From there we went to a lab with thermistors. We're going to make a circuit that does a thing with a thermistor. That thing will be to increase V_out by .5V over a temperature range from 25C to 37C w/ 5V input.
1. Preliminary design (right, top right), calculations (left, and right, center), and comparisons (right). below in picture.
2. Note that cold in our instance is room temperature, and hot is held by Kevin's hot hands.
So the lab:
3. measured thermistor values
So we began by calculating the range of resistor values that will allow for a .5V change or greater in series with a thermistor whose resistance varies from 7k to 11k. We found a range of acceptable values between: 4.36k Ohms and 17.6k Ohms
Voltage drop vs time across thermistor
We then optimized that value by taking the derivative of the voltage out with respect to resistance. We found the ideal resistor value of 8.6kOhms. The closest resistance was 8.2kOhms, so we went with that.
We set up our circuit, and measured the actual resistances of the resistor and thermistor. We found the resistor was actually 8.26kOhms, and the thermistor ranged from 10.9kOhms cold to 7.1kOhms hot.
4. measured voltages
We then measured the voltage out:
Measured Calculated Error
Cold: 2.8+-.1V 2.86V 2%
Hot: 2.2 +-.1 V 2.14V -2%
We think, as engineers, 2% error is pretty good.
5. Post lab
We would set up a series voltage divider with the thermistor in the bottom half of the divider, such that the output would increase in voltage as the temperature increases. To achieve at least .1V/ºC, we observe that the rate of increase of the the thermistor increases in resistance roughly linearly, at a rate of about .3kOhms/degree Celsius (the thermistor increased 3.83kOhms when heated from 75ºF to 98ºF). I elect to use proportionality for simplicity in math.
If we choose a 5V source to match the constraints of the lab, we would need to choose a resistor such that the voltage divider shifted 2% in Voltage out per degree Celsius to satisfy a .1V/ºC requirement. To ensure we always meet 2% shift, we calculate with the maximum thermistor value (10.1kOhms) in our total resistance, as follows:
Marginal increase in Resistance = .3kOhms/ºC
(Marginal increase)/(Total Resistance) > .02
(.3kOhms/ºC) = .02*(10.1kOhms+R_2)
R_2 = 4.9kOhms
Since the next smallest value in E12 resistors is 4.7kOhms, we would suggest a voltage divider across a 4.7kOhm resistor and the Thermistor to guarantee .1V/ºC change in v_out.
BREAKEROO
I went to go talk to Anderson about the research project I'm doing for him for a bit.
When I came back, we were talking about current analysis and ground.
We talked about the utility of nodal analysis.
But the real lesson here was learning how to use MATLAB to solve systems of equations:
1. Start by enumerating variables:syms x y
2. Label equations:
eqn1 = (right) == (left) %noting that the == sign signifies an equation
eqn2 = (right) == (left)
3. Solve, and assign solutions to a structure
S = solve([eqn1 eqn2], [x y]) %equations, variables, both in a list
4. Access solutions by addressing structure elements
S.x
S.y
Supernodes are nodes that are include a voltage source along the node.
Properties of a supernode:
1. The voltage source inside the node provides a constraint equaition needed to solve for the node voltages
2. A supernode has no voltage of its own
3. A supernode requires the application of both KCL & KVL
No comments:
Post a Comment