Autism electronic sensory box

Thread Starter

Deputyduke

Joined Apr 22, 2019
32
Hi folks

I have 3 young kids all with special needs

Twin boys aged 7 both autistic and adhd

Daughter aged 12 with adhd

One of my sons is also non verbal and gets immensly frustrated not being able to tell us what he wants andbgets bored very very quickly which normally ends up with both myself and my wife getting bitten / nipped / scrabbed / slapped etc...

My wife is only 48 and this has led to her having 2 strokes recently due to the stress.

I woyld love to build hom an electronic sensory box all housed within a 380mm x 300mm x 160mm plastic adapatble box with 4 recesessed screws powered by an a23 battery inside it.

Basically i will drill the lid of the box to hold 6 diff coloured 5mm leds and rocker and toggle switches hence whem he flicks a switch the led lights up.

My problem is i know nothing about electronics i have all the bits that have been donated to me but im crap at maths as i believe i need to wire a resister to the anode of each led to drop the voltage to 2v each 6 x 2=12v

I was also donated a 12v 20amp toggle switch well 2 of them but will the a23 power these?

To be honest ive been reading about series and paralell wiring but its double dutch.

Any advice very greatfully received and soory to be a pain but im urgently trying to get this built.

Many thanks in advance

Best Regards

Glenn.
 

dendad

Joined Feb 20, 2016
3,924
Have a look at this..
A 2200 ohm resistor will be ok. Usually refereed to as "2K2". You could go a lot lower but the battery will flatten faster.
Yes, you do need the resistor. It is not to drop the voltage but to limit the current. LEDS run on current and have a fixed voltage drop, sort off.
so, if you want 5mA (0.005 Amps) through the LED, and the supply is 12V, the resistor can be calculated something like this..

Voltage drop across the LED is nominally 2V, so the volts across the resistor is supply - LED volts...
12V -2V =10V
Resistance = Volts / Current
R = 10 / .005
R = 2000 ohms.
Use a 2200 ohms.


LED_SW.jpg
2K2.jpg
LED2.png
The LED has a long lead, it is the + and the - usually has a flat edge on the LED body.
They must be wired the correct way otherwise a popped LED will result.

A better power supply would be a set of AA cells in a battery holder. They will last longer than the 12V one you have.

AA holder.jpg

A 4 cell holder will be ok too, and the volts will be 6V. A lower resistor, like 1K (1000 ohms) could be used then.

If you need more help, just ask :)
 
Last edited:

AlbertHall

Joined Jun 4, 2014
11,534
Does anyone else remember someone on here wanting a box with switches and lights (I don't think it was anything to do with autism) but with a function intended to intrigue?
 

Thread Starter

Deputyduke

Joined Apr 22, 2019
32
Have a look at this..
A 2200 ohm resistor will be ok. Usually refereed to as "2K2". You could go a lot lower but the battery will flatten faster.
Yes, you do need the resistor. It is not to drop the voltage but to limit the current. LEDS run on current and have a fixed voltage drop, sort off.
so, if you want 5mA (0.005 Amps) through the LED, and the supply is 12V, the resistor can be calculated something like this..

Voltage drop across the LED is nominally 2V, so the volts across the resistor is supply - LED volts...
12V -2V =10V
Resistance = Volts / Current
R = 10 / .005
R = 2000 ohms.
Use a 2200 ohms.


View attachment 175804
View attachment 175805
View attachment 175806
The LED has a long lead, it is the + and the - usually has a flat edge on the LED body.
They must be wired the correct way otherwise a popped LED will result.

A better power supply would be a set of AA cells in a battery holder. They will last longer than the 12V one you have.

View attachment 175807

A 4 cell holder will be ok too, and the volts will be 6V. A lower resistor, like 1K (1000 ohms) could be used then.

If you need more help, just ask :)

Hi and many msny thanks for taking the time to reply on your very hekoful and detailed answer.

I fully understand your drawing of the switch to resister to led to battery.

Just a few things i would need to clarify

1. The switch has 6 wires and is rated 12v 20amp with 6 pins numbered 1 - 6 which 2 pins do i wire the led to?

2. The 5mm leds i will be using are red, blue, green, yellow and orange.

3. Ive been informed that the leds are bog standard leds and require 20mA. Between 1.8v and 2.4v dependent on the colour. I believe red requires 1.8v?

4. I will also be using buttons push on push off type pic attached.

5. Which pin is which to go to the cable?

6. Im also using rocker switches with 2 pins pic attached pins labelled 1 + 2 which way around should this be wired?

At the minute ill stick to the 12v A23 battery ive ordered just to play about with as its a prototype for my sun

I have included a link below



Have a look at this..
A 2200 ohm resistor will be ok. Usually refereed to as "2K2". You could go a lot lower but the battery will flatten faster.
Yes, you do need the resistor. It is not to drop the voltage but to limit the current. LEDS run on current and have a fixed voltage drop, sort off.
so, if you want 5mA (0.005 Amps) through the LED, and the supply is 12V, the resistor can be calculated something like this..

Voltage drop across the LED is nominally 2V, so the volts across the resistor is supply - LED volts...
12V -2V =10V
Resistance = Volts / Current
R = 10 / .005
R = 2000 ohms.
Use a 2200 ohms.


View attachment 175804
View attachment 175805
View attachment 175806
The LED has a long lead, it is the + and the - usually has a flat edge on the LED body.
They must be wired the correct way otherwise a popped LED will result.

A better power supply would be a set of AA cells in a battery holder. They will last longer than the 12V one you have.

View attachment 175807

A 4 cell holder will be ok too, and the volts will be 6V. A lower resistor, like 1K (1000 ohms) could be used then.

