circuit review

Thread Starter

bobdxcool

Joined Feb 12, 2012
39
I built an automatic liquid dispenser using an arduino. Now I am planning to build one without an arduino. I am planning to use

1.) IR sensor (https://www.amazon.com/DAOKI-Infrar...d=1&keywords=IR+Sensors&qid=1592302302&sr=8-6).
This sensor outputs a low signal from out pin whenever an obstacle is detected.

2.) Small submersible dc motor pump (3-6volts 200mA).
(https://www.amazon.com/3V-6V-Submer...submersible+water+pump&qid=1592302828&sr=8-12)

3.) 555 Timer IC

4.) LM7805

5.) 4 alkaline 1.5V batteries in series

Please find attached circuit below.

The values of R & C that I plan on using are 100k & 10uf for 1 second delay (or a potentiometer and 10uf capacitor).

Also I plan on integrating a DC float switch (https://www.electronicscomp.com/water-level-sensor-float-switch-p43-india) in the tank. When the tank is empty, the motor should not run. When the tank is empty, there would be a connection between the two wires of the float switch. So I am thinking of connecting one end of the switch wire to positive terminal of motor (instead of +5v as seen in circuit below) and the other end of the wire to +5V.

Also, I was thinking if it would be possible to directly connect sensor output to trigger pin of 555 without any transistors or resistors.

Request the people in this forum to let me know their opinions on this circuit.

nocontroller.png
 
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AlbertHall

Joined Jun 4, 2014
12,345
I see two problems here.
The LM7805 needs at least 7V input. Is the supply voltage coming from a battery (whose voltage will reduce as it runs down)?
Also, you are switching the ground of the '555 circuit so when switched off the '555 will have no ground connection so cannot pull its output down to zero. This will result in the pump running continuously. To fix this you could use a circuit similar to the one below.
1592304277119.png
 

MisterBill2

Joined Jan 23, 2018
18,179
Given that the supply source is a set of batteries, why add a voltage regulator at all? the circuit should function very well without any voltage regulator. That will extend battery life a bit.
If you can arrange for a float switch to open contacts when the tank is empty that will allow a much simpler mode of operation and reduce the power consumption a bit.
 

ElectricSpidey

Joined Dec 2, 2017
2,758
In that circuit I don't see the 555 ground being switched, but I do see a permanent connection from the ground pin to the trigger.

Also without capacitor coupling between the NPN and the trigger you can never get a 1 second delay if the IR sensor stays high for longer.

If your sensor outputs a low when triggered you don’t need the transistor, in fact it won’t work the way you have it drawn, because a low can only turn off a NPN.

So…

Break the connection between pin 1 and 2 on the schematic.

Get rid of the transistor.

Couple the sensor output to the trigger pin with a cap, and place a pullup on either side of the cap. (if you need a pulse shorter than the output of the sensor)
 
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Thread Starter

bobdxcool

Joined Feb 12, 2012
39
Given that the supply source is a set of batteries, why add a voltage regulator at all? the circuit should function very well without any voltage regulator. That will extend battery life a bit.
If you can arrange for a float switch to open contacts when the tank is empty that will allow a much simpler mode of operation and reduce the power consumption a bit.
So my float switch wiring described above in the original post is right ? So you suggest to connect the float switch in parallel with battery instead of motor in order to save power. You are right. I can eliminate the regulator. Since the sensor should most probably work for voltages between 3 and 5V. I can use three AAA batteries in series to get 4.5V. That should be able to power the motor too but at a slower speed.
 

Thread Starter

bobdxcool

Joined Feb 12, 2012
39
In that circuit I don't see the 555 ground being switched, but I do see a permanent connection from the ground pin to the trigger.

Also without capacitor coupling between the NPN and the trigger you can never get a 1 second delay if the IR sensor stays high for longer.
Can I remove the transistor altogether and directly connect the sensor to 555 with a capacitor in between (0.1uf) ?
For the 1 second delay I am planning to use a 10uf capacitor and 10k resistor. Circuit corrected in original post.
 

ElectricSpidey

Joined Dec 2, 2017
2,758
Yes if the sensor outputs a low when active, you can get rid of the transistor, but you will still need the pullups to return the trigger high and discharge the cap.
 

Thread Starter

bobdxcool

Joined Feb 12, 2012
39
Yes if the sensor outputs a low when active, you can get rid of the transistor, but you will still need the pullups to return the trigger high and discharge the cap.
Yes if the sensor outputs a low when active, you can get rid of the transistor, but you will still need the pullups to return the trigger high and discharge the cap.
nocontroller1.pngSo this would work ?
 

Thread Starter

bobdxcool

Joined Feb 12, 2012
39
One more pullup on the other side of the cap.
Thanks for your inputs. Final circuit below. Added a dc float switch to the dc motor, which will turn on only if float switch is connected. Do I need to add any pullup (10k to +5v) or pulldown (10k to ground) to one of the pins of the float switch?

Additionally I am considering eliminating the regulator and directly connect three 1.5V AAA batteries to power the system. As my IR sensor should be able to work between 3 and 5V. Also, my motor (3-6V 220ma) at 4.5V should be fast enough for my application.

nocontroller2.png
 

ElectricSpidey

Joined Dec 2, 2017
2,758
You don't need any pullups at the float switch, just be sure it's rated for that motor.

4.5 volts is pushing the Lower limits of a NE555.

Your circuit should work, but be sure to test before building. (I know that sounds like a nag, so please forgive me)
 

ronsimpson

Joined Oct 7, 2019
2,989
You should think about not using the 7805 but use a "LDO" voltage regulator. The Low Drop Out regulators are harder to use and some require extra thinking on the capacitors. The 6V battery will get below 6V and the 7805 may not be able to regulate.
I noticed the motor is running from 5V which will also stress the 7805. Maybe move the motor to 6V.
 
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