# newbie trying to make a basic monitoring circuit

#### dirkdiggler

Dec 6, 2016
3
Hi guys and girls,

I've been googling for weeks and going crazy trying to teach myself this stuff, but I simply don't have the brain capacity or the patience to find what I'm looking for - so I'm hoping someone out there will find this simple to the point of stupid and can help me out

I'm trying to monitor a circuit of a set resistance, say 100ohms. If the circuit goes above or below this value, then the monitoring device will supply a 12v output (to a warning light). It's basically an alarm circuit that will run off a 12v SLA battery.

I have looked into using a basic comparator as the switching component, 100ohm in and 100ohm on the other/monitoring side - change the value by say +/- 10% and it switches an output. Unfortunately I don't know enough about them to know if they can switch when open circuited on one side, rather than sensing a change of voltage (needing that PD to switch like a relay). Is this the easiest way to do it, with the least components? Or can I use a transistor somehow?

I'd greatly appreciate any help or suggestions on a simple way of making this circuit effective and using the least amount of power as possible. I don't mind if the output, once triggered, switches a relay and draws more power, I just need to work out the switching side of things for now.

Thanks again

Dirk

#### davenn

Moderator
Sep 5, 2009
14,179
hi there
welcome

I'm trying to monitor a circuit of a set resistance, say 100ohms.

so, specifically, what is this 100 Ohms thing and what sort of circuit is it in ?

you will not be measuring a resistor value as such, but more likely the changing voltage drop across the resistance

#### dirkdiggler

Dec 6, 2016
3
hi there
welcome

so, specifically, what is this 100 Ohms thing and what sort of circuit is it in ?

you will not be measuring a resistor value as such, but more likely the changing voltage drop across the resistance

It's just a home made monitoring circuit, I want to make a little switch to go on the garage door so that when it opens the 100 ohms will open circuit, but it also can't be bypassed by bridging it out because then it will short circuit. If either of these things happen then it will bring a light on. I know you can buy things off the shelf but I want to make it

#### pgib8

Jul 26, 2015
107
You said you want to keep it simple so with that in mind here is what I would do...
Use two resistors in series, one is in a safe place (call it R1), the other is where one may tamper with it (call it R2).
If R2 is bridged, R1 will still limit the current so things hopefully don't catch on fire.
Let's assume your battery is 12V and you make each resistor 500 Ohm. Under normal circumstances 12mA will flow through the circuit (12V / 1000 Ohm).
It starts looking like this:
BAT+----->R1-------->R2------>BAT-

The left of R1 will always be at 12V while the right of R2 will always be at 0V. To make this transition something is happening in between, this is that each resistor will cause a voltage drop.
Think of it this way, more current can drain away from the right side of each resistor than what can flow through from the left, so think of it as a pressure drop on the right side of each resistor compared to the left of each resistor, and of course by that I mean voltage.

When you tap right in between R1 and R2 you will see 6V. Let's call this point the center.
What happens if R2 is removed and the circuit left open is that there is no more current that can drain away from the right side of R1. Despite R1 only letting some current come through from the left, since there is no where to go the voltage will build up right away so that the center becomes 12V too.
To visualize this you may imagine a garden hose that is sealed off. If you open the input valve even a slight bit, the water pressure will quickly build up to the same pressure that you have everywhere else in your house.
However open the garden hose up and the pressure will drop right away and you'll get a trickle coming out the other end. It's just like that. (The valve is the resistor).

So in your design let's say someone bridges R2, shorting it out. That means your 0V point moves right up the line towards the left so that at the center point you will have 0V. The little current that R1 lets through drains away to the battery immediately.

That's all I have for now (ran out of time sorry lol).

#### chopnhack

Apr 28, 2014
1,575
Well said pgib8!

Dirk, if you understood the above, its called a voltage divider. You can feed this into your op amp's inputs and use that to trigger your light.

#### AnalogKid

Jun 10, 2015
2,742
Your approach is a classic in alarm sensor loops, detecting both an open and short circuit with what is called a window comparator. 100 ohms is unnecessarily low, doing nothing but discharging the battery more quickly. I recommend 1K as the loop resistor.

The circuit uses two voltage comparators. These look a lot like opamps, but are not optimized for linear amplification and low distortion. Instead, have an open collector output stage that lets them do interesting things in control circuits. You can tie two or more directly together to a common output point, so that if any of the comparators "trips", the common output goes low and can be used to do other stuff. In your case, if you set up the resistor string with three 1K resistors (two with the circuit and one out on the loop), then, when all is normal, the lower node is at 4 V. A short circuit will pull this point to 6 V, and an open circuit will drop it to 0 V. If the two comparators are set to change their outputs at 3 V and 5 V, one of them will go low in either error condition.

Look up an LM393 dual comparator. This (and it's quad brother the LM339) are the most popular and robust comparator chips in the universe. The LM393 can sink enough current to drive an optocoupler, a small reed relay, or drive a transistor that can move even more current. What is it the circuit ultimately is turning on?

Schematic to follow.

ak

#### AnalogKid

Jun 10, 2015
2,742
First pass, rushed, maybe errors. Note that the components are what's in my design libraries. There is no need for a 0.1% R or a 50 V C. I show three different output options.

ak

#### Attachments

• Alarm-Loop-1-c.pdf
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#### dirkdiggler

Dec 6, 2016
3
Thanks heaps @AnalogKid !
I'm trying to decipher it all, like I said I'm very new to the electronics game so leave it with me and I'll try to see how you make it work Good on ya!

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