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Need Assistance on Project

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pwollner

New Member
Need assistance on following project:

I need to design a circut which will be put into a cupboard with 6 drawers. If one drawer is pulled out, a red led should light up, if two are pulled out, a buzzer should buzz and the red led should light up.
Now comes the hard part :) : this has to work with a battery at least 5 to 10 years...

(by the way, this is to prevent the cupboard from falling over if too many drawers are open)

Thanx

pwollner
 

Nigel Goodwin

Super Moderator
Most Helpful Member
pwollner said:
I need to design a circut which will be put into a cupboard with 6 drawers. If one drawer is pulled out, a red led should light up, if two are pulled out, a buzzer should buzz and the red led should light up.
Your main problems are getting a signal from the drawers, there are a great many ways of doing it - but you need to sort it out first. Reed relays and magnets would work, as would mechanical switches.

Now comes the hard part :) : this has to work with a battery at least 5 to 10 years...
It's not hard, it's impossible! - this exceeds the shelf life of most batteries, but the impossible part is that you don't know how often, or how long, the LED and buzzer will be on. The only way to meet the requirement, as you've written it, would to have enough battery power (of a type with a long enough shelf life) to last the full 5 years with the LED and buzzer going continuously for the 5 years!.

To maintain minimum consumption, I would use a PIC processor for the design, putting it in sleep mode and only waking up for very short periods, this is commonly done using the WDT (Watchdog Timer). This sort of design keeps current consumption down in the uA range - but when the LED and buzzer come on consumption will increase many fold.

(by the way, this is to prevent the cupboard from falling over if too many drawers are open)
How about screwing the cupboard to the wall?, or if it's freestanding the floor!.
 

pwollner

New Member
re: need assistance

There are 6 drawers- not only 2...

If any one is open then the led should light up, if any two are open, the buzzer should sound.

Thanx

pwollner
 

Russlk

New Member
OK, 6 drawers is too many switches, try diode and resistor logic:
With one drawer open the bias on the LM393 is 0.2 volts, with two open the bias is about .4 volts and the buzzer sounds.
The switches should have been shown normally closed!
 

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tavib

Member
:)
Use CMOS 4000 circuits for logic, they static supply current is very low(under 1uA).
Simple use controlled gate oscillator for buzzing(the oscillator not oscillate when not need buzz) and a simple transistor for LED.
 

pwollner

New Member
:D
Your responses were all very pleasing, yet I don't know how to program a PIC processor nor what CMOS 4000 circuts are. :oops:


Could anyone give me some more info on these two topics?


Thanx

pwollner
 

john1

Active Member
Hi pwollner,

I feel some of these replies are a little over-done.
You might like to consider something simpler.
I haven't worked out the resistor values,
they depend on the SCR.

The two diodes would be chosen to give appropriate
voltage to the led.

The resistor on the gate would be chosen so that one
drawer would not engage the SCR, but more would.

Any drawer would light the led,
any two (or more) would operate the buzzer,
the intermittent nature of the buzzer would turn it
off when the drawers are closed, or one left open.

SCR's have a very high resistance, almost open circuit
when not operated, not enough to affect battery life.

Two or maybe three U2 type batteries should give a
very long life, its hard to guess but i would expect
a year, more maybe, depends on usage.

Sorry this took so long, i think i had a mental
block.

Best of luck with it, Regards, John :)
 

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tavib

Member
:roll:
john 1
I have some comments.
1. The current will avoid the LED.
2. The SCR is trigger by current not voltage, over 0.6V a very small amount of current will trigger the SCR, this current is more small when pulse is more long.
3. The buzz will never stop.
 

tavib

Member
:)
This is a schematic with CMOS 4000.
When all switchers is open the current sink from battery is very low.
When 1 switch is close, the first D bistable is load with 1 and LED become light.
When another switch is close, while first is close, the second D bistable load 1 from first and allow NOR gate oscillator to start, so the speaker will start buzzing.
This state(LED on and buzz on) is keep until all switchers become opens, when D bistables are reset with a short pulse(50us), and schematic return in standby state.
 

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tavib

Member
:)
This is a schematic with CMOS 4000.
When all switchers is open the current sink from battery is very low.
When 1 switch is close, the first D bistable is load with 1 and LED become light.
When another switch is close, while first is close, the second D bistable load 1 from first and allow NOR gate oscillator to start, so the speaker will start buzzing.
This state(LED on and buzz on) is keep until all switchers become opens, when D bistables are reset with a short pulse(50us), and schematic return in standby state.
 

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pwollner

New Member
thanx for everyone's help


im still workin on this project and post the result once im done

could any of u maybe give me an idea what i could do next as a project - im very interested in any circuts that interact with the computer...


:D Thanx again!
 

pwollner

New Member
just finnished it and it works!!!!!! :D :D :D


I used Russlk's idea: THANK YOU! :D




Thanx for everyone's help,
 

john1

Active Member
Nice going Pwollner,
and good luck with your next project !

Regards, John :)
 
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