Questions on Thermistor Circuit

[bismark99]

Hi, I am working on understanding a simple thermistor temperature switch using a 741 Op Amp for a comparator.

The circuit I have attached does not have all the parts assigned a value, except for the thermistor and related reference resistors.

First comments and questions:
I choose a 100K thermistor because (if I've figured correctly) it will be 9.4k ohms at 212 degrees F. Using this figure, I chose a 10k resistor in series with the therm. This should bring the voltage to approx 1/2 the supply at 212 deg F.

I chose a 20 K Trimmer for the reference adjustment. This would seem to me that when the wiper is in the middle, I would be in the range of my 212 deg F trip point Does this make sense?

Also you will see that I am using a TIP 120 transistor. The sole reason is that I have some already.

The 741 spec sheet says I can have a 20m amp continuous draw at the output. If I take the 1000 amplification factor for the TIP 120, I am way beyond what is needed to close a relay with a 100 ohm coil (.12 amps @ 12 V)

Do I need to put a current limiting resistor between the 741 and Tip120?

Also I am curious on how to handle the offset null pins. Can I let them float?

I would also appreciate any comments on anything else you might see that I have done wrong.

I think that's it for now. Thanks!

Steve

 

[bismark99]

Thermistor, 741 opamp, Tip 120 questions

I have some questions about the attached schematic.

I used a 100k thermistor in series with a 10k resistor for the sensing part of the circuit. I have studies several charts on how to figure the resistance of the thermistor at different temperatures and if I'm figuring things right the thermistor should be about 9.4k ohms at 212 degF.

Is this correct?

If so, I used a 20k trimmer on the reference leg of the circuit, thinking that when the wiper is about midway I should be in the range for the 212 degF tripping point. The 741 should go low above 212 degF.

Do I have this figured correctly?

I attached the 741's output to the TIP 120's base directly. The data sheet says the 741 will supply 20 mA continuously. This is way below the 120 mA max base current for the TIP 120 and way above the current needed to drive a relay with a 100 ohm coil. It should only draw .12 amps.

Is this correct or should I put a current limiting resistor on the base for some reason.

Can I let the offset null pins on the 741 float?

Any other comments are welcome.

Steve

 

[crutschow]

I believe the circuit should work.

It always good policy to limit the output current of an op amp to minimize it's power dissipation. The maximum current as determined by the built in limiter is only designed to prevent damage from faults, not for long term operation in a shorted mode.

Using a current gain of 250 for the transistor to give a low saturation voltage and a .12A load you would need a base current at 0.5mA or so.
Assuming a high output voltage for the op amp of 10V and a 1.3V base-emitter voltage, the op amp output resistor value would be ~17k ohm.

The offset pins can be left unconnected.

Incidentally, if the capacitors in the schematic are polarized electrolytics, they are shown backwards. The straight plate is the positive terminal.

 

[Boncuk]

Hi Bismark,

the circuit seems OK so far, except or the already mentioned wrongly connected electrolytic caps.

If you want to drive a 100 Ohm relay (R8?) you'll fry the LED and the relay won't pull to on. It requires 120mA at 12V and that current is far beyond any chance for any LED to survive.

Better connect the LED parallel with the relay coil and use an appropriate current limiting resistor.

Concerning the base current of the switching transistor I'd be generous and spend 2 to 5 mA. So 6K8 should do.

Boncuk

 

[Diver300]

You do need a current limiting resistor between the output of the 741 and the TIP120 base. That isn't to protect the TIP120 but it is to let the 741 work correctly.

You should also add a resistor from the base or the TIP120 to ground. The 741 output will not drive to ground (use a better OP AMP. 741s were obsolete 20 years ago) so the TIP120 will stay on all the time. The TIP120 needs about 1.4V to turn on.

If you fit a 4.7 khm: in series with the 741 output, and a 1 khm: to ground it should work.

When the ouput is at maximum, about 9 V, then the current will be limited to about 2 mA. When the ouput is at minimum, about 3 V, the base of the TIP120 will be at about 0.52 V so it will be off.

You can let the offset null pins float.

R1 should be about 1 Mhm:. It is there to add hysteresis so that the relay turns on and off cleanly.

 

[wingerr]

Stumbled onto the forum during some search, looks pretty interesting-

I did something similar when the thermostat on a portable refrigerator went out but everything else otherwise worked fine- the thermistor made the thing work better than it ever did before; very tight temp control.

I used a smaller resistor for greater hysteresis to minimize the amount of switching at the control point; it widened up the temp control range, but was easier on the compressor. Don't know if that might be an issue with this application, but it could be tweaked on testing anyway.

Looking at the schematic in the 741 datasheet, shouldn't it be able to drive the output to ground through the Q20 and 50 ohms?


**broken link removed**

 

[bismark99]

Was question about thermistor curcuit, now about LM741

Thanks for your answers Diver300, which leads to more questions. As Wingerr points out there is only a 50 ohm resistor between ground and the output of the 741. My guess is that Q20 offers some resistance even when fully on and that plus the 50 ohm resistor brings the output somewhere above ground.

Looking at a 741 data sheet, where do you get the info to know what the min and max voltage would be on the output.

Also I was thinking that the TIP120 needed 2.5 volts to turn on not 1.4. I was looking at the Vbe value on the sheet. I thought that made sense, since the TIP120 was a darlington with two base-emitter junctions. Is this correct, if not, where do I look on the data sheet to know what the value is?

