# Can someone explain me how to solve following problem:

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#### loki4000

##### New Member
This circuit represent temperature measuring device that light up different LEds as temperature rise.

can someone explain (or at least hint) how to calculate resistors?

#### crutschow

##### Well-Known Member
Here's an article on resistors in series.

One way to calculate the values for the string is to start with an arbitrary total series resistor value for the string (say 10kΩ) and calculate each resistor value to give the required comparator trip voltage, starting from the bottom using the "The Potential Divider Circuit" equation.

For the required resistor in series with the LEDs you just Ohm's law to give the required maximum 15mA LED current.

#### Ian Rogers

##### User Extraordinaire
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273 kelvin = 0°C then it is linearly the same... 373 Kelvin is 100°C, as these things are calibrated at 10mV/K the voltage at the top of the sensor at 0°C should be 2.73V

That should be a good start for you.. Work out you potential divider... Then the LED resistors are easy... (10V) even a worm could do it.

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#### loki4000

##### New Member
Hi.
Thanx for help, but I am still confused. I don't know what should happend at the end. Voltage rise from temp sensor and as V rise it lights up LEDs.Isthis the idea?

This mean that resistors r8 r9 r10 must have resistances at certain fixed rate change so that it allow certain voltage only to pass to LED. why do we need the rest of resistors?
If temperature sensor is voltage source why do we need extra voltage sources at the top?

P.S. We did not study amps yet, I understand they just to strengthen the voltage. Does that mean that resistors r4 r3 r2 r1 should change by fixed rate.

#### loki4000

##### New Member
Hmm.
So, can this circuit be redrawn like this?

#### crutschow

##### Well-Known Member
No. The power supplies must have their negative connection go to common (ground).

U1 is not a linear amplifier, it is a comparator. The output is on (high) when its plus (+) input is more positive then its negative (-) input, and its output is off (low) when its plus input is more negative then its minus input.

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#### loki4000

##### New Member
I understand that they must go to ground, however I leared in circuit analysis that grounds can be replaced by single line connecting all ground nodes...

#### RCinFLA

##### Well-Known Member
List the three LM335 voltages that the will appear at the the three transition points (10, 30, and 50 degs C) then make the divider resistor ratio match these voltages.

I would normally recommend some positive feedback to create hysteresis on each comparator to prevent them from 'chattering' at transition points but not is not part of the question.

#### crutschow

##### Well-Known Member
I understand that they must go to ground, however I leared in circuit analysis that grounds can be replaced by single line connecting all ground nodes...
True. But that's not what you have in you last schematic.

#### loki4000

##### New Member
so,what exactly wrong with schematic? is it that 5v and 12 v sources must not be connected to other grounding?

Like this?

#### Ian Rogers

##### User Extraordinaire
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Sorry............I apologize!! the schematic was fine, the grounds were correct..... The batteries are drawn upside down and it threw me..

Just a little hint ( to say sorry ) R1, R2, R3 and R4.......... 377uA starting with R1 then go upwards

#### crutschow

##### Well-Known Member
Yes, I fell to the same error as Ian. You drew the ground connection at the top of the schematic, which I didn't recognize. Sorry..

#### loki4000

##### New Member
Ok, I understood that I need to calculate this schematic according to voltage changes in lm 335 due to temperature changes.
Only problem is no matter how much I read documentation on lm335 ( http://www.electro-tech-online.com/custompdfs/2012/03/lm135.pdf ) I couldn't find where it says about voltage change per temp change...

#### Ian Rogers

##### User Extraordinaire
Forum Supporter
I told you in post #3

0°C is 2.73V then it's linear.... 100°C is 3.73V... So 10°C = 2.83V.. 30°C = 3.03V... 50°C = 3.23V.
Using 2.83V / 7.5K (available resistor ) = 0.377mA;

0.2V / 0.377mA = 530 Ω (510Ω + 20Ω)... so on and so on.

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#### loki4000

##### New Member
Ok, I submit my work today and it seem to be ok.
So I want to thank you for your help, It is unlikely that I would have solved this problem without it.

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