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Buzzer Alarm at Mains Voltage Cutoff

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Kerim

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Since about a decade ago our houses (including mine) had no more one continuous AC supply (220V, 50Hz). Most people in my city ended up having 4 sources of electricity. Lately, these sources are typically distributed daily as follows: a couple of hours from mains, 8 hours of the neighborhood’s generator, about 4 hours from the in-house DC/AC inverter and, when real necessary, about an hour from an in-house small generator.

In general, and due to unexpected faults, the AC supply of my laptop (whose internal battery reached its end of service since long ago and cannot be replaced due to some world’s regulations) could be off at any moment and I have had to react rather very quickly before the laptop loses its power (I have about 5 minutes only, if not less if certain programs are running). So, although this buzzer alarm is likely not useful to 99% of the world’s population, it saves my work by helping me taking care of my old laptop by keeping it alive :)

The buzzer, I used, works from 6V DC and higher.

Please don’t hesitate if there are questions about its circuit.

Kerim

ACmainsAlarm.png
 

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Can’t you set laptop to shutdown in 1 minute if on battery power? Check your power profile settings in the laptop.
 
Can’t you set laptop to shutdown in 1 minute if on battery power? Check your power profile settings in the laptop.

Your solution is very good. But I am sorry for not being very clear about the situation I face since many years.

When one source is off, I usually can switch to another one. But since this is done manually and I am not sure how long it will take, I just put the computer in sleep mode when I hear the buzzer. (Wow... this happened now while I am writing you!)

And I am afraid that if LTspice is running for about 1 hour, for example, the laptop shutdown will not be able to restore this hour later :(

Also, if I am editing some files, I can't be sure that right before shutdown these files could all be saved properly.

On the other hand, restarting my laptop takes about 4 minutes besides the time I may need to restart what I was working on. This may be tolerable in general, but not in my case since the laptop may need to shut down every 10 to 15 minutes!

Anyway, thank you for your care.

Kerim
 
Can’t you set laptop to shutdown in 1 minute if on battery power? Check your power profile settings in the laptop.

Sorry, I misunderstood what you meant exactly by ‘shutdown’. You are right there is an option to put the laptop to sleep after 1 minute.

Now, I recall that this worked in the early years when the battery wasn't too old till I realized that the battery started losing its power in less than one minute when certain programs were running. Since then and to be on the safe side, I decided to do it manually when I hear the buzzer on my desk (while keeping the option you suggested).
 
Sorry, I misunderstood what you meant exactly by ‘shutdown’. You are right there is an option to put the laptop to sleep after 1 minute.

Now, I recall that this worked in the early years when the battery wasn't too old till I realized that the battery started losing its power in less than one minute when certain programs were running. Since then and to be on the safe side, I decided to do it manually when I hear the buzzer on my desk (while keeping the option you suggested).
Most laptop batteries use 18650 cells, if you can't get a battery then just replace the cells in your existing one.
 
Most laptop batteries use 18650 cells, if you can't get a battery then just replace the cells in your existing one.

I looked at its battery. It has 10 slots (pins).
My laptop was made, about 13 years ago, by Toshiba which, as I heard, doesn't produce laptops anymore.
 
What is the model number of the laptop?
Also, what is the voltage and AH (or mAH or WH) rating of the battery - or the type number on that, if it has one?

The common removable type laptop batteries are usually made with either three or four series 18650 cells, then up to three parallel sets to give larger capacities.

The three cell ones usually have chargers rated around 15 - 16V and the four cell use 19 - 20V or so.
 
The laptop might run from 16 to 21Vdc off external power easily with bulk storage from a car battery solution with a buck/booster. Boost to run, buck to recharge in opposite directions. The internal battery charger and MOBO regulators will be independent as the laptop can operate without the internal battery.
 
I am afraid that all labels on the laptop are erased. If I remember well, it is of the family Toshiba Satellite and on the cover page of its user's manual, it is written L500.
On the battery (a rectangular box, 12cm x 6.5cm x 2 cm), it is written DC 10.8V 3600mAh.
On its Toshiba charger, it is written DC 19V, 3.42A.

For instance, I personally I have no idea about 18650 cells. But I can ask some local retailers about them, perhaps they are available.

By the way, the set of battery (7V2) in my digital scope (aged 15 years from Fluke) is also dead since long (it acts now as a capacitor). So, I added, since many years, 6 similar batteries in series (6*1V2=7V2) which I replace whenever they got tired. But in the case of the laptop there are 10 pins on the battery pack.
 
