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參數(shù)資料
| 型號： | ICL8013 |
| 廠商： | Intersil Corporation |
| 英文描述： | 1MHz, Four Quadrant Analog Multiplier |
| 中文描述： | 1MHz的，四象限模擬乘法器 |
| 文件頁數(shù)： | 5/8頁 |
| 文件大?。?/td> | 132K |
| 代理商： | ICL8013 |
5
This circuit of Figure 5 still has the problem that the input
voltage V
IN
must be small to keep the differential amplifier in
the linear region. To be able to handle large signals, we need
an amplitude compression circuit.
Figure 2 showed a current source formed by relying on the
matching characteristics of a diode and the emitter base
junction of a transistor. Extension of this idea to a differential
circuit is shown in Figure 6A. In a differential pair, the input
voltage splits the biasing current in a logarithmic ratio. (The
usual assumption of linearity is useful only for small signals.)
Since the input to the differential pair in Figure 6A is the
difference in voltage across the two diodes, which in turn is
proportional to the log of the ratio of drive currents, it follows
that the ratio of diode currents and the ratio of collector
currents are linearly related and independent of amplitude. If
we combine this circuit with the voltage to current converter
of Figure 3, we have Figure 6B. The output of the differential
amplifier is now proportional to the input voltage over a large
dynamic range, thereby improving linearity while minimizing
drift and noise factors.
The complete schematic is shown after the Electrical
Specifications Table. The differential pair Q
3
and Q
4
form a
voltage to current converter whose output is compressed in
collector diodes Q
1
and Q
2
. These diodes drive the
balanced cross-coupled differential amplifier Q
7
/Q
8
Q
14
/Q
15
.
The gain of these amplifiers is modulated by the voltage to
current converter Q
9
and Q
10
. Transistors Q
5
, Q
6
, Q
11
, and
Q
12
are constant current sources which bias the voltage to
current converter. The output amplifier comprises transistors
Q
16
through Q
27
.
I
E
R
L
V
IN
-
V
OUT
= 0
V+
I
E
V-
1
/
2
I
E
+
+
1
/
2
I
E
-
R
L
Q
1
Q
2
Q
3
Q
4
1
/
2
I
E
-
1
/
2
I
E
+
I
E
I
E
FIGURE 4A. INPUT SIGNAL WITH BALANCED CURRENT
SOURCES
V
OUT
= 0V
I
E
R
L
V
IN
= 0
-
V
OUT
= 0
V+
2I
E
V-
1
/
2
I
E
+
1
/
2
I
E
R
L
Q
1
Q
2
Q
3
Q
4
I
E
I
E
FIGURE 4B. NO INPUT SIGNAL WITH UNBALANCED
CURRENT SOURCES
V
OUT
= 0V
I
E
R
L
V
IN
-
V
OUT
= 0
V+
2I
E
V-
1
/
2
I
E
+
+
1
/
2
I
E
-
R
L
Q
1
Q
2
Q
3
Q
4
1
/
2
I
E
- 2
I
E
+ 2
3
/
2
I +
3
/
2
I -
FIGURE 4C. INPUT SIGNAL WITH UNBALANCED CURRENT
SOURCES, DIFFERENTIAL OUTPUT VOLTAGE
I
E
R
L
V
IN
V = K (V
X
V
Y
)
V+
I
E
V-
+
-
R
Q
1
Q
2
Q
3
Q
4
R
E
V
IN
FIGURE 5. TYPICAL FOUR QUADRANT MULTIPLIER-
MODULATOR
X x I
D
X x I
E
(I - X) I
E
(I - X) I
D
2 I
E
FIGURE 6A. CURRENT GAIN CELL
ICL8013
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