PRODUCT SPECIFICATION
RC4200
R
R
a
a
R
a
R1
5
8
7
VM
+V
S
R1
R
30K
a
R
a
10K
-VS
100
RO
µ
RC4200
Multiplier
0.1
F
R2
1
2
VY
=
ASIN ω
t
+VS
10K
R2
R
30K
a
4
µ
F
0.1
1/4
RC4156
100
3
6
VO
VM sin ω
=
-VS
t
-VS
+V
µ
0.1 F
R1
S
R*
R2
10K
R3
30K
30K
470
C1
-VS
1/2 R3
15K
R3
30K
R3
30K
p
2
R3 47K
1N914
p
2
t
= A
ω
VH
=
AVG A sin
1/4
RC4156
1/4
RC4156
*R1 R2
R
RO
a
1N914
65-1866
Figure 16. Amplitude Modulator with A.G.C.
Example 1
= Asinωt 2.5V ≤ A ≤ 10V, therefore N = 4
The maximum and minimum values for I and I lead to:
1
2
V
Y
VX(max.) VH(max.)
0V ≤ V ≤ 10V, therefore V (max.) = 10V
M
X
I1(max.) = ------------------------ + ------------------------ = 250µA
K = 1, therefore V = V sinωt
R1
Ra
0
M
VX(max.)
VH(min.)
10V
50µA
R1 = ----------------------- = ------------- = 2 0 0 K
I1(min.) = ----------------------- = 50µA VM(min.) = 0
50µA
Ra
A(max.)
50µA
10V
50µA
VH(max.)
A(max.)
R1 = ------------------- = ------------- = 2 0 0 K
I2(max.) = -------------------- + ------------------------ = 250µA
R2
Ra
A(min.)
2.5V
50µA
Ra = ------------------ = ------------- = 5 0 K
VH(min.)
50µA
I2(min.) = ----------------------- = 50µA
Ra
R1R2
200K × 200K
-------------
---------------------------------
= 800K
RO = K
= 1
Ra
50K
For a dynamic range of N, where
A(max.)
Example 2
= Asinωt 3 ≤ A ≤ 6, therefore N = 2
--------------------
N =
< 5,
A(min.)
V
Y
0V ≤ V ≤ 8V, therefore V (max.) = 8V
M
X
These equations combine to yield:
V (max.)
K = 0.2, therefore V = 0.2 V sinwt
0
M
so:
R = 53.3K, R = 40K
A(max.)
(5 – N)50µA
----------X----------------------
--------------------------------
,
R1
=
, R2
1
2
(5 – N)50µA
R = 60K and R = 7.11K
a
0
R1R2
A(min.)
------------------
-------------
,
Ra
=
and RO = K
50µA
Ra
REV. 1.2.1 6/14/01
17
FAIRCHILD [ FAIRCHILD SEMICONDUCTOR ]
