Need help with a homework (LM335 thermal controller circuit)

JoeJester

Joined Apr 26, 2005
4,130
Reaylly great explanation!! But what about the resistors how can i dermine there values now.
Now that you know the upper and lower "trigger limits" for turning the fan on and off, you need to start calculating values.

R5. R7. and R6 forms a voltage divider. You should be able to get the desired output from the wiper by doing the calculation. Why do you think R5 and R6 are in that circuit?
 

MrAl

Joined Jun 17, 2014
6,949
Reaylly great explanation!! But what about the resistors how can i dermine there values now.
Hi,

Ok that's next really. Do you know how to solve simultaneous equations?
I ask because that's what we would do next.
If not, we'll have to go into that too i guess.

Set up the two equations to equal the mid point value (3.13v) and one equation use 3.11v for vLM and the other use 3.15v for vLM. You have two equations in two unknowns, so you can solve for the two resistors, however they will be in terms of R1 so you would then pick a value for R1 after that. Note we may have to go higher than 1k for that because this is such an unusual circuit.

Does this make sense?
 

MrAl

Joined Jun 17, 2014
6,949
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 + RP2).
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.
vL and vH in your equation are both equal to the mid point value. That reduces it to 3 variables (R10, R11, R1) but we pick and choose R1 later after solving for R10 and R11 simultaneously. As i noted earlier, we may have to change R1 from the presumed 1k to a higher value to get practical values for the two resistors. This is a somewhat unusual circuit.
Later, we can modify ti slightly to get better real world values if necessary. Not sure if that is allowed though because the circuit was given already. There might be a trick or two we can use too though because we have control over the other two resistors too that connect to the MOSFET.

The pot resistor values we can choose later. We want a small adjustment range because the working range is so small to begin with.
 

SotosElec

Joined Apr 13, 2018
19
vL and vH in your equation are both equal to the mid point value. That reduces it to 3 variables (R10, R11, R1) but we pick and choose R1 later after solving for R10 and R11 simultaneously. As i noted earlier, we may have to change R1 from the presumed 1k to a higher value to get practical values for the two resistors. This is a somewhat unusual circuit.
Later, we can modify ti slightly to get better real world values if necessary. Not sure if that is allowed though because the circuit was given already. There might be a trick or two we can use too though because we have control over the other two resistors too that connect to the MOSFET.

The pot resistor values we can choose later. We want a small adjustment range because the working range is so small to begin with.
I worked out that:
vL=3.15*R11/(R10+R11) = vAVG = 3.13
=> R11 = 3.13/3.15*(R10+R11) => R11=0.99*R10 + 0.99*R11 => R11=99R10
that gives a ratio between R10 and R11, so R11 can be replaced by 99*R10 on the vH equation:

vH= ( (12*R10)/(R10+R1+R11) ) - ( (3.11*R11) / (R10+R11) ) =
=( (12*R10)/(R10+R1+99R10) ) - ( (3.13*99R10) / (R10+99R10) ) =
=( (12*R10)/(100R10+R1) ) - ( (3.13*99R10) / 100R10 ) = vAVG = 3.13

At this point I think it's impractical to figure out a relationship of R10 and R1, so I assumed R1=1KΩ, thus:
3.13 = ( (12*R10)/(100R10+1) ) - ( (3.13*99R10) / 100R10 ) which returns a negative value for R10...
 

SotosElec

Joined Apr 13, 2018
19
Now that you know the upper and lower "trigger limits" for turning the fan on and off, you need to start calculating values.

R5. R7. and R6 forms a voltage divider. You should be able to get the desired output from the wiper by doing the calculation. Why do you think R5 and R6 are in that circuit?
I think the purpose of R5 and R6 are to prevent setting the voltage too high or too low. With them in place the voltage range on the inverting input is narrower than the 0-5.6V enforced by the zener diode.
So, if the pot is set to 0% (wiper is all the way up) the voltage should equal 5.6*(R6+10)/(R5+R6+10), while when it's set to 100% (wiper all the way down) it should be 5.6*(R6)/(R5+R6+10). If R5 and R6 weren't there, at 0% it would be 5.6V and at 100% it would be 0V.
 

