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DATA SHEET
KH561
8
REV. 1A February 2001
Figure 4:Voltage Gain Derivation
Note again that if R
i
= 0 this expression would simplify
considerably. Also, if G were very large the voltage gain
expression would reduce to the familiar non-inverting op
amp gain equation. These two performance equations,
shown below, provide a means to derive the design equa-
tions for R
f
and R
g
given a desired no load gain and out-
put impedance.
Performance Equations
Design Equations
Equivalent Model
Given that the physical feedback and gain setting
resistors have been determined in accordance with the
design equations shown above, an equivalent model may
be created for the gain to the load where the
amplifier block is taken as a standard op amp. Figure 5
shows this analysis model and the resulting gain
equation to the load.
Figure 5: Equivalent Model
This model is used to generate the DC error and noise
performance equations. As with any equivalent model,
the primary intent is to match the external terminal
characteristics recognizing that the model distorts the
internal currents and voltages. In this case, the model
would incorrectly predict the output pin voltage swing for
a given swing at the load. But it does provide a simplified
means of getting to the external terminal characteristics.
External Compensation Capacitor (C
x
)
As shown in the test circuit of Figure 1, the KH561 requires
an external compensation capacitor from the output to
pin 19. The recommended values described here assume
that a maximally flat frequency response into a matched
load is desired.
The required C
x
varies widely with
the desired value of output impedance and to a lesser
degree on the desired gain.
simplified internal schematic, that the actual total
compensation (C
t
) is the series combination of C
x
and
the internal 10pF from pin 19 to the compensation nodes.
The total compensation (C
t
) is developed in two steps as
shown below.
Note from Figure 2, the
recognize that [takingV positive]
V
V
Gi
R
solving for V from two directions
V
V
i
R
G 1 i
R
solving for i
from this
i
V
G 1 R
R
then
V
V
V R
G 1 R
R
and,substitutingfor V and i
in the originalV expression
V
V 1
GR
R
o
err
f
i
err
i
err
g
err
err
i
g
i
i
i
g
i
err
o
f
=
+
=
=
(
=
(
+
=
(
+
=
+
i
+
g
i
f
g
v
o
i
f
g
i
f
R
R
i
g
v
f
g
i
G 1 R
R
pullinganR
R
out of the fraction
A
V
V
1
R
R
G
R
R
G 1
note that A
1
R
R
G
G 1
R
0
(
≡
= +
= +
=
R
R
R
R
R
G
R
R
A
R
R
G
R
R
G
R
R
o
f
i
f
g
i
g
v
f
g
i
f
i
g
=
+
+
+ +
1
= +
1
+ +
1
1
R
G
R
A
R
A R
R
R
f
o
i
g
f
o
1
v
=
+
(
)
=
1
R
g
V
i
R
L
V
o
R
o
R
f
- R
o
Classical
op-amp
-
+
V
V
1
R
R
R
R
R
R
substitutingin forR andR with their design
equationyields
V
V
A
R
R
R
A (gain toload)
o
i
f
o
g
L
L
o
o
i
v
L
L
o
=
+
+
=
+
=
C
300
R
1
2.0
R
pFintermediateequation
C
C
1
0.02 C
pF totalcompensation
1
o
g
t
1
1
=
=
+
(