# Circuit Help :) - LDR LED Cube (Urgent)

Discussion in 'LEDs and Optoelectronics' started by depressed_ninja, May 22, 2016.

1. ### depressed_ninja

3
0
May 22, 2016
Hello, My name is Ethan
I am currently doing a B-TEC level 3 extended diploma in electrical/ electronic engineering at City College Coventry

I have my final project that I am doing which needs to be completed and handed in by Friday 27th May.
I am building a circuit that is an LED cube (Lamp) That turns on due to it being dark (using an LDR)
I wanted to know if anyone could see if the circuit I am doing is even feasible. or whether I am completely screwed.
https://gyazo.com/6b5bc9c539c2fb2e94a32b4d36793861 That is a screenshot of the circuit drawing its kinda bad quality and there's an image of the drawing.
I'll give information about each component below.
Potentiometer = 10kΩ
R1 = 22KΩ
R2= 1KΩ
R3 R4 = 390Ω
R5 = 1kΩ (before LED above relay) - not shown on image)
R6,7,8,9,10 390Ω
C1 = 0.01µF
Q1, Q2 (BC108s)
IC1 = NE555(p) Timer
The Relay I am using is really hard to get my head around It's called a:
Zettler Electronics AZ9405-1A-9DEF PCB Mount Relay. Found at http://www.rapidonline.com/zettler-electronics-az9405-1a-9def-pcb-mount-relay-59-5567
If you could provide any information or even redrawing of the circuit it would be really helpful.
The PCB I have designed is also below. However, the relay may have to be drawn differently because it would be reversed.
On the PCB design the Potentiometer and LDR are listed as R1 and R2 respectively so the order may be different

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
2,830
Jan 21, 2010
Apart from design issues with your light detector, 555 circuit, transistors and relay, the left side looks fine.

The right hand side looks fine other than it won't work.

How long have you had to do this?

When did you start?

What testing have you done?

You can fix all of these things, but I hope you haven't got much else to do for the next couple of days.

Do you need to do a write-up on this as well?

I would probably start with just the 555. What do you want it to do? Can you find a circuit that does it? What are the input requirements and the output capabilities?

Then look at how I could drive the relay -- what is actually required?

Next, can I use the photodiode to provide the required input? Is another component more suited to this?

Finally, how can I use the relay to turn the LEDs on and off?

3. ### depressed_ninja

3
0
May 22, 2016
We started the project last year in October, However I only started to really apply to it around January time.
I have attempted simulation softwares' such as Multi-sim and circuitmaker2000, With multi-sim the photo resistor wasn't available and circuitmaker doesn't have sufficient guidelines.
I made the breadboard and had a 9V DC supply to it where it didn't work. I didn't know why at the time.
I have a write up to do but I have done a large quantity of it not including the circuit itself.
The product when turned on would light up and the timer would switch after a certain time period to turn off the LEDs.
The supply voltage for the 555 timer is between 4 & 16v, with an output of current up to 200mA

The IC is mainly used so that after a certain amount of time everything after the relay will turn off.
Would the circuit be better off if it didn't have the IC, as It's not an integral component?

When you say it looks fine but it wont work. Is that because of the other side of the relay?

The relay requires a minimum of 6.3v to the coil up to 9v.

Sorry for my brief description and replies, I'm worrying at this point.

4. ### Alec_t

2,940
799
Jul 7, 2015
Also consider whether you have the pot wired correctly and what are the consequences if it gets twiddled to either extreme position.
And welcome to EP!

5. ### depressed_ninja

3
0
May 22, 2016
thanks . Originally I had the first terminal from my supply and the 3rd terminal to its output towards the LDR and IC etc.

6. ### BobK

7,682
1,688
Jan 5, 2010
That is some creative circuitry.

I especially like the way the relay shorts the power, and the power and ground for the 555 and sensor circuit are routed through reverse biased LEDs. I would have never thought of either of those.

Also, honorable mention for the two redundant inverters between the 555 output and the relay coil.

Bob

7. ### 73's de Edd

3,070
1,299
Aug 21, 2015
.

Ahhhhhh so . . . .Most honnnable depressed_ninja. . . . . . . . . . . .

Circuit . . . circuit . . . how thou doth not work . . . . let me count the ways . . . . . ( Shakespeare . . . . or Confucius . . . . .or Plato . . . .or Pepe Rodriguez and the Tijuana Pimps . . .or someone ? )

I wonder of you did not smell the odor of burnt carbon at one extreme of the left top corners pot in its adjustment, HOWEVER on a 10 K units resistance slurry mix, there just might? be enough series resistance between the short path that either resistance elements end, takes in getting to its companion end terminal, so as to not have that low of a resistance value.
If you burned out one ends connectivity, install the current limiting / range limiting resistor R0 as shown in the YELLOW box for preventing another future boo-boo..

Strip off some parts to make the right end circuitry as is being shown, on the lower plan.
Add in an EMF kickback supression diode as shown in another YELLOW box.
(1N4001 thru1N4007 series . . . . . or 1N914 . . . . or 1N4148 . . .what have you. )

The relay will probably be activated and the LEDS lit unless your left circuitry might actually be working and engaged.
If not, jumper across BJT #1's emitter resistor . . . R9 to ground . . . with a jumper clip.

I'm standing by to see your results . . . . . for an upcoming branch in the thought process road.
With the idea being to to initially establish a working circuit on that right side and then progressively work towards the left side circuitry.

Your circuit . . . . . being QUASI cloned . . . . . .

73's de Edd

.

Last edited: May 23, 2016
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