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Thermistor in Wheatstone's bridge 
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#1
Dec1412, 07:54 AM

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#2
Dec1412, 08:53 AM

Mentor
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#3
Dec1412, 09:08 AM

#4
Dec1412, 09:24 AM

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Thermistor in Wheatstone's bridge



#5
Dec1412, 10:54 AM

PF Gold
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#6
Dec1412, 11:04 AM

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An unpowered circuit that produces zero output under all conditions is hardly practical How will you manage to obtain Vb = 1.0 V when the temp goes to 90C?



#8
Dec1512, 04:13 AM

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Have you determined the equation relating V_{ab} to R, R_{t}, and E? 


#9
Dec1512, 05:32 AM

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Good to see you guys still around :)
So therefor no current in the circuit and Vab must equal 0 as well 


#10
Dec1512, 06:12 AM

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The bridge will not be much use without E being set to some value. ☹ So you have the equation for V_{ab}, and it contains 2 unknowns. You are provided with 2 conditions to be met, so using these you can assign values to the unknowns. You have determined the value for R. Now determine E. 


#11
Dec1512, 07:37 AM

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G'day FP. You've written an appropriate equation to describe the bridge (although you've changed the names of some of the given parameters: Vin for E, Vf for Vab). However, it's not fair setting the voltage supply E to zero when you're looking for R; the circuit stops behaving as intended if it has no power supply and will not meet the other operating conditions. E should be assumed to be a fixed, nonzero value (value to be determined later).
You will have noticed that the bridge is made up of two independent voltage dividers, and that both dividers split the same voltage (Vin or E). It's also been given that the numbered resistors will all have the same value. It's handy to remember that a voltage divider with two equal valued resistors always divides the potential in half. Knowing this allows you to immediately conclude that the potential at node b is 1/2 E, no matter what E eventually turns out to be. If Vab is to be zero, then the other voltage divider containing the thermistor must produce the same voltage division since it is powered from the same voltage source, E. The same principle of equal resistors dividing the potential in half compels us to choose R_{3} to be equal in value to the thermistor (under the specified conditions: temp is 20 C; Vab = 0). 


#12
Dec1512, 08:30 AM

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Yes, you two are right I changed the variables since I looked at an older example exercise and copied it.



#13
Dec1512, 09:07 AM

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So, yes, you're looking for a suitable value for E which will produce Vab = 1 V when R_{t} = 208.5 Ω. 


#14
Dec1512, 10:02 AM

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Great :) That clears it up! Thank you!
They're asking me if there is a linear ratio...does that mean I need to start plugging in thermistor values and drawing them on a graph to see, or is there an easier shortcut? 


#15
Dec1512, 10:24 AM

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I suppose you could start by doing a plot of Vab versus Rt, letting Rt go linearly from 2814 to 208.5 Ohms (so no table lookups required). You might find the resulting curve interesting. You might also plot a handful of resistance values versus temperature from the table to see what that curve looks like. Is the thermistor a linear R vs temperature device? But I suspect that what they want to know is if there's a linear relationship between temperature and Vab, which will require plotting Vab versus temperature. So the table comes into play. I'd think that half a dozen points strategically chosen to make the curve obvious would suffice. 


#16
Dec1512, 11:58 AM

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The question is " whether there is a linear relation between voltage Vab and the temperature at the range (2090 degrees celcius) . Explain your answer."



#17
Dec1512, 12:11 PM

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You could go by the linearity of the thermistor versus temperature curve alone IF the Wheatstone bridge produced linear response to a linear change in resistance. But it doesn't... 


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