• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

could someone help me design a simple circuit please?

Status
Not open for further replies.

FreeBaseBuzz

New Member
Hi,

I have a motion sensor door chime/alarm and would like to alter it slightly to drive a solenoid. The alarm is designed to run off a 9v battery, but the solenoid requires 12v.

Also the signal line from the sensor goes positive at 2.7v going into the IC that generates the chime sounds.

Could someone please help me design a simple circuit to drive the solenoid?

I guess there are two parts to this.

1. a step-up DC-DC converter to get the required 12v from the 9v supply.
(alternatively, I could power the system using 12v wall transformer and somehow drop the 9v for the motion sensor off that)

2. some sort of transistor/relay circuit to switch on the 12v solenoid when the 2.7v is detected on the motion sensor circuit.

I have a couple of PN2222's, 2N3904's and 2N3906's lying around here if they can be used somehow.

I have some basic electronics knowledge, but not enough to design this sort of thing from scratch.



All this effort to try and keep the chickens out of the veggie patch, maybe I should just invest in a fence :)

thanks in advance,
FBB
 

Boncuk

New Member
Hi,

I suppose you want to step up from a 9V battery to 12V. That should work. However it won't work for an extended time period. The battery will deplete very quickly.

A better way is using a 12VDC (unregulated) power supply and use a linear stabilizer for the 9V supply. (LM7809 (1A) or LM78L09 (100mA)).

You would help the forum members to help you if you post a schematic of what you have done so far.

Regards

Boncuk
 

colin55

Well-Known Member
Take the line that goes into the IC and add a 10k and use a Darlington transistor to drive the solenoid from a 12v supply.
 

FreeBaseBuzz

New Member
Thanks for the quick responses,
I haven't done a circuit diagram as yet, just some exploratory surgery with the multimeter, on the motion sensor. However I can put something together with mspaint and post it up.

Using the 12v power supply would be easiest, I was concerned with the battery life if I used 9v, but thought it might be worth a shot, as the solenoid would only be activated for a maximum of 15 seconds at a time (the duration of the alarm on the motion sensor)



don't suppose I could find a LM7809 or similar in a computer power supply? I have a few of those lying around.

where could I find Darlington transistors? what product code? or could I use 2 of the 2n2222's to make one? The only thing I know about darlington transistors now is 5min worth on wikipedia :)
 
Last edited:

KMoffett

Well-Known Member
FreeBaseBuzz,

:eek: Please find another site to host your images. This is a G-Rated Forum.

Ken
 
Last edited:

FreeBaseBuzz

New Member
repost of the image..

Motion Sensor System.jpg

@KMoffett: sorry if I used an unsuitable site, but I don't understand the relevance of the "g-rated forum" comment. AFIK, I didn't swear or post anything that wasn't g-rated. Could you please expand on your comment?
 

FreeBaseBuzz

New Member
drawing as requested

Motion Sensor System.jpg

this may be a really dumb question, but would I be able to use 2 LM7805's in parallel to give 10v?
 
Last edited:

colin55

Well-Known Member
You will find the signal to the chip is a pulse. You will have to extend it.
Put a 10k resistor in series with a 1N4148 signal diode off the pulse line and from the end of the diode put a 100u to 0v. From the join of these two put a 47k to the base of a BD679 Darlington transistor.

The solenoid and chime circuit do not have to run off the same voltage. So long as they have a common 0v rail, the concept will work.
 
Last edited:

FreeBaseBuzz

New Member
Thanks Colin55,

I've been playing around with livewire and trying out different things. to utilise components I already have lying around.

solution.jpg


The bottom circuit seems to work to drive the solenoid. Why though is the 10k resistor needed on the signal line?

THe top circuit however doesn't seem to work. It's a circuit I found on the Fairchild LM7805 datasheet that should allow me to dial in a output voltage somewhere between 7V and 30V. I can't seem to get it to work though.

I'd prefer to use this simply because I have 7805's here and a LM741 Op Amp already.
 

colin55

Well-Known Member
You just need 2 resistors on the output of the 3-terminal regulator to get any voltage you want (from 5v up). Put say 2k2 on the output and connect a 5k pot to it and take the other end of the 5k pot to 0v.
Now take the slider to the "common" or "adjust" terminal. See what voltage you get. You may need lower resistors (such as 1k resistor and 2k pot) depending on the current requirement of the "common" terminal.

I put 10k on the signal line so that the signal is not shunted to deck via the 100u as this may prevent the chip from working.
 

FreeBaseBuzz

New Member
Thanks Colin,

Is this what you mean?


solution2.jpg



One other final question. As I understand it "ground" is just a point of reference from where all other voltages are measured.

In the lower circuit, if I remove ground from the circuit, the positive side of the dc current source is +8v and the negative side is -4v and the system fails.

Could you please explain why it doesn't work when the ground reference is removed?

If I'm using a standard 240Vac transformer that outputs 12Vdc, would it be the same as the lower diagram with ground or without ground? Does that make sense? Just trying to understand the difference between this theory and real life.
 

mneary

New Member
Does the circuit fail, or does the simulation fail? Simulators need to be given assumptions, such as where to begin (e.g. zero volts.)

The "2.4v" provided to SW1 has no reference. You may assume that the reference is the southern most location on the diagram, but a simulator will not.
 
Last edited:

FreeBaseBuzz

New Member
Does the circuit fail, or does the simulation fail? Simulators need to be given assumptions, such as where to begin (e.g. zero volts.)

The "2.4v" provided to SW1 has no reference. You may assume that the reference is the southern most location on the diagram, but a simulator will not.
The simulation fails.

I haven't breadboarded the circuit yet. That'll be next :)

Thank you all for your help. I'll breadboard it and let you know how it goes.
 

FreeBaseBuzz

New Member
It Works!

The chickens are staying out of the veggie patch now, every time they go in, the motion sensor trips and switches on the sprinkler.. they don't stay in there for long :)

I just have to remember to switch off the water before I go in there now
:rolleyes:

I'd like to thank you all for your help, I wouldn't have been able to work it out without it.

Cheers..
FBB
 

Sceadwian

Banned
Chickens are pretty stupid they'll probably never learn to deal with the water. You on the other hand will only forget to turn the water off once or twice before you learn hehe. Please post a photo or three of your setup. Photo's on finished projects are always nice to see. Especially one's that work =)
 
Last edited:

flat5

Member
Is it good practice in a circuit like this to include a back emf diode across the motor? Is it completely unnecessary because the cap discharge causes the motor to slowly lose power?
 
Last edited:

Sceadwian

Banned
If you're suddenly switching the power on and off all the time as in a PWM circuit it is not a good idea to rely simply on a capacitor, it will limit the peaks a bit from the inductive kickback when the power is suddenly switched off and the motor becomes a generator. The diodes are not always required, but it depends on what you have driving the motor and the motor itself.
 

flat5

Member
Thank you, Sceadwian. Would this circuit benefit from a diode across the motor, to protect the transistors?
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top