National Instruments 6527

Chapter3 MakingSignal Connections

©NationalInstruments Corporation 3-11 6527User Manual

Writing a 0 (logic low)to an outpu tb it closes the relay,and writing a

1 (logic high) opens the relay.

Toboth sink and source current with one channel requires an external

resistor.You can use the solid-state relays of a 6527 device with external

resistorsto drive voltages at TTL or non-TTL levels, from –60 to 60 VDC

or3 0 VAC(42 V peak).

Forisolated 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

the6527 as a TTL-level output device—you can use the +5 V line from the

6527 deviceas your voltage source onl yw hen each of the following

conditions is true:

•Non-isolatedpower

•Totalcurrent is less than 1A

•Voltagelevel needed is +5 V

Ifany of the above conditions is not met, use the appropriate external power

supply.

Using the +5 V line from the 6527 deviceallows you to use it as a

TTL-leveloutput device with non-isolated power.

Figure 3-6 showsa signal connection example for both sin king and

sourcingcurrent. The example shows a TTL-level application with a supply

voltageof +5 V. The 6527 provides sinkcurrent when the relay is closed.

Resistor RLprovidessource current when the relay is open .

When the relay is open, little current flows through the resistor and the

outputvoltage is close to 5 V, a logic high. When the relay is closed,

current flowsthrough the load and th eo utput voltageis close to 0 V,

al ogic low.If isolation is not a concern, you can use the +5 V line

from the 6527 devicein place of the external +5 V supply.

Choose a valueof RLsmall enough to provide the source current you

need butlarge enough to avoid reducing sink current or consuming

unnecessarypower. For many TTL-level applications, a value of

approximately RL=5kΩworkswell. Tomaintain 2.8 V at VOUT

the source current is givenby the following equation:

5V 2.8V–

(

)

Ω

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