Simple Hall sensor alarm circuit?

[Alistair Ballantyne]

If I may be so bold... The gentleman is teaching you about reference designators..

View attachment 63017

Thanks for getting back to me but apologies, I have no idea what you mean.
Felt positive I made the right connections but I am obviously doing something wrong?

 

[Delta Prime]

I have no idea what you mean

I know kiddo!
Drain Source Gate the 3 Leeds of a mosfet transistor.
Collector base emitter the three leads of a BJT transistor. The designators are the first letter of each terminal add them to your schematic a notation alongside the lead, this way you can tell the orientation or which direction your transistors are connected in your circuit schematic... you forgot to include them when you submitted a schematic that you have created yourself, it is a lot easier for those reading your schematics to follow with designators. This way it leaves no question as to their connection.
Sorry about the spelling I'm on a stupid phone with a stupid operator me
It must be noted that I am in no way so pretentious as to speak for another member these were only my observations... And do not wish to offend fellow members.
Especially analog kid cuz he will rip me a new one.

 

[Alistair Ballantyne]

The original Darlington circuit actually worked very well.
I have followed your schematic for the MOSFET version but cant get it to work.
Power on and the buzzer is constantly on - the switch has no effect.
Have I got my pinout wrong?

Just for fun, here is the MOSFET-based circuit. We're running out of parts to cut.

Adjusting the input network impedance for a larger resistor reduces the static current draw through R1 when the door is closed. In this circuit it is only 9 microamps, pretty close to the battery's self-discharge current.

ak
View attachment 62959

Perfect - works extremely well, thanks!
I thought I had a problem with the MOSFET wiring initially as the buzzer would not turn off.
Surprised me as all pretty straightforward.
However, seems the actual switch is the problem.
A conventional glass type switch works well - the contact set (below) which I wanted to use is (a) temperamental or (b) just doesn't work.
Do you know if they are known for being temperamental?

 

[AnalogKid]

Usually, what's inside the switch module is a hermetically sealed reed switch. In that application, good for decades - depending on quality. In a security application, the switch is usually closed 99.99% of the time. If the reeds have even a tiny bit of ferromagnetic content they can become permanently magnetized, and not separate when the external magnet is removed.

ak

 

[Alistair Ballantyne]

Perfect - works extremely well, thanks!
I thought I had a problem with the MOSFET wiring initially as the buzzer would not turn off.
Surprised me as all pretty straightforward.
However, seems the actual switch is the problem.
A conventional glass type switch works well - the contact set (below) which I wanted to use is (a) temperamental or (b) just doesn't work.
Do you know if they are known for being temperamental?

View attachment 63023

 

[Alistair Ballantyne]

You kindly supplied me with this Reed switch circuit breaker alarm.
Works well with the bare Reed switch but not with the switch enclosed in a contact set.
I contacted the supplier who told me:
“It is a current handling issue. Reed switches are meant for very low current switching and obviously the bare switch can handle more than the contact set reed switch.”
“Try a 10k resistor in series with the reed switch, so between the junction of the 1M resistor, capacitor +ve and fet gate and the top of the reed switch. The other end of the reed switch remains grounded”.
This actually works. I presume it is a circuit divider?
I purchased my first multimeter and thought I would find a noticeable difference in current and voltage (resistor drop) readings at V1 and V2 - see attached revised schematic.
Could find no real difference so looked at the arithmetic involved:
V1 Current: 7.8v
1,000,000 = 7.8 micro amps
V2 Current 7.8v
1,010,000 = 7.7 micro amps
V1 Voltage 0.0000078 * 1,000,000 = 7.8v
V2 Voltage 0.0000078 * 10,000 = 0.078v
My question is this – is it possible for the addition of a 10k resistor to make such a negligible difference in voltage and current to lower the power sufficiently to make the contact set Reed switch work?

 

[AnalogKid]

My question is this – is it possible for the addition of a 10k resistor to make such a negligible difference in voltage and current to lower the power sufficiently to make the contact set Reed switch work?

Yes.

The 10 K resistor protects the reed switch contacts. When the switch is open, R1 charges up C1 to 9 V. This takes a several seconds because R1 is so large. When the switch closes, all of the charge in C1 is dumped through it in microseconds. Even though it is very brief, this can be a high current that affects the switch operation over time. With the added 10 K resistor (let's call it R2) in series with C1, that peak current is limited to less than 1 mA. That is way below the switch rating.

Note that because R1 is so large, it affects the meter reading. Your meter probably has an input resistance of 10 megohm. When measuring the voltage across the switch contacts, this 10 M resistor forms a voltage divider with R1, reducing the reading by 9%. If the meter input impedance is lower than 10 M, the error will be greater. What does the meter specification page say about DC input resistance or impedance?

Also, is your capacitor ceramic of electrolytic?

ak

 

[Alistair Ballantyne]

Yes.

The 10 K resistor protects the reed switch contacts. When the switch is open, R1 charges up C1 to 9 V. This takes a several seconds because R1 is so large. When the switch closes, all of the charge in C1 is dumped through it in microseconds. Even though it is very brief, this can be a high current that affects the switch operation over time. With the added 10 K resistor (let's call it R2) in series with C1, that peak current is limited to less than 1 mA. That is way below the switch rating.

Note that because R1 is so large, it affects the meter reading. Your meter probably has an input resistance of 10 megohm. When measuring the voltage across the switch contacts, this 10 M resistor forms a voltage divider with R1, reducing the reading by 9%. If the meter input impedance is lower than 10 M, the error will be greater. What does the meter specification page say about DC input resistance or impedance?

Also, is your capacitor ceramic of electrolytic?

ak

Got it - clever.
Electrolytic.
The spec sheet says 'input impedance about 10mΩ'
The left lead plug on the meter states 10 A max.
The right plug states VmaA µA.
My understanding if measuring current to start on the left plug and if under 10A then move lead to right plug.
That right?

 

[AnalogKid]

That is the safe way to start if you have no clues to the approximate current levels you are measuring. But in this case there are lotsa clues. Except for accidentally dead-shorting the batter, you are probably safe starting with the low range.

But . . .

At its heart, what you have is a volt meter. To measure amps or ohms, it converts those to volts and then measures that. For amps, the meter puts a small-value resistor between the + and - terminals, called a current shunt. The circuit current goes through this resistance, and produces a voltage across it per Ohm's Law. The datasheet *might* list the shunt values for the ranges.

The problem is that the shunt appears in series with whatever current you are measuring. For example, if the circuit resistance is 1 K and the shunt is 100 ohms, that is a 10% increase in the overall circuit resistance, introducing a 9% error in the reading. Making the shunt value smaller reduces the error, but then the meter has to measure a smaller voltage across it. This design tradeoff is in every DMM on the planet.

One of the immutable laws of physics is that measuring something changes the thing. All we can do is minimize the effects.

ak