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Thanks very much for the above explanation of R3.
I assume that R1 leaks away current/voltage till it drops to a level below what the chip recognises as an "on" voltage, at which point, the output changes post-haste. Nice
I have a whole lot of pretty coloured LED's including red, orange, yellow, green and UV.
But the one I use mostly, and have many of, is LED539 with its data sheet at http://www.fort777.co.za/images/Datasheets/LED539.pdf
"I assume that R1 leaks away current/voltage till it drops to a level below what the chip recognises as an "on" voltage, at which point, the output changes post-haste"
Yes basically,
put a bit differently:
C is discharge through R1 till the input voltage is below VinL of the inverter ,
causing the output to switch from VoL to VoH .
Yes basically,
put a bit differently:
C is discharge through R1 till the input voltage is below VinL of the inverter ,
causing the output to switch from VoL to VoH .
I have 74LS05 and CD4069UBE and GDF543. I am hoping the first will be suitable.
That is what I meant to say
Your circuit shows that the inverter sinks current when VoL as I understand it.
Dont know what GDF543 is.
The 74LS series draw too much STBY power .
I would test the circuit with the CD4069UBE:
1.with "parallel" INs and OUTs.
2.use a 5ma LED or less( if you have a 2ma one even better).
You will not get the proper illumination you need,
but just test the circuits operation.
order the 74HC14,when you have it ,use it in the real thing...
The 74LS series draw too much STBY power .
I would test the circuit with the CD4069UBE:
1.with "parallel" INs and OUTs.
2.use a 5ma LED or less( if you have a 2ma one even better).
You will not get the proper illumination you need,
but just test the circuits operation.
order the 74HC14,when you have it ,use it in the real thing...
Will try it right now thanks.
Hallo
Collin is not being rude, just being himself. Have a look at some of his other posts.
I have used a 270 Ohms LED resistor with the original 20mA LED which is bright enough in the dark.
The quiet current when off is 1.3mA at 5V = 0.0065Watts resting. That sounds pretty good to me.
The Cap is 1000u and the R1 is 168K Ohms. I paralleled only the left inputs and outputs which was easily bright enough, noticing that the more being // the longer the LED stayed on.
I am using 4 really old AA batteries which add up to 5V. When pressed and while LED glows, the V drops to 3.3V.
Ps. Swoping out the capacitor for a 220u and R1 for a 1M it now is perfect, staying on for 100s.
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I am using the 3 left inputs in // with the 3 left outputs only.Something is still wrong ,the STBY current of 1.3ma is way too high!
You should get no more than a few uA(a factor of 1000 down!).
I guess you didn't tie the unused inputs to GND or VCC...
It does not matter whether I ground or don't ground the 3 right inputs, now I am getting a STBY current of 6mA.
The only thing I changed was the cap and 1M resistor since taking the last STBY current of 1.3mA. I would not have remeasured if you had not said 1.3ma was too high.
Thanks Alec. Swopping out the CD4069UBE for the HEF40106, the HEF snaps off with no fade, and only draws 0.7mA.
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Something is still wrong ,the STBY current of 1.3ma is way too high!
You should get no more than a few uA(a factor of 1000 down!).
I guess you didn't tie the unused inputs to GND or VCC...
Dorke, taking your advice, tying down the 3 unused right part of the chip inverters, and by wiring the Ammeter into the circuit, I see the STBY current is less than 0.001mA using the Hef40106. Now that is impressive!!
With the LED on it draws 7 mA.
The CD4069UBE draws 7-8 mA with the LED on. The current drops slowly to below 6mA at which point it switches the LED off. Even then it draws 5mA and this just keeps dropping slowly.
Therefore the HEF seems to have done it with your circuit!
My original transistor drew 4.5mA when LED on, and this dwindled over 10(??) minutes to zero.
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O.K, that's how it should be !
Not only is it impressive ,
you got yourself 0.7/0.001=700 times the number of battery life STBY days.
That is the power of CMOS...
Furthermore ,If you can use a 2ma LED, it would be the most life you can get from the battery.
You can try lowering the current of the LED you are using now(from 7ma to 5ma or less, depending on illumination) it will improve battery life.
And you can leave it working with the 40106.
Not only is it impressive ,
you got yourself 0.7/0.001=700 times the number of battery life STBY days.
That is the power of CMOS...
Furthermore ,If you can use a 2ma LED, it would be the most life you can get from the battery.
You can try lowering the current of the LED you are using now(from 7ma to 5ma or less, depending on illumination) it will improve battery life.
And you can leave it working with the 40106.
