Controlling the shutdown pin of an IR Transceiver
I am using an IR transceiver in my design.This IC is working at 3.3V supply.This IC need not be on always so I decided to give a control to it's shut down pin (Pin no 5,Shutdown Input. Active low).A transistor switch is used for this purpose.When the switch is ON the transceiver will be active and when it is off the transceiver shutdown pin will be connected to 3.3V. My circuit is given below.The value of R1 is 10K.
The base control signal is coming from a Microcontroller.The uC is also working at 3.3V
May I know this circuit is fine or not.
1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| newbie | (no comment) | Nov 22, 2024 at 09:51 |
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 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.
this circuit is fine or not
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.
in general how we are calculating the value of pull down resistor
See https://electrical.codidact.com/posts/276136.
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