and noise, it is
usually best to use the full
±10V output range. Once an
output range of
±10V has been chosen, then “A” and I
R can
be determined from the Min/Max of the input current, I
S.
E
O = – A log
, where I
MIN < IS < IMAX
The output range of
±10V for an input range of I
MIN to IMAX
means that:
+10 = –A log
and –10 = – A log
Adding these two equations together
log
= 0, or I
R =
I
MAX IMIN
The value for A can be found from:
10 = A log
In terms of the input current range for I
S, the values for IR and
A that will provide a full
±10V output swing are:
I
R =
I
MAX IMIN and
A =
EXAMPLE
Assume that I
MIN is +10nA and IMAX is +100A.
This is an 80dB range.
I
R =
I
MAX IMIN =
(10–4) (10–8) = 10–6, or 1
A.
=
= 100
log
= 2; So, A = 5
For an I
R of 1A and A of 5,
E
O = –5log
CONNECTION DIAGRAMS
Transfer function is E
O = –A log
where I
1 is a positive
input current and I
R
is the resistor-programmed internal
reference current (see Figure 2).
R
T + R2
R
T
0.434
1
10–6
10–4
I
R
I
MAX
1
A
I
S
I
R
I
1
FIGURE 2. Transfer Function When I
1 is Positive.
ADJUSTMENT PROCEDURE
1. Refer to Choosing the Optimum Scale Factor and
Reference Current.
2. Apply |I
1| = IR, adjust R1 such that EO = 0.
3. Apply |I
1| = IMAX, adjust R2 for the proper output voltage.
4. Repeat steps 2 and 3 if necessary.
5. Ignore this step if |I
1MIN| ≥ 10nA. Otherwise, apply |I1| =
1nA, make R
3 = 1kM and adjust R4 for the proper
output voltage. For R
3, a single resistor is recommended.
A voltage divider network is difficult to use due to
amplifier offset voltage.
Transfer function is E
O = –A log
where I
1 is a negative
input current and I
R
is the resistor-programmed internal
reference current (see Figure 3).
I
MIN
I
R
I
MAX
I
R
I
MAX + IMIN
I
R
2
I
MAX IMIN
I
MAX
10
I
R
I
MAX
log
R
4
(1)
10k
R
3
(1)
–15VDC
+15VDC
4
23
22
14
21
1
2
4127
19
18
I
1
Reference
Current
+15VDC
–15VDC
Gain
R
2
E
O
NOTE: (1) Needed only
if I
1< 10nA.
I
R
I
S
I
R
I
S
I
R
|I
1|
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