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Q&A Controlling the shutdown pin of an IR Transceiver

When the switch is ON the transceiver will be active No, it's the other way around. First, you should be able to see the bar over SHDN in the datasheet. That means negative logic, where low is a...

posted 1mo ago by Olin Lathrop‭ · edited 1mo ago by Olin Lathrop‭

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#2: Post edited by

Olin Lathrop‭ · 2024-11-22T11:46:28Z (about 1 month ago)

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<blockquote>When the switch is ON the transceiver will be active</blockquote>
No, it's the other way around.
First, you should be able to see the bar over SHDN in the datasheet. That means negative logic, where low is asserted.
Second, on page 5 of the datasheet the description for the SHDN pin is "Shutdown input. Active low". For the shutdown function to be active (the device shut down), the SHDN pin must be low.
Another way to think of it is that instead of a not-shutdown input, it's a enable input.
<blockquote>this circuit is fine or not</blockquote>
Since a microcontroller is driving the shutdown logic, the inversion mentioned above should be no issue. The firmware can produce either polarity. I'm not sure I'd say this circuit is "fine", but it should work.
However, what's the point? You have a 0 to 3.3 V logic signal you want to drive with a microcontroller that is also running on 0 to 3.3 V. The obvious answer is to connect an output of the micro directly to the shutdown input of the IC. If you care what happens at power up, put a 100 kΩ or so pulldown on the line. That way the IR IC stays off until the micro deliberately enables it.

<blockquote>When the switch is ON the transceiver will be active</blockquote>
No, it's the other way around.
First, you should be able to see the bar over SHDN in the datasheet. That means negative logic, where low is asserted.
Second, on page 5 of the datasheet the description for the SHDN pin is "Shutdown input. Active low". For the shutdown function to be active (the device shut down), the SHDN pin must be low.
Another way to think of it is that instead of a not-shutdown input, it's a enable input.
<blockquote>this circuit is fine or not</blockquote>
Since a microcontroller is driving the shutdown logic, the inversion mentioned above should be no issue. The firmware can produce either polarity. I'm not sure I'd say this circuit is "fine", but it should work.
However, what's the point? You have a 0 to 3.3 V logic signal you want to drive with a microcontroller that is also running on 0 to 3.3 V. The obvious answer is to connect an output of the micro directly to the shutdown input of the IC. If you care what happens at power up, put a 100 kΩ or so pulldown on the line. That way the IR IC stays off until the micro deliberately enables it.
<blockquote>in general how we are calculating the value of pull down resistor</blockquote>
See https://electrical.codidact.com/posts/276136.

#1: Initial revision by

Olin Lathrop‭ · 2024-11-21T19:03:42Z (about 1 month ago)

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<blockquote>When the switch is ON the transceiver will be active</blockquote>

No, it's the other way around.

First, you should be able to see the bar over SHDN in the datasheet.  That means negative logic, where low is asserted.

Second, on page 5 of the datasheet the description for the SHDN pin is <i>"Shutdown input.  Active low"</i>.  For the shutdown function to be active (the device shut down), the SHDN pin must be low.

Another way to think of it is that instead of a <i>not-shutdown</i> input, it's a <i>enable</i> input.

<blockquote>this circuit is fine or not</blockquote>

Since a microcontroller is driving the shutdown logic, the inversion mentioned above should be no issue.  The firmware can produce either polarity.  I'm not sure I'd say this circuit is "fine", but it should work.

<b>However</b>, what's the point?  You have a 0 to 3.3 V logic signal you want to drive with a microcontroller that is also running on 0 to 3.3 V.  The obvious answer is to connect an output of the micro directly to the shutdown input of the IC.  If you care what happens at power up, put a 100 k&Omega; or so pulldown on the line.  That way the IR IC stays off until the micro deliberately enables it.

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