Timer circuit question

[Cormacs]

I have a question about designing a timer circuit. I have two LED's wired in series running at 18 mA on a 12VDC circuit. All I want is a way to have them stay lit for about 5 minutes after they loose their switched power. I know I have to supply them with a steady power and ground. I want to be able to turn them on with a switched power/ground and then when they loose that switched power/ground they stay lit for roughly 5 minutes. I was just wondering if there is some kind of simple chip or way of wiring them so that this is possible.

 

[Grossel]

The device you're looking for is named a capacitor.

Is it very important that light intensity keep at same level, or is it good enough as long as the led's keeping off dying for the next five minutes?

Since I don't know the least voltage you must have in order to keep your LED's brightening, it's very hard to tell how big the cap may bee. Calculate it's size is waste of time before I know more. It's hard enough to calculate because the voltage over led's isn't propotional to the current like resistors.

Dealing with low voltage cut off circuits would make it even harder to calculate time to emit light for any led. But, If you under any circumstances should find a recipe for calculating discharging time where an LED is involved - you MUST share. Too difficult for me to solve.

However, If you can afford using a resident 12V supply, you can simply use an analog comparator to turn off whenever the voltage goes under certain level. That way, the value of the cap isn't that important (unless - voltage over cap should be in the range of the comparator).

 

[Cormacs]

A capacitor would work for sure. The only down fall to a capacitor for me is my space is a little limited. I'm not sure what size of a capacitor I would need but it would most likely be to big. I'm looking into timer modules but my electronics knowledge fails when it comes to chips. My led's are running on 12VDC with a 470 ohm resistor to limit the current to roughly 19 mA. If anyone can calculate the size of capacitor I would need I could see how big it would be. I still think a chip would work better I just don't know what to use.

 

[Dragon Tamer]

If your space is limited, then use a 555 timer configured for monostable operation. For an on time of 5 minutes, you would need to have a capacitor of about 300uF and a resistor of 1M.

 

[Cormacs]

Sorry, but I'm not the best with chips. Would you be able to give me a example of a way to wire a 555 timer? One other question, are there different 555 timers or do i just go in to an electronics store and buy a 555 timer?

 

[Grossel]

Neither I am familiar with the 555 timer. But I do know how nice chematics diagram looks.

Doing some googling around the web, I found **broken link removed** with both schematic, formula and timing diagram.

 

[Reloadron]

Based on your original post:

I have a question about designing a timer circuit. I have two LED's wired in series running at 18 mA on a 12VDC circuit. All I want is a way to have them stay lit for about 5 minutes after they loose their switched power. I know I have to supply them with a steady power and ground. I want to be able to turn them on with a switched power/ground and then when they loose that switched power/ground they stay lit for roughly 5 minutes. I was just wondering if there is some kind of simple chip or way of wiring them so that this is possible.

My understanding is that when power is removed you want these two LEDs in series to remain lit for about 5 min. First, I don't see using a 555 configured as a monostable one shot as a viable solution and here is why. Configured as a one shot the 555 when triggered will output a pulse for a duration determined by a RC (Resistor Capacitor) timing network. However, if power is removed then the output immediately is also removed. If I trigger the 555 the LEDs illuminate but anytime power is removed the lamps extinguish.

The basic circuit suggested using a capacitor would work like the attached circuit. When the switch is closed the capacitor C1 would quickly charge through R1 and the LEDs would illuminate. Resistor R2 is there to limit the series LED current to 18 mA per your numbers. The circuit will sit there with LED 1 and LED 2 glowing as long as power is applied. When the switch is opened the LEDs will remain glowing until the charged capacitor C1 discharges through R2 and the LEDs.

What is not mentioned and needed is the forward voltage drops of the LEDs. Without knowing the forward voltage drop of the LEDs we can't calculate R2. However, you should get the general idea. Again we are looking at power being removed as in no power to anything. That being what I got from your initial post.

In reality R1 could be removed. It is there merely to illustrate a charge path for C1 rather than instantly have C1 charge when 12 volts is applied..

Ron

 

[Grossel]

However, if power is removed then the output immediately is also removed. If I trigger the 555 the LEDs illuminate but anytime power is removed the lamps extinguish.

Cormacs must tell if he does have another DC source avaible. A backup battery t.ex.

 

[Cormacs]

After further review I believe the capictor is difenetly the way to go. I already know the resitance of R2, Since the foward voltage of my LED's are 2.2 and I'm running on a 13 volt circuit with a max current flow of 30 mA I will be going with a 330 ohm resistor.

Now here is my latest battle. What size capacitor would you think I would need to sustain two led's in series with a 330 ohm resistor for roughly 5 minutes?

 

[Cormacs]

For furthur clarification this will be running on a 12.6 to 14.5VDC car power system. That's why I use the 13 volt average number in my previous posts.

 

[Reloadron]

As to discharge time, you can give this a read. However, just winging it for the time you want you will likely need a pretty high capacitance. I would start with maybe a 2,500 uF cap rated for 25 or 50 working volts DC and see what you get. Then maybe parallel another using trial and error till you get about what you want. That is just winging it so read the link.

Ron

 

[MikeMl]

Here is a circuit that comes very close to what you need.

 

[Grossel]

Oh, I can calculate the values very rough, not precise.

Say you have a voltage drop of 4 volts over both led. We must asume that it stay 4 volts, but in reality the voltage will drop when cap is discharging - so that make calculation less presice.

Say that diodes needs at least 15 mA to light. Lets also say that voltage over cap must bee at least 6V then. To get maximum resistance value:
Ur = 6-4 = 2V.
R = U/I = 2V/0,015A = 133 ohms.

Since it's hasrd to get a 133 ohm resistor I choose the closest avaiable E12 value: 150 ohm.
Then again, we must recalculate the lowest voltage over cap neccesary for leds to light:

U_cap_minimum = R_e12*I_led + U_leds = 150*0,015 + 4 = 6,25V

To make discharge time calculation easier, lets add 4 volts to all voltages. So new values become:
'Vcc= 9V
'V_low = 2.25V


Now, all we have to do is to calculate. The discharge formula says:
T = -R*C*ln(1 - Vcc/V_low)

So calculating C:
C = -T / (R*ln(1 - V_low/Vcc))

Inserting values:
C = -5*60 / (150*ln(1 - 2,25/9)) = 6,95 F

Asuming my calculations isn't extremly wrong, you'll ending up with a relly huge cap. You're really sure you're better off with a cap instead of a little extra battery?


[Edit]
I think a jfet connected as a current source would be the best alternative. Still, 5 minutes is a long time for any cap.

 

[Cormacs]

I don't really know what to use. I'm starting to think it will be an impossible feat to have a simple time delay circuit. Having constant power is not an issue, there is always 13 volts available from the battery. I just want when the ignition is turned off the led's will stay lit for another 5 minutes. It can draw from the battery for those 5 minutes. I just don't know how to design a circuit that will shut the led's off in 5 minutes after looseing the switched power.

 

[Grossel]

I don't really know what to use. I'm starting to think it will be an impossible feat to have a simple time delay circuit. Having constant power is not an issue, there is always 13 volts available from the battery. I just want when the ignition is turned off the led's will stay lit for another 5 minutes. It can draw from the battery for those 5 minutes. I just don't know how to design a circuit that will shut the led's off in 5 minutes after looseing the switched power.

That make things more clear. I would advice you use the 555 timer as a monostable.

A really simple trigger mechanism could be a jfet connected to a cap and resistor in paralell. Then negative voltage will arise, and thus make the jfet to not conduct the moment voltage drops.