The D2 is 5.6V/0.25Whi M,
What are the two zener diode voltages.? I cannot see them clearly on that blurry diagram.
E
D1 is LM335
The D2 is 5.6V/0.25Whi M,
What are the two zener diode voltages.? I cannot see them clearly on that blurry diagram.
E
VH i think it's=11.89v i setted r8=10kΩ and r9=100kΩ so VH is =12v*(r8+r9)/(r1+r8+r9)=11.89v (r1=1kΩ)Regarding the potentiometer, we know that the voltage at the R3, R4, R5 junction is going to be 5.6V (right?).
So the vP (voltage pot) at the inverting input will be:
vP = 5.6 (RP2 + R6) / (R5 + R6 +RP1 + RP1), where RP1 is the resistance between the pot's wiper and R5 and RP2 the resistance between the wiper and R6. The expression can be simplified, since RP1+RP2=10, so:
vP = 5.6 (RP2 + R6) / (R5 + R6 +10)
Since we want vP to be the average of the voltage levels of the LM355, then vP=(3.11+3.15)/2 = 3.13V
Thus 3.13 = 5.6 (RP2 + R6) / (R5 + R6 +RP1 + RP1).
3.13V roughly half of 5.6V, so R5 could be equal to R6. Let R5=R6=1K. Then RP2 can be calculated as follows:
RP2 = 3.13 (2*R5+10) / 5.6 - R6 = 3.13 (2*1+10) / 5.6 - 1 = 5.7K
So the pot should be set to 57% (or 43%?).
Regarding the vL and vH, are those the thresholds for the comparator to switch states (aka vLTP and vUTP)?
So far:
vL = 3.15*R11/(R10+R11) and
vH = ( (12*R10)/(R10+R1+R11) ) - ( (3.13*R11) / (R10+R11) ) (corrected with R11=R11+R1)
vL and vH aren't known, though... right? So, that's actually 4 unknown factors.
Sorry if I'm missing something.
It's the application of the superposition theorem. vH will be the two voltages added together. vH1 has the reverse polarity of vH2 though, so it turns out to be vH=vH2-vH1Why do we get the sum of VH1 and VH2?
If im correct 420mV starts working and 380mV stops working so hysterisis should be Vhys=420mV-380mV=40mVhi M,
Lets take it step by step.
You say when it reaches 42C the fan starts working and when it reaches 38C it stops
The datasheet for the LM335 states a range of -40C thru +100C, so what voltage would you measure across the LM335 at 38C.?
Based on 10mV/C what do you calculate the hysteresis band.?
E
Can u please explain the Vp and what do u mean by RP2 and RP1 and why do u use RP1 two times at the equation.Regarding the potentiometer, we know that the voltage at the R3, R4, R5 junction is going to be 5.6V (right?).
So the vP (voltage pot) at the inverting input will be:
vP = 5.6 (RP2 + R6) / (R5 + R6 +RP1 + RP1), where RP1 is the resistance between the pot's wiper and R5 and RP2 the resistance between the wiper and R6. The expression can be simplified, since RP1+RP2=10, so:
vP = 5.6 (RP2 + R6) / (R5 + R6 +10)
Since we want vP to be the average of the voltage levels of the LM355, then vP=(3.11+3.15)/2 = 3.13V
Thus 3.13 = 5.6 (RP2 + R6) / (R5 + R6 +RP1 + RP1).
3.13V roughly half of 5.6V, so R5 could be equal to R6. Let R5=R6=1K. Then RP2 can be calculated as follows:
RP2 = 3.13 (2*R5+10) / 5.6 - R6 = 3.13 (2*1+10) / 5.6 - 1 = 5.7K
So the pot should be set to 57% (or 43%?).
Regarding the vL and vH, are those the thresholds for the comparator to switch states (aka vLTP and vUTP)?
So far:
vL = 3.15*R11/(R10+R11) and
vH = ( (12*R10)/(R10+R1+R11) ) - ( (3.13*R11) / (R10+R11) ) (corrected with R11=R11+R1)
vL and vH aren't known, though... right? So, that's actually 4 unknown factors.
Sorry if I'm missing something.
It is indeed +10mv/C, but at room temperature for example, the voltage output of the LM355 is 2.98V (see attachment).Our teacher gave us this output for LM335 10mV per degree celsiu(see attached file)
I corrected it. It should've been RP1 + RP2. I assume the pot is two resistors in series, RP1 and RP2, and their connection point is the pot's wiper.Can u please explain the Vp and what do u mean by RP2 and RP1 and why do u use RP1 two times at the equation.
Can u please explain me this (sorry for doing so many questions )It is indeed +10mv/C, but at room temperature for example, the voltage output of the LM355 is 2.98V (see attachment).
Hope we cleared this up.
