False triggering of a 555 timer

jacbk6121 · Mar 14, 2017

aardyvarky said:
Try the 2N2222A before you buy anything else, a 555 should source plenty of current. However, refer to Audioguru's comment about the battery - here (UK) we have 9v batteries in the shape of "PP3" and a good quality one has plenty of power - I just tried one (an Energizer) on a 6v motor. Just out of interest, what are you pumping?

Just water.

jacbk6121 · Mar 14, 2017

alec_t said:
.... but not for long. PP3 capacity is only ~500mAh, so a pump current of 150mA would flatten the battery in ~ 3hrs or less.

And that's fine, it only needs to be used for 30 s once a day.

jacbk6121 · Mar 14, 2017

I call the device a dipping bird home. The pump supplies an elevated reservoir (6" high), that slowly fill a lower reservoir by gravity through an orifice in which the bird takes it's dip. When the upper reservoir is empty which takes about a day the magnetic switch will fill up the reservoir. In the picture the lower reservoir is not shown. The dipping bird sits on the bottom reservoir that collects the water. All reservoirs have overflow protection and there is also a priming tube that gets rid of the air in the pump if that were to happen.

aardyvarky · Mar 15, 2017

audioguru said:
But the important problem is that the battery voltage drops so fast that the LM317 regulator does not work.

A 555 is fairly robust - have a biggish cap (1000microF) on that supply side and dispense with the LM317

PS thanks for pic of your project - it always helps to have more info!

jacbk6121 · Mar 16, 2017

aardyvarky said:
A 555 is fairly robust - have a biggish cap (1000microF) on that supply side and dispense with the LM317

PS thanks for pic of your project - it always helps to have more info!

I'd like to get rid of the LM317 but I can't. The pump can take a maximum of 6V. I could use (4) 1.5V AA batteries but if you add the cases around these batteries I run out of space.

audioguru · Mar 16, 2017

I cannot remember why you are not using a "low dropout" regulator. They regulate 5.0V perfectly when the input has dropped to 5.5V or less. An LM317 set to 5V or ordinary 5V regulator needs a minimum input of 8V.

aardyvarky · Mar 16, 2017

One question springs to mind (now that you have told us the whole story) - why can't you just have the magnetic switch turn on the motor (no electronics necessary)? There would be enough "hysteresis" in the system to pump enough water up to the tank and the circuit would consume zero current when the pump isn't operating (you could add the 27R 1W resistor in series to limit the voltage/current if the battery voltage doesn't drop enough).

jacbk6121 · Mar 16, 2017

audioguru said:
I cannot remember why you are not using a "low dropout" regulator. They regulate 5.0V perfectly when the input has dropped to 5.5V or less. An LM317 set to 5V or ordinary 5V regulator needs a minimum input of 8V.

I wasn't aware of an LDO regulator if it helps I'm willing to try it. I found this one on digikey
https://www.digikey.ca/products/en/...t=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25
It is a bit expensive, can you recommend something appropriate at a reasonable cost?

audioguru · Mar 16, 2017

That LDO regulator costs a fortune because it is rated at 5A and it can dissipate a lot of heat. Look for a small low current one. I have a bag full of small LDO 5V regulators that are not made anymore. The search engine at Digikey showed many wrong ones that I did not ask for.

jacbk6121 · Mar 16, 2017

audioguru said:
That LDO regulator costs a fortune because it is rated at 5A and it can dissipate a lot of heat. Look for a small low current one. I have a bag full of small LDO 5V regulators that are not made anymore. The search engine at Digikey showed many wrong ones that I did not ask for.

Here's one:
https://www.digikey.ca/products/en/...t=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25
the Vout voltage can be adjusted with a 10K pot, is this suitable?

audioguru · Mar 16, 2017

You show the same big, adjustable, high current, high dissipation and high cost one. Also you do not need to pay more for one with an adjustable voltage. A fixed 5V one will be fine.

jacbk6121 · Mar 16, 2017

audioguru said:
You show the same big, adjustable, high current, high dissipation and high cost one. Also you do not need to pay more for one with an adjustable voltage. A fixed 5V one will be fine.

