Very nice! I will breadboard that up and see what happens. Thanks!

 

When using transistors as saturated switches, you use a beta/gain of 10. If you try to use the rated beta/gain, the transistor will fall out of saturation, leading to high power dissipation and likely overheating.

The LM393 has a pretty low current sink rating; 6mA if I remember correctly. That means a base current of 6mA if the correct base resistor is selected, and a max collector current of 60mA. Of more is needed, consider using a MOSFET or a Darlington transistor.

 

Marc I noticed there isnt a variable resistor in your drawing. Which one is the pot that controls the thermistors set point? It will be near the end of the thermistors capability at 220 degrees (F).

 

That is were that link I gave you came in.

If you know the PN of the thermistor you have you can select what its resistance should be at whatever temp point you want to use.

As the resistance of the Thermistor is quite low at 100C, you might want to consider using a lower value for the fixed 100K resistor

Also, if you are comfortable with simultaneous equations, and or solving for functions with 3 variables, you can derive the resistance at any point with pretty good accuracy.

Of course you could also put the thermistor in a pot of boiling water and measure its resistance.

I also agree that perhaps you should use a darlington or a MOSFET, but why not try the 2n4401? Its minimum current gain at 150 mA is 100 and you need maybe 60mA. 2 mA of base current should saturate the transistor. I also think that the 'Green' led in the pull up circuit may present problems.

If so, if not 'Yellow" and not "Red" should indicate that you do not have an open circuit and are not too hot.

 

I am not comfortable with ANY of this. I am lost concerning figuring out the exact ohm value for 220 degrees. Why would i have to do that if I want an adjustable number? I want to go from room temp to about 240. This is for a warning system that can also be used for an electric fan control on vintage vehicles. Where to I put the potentiometer to adjust? I like the fact that my current circuit works at room temperature so I can verify it is working. How would you set up a supervisor circuit using the schematic I already have with the change of the therm being on the ground side?

 

I was wrong as usual.. there IS some differences in the circuit when just exchanging the Thermistor position. I will re-write the schematic, but Pin 2 is now the Potentiometer control and Pin 3 is between the two 100K resistor/thermistor. Then wiring the Pot backwards from the way it was puts everything back to normal. It still takes a low on Pin 1 to fire the LED. I do not know enough to make a NPN work, so the PNP does the job. I am trying to keep the component count low without compromising quality. I would add a supervise circuit if I knew enough. I tried breadboarding Marc's design today and nothing happened and me without a clue where to look!

 

Deve:

Sorry, that I was a little carried away.

I have in mind a correction using your original circuit. Later.

 

Thanks Marc.. once I learn something, I pick things up pretty quick. This is all so interesting to me. The plan is to absorb everything I can! Thanks again for the effort!

Latest Schematic:

 

Here, Therm_ver_2.pdf is closer to your original design.

Also please ref the dat sheet for resistance values at selected temps (page 12.pdf)

It would be cool to make the led status indication into a 'traffic light' signal using just one red/green led.

Would require a few more components and some thought.

Actually this circuit would be much more simpler if a uP was used.

 

Thank you again Marc! A few things, I am a using 100K Therm so its okay to sub out the 10k fixed and the Therm to 100K? Also I do not see how you are getting a high at pin 1, all I have ever gotten was a LOW which caused me to go to PNP, but you may have fixed that? I will do this tomorrow! Looking forward to it! So the yellow light will come on if one of the Thermistors leads are cut?

 

The following assumes that the thermistor you are using has the same characteristics as the 100k@25C part in 'page_12.pdf (highlighted)'. If not, review the spec sheet for your device and use its parameters. However, the method outlined below should not have to change.

You can keep the pot at 100k, I did not mean to change it to 10k, but the other fixed resistor needs to be changed from 100k to 10k.

Because the thermistor has a negative temperature coefficient, the voltage at the inputs of pins 2 and 5 will increase as the temperature goes down and decrease as the temperature increases

The resistance of the thermistor at 100C is 5.8k and the voltage at pins 2 and 5 will be 4.4v.

