Need help with a homework (LM335 thermal controller circuit)

Thread Starter

Matherios

Joined Feb 25, 2018
74
So i have never done a homework like that where it asks to find the resistances of the circuit.LM335 has output 10mV/C
So if i could have some help i would appreciate it!
MOD: Rotated your image the correct way up.

askisihlk.png
 
Last edited by a moderator:

WBahn

Joined Mar 31, 2012
24,852
Find the resistances in order to achieve what goal?

You need to show your best attempt. Just do the best you can. That's what gives us a starting point from which to try to help you.
 

Thread Starter

Matherios

Joined Feb 25, 2018
74
We are trying to make a circuit where when it reaches 42C the fan starts working and when it reaches 38C it stops.
I have found until now only R9=100kΩ so we can make sure that gate-source voltage will be 0V in case LM311 gets destoryed or removed.
And R8=so only a few mA will pass when ORF1104 is ON.
Also i have found those but im not 100% sure about them.
 

Attachments

MrAl

Joined Jun 17, 2014
6,611
Here is the circuit rotated so we can read it.
Circuit-1.gif

You have to calculate the voltages for output of the comparator high and low, then figure out what values will get you there.

What we dont know yet is what temperature you want to turn on at and what temperature you want to turn off at. We need both.

You could probably assume that the pullup resistor is 100 times smaller than the loads on it, so the voltage high output is nearly equal to +Vcc which is 12v here. If you dont want to assume that, you have to do the full network calcualtions for both comparator output states, on and off. You then calculate the trip points and from there you find the resistances to get the required hysteresis.
 
Last edited:

Thread Starter

Matherios

Joined Feb 25, 2018
74
I have already calculated output volt.(see attached file)
I also putted on the pull up resistor 1k.
The last thing that you said with the trip points i don't understand it.
I have said also the temperature that it turns on and off(ON at 42 degrees celsius and OFF at 38 degrees celsius
 

Attachments

Thread Starter

Matherios

Joined Feb 25, 2018
74
LM335 V's are 420mV(Starts working 42 degrees Celsius) and 380mV (stops working at 38 degress Celsius)(Because of 10mV per degree celsiu)
But how will i find the Vref,i don't know R10 neither R11 (and i can't find a way to calculate them)
 

MrAl

Joined Jun 17, 2014
6,611
LM335 V's are 420mV(Starts working 42 degrees Celsius) and 380mV (stops working at 38 degress Celsius)(Because of 10mV per degree celsiu)
But how will i find the Vref,i don't know R10 neither R11 (and i can't find a way to calculate them)
Hello again,

First you have to understand how the circuit works, intuitively as they say.
You then analyze the circuit in the usual way, whatever you used in the past.
Once you analyze the circuit you will find that as the voltage of the LM device rises, it will reach a level where the comparator switches states. When the voltage of the LM device falls again, it will reach a second point where the comparator flips back to the original state. Those two points are sometimes called trip points or threshold points.
In any case, the result is two distinct output states, low and high. I would suggest that as a first try, you assume that the output goes to 12v when high and 0v when low. You then write two equation one for each trip point (we might also call them set points in this circuit). You have two equations and two unknowns which are the two resistor values, so you can set up a set of two simultaneous equations that lead to the solution for the trip points, and from there you can calculate the values of the resistors needed or sometimes the ratio of one resistor to the other. Once you get the right values, your circuit turns on at 42 and off at 38 degrees C.

When you assume that the output goes to +12v it gets a lot easier, so i would suggest that you start there, find the solution, and then later purify the results by including the effects of the other resistors too if you wish. You may be satisfied with this simpler approximation though anyway.

So to start, assume two different circuits:
1. Comparator output is low (0v)
2. Comparator output is high (+12v)

Analyze the two circuits for the point where the non inverting input equals the inverting input for each case.
Can you do that? If not, we have to go into some circuit analysis methods.
 

Thread Starter

Matherios

Joined Feb 25, 2018
74
I have understand the philosophy of the circuit but i can't analyze it so i can't do that.
If u can show me a way to start analyzing it.
 

Dodgydave

Joined Jun 22, 2012
8,469
Your circuit is wrong, your hysteresis needs to be on the resistor network, coming from the Drain of the mosfet, that way when the mosfet is off the hysteresis will pull the temperature up, and when the mosfet is on the hysteresis will be pulled low.

Or better still use a Window comparator circuit .
 
Last edited:

MrAl

Joined Jun 17, 2014
6,611
I have understand the philosophy of the circuit but i can't analyze it so i can't do that.
If u can show me a way to start analyzing it.
Hello,

Ok, it is a basic voltage divider with one source when the output is low, and a voltage divider with two
sources when the output is high. We agreed that for now we will call the output +12v when high and 0v when low.

