Over temprature sencing circuit - Page 1
Forum section: electronics / beginners

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 23, 2013, 08:16:23 am
content: 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?

author: Psi
author_rank: Super Contributor
timestamp: May 23, 2013, 09:09:05 am
content: Use a comparatorhttp://pfnicholls.com/electronics/comparator.html

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 23, 2013, 06:43:59 pm
content: Ok so using that circuit to sence if the temp is over 100°CIf 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#specificationThis should turn the one of the led's on at aproximatly 100°C ?

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 23, 2013, 07:21:48 pm
content: 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?

author: c4757p
author_rank: Super Contributor
timestamp: May 23, 2013, 07:51:23 pm
content: Quote from: carbon dude oxide on May 23, 2013, 06:43:59 pm1) Yes, the red one.2) Most common, cheap comparators don't output anything, they onlysinkcurrent. 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.Quote from: carbon dude oxide on May 23, 2013, 07:21:48 pmUse your imagination...

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 23, 2013, 07:59:26 pm
content: QuoteWhat 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

author: Corporate666
author_rank: Unknown
timestamp: May 23, 2013, 08:38:52 pm
content: 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.

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 23, 2013, 08:44:15 pm
content: 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?

author: c4757p
author_rank: Super Contributor
timestamp: May 23, 2013, 09:02:06 pm
content: Quote from: carbon dude oxide on May 23, 2013, 07:59:26 pmThat circuit has V+, V- and ground. There's no need for V- except for the incapability of the designer to understand modern circuit design...Quote from: carbon dude oxide on May 23, 2013, 08:44:15 pmMOSFETs 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.

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 23, 2013, 09:10:05 pm
content: Ok so how would i work out what i would beed to do with the bjt transistor?

author: c4757p
author_rank: Super Contributor
timestamp: May 23, 2013, 09:17:58 pm
content: Calculate ICbased 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 IB(it's ICdivided by the transistor's DC current gain / hFE). 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.

author: c4757p
author_rank: Super Contributor
timestamp: May 23, 2013, 09:25:05 pm
content: Note that I accidentally put the thermistor in the wrong place; it should be on the top.

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 23, 2013, 10:56:37 pm
content: 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 = 4.77kIc = 10mA = 0.01AIb = Ic/Hfe = 0.01/200 = 50uA10xIb = 0.5mAR2 (of potential divider) = R1 / ((Vin/Vout)-1)R2 = 4770 / ((5/0.65)-1) = 712.7Nearest value = 680 ohms = R2I = V/R = 0.65/4770 = 0.14mADoes 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 ^.^

author: Corporate666
author_rank: Unknown
timestamp: May 23, 2013, 11:22:39 pm
content: Quote from: carbon dude oxide on May 23, 2013, 10:56:37 pmYou 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!

author: c4757p
author_rank: Super Contributor
timestamp: May 23, 2013, 11:26:30 pm
content: Yup, there are many ways to approach this. I'd go with theDarlingtonmethod. 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. Openthis datasheetand drool at the gain - but they are 30-40 cents each. They go higher, too.

author: c4757p
author_rank: Super Contributor
timestamp: May 23, 2013, 11:34:01 pm
content: Quote from: Corporate666 on May 23, 2013, 11:22:39 pmWho 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...)

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 23, 2013, 11:39:30 pm
content: Ok so a darlington pair like this :http://www.maplin.co.uk/power-darlington-transistors-33874?action=noagentcheck#specificationShould work as ive calculated it to have a base current of 1.3mAi shall have alook as some better darlington pair transiators in the mornig, thank you very much for your help

author: Corporate666
author_rank: Unknown
timestamp: May 23, 2013, 11:40:18 pm
content: Quote from: c4757p on May 23, 2013, 11:34:01 pmWell, OBVIOUSLY I wasn't talking about you! Wait until he comes along - he'll be pretty angry that you are stealing his thunderI 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 itYeah, that's my excuse and I'm sticking with it!

author: c4757p
author_rank: Super Contributor
timestamp: May 23, 2013, 11:41:46 pm
content: You don't need a special Darlington device... it's called a "pair" for a reason! Look at its symbol!

author: c4757p
author_rank: Super Contributor
timestamp: May 23, 2013, 11:47:53 pm
content: Quote from: Corporate666 on May 23, 2013, 11:40:18 pmExcuse 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

author: carbon dude oxide
author_rank: Frequent Contributor
timestamp: May 24, 2013, 07:51:10 am
content: 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?