I would suggest using a 10k rather than a 100k for the fixed resistor as the ratio of 5.8 k to 100k is 0.058 as the ratio of 5.8k to 10k is 0.58 which puts your voltage at 5.8/(5.8+100) * 12 = 4.4V rather than 5.8/105.8 * 12 = 0.657V if you use the 100k

Because the input of pin 2 is inverting and the pin 3 connected to the reference voltage established by the pot is the non inverting input, when the voltage at pin 2 is greater than the reference voltage at pin 3 the output at pin 1 will be low and the NPN transistor connected to pin 1 will not be able to saturate (is turned off) as the base emitter voltage is 0. When the input at pin 2 is lower than the reference voltage the output at pin 1 will try to go high. At this point the open collector transistor in the LM393 (see spec sheet) stops conducting allowing 2.7 ma of current from the 4.7k pull up resistor (12v/4.7k) to flow out of the base to emitter of the NPN potentially enabling 270ma of collector current (assuming a current gain of 100). 270ma of collector current will not be allowed as the current is limited by the 680 ohm resistor (12v/680) = 17.6 ma and the current requirements of the relay(~ 30 ma) for a total of ~50ma . When this is happens, the Red LED will light and the Relay will energize.

If this does not happen, try a transistor with a higher current gain, or a MOSFET.

How the transistor boosts the current from 2.7 ma to 0.27 Amps is the subject of another discussion.

Also be aware that when I say the output of the comparators goes hi, the voltage that you would measure would only be about 0.7V. This indicates that the base – emitter junction is forward biased and that the Collector – Emitter are saturated. If you were to use a darlington, this voltage would be about 1.4V. If you were to use a N Channel MOSFET the voltage could go to 12v. Watch out for this and pay attention as to the correct gate to source voltage. (Contained in the spec sheet) If Vgs is, say 3v, then use a voltage divider to limit the voltage to this level.

Also if the relay fails to operate the piezo, check the minimum current rating for the relay. If this tuns out to a problem, operate the piezo in series with the LED and use the contacts of the relay for higher current requirements.

In the other half of the LM393, a similar action takes place, but because the voltage at pin 5 is connected to the non inverting input (pin 5), the output at pin 7 is the opposite of the voltage at the output at pin 1, (It follows the input at pin5) thus the NPN connected at pin 7 will not be ON when the voltage at pin 5 is greater than the reference voltage established by the voltage divider at pin 6, but will be ON when it is less (Cold Temp or open ckt).

When the output open collector transistor in the LM393 at pin 7 is OFF (colder or open cat), the led will activate via the pull up resistor connected to pin 7.

By using the resistance vs temp data in the thermistor data sheet (Page_12.pdf) and the voltage divider equation Ra/Ra+Rb) * 12V, you should be able set the low temp and high temp points to whatever you want them to be.

 

Marc I really appreciate your help and your very detailed explanations. The first half of your circuit does not work any better than it did before. I am sure it has to do with trying to forward bias pin 1, but I do not know 1/10th of what you do about this. The circuit IS working up until the point where the relay doesn't fire. My circuit has zero problems, other than the distasteful concept of using a PNP. The relay works, the entire circuit works great. I would LOVE it if someone could be critical of my work and tell me if I am pushing too hard (applying too much current or voltage) to a certain part of the circuit, but if it passes muster, I am just sticking with what I already have on the first part....

THAT BEING SAID.. Congrats on the second half of your circuit! That works flawlessly as long as I connect that half to pin 3 of my half. I am very impressed and elated that I will have a supervisor circuit on board. Removing either lead of the thermistor causes the yellow LED to come on. So, thanks for taking the time to not just provide a solution, but the teaching aspects as well. I need all the help I can get! I will have a detailed schematic drawn up later today. Thanks again!

 

Deve
Glad to hear you got it working. I am looking forward to seeing your schematic.

I probably made some dumb mistake, like I said I never did bread board this thing. I would really like to see where I went off the rails.

 

Here is the latest schematic. This works really well, but if there are any anomalies that could pop up, please let me know. Thanks to everyone for the help!

 

Deve

Glad to hear it is working.

I found out what was wrong with my version. The input voltages need to be 1.5 Volts less than the positive supply voltage (Vcc).

This is in the spec sheet under 'Input Common Mode Voltage Range' .

I changed the 910 ohm resistor to a 2k and it worked perfectly.

I also have an idea for the 'traffic' light display if you would be interested. It would use a tri color led and a quad comparator LM339.

 

It sounds cool Marc, but I don't need it for this project. The box will be stuffed under the dash somewhere and the driver will be focusing on the 4 alarm lights already in play, but thank you for the help.