Datasheet
TC429
DS21416D-page 8 2002-2012 Microchip Technology Inc.
4.0 APPLICATIONS INFORMATION
4.1 Supply Bypassing
Charging and discharging large capacitive loads
quickly requires large currents. For example, charging
a 2500 pF load to 18V in 25 nsec requires a 1.8A
current from the device's power supply.
To ensure low supply impedance over a wide frequency
range, a parallel capacitor combination is recom-
mended for supply bypassing. Low-inductance ceramic
disk capacitors with short lead lengths (< 0.5 in.) should
be used. A 1 µF film capacitor in parallel with one or two
0.1 µF ceramic disk capacitors normally provides
adequate bypassing.
4.2 Grounding
The high-current capability of the TC429 demands
careful PC board layout for best performance. Since
the TC429 is an inverting driver, any ground lead
impedance will appear as negative feedback that can
degrade switching speed. The feedback is especially
noticeable with slow rise-time inputs, such as those
produced by an open-collector output with resistor pull-
up. The TC429 input structure includes about 300 mV
of hysteresis to ensure clean transitions and freedom
from oscillation, but attention to layout is still
recommended.
Figure 4-3 shows the feedback effect in detail. As the
TC429 input begins to go positive, the output goes
negative and several amperes of current flow in the
ground lead. A PC trace resistance of as little as 0.05
can produce hundreds of millivolts at the TC429 ground
pins. If the driving logic is referenced to power ground,
the effective logic input level is reduced and oscillations
may result.
To ensure optimum device performance, separate
ground traces should be provided for the logic and
power connections. Connecting logic ground directly to
the TC429 GND pins ensures full logic drive to the input
and fast output switching. Both GND pins should be
connected to power ground.
FIGURE 4-1: Inverting Driver Switching
Time Test Circuit.
FIGURE 4-2: Switching Speed.
C
L
= 2500 pF
0.1 µF
1µF
Input
V
DD
= 18V
Output
Input: 100 kHz,
square wave,
t
RISE
= t
FALL
10 nsec
4, 5
2
6, 7
1, 8
TC429
t
R
Output
Input
t
D1
t
F
t
D2
+5V
10%
90%
10%
90%
10%
90%
18V
0V
0V
TIME (100ns/DIV)
VOLTAGE (5V/DIV)
C
L
= 2500pF
V
S
= 18V
5V
INPUT
OUTPUT
100ns
TIME (100ns/DIV)
VOLTAGE (5V/DIV)
C
L
= 2500pF
V
S
= 7V
5V
INPUT
OUTPUT
100ns