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Can someone explain me how to solve following problem:

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loki4000

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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

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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

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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

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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.
 

crutschow

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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

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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

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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

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

Ian Rogers

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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

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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..:eek:
 

Ian Rogers

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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

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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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