Yet another Wheatstone bridge with thermistor

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Discussion Overview

The discussion revolves around a Wheatstone bridge circuit involving a thermistor, focusing on the calculation of voltages V1 and V2 at a temperature of 80 degrees Celsius. Participants explore the implications of thermistor resistance values and op-amp behavior within the circuit.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a thermistor resistance equation and asks for the voltages V1 and V2 at 80 degrees Celsius.
  • Another participant questions the initial resistance value of the thermistor, suggesting it should be 100 kΩ instead of 100 Ω, citing physical impossibilities with the given parameters.
  • Some participants agree on the need to correct the thermistor's starting resistance to 100 kΩ.
  • There is a discussion about the current draw of op-amps, with one participant noting that they draw zero current due to infinite internal resistance.
  • A participant suggests using the Wheatstone Bridge equation to solve for the voltages.
  • One participant emphasizes that V1 and V2 should be determined separately as they are independent, except under specific conditions.
  • Another participant points out an error in the analysis regarding the shifting of input voltages and its effect on the output voltage.
  • One participant shares their calculations for R2 and expresses confusion about the sign of the result, questioning whether the configuration is inverting or non-inverting.
  • A later reply provides a simplified approach to calculating R2, indicating a realization of basic electronics principles.

Areas of Agreement / Disagreement

Participants generally agree on the need to correct the thermistor's resistance value, but there are differing views on the analysis of the circuit and the calculation methods for V1, V2, and R2. The discussion remains unresolved regarding the implications of the circuit configuration.

Contextual Notes

Some participants express uncertainty about the assumptions made in their calculations, particularly regarding the behavior of the op-amps and the implications of voltage shifts in the circuit analysis.

Who May Find This Useful

This discussion may be useful for students or individuals interested in circuit analysis, thermistors, and op-amp configurations, particularly in the context of Wheatstone bridge applications.

Femme_physics
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Homework Statement



So I got a thermistor here and am told that in the range 0-80 celsius the thermistor's resistance is given by Rt = R0 - kT ...whereas R0 = 100 ohms which is the thermistor resistance at 0 celsius. k = 0.5 kohms/celsius. T is of course temperature. The op-amps are ideal and their supply voltages are 15 volts

Question: The measured temperature is 80 deg celsius. What re the voltages in V1 an V2.


http://img854.imageshack.us/img854/80/circuit1.jpg

The Attempt at a Solution



According to my logic, V1 has to be 12 volts

Since V+ = V- = V1

http://img837.imageshack.us/img837/8282/loopex.jpg
 
Last edited by a moderator:
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Hi FP.

Can you check the values you've posted for Ro and k for the thermistor? 100 Ω seems a bit low for a starting resistance, and .5 kΩ/C seems awfully large if Ro is so small -- the thermistor would be showing negative resistance after about 0.2 °C. This is not physically possible. I think your Ro should be something like 100 kΩ.

Next, you should recall the "rules of thumb" for op-amps. How much current does an op-amp input draw?
 
I agree with Gneill. Your equation indicates that over the range 0 - 80 the resistance of the thermistor decreases by 500ohms/degree. I also think that your starting resistance is probably 100kohms, not 100ohms
 
You're right, it's 100k, I miswrote.

The op-amp draw 0 current, because of an infinite internal resistance! :) which is why I allowed myself to make a loop there. Perhaps I've forgotten elementary electronics already. I think I should just use the Wheatstone Bridge equation to solve it though

http://img577.imageshack.us/img577/672/solutttt.jpg
 
Last edited by a moderator:
You should determine V1 and V2 separately. After all, they are independent of each other, and determined by separate voltage dividers. They won't be equal in value except for a very particular value of RT.
 
Last edited:
Femme_physics said:
The op-amp draw 0 current, because of an infinite internal resistance! :)

That's right, which means it's like they aren't there :) They sample the voltage but don't affect the circuit in any way.
 
For your analysis you've offset the input voltages by subtracting 6V from each, yet you've left the output unshifted at 12V. That's going to produce incorrect results.

Rather than trying to force the circuit into a form for which you've memorized the gain formula, you can directly analyze the circuit as given without too much trouble.

attachment.php?attachmentid=54922&stc=1&d=1358862041.gif


Notice that you have the potentials at either end of the 1 kΩ resistor. So what is I? You've also got the potentials at either end of R2...
 

Attachments

  • Fig1.gif
    Fig1.gif
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ah...easier than I thought.
I1 = 1.5/1000
VR2 = 12-7.5 =4.5V
R2 = VR2 / I1

I forget my basic electronics sometimes, I didn't think it would come in handy in control systems! Who'd thunk it.

Much obliged gneill, sorry for the long delay :)
 
  • #10
Femme_physics said:
ah...easier than I thought.
I1 = 1.5/1000
VR2 = 12-7.5 =4.5V
R2 = VR2 / I1
That'll do it :approve:
I forget my basic electronics sometimes, I didn't think it would come in handy in control systems! Who'd thunk it.
Devilishly pervasive this electronic stuff :smile:
Much obliged gneill, sorry for the long delay :)
No worries. Always happy to help.
 

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