If you need more help, just ask :)
 

Attachments

Thread Starter

Deputyduke

Joined Apr 22, 2019
32
If you want to make something more versatile, have a read about the Arduino micros controllers.
Here is a good introduction to them.

http://math.hws.edu/vaughn/cpsc/226/docs/askmanual.pdf

Then you could make a more interactive device.
Many thanks for all of your kind help

Here is a video attached to same idea as to what i had in mind the box must be powered by inertnal batteries to keep the risk of shock down

 

dendad

Joined Feb 20, 2016
3,924
1. The switch has 6 wires and is rated 12v 20amp with 6 pins numbered 1 - 6 which 2 pins do i wire the led to?
Use the center pin on one side, and one of the end ones on the same side. There are 2 changeover switches in the one. It is a "Double Pole Double Throw" DPDT switch. You only need a part f the switch.
If you like, connect one LED to each end and feed the power to the center. Then one LED will be on OR the other one.
3. Ive been informed that the leds are bog standard leds and require 20mA. Between 1.8v and 2.4v dependent on the colour. I believe red requires 1.8v?
The 20mA is not "required", but is often a max current and used to have a standard current to specify the brightness at. LEDs can be well run at a lot lower current.
5. Which pin is which to go to the cable?

6. Im also using rocker switches with 2 pins pic attached pins labelled 1 + 2 which way around should this be wired?
Switches do not matter which way around they go.
 

Thread Starter

Deputyduke

Joined Apr 22, 2019
32
Top man in your field! thanks so much and sorry for all the questions.

So to confirm resistor vslues i need for future is the following calculation.

Add together Resister voltage
+Battery voltage
Then divide by current?
 

djsfantasi

Joined Apr 11, 2010
7,870
Sorry i meant led voltage and battery voltage added together
Actually, you subtract the led voltage from the battery voltage and divide that by current.

You start with the battery voltage. Then the LED needs its voltage, so that’s why we subtract it. The remaining voltage then must be “used up” by the resistor. So we divide what’s left by the current to get the resistor value.
 

dendad

Joined Feb 20, 2016
3,924
Top man in your field! thanks so much and sorry for all the questions.

So to confirm resistor vslues i need for future is the following calculation.

Add together Resister voltage
+Battery voltage
Then divide by current?
Not quite.
Take LED voltage (about 2V ) from the battery voltage.
Eg, 6V battery - 2 V LED = 4V.
Then, select a current for the LEDs. make it 10mA. That will be plenty bright.

So, Resistance = Volts / Current.
R = 4/.01
R = 400 Ohms.
The closest "standard" value will be 390 Ohms.
 

Thread Starter

Deputyduke

Joined Apr 22, 2019
32
Actually, you subtract the led voltage from the battery voltage and divide that by current.

You start with the battery voltage. Then the LED needs its voltage, so that’s why we subtract it. The remaining voltage then must be “used up” by the resistor. So we divide what’s left by the current to get the resistor value.
Excellent!

So am i right in ststing the following assumption based on my limited maths.

Battery =12v
Led =2.0v
Current = 0.02mA

Resister required is 500ohms?
 

djsfantasi

Joined Apr 11, 2010
7,870
I
Excellent!

So am i right in ststing the following assumption based on my limited maths.

Battery =12v
Led =2.0v
Current = 0.02mA

Resister required is 500ohms?
You got it!

Just note that there are standard resistor values. You won’t find a 500Ω resistor. But the resistor value isn’t critical. 470Ω and 510Ω are standard. I’d use the higher value.

You can double check by calculating the current for the different resistors. 10V divided by 470Ω is 21mA. Since 20mA is the maximum, I’d pick the 470Ω resistor.
 

Thread Starter

Deputyduke

Joined Apr 22, 2019
32
I


You got it!

Just note that there are standard resistor values. You won’t find a 500Ω resistor. But the resistor value isn’t critical. 470Ω and 510Ω are standard. I’d use the higher value.

You can double check by calculating the current for the different resistors. 10V divided by 470Ω is 21mA. Since 20mA is the maximum, I’d pick the 470Ω resistor.
Guys you lot are terrific thank you so much again for all of your kind help.

One other question if i am using 1 x 12v a23 battery in its holder how do i connect all of the 6 led wires to it?

Do i wire them all in series for example each led wired positive to negitive?
 

djsfantasi

Joined Apr 11, 2010
7,870
Guys you lot are terrific thank you so much again for all of your kind help.

One other question if i am using 1 x 12v a23 battery in its holder how do i connect all of the 6 led wires to it?

Do i wire them all in series for example each led wired positive to negitive?
No

Do you want the switches to turn the LEDs on and off? I’ll assume that you do in my explanation.

First, let’s start by making a set of sub-assemblies. From each switch, wire from one terminal to the appropriate resistor. From the other end of the resistor, wire to the long lead of the LED. From the short LED lead, run a wire to the negative lead of the battery.

Now, run a wire from the positive lead of the battery to the other terminal of each switch.

A couple tips. First, on an LED, the long wire always connects to positive and the short lead always connects to negative. I use red wire for the positive connections. And black wire for the negative or ground connections.

If you don’t want to switch the LEDs, wire all of the resistors to the positive lead of the battery holder.
 

dendad

Joined Feb 20, 2016
3,924
Just understand, an A23 battery is rated at 55mAH, so that means if you run a 55mA load, it will last on hour.
As you look to be wanting to run 20mA per LED, and a number of LEDs, the time will go down rapidly.

Using 12V of AA cells will run at least 10 times longer, and depending on the cell type, maybe even 40 times longer.
That is one reason I urge you to see if 5mA is bright enough. Test and see.
 
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