HA5022EVAL INTERSIL [Intersil Corporation], HA5022EVAL Datasheet - Page 5

• Manufacturer Part Number: HA5022EVAL
• Description: Dual, 420MHz, Low Power, Video, Current Feedback Operational Amplifier with Disable
• Manufacturer: INTERSIL [Intersil Corporation]

Application Information
Relevant Application Notes
The following Application Notes pertain to the HFA1245:
These publications may be obtained from Intersil’s web site
(http://www.intersil.com) or via our AnswerFAX system.
Optimum Feedback Resistor
Although a current feedback amplifier’s bandwidth
dependency on closed loop gain isn’t as severe as that of a
voltage feedback amplifier, there can be an appreciable
decrease in bandwidth at higher gains. This decrease may
be minimized by taking advantage of the current feedback
amplifier’s unique relationship between bandwidth and R
All current feedback amplifiers require a feedback resistor,
even for unity gain applications, and R
the internal compensation capacitor, sets the dominant pole
of the frequency response. Thus, the amplifier’s bandwidth is
inversely proportional to R
optimized for a 750 R
decreases stability, resulting in excessive peaking and
overshoot (Note: Capacitive feedback will cause the same
problems due to the feedback impedance decrease at higher
frequencies). At higher gains the amplifier is more stable, so
R
The table below lists recommended R
gains, and the expected bandwidth. For good channel-to-
channel gain matching, it is recommended that all resistors
(termination as well as gain setting) be 1% tolerance or
better. Note that a series input resistor, on +IN, is required for
a gain of +1, to reduce gain peaking and increase stability.
Channel-To-Channel Frequency Response Matching
The frequency response of channel 1 and channel 2 aren’t
perfectly matched. For the best channel-to-channel
frequency response match in a gain of 2 (see Figure 1), use
R
F
F
• AN9787-An Intuitive Approach to Understanding
• AN9420-Current Feedback Amplifier Theory and
• AN9663-Converting from Voltage Feedback to Current
GAIN
can be decreased in a trade-off of stability for bandwidth.
= 650 for channel 1 and R
(A
+10
+1
+2
+5
-1
Current Feedback Amplifiers
Applications
Feedback Amplifiers
V
)
TABLE 1. OPTIMUM FEEDBACK RESISTOR
560 (+R S = 650 )
R
475
750
200
180
F
F
( )
at a gain of +2. Decreasing R
F
. The HFA1245 design is
5
F
= 806 for channel 2.
F
F
values for various
, in conjunction with
BANDWIDTH
(MHz)
280
260
420
270
140
F
F
.
HFA1245
Non-inverting Input Source Impedance
For best operation, the DC source impedance seen by the
non-inverting input should be 50 . This is especially
important in inverting gain configurations where the
non-inverting input would normally be connected directly to
GND.
Pulse Undershoot and Asymmetrical Slew Rates
The HFA1245 utilizes a quasi-complementary output stage
to achieve high output current while minimizing quiescent
supply current. In this approach, a composite device
replaces the traditional PNP pulldown transistor. The
composite device switches modes after crossing 0V,
resulting in added distortion for signals swinging below
ground, and an increased undershoot on the negative
portion of the output waveform (see Figures 7, 11, 15, and
19). This undershoot isn’t present for small bipolar signals,
or large positive signals. Another artifact of the composite
device is asymmetrical slew rates for output signals with a
negative voltage component. The slew rate degrades as the
output signal crosses through 0V (see Figures 7, 11, 15, and
19), resulting in a slower overall negative slew rate. Positive
only signals have symmetrical slew rates as illustrated in the
large signal positive pulse response graphs (see Figures 5,
9, 13, and 17).
DISABLE Input TTL Compatibility
The HFA1245 derives an internal GND reference for the
digital circuitry as long as the power supplies are symmetrical
about GND. With symmetrical supplies the digital switching
threshold (V
ensures the TTL compatibility of the DISABLE input. If
asymmetrical supplies (e.g., +10V, 0V) are utilized, the
switching threshold becomes:
and the V
V
TH
FIGURE 1. CHANNEL 1 AND CHANNEL 2 MATCHED
-1
-2
-3
-4
2
1
0
1
=
A
V+
-------------------
V
= +2
IH
2
+
V-
TH
and V
FREQUENCY RESPONSE
+
= (V
1.4V,
IL
IH
levels will be V
10
+ V
FREQUENCY (MHz)
IL
)/2 = (2.0 + 0.8)/2) is 1.4V, which
R
F
= 650 , CH1
TH
100
0.6V, respectively.
R
F
= 806 , CH2
1000