This circuit is usable as a Night Lamp when a wall mains socket is not available to plug-in an ever running small neon lamp device. In order to ensure minimum battery consumption, one 1.5V cell is used, and a simple voltage doubler drives a pulsating ultra-bright LED: current drawing is less than 500µA.
An optional Photo resistor will switch-off the circuit in daylight or when room lamps illuminate, allowing further current economy.
This device will run for about 3 months continuously on an ordinary AA sized cell or for around 6 months on an alkaline type cell but, adding the Photo resistor circuitry, running time will be doubled or, very likely, triplicated.

circuit and component list link:-
Battery-powered Night Lamp - RED - Page44

i made it and its not working can any one help me plzzzzzzzzz

 

What is the EXACT part number of the timer chip you are using? An ordinary LM555 will NOT work. It needs to be the CMOS version and even then, it is being operated out of spec, since most of them, as far as I know, need a minimum supply voltage of 2V.

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Inside every little problem, is a big problem trying to get out.

 

yeah i used 7555 number chip, and 1m ohms 10% tolerance resistors, i used polyester cap 100nf 100V instead of 63V, it is ok or its need exactly 63Vcap ????????

 

The 100V part will be fine.
Which manufacturer of the 7555 did you use? Some are only spec'd down to 3V.
Also, try removing the CDS light sensor or going into a very dark room with the circuit to see if it works.
If all else fails, post a clear picture of the circuit. (Top and bottom)

__________________
Inside every little problem, is a big problem trying to get out.

 

Hello there,


The circuit, although interesting, is a bit superfluous. All the circuit is doing is
increasing the voltage so that it can run an LED that has a forward voltage that
is higher than a single battery. Why does the circuit have to do this? It's because
the battery voltage isnt high enough to run the LED by itself. If the battery
voltage WAS high enough, we wouldnt need a circuit anymore.
Enter a *second* battery here, in series with the first.
With TWO batteries in series we can easily drive a red LED and we dont need a
circuit to do that, just a single resistor.

What's the run time for two cells instead of one? Im glad you asked :-)
The run time will be longer because the current draw from the two cells
will be approximately one half of what it would be with a voltage double
circuit. This means the run time will be twice as long as with only one cell.

So in the long run it is better to use two cells in series than to bother with
a circuit unless there is some real need to run off of one single cell.
It's true we would need two cells then, but they will last twice as long so
that's about the same average battery usage meaning over time we will use
the same number of batteries.

Besides the circuit looking like it could use a lot of improvement, one drawback i
see right away is that it doesnt look like it can run a white LED. That's a big
drawback because we often dont want to have to navigate down a hallway
with a dim red light, we would rather have white light. I've been there and done
that so i have a good idea about the difference here. Try it if you want to see
the difference.

Using three cells and a resistor we can drive a small while LED now, and we can
find white LEDs that run on very low current these days and are very efficient
at low currents too. The run time will be about three times that of a single
cell, so we wont be wasting batteries, and we get a nice white light to see by.
Try it and you'll see the difference.

 

[IMG]C:\Documents and Settings\COMPUTER\Desktop\NIght lamps\LED 1.5V_files\NightLamp.gif[/IMG]

 

ooops sorry how to post circuit image ??????? plz help
if it is not possible use given link to see circuit and component list ppllllzzz
Battery-powered Night Lamp - RED - Page44

 

"7555", "IPA" company, in the bottom i think this is batch number or not I don't know
"H 9340"

 

The circuit blinks the red LED at 4Hz and saves battery power because the LED is dim and is turned off for most of the time.
C2 is charged to the battery voltage less the voltage of D2 (0.3V) then the output of the Cmos timer goes low and applies the battery voltage plus the C2 voltage to the LED at a current of only about 2mA which is limited by the Cmos timer IC.

The LED is dim with a current of only about 2mA.

__________________
Uncle $crooge

 

What I meant was a real picture of the circuit you built (Top and bottom). Not the schematic. That way we can spot any mistakes you might have made.
Sounds like it may be the 7555 from Intersil which should be OK as it is guaranteed to work down to 2V. Still out of spec, but less so than a 3V part.

__________________
Inside every little problem, is a big problem trying to get out.

 

The circuit you have posted will not work because the electrolytic will not charge when the output of the chip is HIGH as the output will be about 1v lower than rail and the diode removes another 0.6v.
The output will be 1v lower than rail due to the electrolytic being a DEAD SHORT and the chip will not deliver a charging current.
What you need is the circuit I designed using two transistors. This circuit will even flash a white LED even though a white LED requires about 3.2v to 3.6v under normal conditions.

 

How about using one of those "funky"
little (I believe) 6V batteries, that are
about the diameter of a AA and 2/3
it's length ?

 

No.
The 7555 is a Cmos 555 that has an output that goes rail to rail (as high as the power supply voltage).

No.
The 1N5819 Schottky diode has a voltage drop of only 0.25V or less at the low current used for charging the capacitor.

Therefore the capacitor charges to 2.75V which is plenty to light a 1.8V red LED.
When the 1.5V battery drops to only 1.0V then the LED will still be driven with 1.75V which will make it blink dimly.

__________________
Uncle $crooge

 

A little battery does not have much capacity so it will not last long.

__________________
Uncle $crooge

 

This is entirely wrong:
The 7555 will only source 1mA when the supply is 5v and the output voltage is 1v lower than rail. The capacitor provides a Dead-Short across the output and thus the capacitor will never charge. Secondly, any slight voltage produced by the capacitor will be lost when the output goes low as the output is about 0.4v above 0v rail.
How did you get the impression that the charge-rate will be low?
How did you get the idea that the capacitor will charge to 2.75v?
You should come and work for me. You will make me a fortune with your technical specifications.