12 Volt PIR to Timer 210705

[AllenPitts]

Hello ETO forum,

Pommie, KISS, and rjenkinsgb, the feedback on the post titled 12 Volt PIR to Timer
was greatly appreciated.
Link to previous post: 12 Volt PIR to Timer

Have learned a lot about the PIR/555 combination
All have developed a some questions.

Not sure if there is a protocol about asking more than one
question in a post. If there is, let me know and these two
questions will bifurcated.

Question 1.
Based on feedback, especially from Pommie, a 100 ohm
resistor was added between the PIR output and Q1.
This is the way it looked before the addition.

This is the schematic with R5 added:

Have been running this circuit in the closet using a prototype board with a two amp load for ten days
giving me enough confidence to order PCBs.

When experimenting with the circuit it was desired that an indicator LED be placed between the PIR
and Q1 to see the signal coming out from the motion detector. This would, for instance, show the timer
IC continue to light the load after the PIR signal had ceased. So a LED, D2, with R6 was added:

But even though the meter says that the PIR is putting out 3.3 volts and the 555 seems to be working because
D1 is lit, D2 does not come on. Is the electronic logic incorrect? What is a method for putting an indicator
light showing the signal from the HC SR505?

Question 2.
This experimenter is plagued by being stuck in trial and error mode. Would like to move away from the empirical
to the deductive. Somewhere short of getting BS in EE there must a way of getting more knowledge than Ohm's
law so reason could be used build electronic systems instead being a blind dog in a meat house.
Presently, a circuit is found on the 'net and breadboarded.
Then adaptations of the working circuit is made to the needed purpose by making a change, seeing if it works, and if it doesn't
go into a very long trial and error to get the adaption to work. When a working system is finally found it
is not known why or what was done to find success.

Specifically, Pommie suggested that a resistor be placed between the PIR and Q1 as shown in the schematic dated 210705,
above. The implication being that the BC547 would fail taking the signal from the PIR. So the datasheet for the BC547
was studied to see if by reading the datasheet it could have been deduced that the PIR signal would blow the BC547.

The data sheet is attached herewith. In the table marked 'Absolute Maximum Ratings' it seen that the BC547
will take a max of 50 volts so the 12 volts used in the system should be ok. But it can not be discovered what
that maximum current or amperage that the transistor will take. It is conjectured that it was the current
not the voltage that was making the BC547s fail. Where could that be seen in the datasheet?

Thanks.

Allen in Dallas

 

[Diver300]

The motion sensor output is 3 V when high, and has a 1 kOhm resistance in series. **broken link removed**
I don't really see the point of the 100 Ohm resistor when the 1 kOhm resistor is there.

The 1 kOhm will limit the output current, and the base-emitter junction of Q1 will stop the voltage on the base rising above about 0.7 V, and with the 100 Ohm resistor the voltage on the output of the motion sensor won't get above about 1 V, which is not enough to light the LED.

I suggest that you put the LED in series with R1. R6 can be omitted.

 

[Diver300]

On question 2, the base-emitter saturation voltage of Q1 is rated at 0.7 V when the current is 10 mA.

The motion sensor output is 3 V when high. The base-emitter junction is basically shorting out the output of the motion sensor, so there has to be a resistor in series. If not the current would be too large and could damage the IC inside the motion sensor, or Q1. It would be running the IC outside of its maximum rating, which is never a good idea.

The IC inside the motion sensor is a BISS0001 and the maximum output current is 10 mA. I would have suggested a resistor of 330 Ohms which would keep the current to less than 10 mA if the output were shorted to ground. The current will be a bit less with the 0.7 V from Q1.

However it seem likely that the motion sensor contains a 1 k resistor, which makes it a bit pointless to have another resistor externally.

It is normal to have a resistor in series with the transistor base on a circuit like that. Many of us on this forum have seen that sort of circuit a lot of times and would expect a resistor. All transistors have a base-emitter voltage around 0.7 V, and it becomes second nature to put resistors in series, the same as for LEDs and zener diodes.

A lot of electronics is trial and error. I destroyed an IC today, and tomorrow's job is working out why, and how to stop it. I've been working on electronics for nearly 50 years, and I still make mistakes.

https://www.electro-tech-online.com/threads/regulators-failing.91068/#post-717777 shows how easy I've found it to make mistakes.

https://www.electro-tech-online.com/threads/self-made-components.158857/#post-1376036 should always be remembered.

 

[KeepItSimpleStupid]

Use stuff like @Pommie so it shows up like Pommie . KeepItSimpleStupid and rjenkinsgb
You only have to type about 4 letters. that way we get notifications.

 

[KeepItSimpleStupid]

Diver300 From his other post, it doesn;t work without the 100 ohm resistor. It burns out Q1.

A LED in series with R1, probaby not.

If you put a diode, cathode to Q1 between R1 and Q1 nothing really changes,
Add another diode cathode to Q1

Add a resistor to a LED to +12 as if when you ground the other end, the LED comes on.
Attach this to the free end of the just added diode.

The diodes isolate the two circuits. Only difference is you need to do a (12- 0.7V-0.7-Vled)/ILED
There are two 0.7 drops to go through,

The extra 0.7V for the trigger won't matter.

 

[rjenkinsgb]

But even though the meter says that the PIR is putting out 3.3 volts and the 555 seems to be working because
D1 is lit, D2 does not come on. Is the electronic logic incorrect?

I doubt the PIR is outputting 3.3V when connected to the transistor.

And if that 100 Ohm stops things failing, then the PIR modules are not the same as the public schematic and do not have an internal 1K output resistor. They may be a clone copy or just a different version; there are options on the PCB for different components, but the look of it.

If the 1K resistor was present, that would be forming a voltage divider with the 100 Ohm and giving about 1/11th of the total voltage across the 100 Ohm.

Without a 1K, the current will be over 20mA and probably pulling down the voltage on whatever drives it, so it's too low to drive the LED as well.

1K should work better, but its a bad place for the LED, especially if the PIR modules have different output configurations at random..

Put the LED and its resistor across R1, so it is driven by the transistor rather than loading the PIR.

 

[rjenkinsgb]

Re. the BC547 failing: That would have been due to "unlimited" base current if the PIR is lacking a series resistor & the base was directly connected to an IC output pin.

The 10mA limit in the IC data sheet for the IC is it's maximum continuous design rating, not what it can give under fault conditions if held at 0.7V by a transistor base-emitter junction!

The BC547 datasheet does not explicitly give a maximum rating for the transistor base current.

The current is normally calculated by the circuit designer to be appropriate for the application; the highest value in the datasheet is 5mA, shown as sufficient for saturated switching with a 100mA load, the maximum the transistor is rated for.