Will someone help a math challenged senior citizen.

Discussion in 'Math' started by tracecom, Jan 14, 2011.

  1. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
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    Please see page 18, sections 5.1 and 5.2 of the attached data sheet. I have read and understand in principle everything that is there - except how to do the math. On page 19 in the last of section 5.2, they present an example complete with the answer, but they leave out the calculations that derive the answer. Could someone please go through the calculations step by step and post them here. (I can do basic algebra.)

    Thanks.
     
  2. Georacer

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    Nov 25, 2009
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    It got me thinking, but it's quite simple actually:

    For example 1:
    \text{time for 1 revolution}=\frac{T}{1}= \frac{1}{\frac{1}{T}}=\frac{1}{\frac{N}{min}}= \frac{1}{RPM}\ \text{in minutes}\\<br />
\text{time for 1 revolution}=\frac{60}{RPM}\ \text{in seconds}\\<br />
\text{time for 1 revolution}=\frac{60 \cdot 1000}{RPM}\ \text{in miliseconds}
    So in fact only the RMP is a variable here.

    For the second example:
    You need to choose a pair of values for Duty Cycle/Temperature for your fan. Then, convert that into Input Voltage/Temperature. Finaly, select a thermistor and extract (experimentaly) its values in the previously selected temperatures.
    What you have to do now, is to input those values you found in the set of equeations given, and solve for R1 and R2. The solution pair, will determine the values you will need to use in the resistors in order to achieve the behaviour you started with.
     
  3. tracecom

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    You give me too much credit. I really don't understand how to do the math to complete the equations. If you could just show me how the calculations are made to arrive at the answers (R1 = 34.8k, R2 = 14.7k) in the Example at the top left of page 19, I would be grateful, and then I could use my own data to calculate the answer to my own situation.

    Thank you.
     
  4. Georacer

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    Nov 25, 2009
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    To my surprise, this is turning to be much harder than I though it would. Below I post a solution that I have come by not by hand calculations but by newly acquired Matlab skills, so I don't guarantee anything yet. Use them with much caution. I will try it again tomorrow and will return with verified results.

    <edit> The following equations are wrong! Please refer to post #12 for the correct ones. Sorry for the inconvenience.

    R1=\frac{(V(T2) - V(T1)) \cdot R_T 1 \cdot R_T 2 \cdot V_{DD}}{(V_{DD}-V(T2)) \cdot (R_T 1 \cdot V(T1) - R_T 2 \cdot V(T2))}\\<br />
R2=\frac{(V(T2) - V(T1)) \cdot R_T 1 \cdot R_T 2 \cdot V_{DD}}{(R_T 1 -R_T 2) \cdot (V_{DD} -V(T1)) \cdot (V_{DD} - V(T2))}

    This time it was a freebie. But in your next thread we will expect you to do your own math and algebra. This forum is here to provide guiding and directions, not solve your problems for you.

    Have a nice day, talk to you later.
     
    Last edited: Jan 27, 2011
  5. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
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    Thanks, but don't trouble yourself further on my account

    Sorry if I asked for too much, but since responses from forum members are voluntary, I don't see the problem.
     
  6. Georacer

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    Actually, there is. Homework Forum rules specify that the Original Poster must first provide along with the problem, all of his thoughts and work he has done so far to tackle to problem. Usually we don't solve the problems ourselves but help the OP to solve the problem himself. It is more educational this way.
     
  7. Wendy

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    Mar 24, 2008
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    I don't think he is a student, and this isn't homework. It is a fellow member asking for help. Could be wrong, but that is how I interpret it.
     
  8. Georacer

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    I thought about that too, and I kinda have a soft spot for the elderly, that's why I gave the solution to begin with.
    But in the end this is the Homework Help forum and I had to mention its rules.
     
  9. Wendy

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    Mar 24, 2008
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    No, it is in the Abstract Forums/Math Forum. We get some homework here, but not always. Different rules.
     
  10. Georacer

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    Oops! I spend so much time over there that I actually forgot that this is the Math forum. Sorry tracecom!
     
  11. CaryInVancouver

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    Jan 25, 2011
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    Though I'm new to electronics, and a fellow senior citizen, I think I may understand what's going on here. Firstly, look at the sentence under the box labelled "EQUATION". I think that it should read something like this: "In order to solve for the values of the two resistors, R1 and R2, two equations must be solved, corresponding to two values of temperature that are to be selected, T1 and T2."

    The two equations are a little misleading, as there are three unknowns: R2, Rtemp(T1) (the resistance of the thermistor and R1 in parallel, at temperature T1) and Rtemp(T2). However, for any temperature t, Rtemp(T1) and Rtemp(T2) each depends on only one unknown: R1. To solve these two equations (assuming they are correct), on must first replace Rtemp(T1) and Rtemp(T2) with R1*RT(T1) / (R1+RT(T1)) and R1*RT(T2) / (R1+RT(T2)), respectively, where RT(t) is the (known) resistance of the thermistor at temperature t.

    I decided to try to calculate R1 and R2 from first principles.

    Look at Figure 5-1. The thermistor provides a resistance that varies with temperature. That relationship can be expressed as a function, RT(t), where t equals temperature. I assume that function must be known. Perhaps it is in the data sheet. If not, I expect it would be in the data sheet for that part. So, for a given temperature, t0, RT(t0) can be determined and is a constant (Ohms).

