NTC sensor voltage divider

Discussion in 'General Electronics Chat' started by crash563, Jun 26, 2013.

  1. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
    Moderator edit: This thread was split from here: http://forum.allaboutcircuits.com/showthread.php?t=84219

    My question is about the NTC sensor voltage divider. Is it that when the sensor resistance is lowered due to an increasing temperature, allows the voltage across R4 to increase and charge through C3. But is C3 used as a time delay to compensate for the changes in the temp? Also, is somewhat of this explation correct?
     
    Last edited by a moderator: Jun 28, 2013
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Yes, you are right. When temperature start to increasing. Then NTC resistance is start to drop. So to voltage across R4 start to increase. And until voltage at inverting input is lower then voltage at non-inverting. The op amp output is at positive saturation and Q3 is in saturation and relay will be energize. And as soon as voltage at inverting input goes above the voltage at non-inverting (because NTC resistance is dropping due to temperature start to increasing). The op amp output switch to "negative saturation" to GND (LOW state). And this will cut-off the BJT and relay becomes de-energize (OFF).

    As for the C3. I think that C3 like a filter. NTC can be connected via long wires.
    And this is why they add C3 to filter any RF noise included in the long wires.
     
  3. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
    Now with the schmitt trigger section, I am assuming that the 100k pot and the 56k in series form a reference voltage and it is a part of Rin of the trigger. Or is this different?
     
  4. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
    I understand the full circuit operation, I meant in the previous post if someone can provide the calculations for the trip points for the trigger.
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    656
    Are you hijacking this thread, or are you latitude in sheep's clothing?:rolleyes:
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    If we assume that we set RV1 at 50% (RV_up = 50kΩ and RV_down = 50kΩ) we now can find a threshold voltage.

    Now we need to find the voltage at non-inverting input when op amp is at positive saturation.

    (Vin - Vp)/( R5 + RV_up ) + (Vout - Vp)/(R7) - Vp/(RV_down + R6) = 0V

    Where
    Vin - IC1C - op amp output voltage is equal to ZD1 Zener diode voltage
    Vin = 5.1V

    Vout - op amp positive saturation voltage
    Vout - Vcc - Vd2 - Vsat = 12V - 0.65V - 1.2V = 10.3V

    And we plug those values into equation

    (5.1 - Vp)/( 120 + 50 ) + (10.3 - Vp)/(6800) - Vp/(50 + 56) = 0


    And if we solve this for Vp (voltage at non-inverting input) we get

    Vp = 2.038V.

    So to turn - OFF the relay (and the heater) voltage from voltage divider formed by NTC resistor and R4 must be higher than 2.038V.

    And now we have a op amp at negative saturation voltage.
    So the threshold voltage is now equal to

    (Vin - Vp)/( R5 + RV_up ) - Vp/((RV_down + R6)||R7 )

    Vp = Vin * (RV_down + R6)||R7 )/ ( R5 + RV_up + (RV_down + R6)||R7 )

    Vp = 5.1 * ((50+56)||6800)/(120 + 50 + (50+56)||6800) = 5.1*(104)/(170 + 104) = 1.93V

    As you can see voltage at inverting input must now drop below 1.93V to switch ON the relay and the heater.
     
    • 3.PNG
      3.PNG
      File size:
      60.9 KB
      Views:
      29
    Last edited: Jun 28, 2013
    crash563 likes this.
  7. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
    Where did that 1.2V come from
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    This 1.2V is a op amp LM324 saturation voltage. Internal voltage drop across LM324 output transistors (Q13 , Q14 , Q15 - figure 1 page 5).
    Look at datasheet
    http://www.onsemi.com/pub_link/Collateral/LM324-D.PDF page 4 Output Voltage − High Limit Voh
    For Vcc = 5V and RL = 2K we have 3.5V typical.
    So Vsat = 5V - 3.5V = 1.5V. But I assume 1.2V.
     
  9. crash563

    Thread Starter Member

    Feb 25, 2013
    47
    1
    So the top equation is really the superpostion of the equivalent network and the bottom is also superpostion when Vo is low. Really interesting circuit analysis
     
Loading...