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"1.5V AA" 3 months continuously night lamp

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TDA2030

New Member
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
 

kchriste

New Member
Forum Supporter
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.
 

TDA2030

New Member
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 ????????
 

kchriste

New Member
Forum Supporter
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)
 
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MrAl

Well-Known Member
Most Helpful Member
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.
 

audioguru

Well-Known Member
Most Helpful Member
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.
 

kchriste

New Member
Forum Supporter
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
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.
"7555", "IPA" company, in the bottom i think this is batch number or not I don't know
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.
 

colin55

Well-Known Member
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.
**broken link removed**
 

audioguru

Well-Known Member
Most Helpful Member
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
No.
The 7555 is a Cmos 555 that has an output that goes rail to rail (as high as the power supply voltage).

the diode removes another 0.6V.
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.
 

colin55

Well-Known Member
This is entirely wrong:
7555 is a Cmos 555 that has an output that goes rail to rail. Therefore the capacitor charges to 2.75V

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.
 
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audioguru

Well-Known Member
Most Helpful Member
The 7555 will only source 1mA when the supply is 5v and the output voltage is 1v lower than rail.
No.
Intersil's datasheet shows a sourcing current typically about 7mA.

The capacitor provides a Dead-Short across the output and thus the capacitor will never charge.
No.
The capacitor never discharges to zero volts. It discharges to about 0.3V when the LED stops conducting.

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.
No.
The output of the ICM7555 sinks about 2mA into the LED and discharges the capacitor from 1.25V to 0.3V.

How did you get the impression that the charge-rate will be low?
The sourcing current with a 2V supply is shown as typically 1mA. It is probably 0.8mA with a 1.5V supply. The capacitor needs to charge from 0.3V to 1.25V in 167ms. It cannot do it so the capacitor will charge to only about 0.6V.


How did you get the idea that the capacitor will charge to 2.75v?
I meant that the voltage available to the LED is the 1.5V battery plus the 1.25V of the capacitor's charge.

I would make one to prove that it works but I already have an LM3909 LED flasher working from a 1.0V to 1.5V battery. Its flash is much brighter than this weak circuit.
 

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colin55

Well-Known Member
The capacitor never discharges to zero volts. It discharges to about 0.3V when the LED stops conducting.
This is incorrect. You are assuming the voltage on the capacitor is 1v or more. But the electro never charges.
From the graph you provide it is clear that the current at 2v Vdd is only about 1 mA. The other 7555 chips state 1mA for 5v not 7mA so the current capability at 1.5v will be even less than 1mA.
The whole concept of the circuit is a failure. If one person could not get it to work, then it's a failure. I have been shown these circuits many times and none of them work at 1.5v.
The circuit I designed puts a high current into the LED to produce a bright flash. It's not the most efficient design but it's cheap and works well. The 47R is wasteful but cannot be avoided.
 
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audioguru

Well-Known Member
Most Helpful Member
Colin,
Which manufacturer of a 7555 has an output high current of only 1mA?
Intersil, Harris, Philips and Maxim are all the same with about 5mA to 8mA. National and Texas Instruments do not show typical current graphs.
 

audioguru

Well-Known Member
Most Helpful Member
That is correct.
The capacitor charges with a little less than 1mA for a duration that is twice as long as its discharge a little into the LED.

It is a horrible weak circuit and the 7555 oscillator might need to make many cycles for the capacitor to become charged enough to dimly light the LED.
It probably won't work if the battery voltage is less than 1.4V.
 
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