twin dividers problem

[mdmcs1]

hello all!

i am working with a temp measurement circuit using a 10k NTC thermistor. i will post a pic of what i am doing with this circuit, but my problem is that i am not getting the voltage i expect to my processor.

see attatched diagram

any help is wonderful - is there a formula i need to calculate the expected voltage to my processor?

thanks a ton guys!!

 

[BeeBop]

hello all!

i am working with a temp measurement circuit using a 10k NTC thermistor. i will post a pic of what i am doing with this circuit, but my problem is that i am not getting the voltage i expect to my processor.

see attatched diagram

any help is wonderful - is there a formula i need to calculate the expected voltage to my processor?

thanks a ton guys!!

That looks like a bridge circuit. You need to use precision resistors, and balance your bridge (use a trim pot on one leg.) If I remember, should give you 0V at 25 deg. C.

 

[Nigel Goodwin]

That looks like a bridge circuit. You need to use precision resistors, and balance your bridge (use a trim pot on one leg.) If I remember, should give you 0V at 25 deg. C.

It's not a bridge, it's just a potential divider, two 10K's across 5V, with a thermistor in parallel with the bottom resistor.

Perhaps it's supposed to be a bridge?, but it's drawn wrong, and needs a differential amplifier to provide the output.

 

[BeeBop]

It's not a bridge, it's just a potential divider, two 10K's across 5V, with a thermistor in parallel with the bottom resistor.

Perhaps it's supposed to be a bridge?, but it's drawn wrong, and needs a differential amplifier to provide the output.

Ooops! Guess I should have taken more time or redrawn it.

Oh, yes, I see; it is close, perhaps supposed to be...

 

[mdmcs1]

thats the tricky part - this is on existing equipment that cannot be changed or modified at this time. and the drawing is exactly what i have.

the reason it is connected the way it is, is #1 to provide a varying voltage based on temperature, and then #2 to divide that voltage in half after that to limit the voltage that gets sent to my processor (so i dont let the smoke out of the processor)

what i am really interested in is what voltage i should expect to my processor based on the temp.

maybe there is no simple way to figure this out?

 

[BeeBop]

thats the tricky part - this is on existing equipment that cannot be changed or modified at this time.

the reason it is connected the way it is, is #1 to provide a varying voltage based on temperature, and then #2 to divide that voltage in half after that to limit the voltage that gets sent to my processor (so i dont let the smoke out of the processor)

what i am really interested in is what voltage i should expect to my processor based on the temp.

maybe there is no simple way to figure this out?

Well, at 25 deg. C. your thermistor should have a resistance of 10k ohms. so xxxxxx

 

[mdmcs1]

thats what i was thinking it should be, but i seem to be .5 volts low at room temp - i will have to look and see if my processor is somehow sinking some current on its input instead of floating maybe?

 

[steveB]

thats what i was thinking it should be, but i seem to be .5 volts low at room temp - i will have to look and see if my processor is somehow sinking some current on its input instead of floating maybe?

Assuming the input impedance into the microprocessor is large (>1M), you should see about 1.67 V at the input pin. The way to look at this is to realize that the thermistor has a 10 K resistor in paralel. So at room temperature, that combination is 5K. This 5K forms a voltage divider with the 10K resistor connected to the 5V supply. Hence the voltage is divided by 3. The 10K resistor connected to the micro input pin has no effect on the voltage unless current is flowing in.

Note that the 5V should be a very stable voltage reference if you want to get accurate temperature readings.

 

[BeeBop]

Assuming the input impedance into the microprocessor is large (>1M), you should see about 1.67 V at the input pin. The way to look at this is to realize that the thermistor has a 10 K resistor in paralel. So at room temperature, that combination is 5K. This 5K forms a voltage divider with the 10K resistor connected to the 5V supply. Hence the voltage is divided by 3. The 10K resistor connected to the micro input pin has no effect on the voltage unless current is flowing in.

Note that the 5V should be a very stable voltage reference if you want to get accurate temperature readings.

That sounds correct to me. Forgot about that one.

 

[Nigel Goodwin]

thats the tricky part - this is on existing equipment that cannot be changed or modified at this time. and the drawing is exactly what i have.

the reason it is connected the way it is, is #1 to provide a varying voltage based on temperature, and then #2 to divide that voltage in half after that to limit the voltage that gets sent to my processor (so i dont let the smoke out of the processor)

#2 - there is no potential divider to further divide the output, the two resistors are connected wrongly for that.

what i am really interested in is what voltage i should expect to my processor based on the temp.

maybe there is no simple way to figure this out?

Simple ohms law and parallel resistors, and knowing the range of resistance of the thermistor - it's very basic.

 

[arhi]

here's the simulation if you are still not clear

edit: I apologise for the dots, forgot to turn them off

 

[duffy]

If R2 and R3 are suppose divide the voltage in half, they are connected wrong.

 

[BeeBop]

Well, we have discerned that it is not a divide by two...

the divider is R1 and R2 || Rtherm (I thought the thermistor was 10k?) R3 is only going to limit the current into the controller.

I can see the reason for putting the 10k in parallel with the thermistor - to linearize the thermistor around 25 deg. mark