# Simple LED Timer Circuit

Joined Mar 2, 2008
5
Hello,

This is my first post here and at initial glances, this website seems like an amazing resource.

I need help with a very simple LED timer circuit.

I am making a collection box for my GCSE Design & Technology course. I want to make it so that when money is inserted into the hole, it will press down a copper contact that will complete the circuit and light up six blue LEDs.

Because the money falls into the box right away, the circuit will be broken in less than a second as the copper contact will spring back into place. However, I want the LEDs to remain turned on so that they illuminate the 'thank you' sign for at least 10 seconds in order for the donor to notice it.

I do not know where to take it on from here. I want to use a 9 or 12 V battery as the power supply and 6 blue LEDs (that I am told require 5mA).

I initially thought I could use a capacitor to store the energy and release it to the LEDs for 10 seconds. Do you'll think this is viable? If so would it be possible to point me in the right direction in terms of calculating the measurements because I haven't had any luck? (I've been told I may also need a resistor to protect the capacitor.)

Alternatively, I was told that I could use a time-delay circuit but I have no idea where to begin with that. Would you advice me to take this option, if so can you guide me to any good resources?

Furthermore, are there any other routes I can use to complete this circuit?

Thank You So Much,

Joined Mar 2, 2008
5
Hmm... The circuit seems quite complex at a first look because the timer has 8 pins. I would prefer something a little bit simpler, but I'll give it a shot. Thanks!

Just to clarify, if I used the formula R=T/1.1*C and I wanted 10 seconds using a 2200 microfarad capacitor, I would do: 10/(1.1*2200)=4.1*10(to the power of -3). Multiply by 1000 to get 4.1 K Ohms.

Would this work with a 9 V battery and LEDs that require 30mA? Sorry, I am little but confused here.

#### thingmaker3

Joined May 16, 2005
5,084
2200 uF is an awfully large capacitor for a timing circuit. It will draw a lot of current and your battery will not last as long. Better to use a 22uF capacitor and a 410 Kohm resistor.

Do you know the forward voltage drop for your LEDs?

Joined Mar 2, 2008
5
Unfortunately I am not too sure. I will try and ask the electronic shop where I purchased them.

Thanks.

Joined Mar 2, 2008
5
I couldn't find out the forward voltage drop because the store keepers did not know it, is it really essential because it vary by a lot?

I also just wanted to check one more thing. For the monostable 555 timer circuit, what do the values of the other resistors and capacitors do because they vary from diagram to diagram (-I am guessing that they vary by the input voltage by the battery)?

I started following this diagram: http://www.eleinmec.com/article.asp?4* . Do you think the values for the other capacitor (220 uF & 0.01 uF) and resistor (10 kohms) will fit my needs? Or do I need to change those as well?

Thanks,

Edit: *Note: Referring to figure 2 in the link.

#### thingmaker3

Joined May 16, 2005
5,084
I couldn't find out the forward voltage drop because the store keepers did not know it, is it really essential because it vary by a lot?
We use the LED forward voltage drop and the current specification to select the correct current limiting resistor. We could have also compared the forward voltage drop with 1/6 of 9V to know if we could string the LEDs in series or not.

Blue LEDs at 5mA will drop anywhere from 2.9V to 3.6V. If you have a good voltmeter, you can test yours by putting one in series with a 120 Ohm resistor and measuring across the LED. This will put us in the right "ball park."

I started following this diagram: http://www.eleinmec.com/article.asp?4 . Do you think the values for the other capacitor (220 uF & 0.01 uF) and resistor (10 kohms) will fit my needs? Or do I need to change those as well?
Those values will work just fine.

#### SgtWookie

Joined Jul 17, 2007
22,210
I'm afraid that using a 120 Ohm resistor in series with a blue LED (with a Vf of 3v) across 9V will rapidly result in a burned-out LED, as it will be passing 50mA instead of 5mA.

I suggest that for a 9v supply, a 510 Ohm resistor in series with the LED would be a much better value for current limiting purposes during test. It should safely pass sufficient current to light up most diodes, from a standard red LED (resulting in 14.3mA @ Vf=1.7v) to a white LED (9.8mA @ 4.0V).

The use of a simple current limiting resistor won't ensure that the current remains constant when the LED is changed. A better method would be to use/construct a constant current supply.

Have a look at the attached schematic. It's a constant current supply that is easy to build from a few components. Note that you can use other PNP transistors than a 2N2907; a 2N3906, 2N4403, etc will work as well.
D1 and D2 can be replaced with a single standard red LED.
R1 controls the current; increasing R1 will decrease the current through Dtest.

As shown, the circuit produces a constant 11.94mA source, regardless of the Vf of Dtest (in the range of 1.2v to 5v). However, semiconductors do vary somewhat in their Vf; so construction of the circuit will require adjustments to R1 to obtain the desired current. It's better to start out with a higher value (say, 100 Ohms) and work down. Values below 50 Ohms will likely result in current flow high enough to rapidly burn out standard LEDs.

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Joined Mar 2, 2008
5
Cool, thanks!

I'll try the monostable 555 circuit timer first, if it doesn't work then I'll give a go at your method SgtWookie.

#### thingmaker3

Joined May 16, 2005
5,084
I'm afraid that using a 120 Ohm resistor in series with a blue LED (with a Vf of 3v) across 9V will rapidly result in a burned-out LED, as it will be passing 50mA instead of 5mA.
Has anyone seen a decimal point laying around? I seem to have misplaced one.