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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： | 6/8页 |
| 文件大小： | 107K |
| 代理商： | 4127 |
6
4127
FIGURE 3. Transfer Function When I
1
is Negative.
ADJUSTMENT PROCEDURE
1. Refer to Choosing the Optimum Scale Factor and Refer-
ence 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 E
1 is a
positive input voltage and I
R is the resistor-programmed
internal reference current (see Figure 4).
4. Apply E
1 = EMIN, adjust R3 for the proper output.
5. Repeat steps 2 through 4 if necessary.
Transfer function is E
O = –A log
, where E
1 is a
negative input voltage and I
R is the resistor-programmed
internal reference current (see Figure 5).
R
4
(1)
10k
R
3
(1)
–15VDC
+15VDC
7
23
22
14
21
4
2
51
4127
19
18
I
1
Reference
Current
+15VDC
–15VDC
Gain
R
2
E
O
NOTE: (1) Needed only
if I
1< 10nA.
E
1
R
4 IR
FIGURE 4. Transfer Function When E
1 is Positive.
R
3
10k
100M
–15VDC
+15VDC
4
23
22
14
21
4127
19
18
Reference
Current
+15VDC
–15VDC
Gain
R
2
E
O
E
1
10k
±1%
R
1
2
ADJUSTMENT PROCEDURE
1. Refer to Choosing the Optimum Scale Factor and
Reference Current.
2. Apply E
1 = IR (10k), adjust R1 such that EO = 0.
3. Apply E
1 = EMAX, adjust R2 for the proper output voltage.
ADJUSTMENT PROCEDURE
1. Refer to Choosing the Optimum Scale Factor and
Reference Current.
2. Apply |E
1| = IR (10k), adjust R1 such that EO = 0.
3. Apply |E
1| = EMAX, adjust R2 for the proper output voltage.
4. Apply |E
1| = EMIN, adjust R3 for the proper output.
5. Repeat steps 2 through 4 if necessary.
Transfer function is E
O
= –A log
with I
1 and
I
2
negative; |I
1| ≥ 1nA, |I2| ≥ 1A (see Figure 6).
FIGURE 5. Transfer Function When E
1 is Negative.
R
3
10k
100M
–15VDC
+15VDC
7
23
22
14
21
4
2
5
1
4127
19
18
Reference
Current
+15VDC
–15VDC
Gain
R
2
E
O
E
1
R
4
10k
±1%
R
1
|E
1|
R
4 IR
FIGURE 6. Transfer Function When I
1 and I2 are Negative.
R
4
(1)
10k
R
3
(1)
–15VDC
+15VDC
5
7
22
14
21
10
2
11
9
4127
19
18
I
1
+15VDC
–15VDC
Gain
E
O
I
2
4
R
5
(2)
R
6
(2)
NOTES:(1) Needed only
if |I
1| < 10nA.
(2) R
5 = R6 ±1%
|I
1|
|I
2|
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