Yes you're right that was the wrong link.
And as you say a fixed voltage Vout would be preferable, I need 6V out.
This one might be suitable:
https://www.digikey.ca/product-detail/en/fairchild-on-semiconductor/LM7806CT/LM7806CT-ND/1954727

audioguru · Mar 16, 2017

An LM7806 is not a low dropout regulator, its datasheet shows that it works properly with an 11V supply but with an 8V or 9V supply its performance is not too bad.

jacbk6121 · Mar 16, 2017

aardyvarky said:
One question springs to mind (now that you have told us the whole story) - why can't you just have the magnetic switch turn on the motor (no electronics necessary)? There would be enough "hysteresis" in the system to pump enough water up to the tank and the circuit would consume zero current when the pump isn't operating (you could add the 27R 1W resistor in series to limit the voltage/current if the battery voltage doesn't drop enough).

First the pump can only run max. 6V, so I need the LM317 because I need a 9v battery. Not sure what you mean by hysteresis in the system, I need the timer so that the pump runs only long enough to fill the upper reservoir because I don't have an upper level detector to turn the pump off.

jacbk6121 · Mar 16, 2017

aardyvarky said:
Try the 2N2222A before you buy anything else, a 555 should source plenty of current. However, refer to Audioguru's comment about the battery - here (UK) we have 9v batteries in the shape of "PP3" and a good quality one has plenty of power - I just tried one (an Energizer) on a 6v motor. Just out of interest, what are you pumping?

Just tried it, no luck.

jacbk6121 · Mar 16, 2017

audioguru said:
An LM7806 is not a low dropout regulator, its datasheet shows that it works properly with an 11V supply but with an 8V or 9V supply its performance is not too bad.

Can you recommend one?

audioguru · Mar 16, 2017

jacbk6121 said:
Can you recommend one?

I tried Digikey's search engine but it gave me what I did not ask for. You should try Texas Instruments www.ti.com website.

aardyvarky · Mar 17, 2017

jacbk6121 said:
First the pump can only run max. 6V, so I need the LM317 because I need a 9v battery. Not sure what you mean by hysteresis in the system, I need the timer so that the pump runs only long enough to fill the upper reservoir because I don't have an upper level detector to turn the pump off.

From Dictionary.com:
"noun, Physics.
1. the lag in response exhibited by a body in reacting to changes in the forces, especially magnetic forces, affecting it."

In electronics this is usually where a system has positive feedback - see EE vid on the Schmitt trigger.
In answer to the 6v requirement, that is the purpose of the 27R 1W resistor:
R=V/I so =Vbatt-Vmotor/Imotor = 9-6/.15 = 3/.15 = 20 ... (I suggested 27R because I thought you said that the motor was 5v)
so lets choose 20R instead:
W=I x I x R = .15 x .15 x 20 ≈ .0.45W so a 0.5W resistor will handle that easily seeing as how the pump will only be on for a few seconds.

If you raise the magnetic switch (I can't really see it on your picture) to the minimum point at which the dipper thing still works then problem solved - there will always be enough water in the upper tank.

Why make a complicated circuit when a simple one will do?

aardyvarky · Mar 17, 2017

Try this if you really must make a complicated circuit:
https://www.mouser.co.uk/ProductDetail/Taiwan-Semiconductor/TS2937CZ50/?qs=sGAEpiMZZMug9GoBKXZ752qL35/R8uEkHW/aXIN2ZVA=

I'm sure they are stateside originally and will stock that part.

PS They do.

jacbk6121 · Mar 17, 2017

aardyvarky said:
From Dictionary.com:
"noun, Physics.
1. the lag in response exhibited by a body in reacting to changes in the forces, especially magnetic forces, affecting it."

In electronics this is usually where a system has positive feedback - see EE vid on the Schmitt trigger.
In answer to the 6v requirement, that is the purpose of the 27R 1W resistor:
R=V/I so =Vbatt-Vmotor/Imotor = 9-6/.15 = 3/.15 = 20 ... (I suggested 27R because I thought you said that the motor was 5v)
so lets choose 20R instead:
W=I x I x R = .15 x .15 x 20 ≈ .0.45W so a 0.5W resistor will handle that easily seeing as how the pump will only be on for a few seconds.

If you raise the magnetic switch (I can't really see it on your picture) to the minimum point at which the dipper thing still works then problem solved - there will always be enough water in the upper tank.
View attachment 104981
Why make a complicated circuit when a simple one will do?

Thanks again for your help and comments. I think you are saying that I could raise the mag. switch to a higher level forcing the pump to stop when the water reaches this high level and start whenever the switch is not activated. This means that the pump will stop and start all the time with only a small movement of the float.

What I want to achieve is that when the reservoir is empty, the magnetic switch provides the signal to start the timer which starts the pump to fill the reservoir. This should occur about once a day. The magnetic switch is located at the bottom of the reservoir, the magnet that triggers it is on a float that rises with the water level.

False triggering of a 555 timer

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