JoeJester

Joined Apr 26, 2005
4,130
I think the purpose of R5 and R6 are to prevent setting the voltage too high or too low.
That is the reasoning. I'm sure the thread starter will appreciate the heads up.

It is going to be tough for the TS to learn something if the answers keep popping up in the thread.

I was playing around earlier with the circuit and preparing for an oral history and movie that I'll be posting on another site, so I created this file that is germane to this topic, just for fun.

 

SotosElec

Joined Apr 13, 2018
19
That is the reasoning. I'm sure the thread starter will appreciate the heads up.

It is going to be tough for the TS to learn something if the answers keep popping up in the thread.

I was playing around earlier with the circuit and preparing for an oral history and movie that I'll be posting on another site, so I created this file that is germane to this topic, just for fun.

Yes, you are right, I got ahead of myself. Your simulation seems to be working very well! Looking forward to seeing your finished work.
 

Thread Starter

Matherios

Joined Feb 25, 2018
74
That is the reasoning. I'm sure the thread starter will appreciate the heads up.

It is going to be tough for the TS to learn something if the answers keep popping up in the thread.

I was playing around earlier with the circuit and preparing for an oral history and movie that I'll be posting on another site, so I created this file that is germane to this topic, just for fun.

[/QUR

Really great video!!
 

SotosElec

Joined Apr 13, 2018
19
hi M,
This PDF explains in detail how to calculate the Hysteresis resistors.
E
I want to clear something up. vREF keeps showing up and I'm not sure what its value is in this occasion. Firstly, let's agree on this:
vREF is the voltage on the inverting input. Is that right?
Secondly, we've discussed that it should be the average of the trip voltages, so 3.13V. Correct?
I tried using vREF = 3.13V on the calculators at the okawa pages, but I get an error suggesting that vREF should be greater than V1 and V2 (can't tell for sure, because there's some Japanese text)
 
Last edited:

ericgibbs

Joined Jan 29, 2010
9,320
Hi Soto,
Set Vref at 3.13v
Vthr =3.15v
Vthf = 3.11v

I make the Rhyst= 3.13meg and Rin 9k6

Have written a VB program to do the calcs, will now try in LTSpice, post shortly.
E
 

SotosElec

Joined Apr 13, 2018
19
Hi,
These images are from the VB program and the check is with LTSpice
What do you calc the values to be.?
E

EDIT:
I also get an error message on the denshi link.???
My previous attempt on figuring out the system of equations as MrAI suggested didn't yield any results.
I'm now trying to understand your circuit. On the pdf you've shared, the first component to be calculated is R3, which in this application is absent. On its place lies the LM335. So, how can one proceed in calculating the rest of the values?
You have added a pot to the sensor. Is it there for the purpose of running the simulation?
 

ericgibbs

Joined Jan 29, 2010
9,320
hi,
If you look at the last 2 pages of the PDF. Step 1
For the OPA it is referring to has a input leakage current of 2nA, so to swamp out the effect of this 0.2nA they have assumed a R3 current of 0.2uA .
The LM339 has an input leakage on 100nA so I have assumed a current of 1uA thru R3.
So R3 = Vref/IR3 = 3.13/1uA = ~3.0meg Ohms.

Ask if you have a problem.
E
 

SotosElec

Joined Apr 13, 2018
19
For the OPA it is referring to has a input leakage current of 2nA
Alright that's clear now. Wouldn't that give us a value of ~100KΩ for R1 (R4 in your schematic) though? According to the formula, R1 = (R3+R4) * (vHB/vCC) = (3MΩ+R4) * (40mV/12V) and if R4 is considered neglectable, then R1=3MΩ/30=100KΩ.
 
Top