HA17301P
Quad Operational Amplifier
Description
The HA17301P is an internal-compensation quad operational amplifier that operates on a single-voltage
power supply. Typical applications for the HA17301P include waveform generators, voltage regulators,
logic circuits, and voltage-controlled oscillators.
Features
•
•
•
•
Wide operating temperature range
Single-voltage power supply operation
Internal phase compensation
Low input bias current
Pin Arrangement
Vin(+)2
Vin(+)1
Vin(−)1
Vout1
Vout2
Vin(−)2
GND
1
2
3
4
5
6
−
−
+
14 V
CC
13 Vin(+)3
1
+
−
12 Vin(+)4
11 Vin(−)4
10 Vout4
4
2
+
+
3
9 Vout3
−
7
8 Vin(−)3
(Top view)
HA17301P
HA17301P Application Examples
The HA17301P is a quad operational amplifier, and consists of four operational amplifier circuits and one
bias current circuit. The HA17301P features a wide operating temperature range, single-voltage power
supply operation, internal phase compensation, a wide zero-cross bandwidth, a low input bias current, and a
high open-loop gain. Thus the HA17301P can be used in a wide range of applications. This section
describes several applications using the HA17301P.
HA17301 Circuit Operation
V
CC
Q
5
C
1
3 pF
Inverting input
3
Non inverting
input
2
D
1
GND
Bias circuit
Q
3
Q
10
Op amp 1
Q
1
Q
2
Q
4
4
Output
Figure 1 HA17301 Internal Equivalent Circuit
Figure 1 shows the internal equivalent circuit for the HA17301P bias circuit and one operational amplifier
circuit (Op amp 1).
Op amp 1 is basically an emitter ground type operational amplifier in which the input transistor Q
1
, the
buffer transistor Q
4
, the current source transistor Q
5
, the output emitter-follower transistor Q
2
, and the
current source transistor Q
10
form an inverting amplifier. The voltage gain of this circuit is all given by the
transistor Q
1
, and the adoption of the current-supply load Q
5
allows this circuit to provide a large open-loop
gain even at low power-supply voltages. Next, the emitter-follower transistor Q
2
lowers the output
impedance of this circuit. The use of the power-supply transistor Q
10
as the load for Q
2
gives this circuit an
extremely large dynamic range, and essentially an amplitude from ground to (V
CC
– 1) can be acquired.
Also, the buffer transistor Q
4
is used to reduce the input current without increasing the DC input voltage
level. Since the capacitor C
1
is used to preserve stability when this inverting amplifier is used as a closed
circuit, no external compensation is required.
Now consider the non inverting circuit. Assuming that the current amplification ratio provided by Q
3
is
adequately large for the current flowing into the non inverting input, then all that current will flow through
diode D
1
and the voltage drop induced in the diode D
1
by this input current will be applied to the Q
3
base-
emitter junction. Therefore, if D
1
and Q
3
are matched, a current equal to the input current will flow in the
Q
3
emitter. Assuming that the current amplification ratio provided by Q
3
is adequately large, a current equal
to the input current will flow in the Q
3
collector. This is called a “current mirror”, and when an external
feedback resistor is used, a current equal to the non inverting input current will flow in this resistor and thus
determine the output voltage.
4
HA17301P
Inverting Amplifier
There are three bias techniques for biasing the inverting amplifier, the single power supply bias technique,
the NV
BE
bias technique, and the load voltage bias technique.
1. Single Power Supply Bias Technique
Figure 2 shows a common AC amplifier that is biased by the same power supply as the supply that
operates the amplifier.
R
2
Cin
0.1
µF
Vin
R
1
50 kΩ
I
R
3
= 2R
2
+
500 k
V
D
V
D
−
−
Vout
+
+
1 MΩ
V
+
Figure 2 Single Power Supply Bias Technique
R
Vout
=
−
2
Vin
R
1
(1)
2. NV
BE
Bias Technique
R
2
Cin
R
1
V
RE
82 kΩ
1 MΩ
−
Vout
+
I
−
0.1
µF
100 kΩ I
Vin
R
3
Figure 3 NV
BE
Bias Technique
This is the most useful application of an inverting AC amplifier. In this circuit, the input bias voltage
V
BE
for the inverting input is determined by the current that flows to ground through the resistor R
3
.
R
Vout
=
−
2
Vin
R
1
(2)
5
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