AD8519 Analog Devices, AD8519 Datasheet - Page 12

• Manufacturer Part Number: AD8519
• Description: 8 MHz Rail-to-Rail Operational Amplifier
• Manufacturer: Analog Devices

Specifications of AD8519

• -3db Bandwidth: 8MHz
• Slew Rate: 2.9V/µs
• Vos: 600µV
• Ib: 300nA
• # Opamps Per Pkg: 1
• Input Noise (nv/rthz): 10nV/rtHz
• Vcc-vee: 2.7V to 12V
• Isy Per Amplifier: 1.2mA
• Packages: SC70,SOIC,SOT

AD8519/AD8529
APPLICATIONS INFORMATION
MAXIMUM POWER DISSIPATION
The maximum power that can be safely dissipated by the
AD8519/AD8529 is limited by the associated rise in junction
temperature. The maximum safe junction temperature is 150°C
for these plastic packages. If this maximum is momentarily
exceeded, proper circuit operation is restored as soon as the
die temperature is reduced. Operating the product in an
overheated condition for an extended period can result in
permanent damage to the device.
PRECISION FULL-WAVE RECTIFIER
Slew rate is probably the most underestimated parameter when
designing a precision rectifier. Yet without a good slew rate,
large glitches are generated during the period when both diodes
are off.
The operation of the basic circuit (shown in Figure 23) should
be examined before considering the slew rate further. U1 is set
up to have two states of operation. D1 and D2 diodes switch the
output between the two states. State one is an inverter with a
gain of +1, and state two is a simple unity gain buffer where the
output is equal to the value of the virtual ground. The virtual
ground is the potential present at the noninverting node of the
U1. State one is active when V
ground. D2 is on in this condition. If V
ground, D2 turns off and D1 turns on. This causes the output of
U1 to simply buffer the virtual ground and this configuration is
state two. Therefore, the function of U1, which results from
these two states of operation, is a half-wave inverter. The U2
function takes the inverted half wave at a gain of two and sums
it into the original V
This type of rectifier can be very precise if the following
electrical parameters are adhered to:
•
•
•
All passive components should be of tight tolerance, 1% for
resistors and 5% for capacitors.
If the application circuit requires high impedance (that is,
direct sensor interface), then an FET amplifier is a better
choice than the AD8519.
An amp such as the AD8519, which has a great slew rate
specification, yields the best result because the circuit
involves switching.
V
OUT
=
V
IN
−
2
V
IN
IN
wave, which outputs a rectified full wave.
−
1
<
0
IN
is larger than the virtual
IN
drops below virtual
Rev. D | Page 12 of 16
Switching glitches are caused when D1 and D2 are both
momentarily off. This condition occurs every time the input
signal is equal to the virtual ground potential. When this
condition occurs, the U1 stage is taken out of the V
and V
Node A should be V
condition, Node A is simply tracking V
input centered around virtual ground, glitches are generated
at the sharp negative peaks of the rectified sine wave. If the
glitches are hard to notice on an oscilloscope, raise the fre-
quency of the sine wave until they become apparent. The size
of the glitches is proportional to the input frequency, the diode
turn-on potential (0.2 V or 0.65 V), and the slew rate of the op amp.
R6 and R7 are both necessary to limit the amount of bias
current related voltage offset. Unfortunately, there is no perfect
value for R6 because the impedance at the inverting node is
altered as D1 and D2 switch. Therefore, there is also some
unresolved bias current related offset. To minimize this offset,
use lower value resistors or choose an FET amplifier if the
optimized offset is still intolerable.
The AD8519 offers a unique combination of speed vs. power
ratio at 2.7 V single supply, small size (SC70 and SOT-23), and low
noise that makes it an ideal choice for most high volume and
high precision rectifier circuits.
V
IN
OUT
10kΩ
is equal to V
R1
R6
5kΩ
Figure 23. Precision Full-Wave Rectifier
1N914
VIRTUAL GROUND =
D1
U1
IN
IN
AD8519
inverted or virtual ground, but in this
10kΩ
10kΩ
R4
R2
× R5 × (R4 || R1 + R2 + R3). Note that
NODE A
1N914
D2
V
4.99kΩ
CC
2
R3
IN
. Given a sine wave
R7
3.32kΩ
10kΩ
U2
R5
AD8519
OUT
equation
V
OUT