I thought the formula Vbe minus the min turn on (2.5?) divided by the limiting resistor (4.7k) would be (9v - 2.5v)/4700= 1.4mA base current.

How do you plug the 1k resistor (the one between emitter and ground) into the formula?

My relay is 100 ohm coil FYI.

Thanks!

 

[bismark99]

Thermistor,741 opamp,TIP120 questions

FYI everyone, I have double posted by accident here and in Electronic Projects.
I thought my first attempt at posting didn't go through.

Thanks for the replies. Crutschow, I never considered the dissipation of the 741 until you pointed it out. I'll fix my caps.

Boncuk, I was not clear in my post about the relay. J1 and J2 are the connections for the relay. The LED and resistor is in parallel. Sorry. Thanks for your reply.

Everyone is telling me the TIP120 needs 1.4 volts. Isn't it 2.5. It is a darlington with two base emitter junctions, and the spec sheet says Vbe(turn on) is 2.5.

Also am I anywhere close to being correct on my assuming the value of the 100k thermistor will be around 9.4k at 212 degF?

Thanks!

 

[audioguru]

The TIP120 is a power darlington transistor with a max allowed continuous collector current rating of 5000mA. But your relay uses 120mA and the LED uses 20mA so your load is only 140mA.

The TIP120 switches 140mA well when its base current is only 0.5mA. But the opamp can supply 20mA.

The max base-emitter voltage of the TIP120 is 2.5V when its collector current is 3A and its base current is 12mA. A graph in its datasheet shows a typical base-emitter voltage of 1.26V when its collector current is 140mA.

Use a little 2N4401 transistor instead. Its max allowed continuous collector current is 600mA and it switches 140mA well with a base current of 10mA.
The output high voltage of the opamp when it has a 12V supply is 10V and the base-emitter voltage of the 2N4401 is 0.8v so use a 910 ohm or 1k series base resistor.

The output low voltage of a lousy old 741 opamp is not low enough to turn off a transistor. So connect a 220 ohm resistor from base to emitter of the transistor and use a series 680 ohm base resistor.

Why is there a capacitor and a diode in parallel with the transistor?? The transistor will be stressed by discharging the capacitor. The diode will do nothing.

Both your capacitors are drawn upside down.

 

[Roff]

FYI everyone, I have double posted by accident here and in Electronic Projects.
I thought my first attempt at posting didn't go through.

Thanks for the replies. Crutschow, I never considered the dissipation of the 741 until you pointed it out. I'll fix my caps.

Boncuk, I was not clear in my post about the relay. J1 and J2 are the connections for the relay. The LED and resistor is in parallel. Sorry. Thanks for your reply.

Everyone is telling me the TIP120 needs 1.4 volts. Isn't it 2.5. It is a darlington with two base emitter junctions, and the spec sheet says Vbe(turn on) is 2.5.

Look at Fig. 2 on the datasheet. The Vbe(sat) curve shows the typical values. 2.5V is the worst case max at Ic=3A.

Also am I anywhere close to being correct on my assuming the value of the 100k thermistor will be around 9.4k at 212 degF?

Thanks!

You'll need to post a datasheet for the thermistor. There are countless different types.

I always shudder a little when I see someone using a 741 for a comparator. It will probably work for you in this case (with the modifications AG suggested), but I like to use a real comparator, like an LM393.

 

[audioguru]

Moderator:
Please join his two threads!

 

[bismark99]

Well, I guess the little 741 I have laying around served it's purpose with out even using it. Your comments on how it isn't the best and the reasons and solutions to it's short comings help me to understand.

I will turn the caps right side up.

The Diode in parallel with the coil is there because a fellow years ago told me he put them there on the theory that when the coil's field collapses, there would be reverse current that would be shunted around the TIP120. I know there is another diode between the coil and the transistor to block the current. He seemed to think the parallel one would help.

I don't know the reason for the cap in parallel. I looked at many examples of this type of circuit and saw it there from time to time. It was on page two of my questions.

Thanks everybody for your time. On to the LM 393 and 2N4401 transistor.

 

[Roff]

The LM393 has an open collector output, so you will need a pullup resistor from the output to Vcc.

 

[crutschow]

The Diode in parallel with the coil is there because a fellow years ago told me he put them there on the theory that when the coil's field collapses, there would be reverse current that would be shunted around the TIP120. I know there is another diode between the coil and the transistor to block the current. He seemed to think the parallel one would help.

There is no "reverse" current as such. It's the problem of switching any inductive load. The relay has an inductance that will keep its current flowing until the inductive energy is dissipated after you turn off the transistor and remove the relay's voltage. If you don't have a path for the current, the voltage on the transistor will become high enough to likely destroy it. The diode in parallel with the relay provides a path for that current. A diode in seres with the relay would have no effect on this.

The disadvantage of using a diode is that it increases the turn-off time of the relay since the energy is slowly dissipated (most is dissipated in the relay winding resistance). For most applications this is not a problem but, if it is, you can add a resistor or zener diode (cathode towards transistor) in series with the diode to increase the dissipation rate of the inductive energy. The resistors maximum value would be (Vce-Vsupply)/Irelay where Vce is the maximum transistor voltage and Irelay is the relay coil on-current. The zener's maximum voltage would be (Vce-Vsupply).

 

[bismark99]

Thanks! This helps a lot.

Steve