What is the purpose of the 24V supressor D1 on the PIR Breakout board (see diagram below)? Why is 24V used, but not less (12V) or more (100V)? What happens if you remove the supressor D1?
To suppress transient voltage spikes. I’m not real sure but 24v could be chosen due to a lot of reasons…all depends on stock/supply/datasheet specs/pricing, etc. The operating voltage is 4-6v though, so 12v would probably be fine?
Removing it would make the circuit susceptible to any transient voltage spikes
Thanks for the reply. The EKMC460711xK PIR sensor allows a maximum voltage of 7V. Do I understand correctly that the D1 suppressor will not protect the PIR sensor from a transient voltage spikes of 7-23V? Then the circuit should use a 7V D1 suppressor or no?
And another related question. What does the text “VIN: 2.3V-4.0V” refer to (see below)? The working voltage of the PIR sensor is 3…6V. Does it mean that 2.3V power VCC
cannot be supplied?
I mean, it certainly could - are you having any actual issues?
The VIN listed is directly from the specs https://cdn.sparkfun.com/assets/3/f/8/8/1/4541_fileversion.pdf ; it is designed for 3.3v nominal
Well, but you gave a link to a pdf-file, that does not have EKMC4607112K sensors. I am sending you the datasheet page, which indicates a voltage Vdd (not VIN) of 3…6V by EKMC4607112K. Perhaps there is an error in the circuit and it should be EKMC1607112K? Or is there an error in the text and it should be Vdd=3.0V…6.0V?
Regarding the D1 suppressor. This is my curiosity. But is it possible that the purpose of the suppressor is to protect the human hand from electrostatics?
As far as I understand, there will be no correction in the PIR Breakout schematics. Since the circuit is distributed under the terms of Open Hardware, I will allow myself to give advice. It is better to replace the 24V suppressor with a 5.5V suppressor SMD (used to protect the VBus circuit in the USB path, for example, ESL100503 SEMITEH).
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