The 65 W Toshiba charger is about twice what the laptop needs to run as it was designed to charge the battery and run at the same time but at a slower rate. Turning the backlight brightness to dim as possible in the shade, is perhaps the biggest power save. If you have a variable lab supply, you might be lucky to find it can run off 12~14.2V
 
The 65 W Toshiba charger is about twice what the laptop needs to run as it was designed to charge the battery and run at the same time but at a slower rate. Turning the backlight brightness to dim as possible in the shade, is perhaps the biggest power save. If you have a variable lab supply, you might be lucky to find it can run off 12~14.2V

A year ago, I run the laptop in the morning to find out that its screen is also dead with no clear reason!
So now I use an external monitor which turns on and off with the AC supply :) But, at least, it is more comfortable since its surface is about twice of the laptop's screen.
 
Back to your circuit . . .

What is the purpose of Q1 and its parts? It looks like a 50 uA current sink. If so, why is this necessary?

Also, do you have a manufacturer/part number for the buzzer? Better still, can you post a datasheet?

ak
 
A year ago, I run the laptop in the morning to find out that its screen is also dead with no clear reason!
So now I use an external monitor which turns on and off with the AC supply :) But, at least, it is more comfortable since its surface is about twice of the laptop's screen.
Most likely the edge CCFL vacuum tube has expired or the supply which is normally stocked by Digikey by diameter and length. My Win7 tower is going to be 10 yrs old and still running fast with an ever-growing collection of apps and files.
1678386363298.png

(Intel(R) Core(TM) i7-4770S CPU @ 3.10GHz)
1678386275468.png

My old laptop with a dead battery now has a power jack that needs solder service collecting dust.
 
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The battery is presumably one of these, then:-

If so it's a six cell, two parallel sets of three cells.

(Ignore the supposed voltage difference; different makers rate each cell at different voltages - 3.6V, 3,7V and sometimes 4V - all of which are within the voltage range for 0 - 100% charge of a lithium cell.)

Edit - someone doing a cell replacement on one of those exact batteries!
 
Back to your circuit . . .

What is the purpose of Q1 and its parts? It looks like a 50 uA current sink. If so, why is this necessary?

Also, do you have a manufacturer/part number for the buzzer? Better still, can you post a datasheet?

ak

If Q1 is removed, V(CC) on C1 will be relatively very high since there is no load (Q3 is turned off by Q2 when the mains voltage is on). We may see Q1 as a Zener which starts to draw current thru R4 (12K) when V(CC) reaches about 45V. This lets the voltage on the capacitor C1 be limited to a certain level (as about 55V at 250Vac).

When the mains is off, Q1 and Q2 turn off too, this lets Q3 and its load, the buzzer, be on.

By increasing the capacitance of C1, the decay of the buzzer sound will be slower.
C3 lets the buzzer work at relatively low voltages.
R9 is added (with two LEDs at the base) to limit the collector current of Q3 in order to avoid excess voltage on the buzzer when on.
 
Limiting the Q3 collector current by starving its base is called dangle-biasing, and is *very* unreliable. The beta of a transistor changes with temperature, atmospheric pressure, collector current, base current, and just about everything except the phase of the moon.

You already have R9 in there, which will limit the buzzer current. I'm curious - why is it in the emitter rather than the collector?

ak
 
Limiting the Q3 collector current by starving its base is called dangle-biasing, and is *very* unreliable. The beta of a transistor changes with temperature, atmospheric pressure, collector current, base current, and just about everything except the phase of the moon.

Thank you for discussing this circuit. I built it many years ago and it is still working in these days.
Now, I wonder why I added the 2 LEDs. Their current is too low to let them emit a practical visible light when Q2 is off.

Here, when the mains is on, the V(CC) limit is about 54V. When Q2 is off, (mains is off) and Q3 starts to trun on, Q3 Vce is relatively high which lets its current gain, Ic/Ib, be high too. In case of BC337, this gain is about 200-400. Therefore, for a 3 mA collector current, the base current needs 15 uA. And the base voltage = Vbe + Ie*R9 = 0.6 + 3*0.220 = 1.26V.
At this moment, V(CC) should be (15uA*1megaR + 1.26) = 16.26 V which is much lower than the starting 54V. This means, the margin is rather big.
For instance, the collector current is distributed between C3 and the buzzer. The charged C3, besides supplying the pulsing currents of the buzzer, keeps the buzzer sounding a bit more, though not with a continuous sound.

You already have R9 in there, which will limit the buzzer current. I'm curious - why is it in the emitter rather than the collector?

In real life, a new final circuit is the end fruit of many changes, if not the fruit after trying several different methods.

R9, at the emitter, was supposed to reduce the effects of the temperature sensitivity of the current gain and Vbe. But in this final circuit in which the two LEDs have no practical role (they can be removed) and a wide range of gain is tolerable, R9 could be moved to be at the collector instead of being at the emitter.
 
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