PA162
Product Innova tionFrom
QUIESCENT CURRENT
BIAS CURRENT
75
20
120
16
80
65
12
40
VOS
3
2
8
0
55
1
4
-40
0
-80
2
4
6
8
10
AVERAGE QUIESCENT CURRENT, IQ (mA)
PHASE MARGIN vs. OUTPUT LOAD CAPACITANCE
65
45
-40 0 40 80 120
CASE TEMPERATURE, TC (°C)
VOLTAGE GAIN & PHASE vs. FREQUENCY
60
90
55
40
100
45
20
110
35
0
120
25
00 04 08 12 16 20
OUTPUT LOAD CAPACITANCE, CL (nF)
PULSE RESPONSE
10
8
6
+VS = +15V
4
AV = +1
VIN = 10Vp
2
0
-2
-4
-VS = -15V
-6
RL = 20Ω
fIN = 20kHz
-8
-10
0
10 20 30 40 50 60 70
TIME, t (s)
-20
1
10
100
1K
FREQUENCY, ƒ (KHz)
130
10K
PULSE RESPONSE
10
8
6
+VS = +15V
4
AV = +1
VIN = 10Vp
2
0
-2
-4
-6
-8
-10
0
-VS = -15V
fIN = 1kHz
200 400 600 800 1000 1200 1400
TIME, t (s)
0
-40 0 40 80 120
CASE TEMPERATURE, TC (°C)
OUTPUT VOLTAGE SWING
2
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
OUTPUT CURRENT, IO (A)
TYPICAL APPLICATION
R1 and R2 set up Amplifier A as non-inverting. Ampli-
fier B is set up as a unity gain inverter driven from the
output of Amplifier A. Note that Amplifier B inverts the
signals about the reference node, which is set at mid-
supply by R5 and R6. When the command input is
midrange, so is the output of Amplifier A. Since this is
also equivalent to the reference node voltage, the out-
put of Amplifier B is the same resulting in 0V across
the motor. Inputs more positive than 5V result in motor
current flow from left to right (see Figure 3). Inputs less
than 5V drive the motor in the opposite direction.
R2
R1
_
A
1/4 PA162
M
INPUT
0-10V
+
R3
R4
_
B
1/4 PA162
+
+28V
R5
R6
FIGURE 3. Bi-directional speed control
from a single supply.
4
PA162U