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Topic: Problem with/purpose of diode and voltage regulator on PIR HC-SR501 (Read 104 times) previous topic - next topic

terraduino

Hi,

I've some HC-SR501 PIR sensors, see http://mpja.com/download/31227sc.pdf for schematic and facts.
They all work fine when powering them as usual using 5V. They even work fine when powering them with 3.3.V or less directly through "H: Repeat trigger pin" (see schematics, JP1, pin #3 right below BISS0001). The latter only works if no other consumer load is attached and is not drawing power, respectively. See stability problems I reported here. Now I aimed at building a circuit to overcome the stability issue.

First, however, I'm trying to understand why the PIR does not work at all, i.e. data pin arbitrarily changing from LOW to HIGH and back, when powered through VCC pin (see schematic, right most pin #3) using 3.3V or less. First, I thought the diode drops the voltage too much because data sheet says "input voltage 5V-..."). I measured it and diode D1 droped the voltage about 0.4V. When applying 3.3V that's still more than the 2.8V--and I have successfully powered the PIR through "H: Repeat trigger pin" even with 2.6V.

What is the difference in applying, for example, 2.8V directly to "H: Repeat trigger pin" and getting 2.8V VOUT pin of the HT7133? Both are connected to the same rail, right? First works, second does not.

Thanks a lot & best
--
Just a curious hobbyist. Only basic knowledge about electronics, microchips and the like.
Non-native speaker, please bear with me.

androidfanboy

I believe that it's because the 3.3V voltage regulator (shown at the top right of the schematic) has a certain dropout voltage, and if you supply 3.3V or less, it won't really operate properly. The schematic doesn't give a definite part number, but for example, if the dropout voltage of a 3.3V regulator is 0.4V at 65mA (the current consumption of the HC-SR501), that means it won't operate properly unless the input voltage at VCC is 3.3V + 0.4V = 3.7V. Anything below that is fingers crossed (as in, the output voltage will dip below the rated 3.3V).
"Genius: one percent inspiration and 99 percent perspiration." - Thomas Edison

terraduino

Thanks for the reply. :)
The voltage regulator has a drop out of about 0.1V@3.48V like the datasheet claims. Also, when less than 3.3V are applied the HT7133 just seems to pass through, i.e. I applied 2.8V on VIN of the regulator and I measured 2.8V on VOUT, hence, no dropout.

I assume it has to do with the load and the behavior of the HT7133. Too bad I cannot figure it out.

Btw. any idea why the diode in front of the HT7133?

Best
--
Just a curious hobbyist. Only basic knowledge about electronics, microchips and the like.
Non-native speaker, please bear with me.

androidfanboy

The diode is for reverse-polarity protection so that if you apply VIN to GND and GND to VIN, the diode won't conduct.

Now that you mention it, this could've been what's causing your problem, because since it's not a Schottky diode with low forward voltage, normal diodes typically have a larger Vf, around 0.7V, which would drop your input by that amount. However, you said that you supplied 2.8V to VIN of the regulator and got 2.8V out, so I'm not sure what's happening there.
"Genius: one percent inspiration and 99 percent perspiration." - Thomas Edison

terraduino

Reverse-polarity protection, simple as that, ha :)

Thanks again & best
--
Just a curious hobbyist. Only basic knowledge about electronics, microchips and the like.
Non-native speaker, please bear with me.

MarkT

If the voltage regulator isn't regulating, there will be a lot more noise potentially.  This kind of circuit is
very sensitive to noise as PIR sensors produce only tiny signals measured in microvolts.  You may be
injecting noise direct onto the circuit board with your test probes even.

Operating a chip below its minimum design supply voltage and finding it doesn't work is unsurprising,
trying to make it work outside its limits is probably a way to waste much time and energy.  And you
are likely to encounter various different behaviours in different chips due to manufacturing spread.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

terraduino

@MarkT
You're probably right. And I did not think about the additional noise when the ht7133 passes through instead of regulating.

Still, I'm curious how and why things work or do not work as I thought they should.
Can you see an explanation why 5V is the minimum supply voltage? IMO the diode and regulator have a combined drop out of about 0.5V.

Thanks & best
--
Just a curious hobbyist. Only basic knowledge about electronics, microchips and the like.
Non-native speaker, please bear with me.

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