    Because RT and R1 are in parallel, the combined resistance equals:

    Rtemp(t) = R1 * RT(t) / (R1+RT(t)) (1)

    For example, if t = 10 degrees (C or F), and one knows RT(10) (RT at 10 degrees) and R1, one can compute:

    Rtemp(10) = R1 * RT(10) / (R1+RT(10))

    However, R1 is not known. We are to determined its value. For the time being, forget about R1. We'll just deal with the unknown Rtemp()'s and come back to calculating R1 later.

    Ohms Law tells us that, for a given temperature t, the current flowing from Vdd to ground equals:

    Idiv(t) = Vdd / (Rtemp(t) + R2)

    Hence, the voltage at Vin (voltage drop over R2) is:

    Vin(t) = R2 * Idiv(t)
    or
    Vin(t) = R2 * (Vdd / (Rtemp(t) + R2))
    or
    Vin(t) * (Rtemp(t) + R2)) = R2 * Vdd
    or
    Vin(t) * Rtemp(t) + Vin(t) * R2 = R2 * Vdd

    Solving for R2, we obtain:

    R2 = Vin(t) * (Rtemp(t) / (Vdd - Vin(t))

    Now, for any give temperature t, we may specify what we want to be Vin(t) to be (a constant). So, we are left with two unknowns: R1 and Rtemp(t).

    As I understand it, we are now to select two temperatures, which I'll call Ta and Tb. For example, perhaps Ta = 10 and Tb = 20 (degrees). Maybe the two temperatures should normally span the product's temperature operating range.

    Next we decide what we want the desired values of Vin to be for each of the two chosen temperatures. Call these values Va (when temperature = Ta) and Vb (when temperature = Ta). For example, we might specify Va = 5v and Vb = 8v, or Va = Vb = 12v.

    We now plug these values into the equation above and obtain:

    R2 = Va * (Rtemp(Ta) / (Vdd - Va) (2)
    and
    R2 = Vb * (Rtemp(Tb) / (Vdd - Vb) (3)

    Since both equations have R2 on the left, we can eliminate R2:

    Va * (Rtemp(Ta) / (Vdd - Va) = Vb * (Rtemp(Tb) / (Vdd - Vb)

    Hence,

    Rtemp(Ta) = k * Rtemp(Tb) (4)

    where the constant k (which we can compute) equals:

    k = (Vb / (Vdd - Vb) / (Va / (Vdd - Va))

    = (Vb/Va) * (Vdd - Va)/(Vdd - Vb)) (5)

    Now, in equation (4), let's substitute for Rtemp() from equation (1):

    R1 * RT(Ta) / (R1+RT(Ta)) = k * R1 * RT(Tb) / (R1+RT(Tb))

    where RT(Ta) and RT(Tb) are the resistance of the thermistor at temperatures Ta and Tb, respectively, which we can determine.

    We can divide both sides by R1 to simplify:

    RT(Ta) / (R1+RT(Ta)) = k * RT(Tb) / (R1+RT(Tb))

    which allows us to solve for R1:

    RT(Ta) * (R1+RT(Tb)) = k * RT(Tb) * (R1+RT(Ta))
    or
    R1 * (RT(Ta) - k * RT(Tb)) = (k - 1) * RT(Tb) * RT(Ta)
    or
    R1 = (k - 1) * RT(Tb) * RT(Ta) / (RT(Ta) - k * RT(Tb)) (6)

    Knowing R1, R2 can be computed from equation (2) or (3). Make sense?
     
    Last edited: Jan 26, 2011
  12. Georacer

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    Nov 25, 2009
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    When I tried to solve the system by hand it looked much more daunting. But you put the process in a clean order and now it all looks so clean. Maybe I wasn't paying enough attention, and tried to simplify for the wrong variable.

    Anyway, let me re-write all that for you:
    <br />
k=\frac{V_{in} (T_2) \cdot (V_{DD} - V_{in} (T_1)}{V_{in} (T_1) \cdot (V_{DD} - V_{in} (T_2)}\\<br />
R_1=\frac{ R_T (T_1) \cdot R_T (T_2) \cdot (k-1)}{R_T (T_1)-k \cdot R_T (T_2)}\\<br />
R_{temp} (T_1)=\frac{R_1 \cdot R_T (T_1)}{R_1 + R_T (T_1)}\\<br />
R_2=\frac{V_{in} (T_1) \cdot R_{temp} (T_1)}{V_{DD} - V_{in} (T_1)}<br />

    Of course k and Rtemp are intermediate variables that facilitate the calculation of R1 and R2.
    Once again, kudos to CaryInVancouver for solving the system.
     
  13. CaryInVancouver

    New Member

    Jan 25, 2011
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    Thanks for the kudos, Georacer. Glad to help. I kinda have a soft spot for the young-uns.

    Cary

    PS Did you check my results against Matlab's? I didn't think to do so and am not quite up to it at this moment. We old folk need our sleep.
     
  14. Georacer

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    Nov 25, 2009
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    It seems I had done a typo when setting up the equations on Matlab the first time. I gave it another shot today. I am using the Symbolic Math Toolbox.

    I reached on the correct formula for R1, but it insists on giving me the formula of post #4 for R2. I don't know what's going wrong and I think I 'll drop it for now.

    If someone knows how to come up with the correct results on Matlab, I 'd love to see their code.
     
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