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We are using AM26LS32AIDR in our design, as per the datasheet when input is open/floating, output should be high, but the output is in low state.

We are unable to trace out the issue.

We followed the schematic as below.

enter image description here

enter image description here

Edit:-

We are using this chip for controlling active low device. Default state should be high in order to disable the device.

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  • \$\begingroup\$ How are the enables connected? \$\endgroup\$
    – Justme
    Nov 18, 2021 at 7:48
  • \$\begingroup\$ @Justme enables are tied to GND(ground) \$\endgroup\$ Nov 18, 2021 at 7:51
  • \$\begingroup\$ What is the supply voltage? Is the input terminated with resistor, but with no connections, like in the picture? What is the resistor value? \$\endgroup\$
    – Justme
    Nov 18, 2021 at 8:06
  • \$\begingroup\$ Supply voltage is 5V, input is terminated with 100R \$\endgroup\$ Nov 18, 2021 at 9:16

2 Answers 2

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Since the input is terminated with 100 ohms resistance, the difference between input pins is in the order of 2.5 mV due to internal pull-up and pull-down resistors in the chip. Both inputs will have about 2.5V of common mode voltage.

The voltage is not high enough to cross the hysteresis threshold to neither positive or negative direction, so output will be whatever it previously was, or rather, undefined after powering the chip up.

Even the datasheet table 1 confirms this, if voltage is between high and low thresholds, output is indeterminate.

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  • \$\begingroup\$ Its working after removing 100R termination resistor \$\endgroup\$ Nov 18, 2021 at 10:55
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In your schematic, the 2/3/4A/B inputs are open, so the 2Y/3Y/4Y outputs are high. The 1A/1B inputs are shorted through the termination resistor, so the output state is undefined.

This device does not have a failsafe feature for an idle bus. If you want that, you need to add pull-up/-down resistors to the bus lines.

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