Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Total Newbie Here, where to start

Status
Not open for further replies.

newbiebuntu

Newbie level 6
Newbie level 6
Joined
Jul 22, 2011
Messages
13
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,409
I would like to build an on/off switch for a science project. I am new to boards and electronics but have an interesting idea.

I'm looking for the switch to turn on when a ohm rating above 40 gets sent through a ground wire. It would then be off with any ohm rating from 0-39 (+/- 5 or 10 ohms is okay).

0-39 Ohms = good
40+ ohms = bad and the negative after the switch is turned on and a current is allowed through to turn on an LED light which is always plugged into positive. A piezo buzzer for an audible alarm could be added along with the LED.

Such an ohm reading would be read from a thermal sending unit (thermistor) to a gauge showing how cold or hot something is based on the ohm rating received from the thermal sending unit. Fridges, cars, etc have these setups for their gauges.

Thank you for your time and effort. I'm not sure if I was clear enough or not, and if I did anything wrong by posting this here. Again I'm a complete beginner and hope to get some info and use this as a starting point.

I read somewhere yesterday here someone was using a transistor as a switch?

Update: I'll be using a car battery to power everything. Working on the blinking LED, I take Supply Voltage minus LED Voltage divided by LED current. So in my case (14.4V battery - 3.0V LED) / .035A which is 35mAmps max = 325ohms, rounded up to next highest value of 330ohms (I'm using a 1/2W resistor).
 
Last edited:

I don't understand how the ohm rating is getting sent through the wire?

Yes a MOSFET is the easiest type of switch you can use. A high Voltage on its Gate will short the switch. A low voltage on the Gate will open the Switch.
 

So how would I calculate for the size of mosfet. based on 14.4v supply to + side of Alarm (LED and buzzer).

Ohms are getting through based on the regulation of the sending unit. Ground sent through the unit which is regulated by heat. Heat makes the ohms fluctuate. Lets say 5 ohms at room temp and 51.2 ohms +/-4.3 ohms at 90 deg C

Thermistor - Wikipedia, the free encyclopedia


Ah, another ohm gauge/ system:
Check out the puck that slides up and down which regulates 30-240 ohms of the negative side of current as is slides- Amazon.com: Moeller Marine Electrical Reed Switch Fuel Sending Unit (13" Deep Tanks): Sports & Outdoors

---------- Post added at 18:07 ---------- Previous post was at 17:58 ----------

The temp gauge in a car also runs off of a sensor that limits ohms that is read on the gauge.
 
Last edited:

A transistor can be used to sense the varying resistance but the circuit will be sensitive to temperature.
The sense error would be about 1% for 3 deg change.
Let us consider the 2-wire input as a resistor (actually it is like a variable resistance).
This resistor (two wire terminals) will be connected between the base and emitter of an npn transistor.
Then a constant current (calculation is approximated to have a practical starting):
Isense= Vbe / Rsense = 0.65 / 40 = 16 mA
So when Rsense < 40 Ohm , the voltage across it will be less than Vbe and the transistor turns off.
And when Rsense > 40 Ohm , the voltage across it will be greater than Vbe and the transistor turns on.

Unfortunately, the transistor will change from off to on (and on to off) gradually.
And as you noticed, the output state is the inverse of what we expect. One possible solution is to add another npn transistor to reverse it at the LED side.
And since we have now two transistors, it is possible to add a small positive feedback to let the transient region narrower.

Did you get the general idea? Could you try drawing a possible circuit so that we can discuss it together :)

Kerim
 
Last edited:

what is the exact device that you want to connect ?
the ohm rating , as specified by you , is not giving the full information.
 

The gauge and temp sensor are already in the fridge system and not changeable. I need to make from switch to the right of the paper. I have the LED/buzzer done. what about the "?" (switch)

2011-07-22_13-42-41_402.jpg





---------- Post added at 18:50 ---------- Previous post was at 18:49 ----------

One possible solution is to add another npn transistor to reverse it at the LED side.
And since we have now two transistors, it is possible to add a small positive feedback to let the transient region narrower.

This is where I got lost

Also I guess "On" could just be <45 ohms

It's a warning system for when the Temp gets too high :)
 
Last edited:

from the diagram , there is a temeprature sensor connected to a gauge(a meter basically).

and you want to connect an electronic circuit to light an led at a proper temperature ?
 

I think that is it in a nut shell. Oops I forgot the positive 14.4V side of the gauge.
 

I draw this one ... hope it helps.

The horizontal axis is actually 0 to 60 Ohms (though it shows seconds).

Below 40 Ohms, the LED is on (current is high).

I guess you did see the switch in the schematic :wink:

If you like we can talk about each part of it... so you would know how to change it if you like.
 

Attachments

  • OhmSns_01.png
    OhmSns_01.png
    104.7 KB · Views: 110
Last edited:

I'll have to do some research and figure out what all the labels/ symbols are and put it together. Thank you
 
Last edited:

Start with the 1st one in your mind :)

I mean which one ... not known to you?

For instance try to save the pic on your PC so that you can see it clearer (with Preview for example). Its resolution here is reduced.

I guess you recognized the resistors.

Which transistors you may have? 2N2222 for example?

The big yellow one is just the well know 5V regulator (having 3 pins). I use it here to generate a constant current to pass in the guage resistance about 13mA
 
Last edited:

Last edited:

Oh... you mean you have two resistances to sense ?

D1 is the LED.

Perhaps I couldn't understand fully your question... I thought one switch for one input (variable resistance).
 
Last edited:

If you left meter in circuit, you should measure voltage Ux to ground,
at point led and buzzer triggers on
and calculate the prestage component values based to voltage Ux.

Meter_Alarm_01.png
 

Kak111... Thank you... You are in charge here from now :grin:

I mean I couldn't understand what he likes to do.
 

Confused
Meter_Alarm_03.png


---------- Post added at 20:28 ---------- Previous post was at 20:14 ----------

What I need to figure out is the "?" in my first diagram. How to turn on the LED once 45 ohms has been sent through the thermistor on it's way to the gauge. When my "?" (built circuit) hits 45 ohms it would turn on the LED and buzzer until the temp/ohms goes below 45ohms at which point the LED/buzzer would turn off. I'm trying to explain it, but it's hard. This is my first try at this stuff. Hence the Elementary Electronics forum :)

Another example without heat involved.
Take the fuel sender I linked above. You send the ground through one side and the puck returns the resistance (the ohms value) to the gauge and the gauge has this "variable ground" adjusting the needle while it is fed by a positive line on the other end. Lets say I want to create a buzzer that alarms when the gas is low which would be 30 ohms. between the sending unit and the gauge I splice in and put a switch of some sort telling the buzzer to turn on at 30ohms at which point the needle would also show empty. I hope this helps.

A new version with the positive side to the gauge.
2011-07-22_16-09-59_688.jpg
 
Last edited:

So should I build the circuit and get a hand full of transistors and put them in for the lower value and work up to the hottest point I want that sets off the alarm = trial and error.

How would I find the voltage? Maybe heat up the original system to the point I want it to buzz and get the voltage?
 
Last edited:

No, we do some calculations.
and then draw needed circuit.
Do this..............

Temp_alarm_meas_01a.jpg

or best is if you get value Ux at trigger point.

KAK

Pict. reloaded 23.07.11...................................
 
Last edited:

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top