MrAI suggested that the pot should be set so the voltage at the inverting input will be the average of the LM355 voltage levels of interest.Can u please explain me this (sorry for doing so many questions )
vP=(3.11+3.15)/2 = 3.13V
We also have to keep in mind for later that this circuit contains a potentiometer too, that adjusts the set point. We have to look at that too later. A good place to start though is to force it to be equal to the average value of both set points, the 420mv and 380mv points. You should calculate the average of those two also.
Also, since as Joe pointed out the real voltage relationship for the sensor, the average input will be different too. The average of 3.11 and 3.15v.
But why do we have (3.11+3.15)MrAI suggested that the pot should be set so the voltage at the inverting input will be the average of the LM355 voltage levels of interest.
Ok i understood it!hi M,
As Sotos points out the LM335 is rated in Kelvin, so that is -273C degrees, ie: 2.73V at 0C, so you must add your temperatures/mV to that value of 0C.
E
Can u please tell me why do we want pot voltage to be the average of the LM335hi M,
As Sotos points out the LM335 is rated in Kelvin, so that is -273C degrees, ie: 2.73V at 0C, so you must add your temperatures/mV to that value of 0C.
E
Hi,Can u please tell me why do we want pot voltage to be the average of the LM335
That's what im also thinking.MrAi told about in a previous post.Hi,
If I understand your question correctly.?
Its because the pot Ref voltage and the LM335 voltage are connected to the inputs of a Comparator.
Is that what you mean.?
E
Hello again,Can u please tell me why do we want pot voltage to be the average of the LM335
Reaylly great explanation!! But what about the resistors how can i dermine there values now.Hello again,
Let me see if i can clear up a few things here.
First, the voltage is the average of the high sensor and low sensor voltage extremes because the average voltage is the midpoint voltage and we want the hysteresis to pull this higher and lower as the circuit is in operation and it makes sense to pull it just as low as we do high, hence the midpoint voltage accomplishes this.
It's also best for noise immunity because we get equal upper and lower threshold change as the circuit switches back and forth.
We are not clued to this value as we can always try other values later, but it makes sense to start there. I'll show a graph of a working system so we can see how well this works out in practice.
Second, the output of the LM device is 10mv/K but that is also sometimes referred to as 10mv/C because they are both really true, it's just that the 10mv/C spec needs to have an offset added to it. This is because Celsius is a relative scale, hence we see "degrees" C, while the Kelvin scale is an absolute scale, so we dont see "degrees" K we just see K (unless the author makes a mistake in writing). So Kelvin is based on absolute zero while degrees C is based on freezing temperature of water.
The only point left then is what the instructor really wants you to use. If we wants everyone to use 390mv/420mv then you have to use that, but if he wants you to look up the data sheet and get the full voltage spec, then he'd want you to use 3.11v and 3.15v. The latter are the more real life values so i would expect those to be used.
Third, when calculating the upper transition point (at 3.15v) the output of the comparator is zero, so we only have one voltage source to deal with so we dont need superposition. When calculating the lower transition point howefver (at 3.11v) the output is high so we have TWO voltage sources to deal with along with the voltage divider, so we can use superposition to help calculate the voltage trip point and superposition is the addition of two or more effects from two or more sources, and that's why we add the two. Going left to right, we use the LM voltage, and going right to left we use the output voltage.
Another note is that we should include the pullup resistor in the calculation using superposition. This just means that R11 becomes R11 plus R1, but only for that one calculation because when the output is zero the pullup resistor is not in the circuit anymore.
A final note is that we may end up with resistor values that are not practical, so we may have to change the circuit a little bit. This is because the input range is so low (40mv difference between high trip and low trip points). Note the difference between 38C and 42C is only 4 degrees C also.
The attachment i will add to this post in a few minutes. It will show a working system so we can see what the voltages would be. I have to prepare the drawing so it will take a few minutes to do.
[LATER]
Here is the diagram. It shows a system with the 380mv/420mv standard for the LM335 which as we know now should be 3.11v and 3.15v.
The LM335 voltage is shown in RED. We see as the temperature goes up the LM voltage goes up, then the fan turns on and the object starts to cool so the red graph voltage starts to decrease, then the fan turns off and it starts to heat up again, etc., etc.
We can see that the voltage at the non inverting input gets pulled up and down in equal amounts.
Also note that the blue wave (non inverting input) always changes when it reaches 400mv and that was done by design. The 400mv line is the green horizontal line centered between the top green line (420mv) and bottom green line (380mv).
Also, the square wave is just there to show the on/off state of the output of the comparator. It was really 12v but it was reduced for graphing purposes. When on it is really 12v and when off it is 0v.
Thread starter | Similar threads | Forum | Replies | Date |
---|---|---|---|---|
Y | NEED HELP TO MY HOMEWORK | Homework Help | 2 | |
N | In need for help for my homework | Homework Help | 0 | |
W | Need guidance on this homework problem | Homework Help | 32 | |
M | Need help with a homework! | Homework Help | 11 | |
M | Hi, I need help with my homework on circuit diagrams! | Homework Help | 12 |
by Jake Hertz
by Jake Hertz