Over temprature sencing circuit

[carbon dude oxide]

Hello, if i want to make an over temprature circuit to turn an led on to let you know its over x°C would it be done by:

Having a thermostor in a potential divider and having the output from that put into a smitt trigger to stop flickering and maby to make the LED stay on for a bit of time have a pulse lengthener?

Or is there a better way of doing it?

[Psi]

Use a comparator

http://pfnicholls.com/electronics/comparator.html

[carbon dude oxide]

Ok so using that circuit to sence if the temp is over 100°C



If i set the bottom two resistors to 100k and the top left resistor to 4.7k. Acording to this thermistor :
http://www.maplin.co.uk/100k-bead-thermistor-2220?action=noagentcheck#specification

This should turn the one of the led's on at aproximatly 100°C ?

[carbon dude oxide]

Also what type of thermistor would be best to put onto an IC or a heatsi k and how should it be attached to it?

[c4757p]

This should turn the one of the led's on at aproximatly 100°C ?

1) Yes, the red one.
2) Most common, cheap comparators don't output anything, they only sink current. If you use this circuit, use an op amp instead of a comparator unless you rearrange it with that in mind. (Watch Dave's video on the overload detector if you don't understand what I mean.)
3) A bipolar supply is pointless for a temperature comparator.

Also what type of thermistor would be best to put onto an IC or a heatsi k and how should it be attached to it?

Use your imagination... 

[carbon dude oxide]

3) A bipolar supply is pointless for a temperature comparator.

What do you mean by this?

And i was thinking of adding a dab of thermal paste between the IC/heatsink and use the solid core wire on the thermistor to hold it there, maby some hot snot

[Corporate666]

How precise are you trying to be?

You could make this ridiculously simple by just using a thermistor and resistor set up as a voltage divider, driving the base of a transistor.

[carbon dude oxide]

What type of transistor would it need to be? Mosfet?

And then just set up the potential dividers output to go higher than the minimum required voltage of the fet in order to turn it on?

[c4757p]

3) A bipolar supply is pointless for a temperature comparator.

What do you mean by this?

That circuit has V+, V- and ground. There's no need for V- except for the incapability of the designer to understand modern circuit design...

What type of transistor would it need to be? Mosfet?

MOSFETs don't have bases, do they?  You could use a MOSFET, but it will be more predictable with a BJT. You have the right idea, though you only need one divider.

[carbon dude oxide]

Ok so how would i work out what i would beed to do with the bjt transistor?

[c4757p]

Calculate I_C based on how bright you want the LED to be. (I'd say a hefty 10mA or so - it's a warning LED, you want it to be bright). Then calculate I_B (it's I_C divided by the transistor's DC current gain / h_FE). Now, you want around 0.65V on the base (the usual threshold voltage), and you want the current through the divider to be around ten times the base current to avoid that affecting your calculations. Calculate the resistor to give you 0.65V, then make sure you'll have enough current. (If you won't, you will have to modify the design a bit, but I don't think it will be an issue.)

Of course, if you do find you don't have enough current, you could just choose a lower resistance thermistor, unless you've already bought yours. You also need to make sure you're not putting enough current through it to warm it up, of course.

[c4757p]

Note that I accidentally put the thermistor in the wrong place; it should be on the top.

[carbon dude oxide]

Ok you will have to forgive me if i get this wrong, i have not gotten threw transistor theory and stuff like that yet but this is what i have calculated for a BC548B (one i had in my draw)

Thermistor resistance at 100°C = 4.77k

Ic = 10mA = 0.01A
Ib = Ic/Hfe = 0.01/200 = 50uA
10xIb = 0.5mA

R2 (of potential divider) = R1 / ((Vin/Vout)-1)
R2 = 4770 / ((5/0.65)-1) = 712.7
Nearest value = 680 ohms = R2

I = V/R = 0.65/4770 = 0.14mA

Does this mean that my thermistor is too high of a resistance for this to work and i will need to get a smaller one? That is if i have calculated this all correctly ^.^

[Corporate666]

Ok you will have to forgive me if i get this wrong, i have not gotten threw transistor theory and stuff like that yet but this is what i have calculated for a BC548B (one i had in my draw)

Thermistor resistance at 100°C = 4.77k

Ic = 10mA = 0.01A
Ib = Ic/Hfe = 0.01/200 = 50uA
10xIb = 0.5mA

R2 (of potential divider) = R1 / ((Vin/Vout)-1)
R2 = 4770 / ((5/0.65)-1) = 712.7
Nearest value = 680 ohms = R2

I = V/R = 0.65/4770 = 0.14mA

Does this mean that my thermistor is too high of a resistance for this to work and i will need to get a smaller one? That is if i have calculated this all correctly ^.^

You could use a different thermistor.

Or you could use a different transistor (with a higher beta)

Or you could add another transistor and set it up as a darlington feeding from the first transistor.


c4567p is a good dude - he's given you excellent advice and obviously knows his stuff!

[c4757p]

Yup, there are many ways to approach this. I'd go with the Darlington method. Read that whole article and you should be able to figure out how to adapt it.

Alternately, use a different thermistor, yes.

You can always use a higher gain transistor, but above 200 or so they start getting more expensive. Open this datasheet and drool at the gain - but they are 30-40 cents each. They go higher, too.

[c4757p]

c4567p is a good dude - he's given you excellent advice and obviously knows his stuff!

Who is this c4567p of whom you speak?

Alternately - hell yes I do, I'm a Super Contributor! 

(Yes, I'm easily amused, and those new labels seem funny to me...)

[carbon dude oxide]

Ok so a darlington pair like this : http://www.maplin.co.uk/power-darlington-transistors-33874?action=noagentcheck#specification

Should work as ive calculated it to have a base current of 1.3mA i shall have alook as some better darlington pair transiators in the mornig, thank you very much for your help

[Corporate666]

c4567p is a good dude - he's given you excellent advice and obviously knows his stuff!

Who is this c4567p of whom you speak?

Alternately - hell yes I do, I'm a Super Contributor! 

(Yes, I'm easily amused, and those new labels seem funny to me...)

Well, OBVIOUSLY I wasn't talking about you!  Wait until he comes along - he'll be pretty angry that you are stealing his thunder 


I can't believe I typed that wrong - I even double checked it... still at work now though, and one of the CNC machines was making a funny sound so I clicked "send" and ran to check on it    Yeah, that's my excuse and I'm sticking with it!

[c4757p]

You don't need a special Darlington device... it's called a "pair" for a reason! Look at its symbol!

[c4757p]

I can't believe I typed that wrong - I even double checked it... still at work now though, and one of the CNC machines was making a funny sound so I clicked "send" and ran to check on it    Yeah, that's my excuse and I'm sticking with it!

Excuse accepted, partially out of jealousy. I'm just a student right now, the closest I have to a CNC machine to distract me is a boring old dog. And they (usually) know what "shut up" means

[carbon dude oxide]

Ok to make sure a darlington pair is two transistors arranged in such a way that the current comming out of the emmitter of the first transistor is used to turn on the base of the second transistor. So you can have a low base current on the first transistor which normaly would not fully saturate a single transistor or it would fully saturate but not enough for the current that is required to go through the transistor, so the emmitter current comming out of the first transisor which is significantly higher than the base current due to the gain is used to fully saturate the second transistor allowing a significantly higher current to be switched with a relativly low base current.

Am i correct in saying this?