National Instruments 6527 User Manual
Page 27
Chapter 3
Making Signal Connections
© National Instruments Corporation
3-11
Writing a 0 (logic low) to an output bit closes the relay, and writing a
1 (logic high) opens the relay.
To both sink and source current with one channel requires an external
resistor. You can use the solid-state relays of a 6527 device with external
resistors to drive voltages at TTL or non-TTL levels, from –60 to 60 VDC
or 30 VAC (42 V peak).
For isolated power, total current on all channels exceeding 1 A, or voltages
other than +5 V, you can provide an external power supply. For driving
non-isolated +5 V outputs totaling less than 1 A—for example, when using
the 6527 as a TTL-level output device—you can use the +5 V line from the
6527 device as your voltage source only when each of the following
conditions is true:
•
Non-isolated power
•
Total current is less than 1A
•
Voltage level needed is +5 V
If any of the above conditions is not met, use the appropriate external power
supply.
Using the +5 V line from the 6527 device allows you to use it as a
TTL-level output device with non-isolated power.
Figure 3-6 shows a signal connection example for both sinking and
sourcing current. The example shows a TTL-level application with a supply
voltage of +5 V. The 6527 provides sink current when the relay is closed.
Resistor R
L
provides source current when the relay is open.
When the relay is open, little current flows through the resistor and the
output voltage is close to 5 V, a logic high. When the relay is closed,
current flows through the load and the output voltage is close to 0 V,
a logic low. If isolation is not a concern, you can use the +5 V line
from the 6527 device in place of the external +5 V supply.
Choose a value of R
L
small enough to provide the source current you
need but large enough to avoid reducing sink current or consuming
unnecessary power. For many TTL-level applications, a value of
approximately R
L
= 5 k
Ω works well. To maintain 2.8 V at V
OUT
,
the source current is given by the following equation:
5V
2.8V
–
(
)
5 k
Ω
-----------------------------
440
µA
=