Notes:
The LM device is viewed as a voltage source that changes with temperature.
The output of the comparator is viewed as a voltage source, either 0v or +12v.
These two mean again we have a voltage divider made up of vLM and vOut and two resistors
R10 and R11.
What we need to do is calculate the voltage at the non inverting terminal for each outptu state, 0v or 12v.

I'll do the 0v case but then you do the 12v case ok? You should seriously try this at least.

With 0v out of the comparator, we have just one source the LM device, and it's voltage iwe will call vLM.
We have the right side of R11 going to ground, so the right side is 0v and the voltage at the junction of R10 and R11 using the voltage divider formula is:
VL=vLM*R11/(R10+R11)
i am calling this VL so we know it is the voltage at the non inverting input when the output is low hence the "L" in "VL".
Note we dont care what any values are yet for either resistor or for vLM, we just need the expression for VL.

Ok, so we have an expression for VL which is the voltage at the non inverting input when the output is low, so now we need the expression for the voltage at the non inverting input when the output of the comparator is 12v. Can you find this expression? It's a voltage divider again, but this time we have a non zero voltage at both ends of the divider. That means you may want to use superposition to find the expression.
Can you at least attempt to get this?
[Hint: use the voltage divider formula twice, killing just one source for each solution, then add the results].

So we can call Part 1 where we find the expression with 0v output, and Part 2 where we find the expression with 12v output. Once we have both of these we will be in a very good position to find all the unknowns.

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.
If we have to determine R5 and R6 too, we can do that later. We will have to pay some attention to the adjustment sensitivity since the working range here is less than 50mv which is quite small.
 

SotosElec

Joined Apr 13, 2018
19
Ok, so we have an expression for VL which is the voltage at the non inverting input when the output is low, so now we need the expression for the voltage at the non inverting input when the output of the comparator is 12v. Can you find this expression? It's a voltage divider again, but this time we have a non zero voltage at both ends of the divider. That means you may want to use superposition to find the expression.
Can you at least attempt to get this?
[Hint: use the voltage divider formula twice, killing just one source for each solution, then add the results].
So, I gave it a shot. For Part 2 I think it is:
vH1 = vLM * (R11/(R10+R11) killed the 12V of the comparator output
vH2 = 12 * (R10/(R10+R11) killed vLM
So, overall vH should be the the algebraic sum of vH1 and vH2, which should equal vH = vH2 - vH1 = ((12*R10)-(vLM*R11)) / (R10+R11).
At this point there are two equations and three unknowns. Should we assume vLM=380mV at Part 1 and vLM=420mV at Part 2 ?
 

JoeJester

Joined Apr 26, 2005
4,126
I don't want to rain on your parade, but, here is a picture from the LM335 datasheet. You might want to rethink your voltage levels.

LM335-.jpg
 

MrAl

Joined Jun 17, 2014
6,611
So, I gave it a shot. For Part 2 I think it is:
vH1 = vLM * (R11/(R10+R11) killed the 12V of the comparator output
vH2 = 12 * (R10/(R10+R11) killed vLM
So, overall vH should be the the algebraic sum of vH1 and vH2, which should equal vH = vH2 - vH1 = ((12*R10)-(vLM*R11)) / (R10+R11).
At this point there are two equations and three unknowns. Should we assume vLM=380mV at Part 1 and vLM=420mV at Part 2 ?
Hi,

Yes that's the basic idea. For the second expression though we have to include R1 too. So make R11=R11+R1 but only for that one expression. That allows us to solve for the two resistors.

We might have to go higher with R1 too, higher than 1k, but also we might have to modify the circuit just a little for practical use. No big deal right now though as we can solve for the resistors first, then look at that next.
One catch about this circuit is that the input range (3.11v to 3.15v) is so narrow.

Oh yeah BTW as Joe pointed out the voltage is not 380mv to 420mv because the sensor response is 10mv per K not 10mv per degree C. So the LM device voltage range is 3.11v to 3.15v.

Yes the equations have only two unknowns. The voltage of the LM device is used as you thought. When the output is low we use the higher voltage, when the output is high we use the lower voltage, so we always use the LM voltage that will cause the opposite state to what we have at the present time.

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.
 

Thread Starter

Matherios

Joined Feb 25, 2018
74
So, I gave it a shot. For Part 2 I think it is:
vH1 = vLM * (R11/(R10+R11) killed the 12V of the comparator output
vH2 = 12 * (R10/(R10+R11) killed vLM
So, overall vH should be the the algebraic sum of vH1 and vH2, which should equal vH = vH2 - vH1 = ((12*R10)-(vLM*R11)) / (R10+R11).
At this point there are two equations and three unknowns. Should we assume vLM=380mV at Part 1 and vLM=420mV at Part 2 ?
Why do we get the sum of VH1 and VH2?
 
Top