Stuck with an automatic temperature controller

Thread Starter

MrP90

Joined May 19, 2013
12
Hello!

Can someone please help me understand how to analyse this circuit? I just can't seem to get it...



I need to find the resistor RA value when I want to impose a 45ºC temperature on resistor R5.
Also, how can I find the current that heats the resistor R5 when both switches are open or when just one of them is open.

I just need some advice to see if I can understand this and do it. Thank you so much!
 

Thread Starter

MrP90

Joined May 19, 2013
12
Sorry, I totally forgot.

Vcc=12V;
R1=330K;
R2=110K;
R3=10K//8,2K;
R4=R6=R7=1K;
R5=47ohm/5W;
R9=R11=10k;
R10=33K;
R12=R21=100K;
R13=R14=R15=910 ohm;
R20=2,7K;
Ra=Rb=22K;
Rp=Rn=10M;
Rntc=10 k (at 25ºC)
C1=22microF/25V;
C2=C3=100microF/25V

Variation of the resistance with T: R(T)=R(T0)exp[B(1/T)-(1/T0)]

T0 is usually 298,15K
B is not given, it just says that it's a constant and it depends on the material properties.

And that's all I got.

Thank you!
 

Dodgydave

Joined Jun 22, 2012
10,139
Assuming the transistor is fully saturated then the current through R5 is 12V/ 47r = 0.255 Amps.


so what is the voltage out of the regulator Vcc2?
 

Dodgydave

Joined Jun 22, 2012
10,139
Vcc2 = Vcc = 12V

Thank you for your help.
i get the feeling you don't understand the circuit,

The resistor R5 is on the Vcc supply and the reference voltage is from the regulator is Vcc2 what are these voltages , they can't both be 12 Volts ,
otherwise why have a regulator???
 

Thread Starter

MrP90

Joined May 19, 2013
12
Yeah, I'm having some trouble getting the circuit. That's my main problem.

The questions says: Determine the resistor RA value (potentiometer as rheostat) so that the controller imposes a 45ºC temperature on resistor R5. Potentiometer is linear, so resistance is proporcional to rotation angle. Admiting that potentiometer maximum rotation is 270º, how much should you rotate the potentiometer? Considere Vcc2=Vcc=12V
 
Last edited:

LDC3

Joined Apr 27, 2013
924
An exercise!

Is the amplifier is driven with +12 and -12 V ?
What is the value of the zener diode ? (it's red) I am guessing that it is used to measure the temperature. For one of the diodes I looked up (1N5264B), it has a 0.097 %/k temperature deviation. So for a change from 20 ºC to 45 ºC, the zener voltage changes by 24.2 mV.

The loops with Rp and Rn (C1) provides some hysteresis to the op-amp. It will not effect the temperature or the setting of RA. Also, the red LED will be on when heating and the green LED will be on when the the temperature is above the set point.

The voltage between R1 and R2 varies from Vcc2 * (110K / (110K + 330K)) to Vcc2 * ((110K + 22K)/ (110K + 22K + 330K)) or 3.0000 V to 3.4286 V.

Without knowing B, we would need to know the temperature range of the controller.
 

Thread Starter

MrP90

Joined May 19, 2013
12
The diode is the 1N4001, I can't find it's temperature deviation. The LM355 temperature sensor's temperature range is -50ºC to +150ºC. The only information about B is that it depends on the properties of the material.

Using LM335, it gives a voltage proportional to the temperature (vT=0.01T , Temperature in K)

Thank you!
 

LDC3

Joined Apr 27, 2013
924
If the temperature range is -50 ºC to 150 ºC, then 45 ºC would be
(45 - (-50))/(150 - (-50)) x 22k
= 95/200 x 22k
= 10.45k

Since the transistor has 0 voltage drop when it is fully on, the current through R5 is
12 V / 47 ohms
= 0.255 amps
The state of the switches does not change the current in R5 since VF is considered to be at ground potential (that is why Dv is off).
 

Thread Starter

MrP90

Joined May 19, 2013
12
These 0.255A is the current that heats Ra when both the switches are closed, right?

I'm sorry, but I'm really confused...
 

LDC3

Joined Apr 27, 2013
924
No, that is the current through R5. It does not matter what the current through RA is, only the voltage level at the op-amp input.
 

Thread Starter

MrP90

Joined May 19, 2013
12
Yes, I meant R5.

To find VR, can I use a voltage divider between R2+RA and R1//R3 using the output voltage of the regulator?
 

LDC3

Joined Apr 27, 2013
924
Well, it is actually (R2 + RA) / (R2 + RA + R1) x Vcc2
where R1 is 330k, R2 is 110k, RA is 10.54k and Vcc2 is 12 V
 

Thread Starter

MrP90

Joined May 19, 2013
12
Just another question. I can't understand how the K-ºC converter works... I know it will subtract 0.273V to the voltage, but how does it happen?

Thank you very much for your help!
 

LDC3

Joined Apr 27, 2013
924
I'm not certain what you mean.

If you are talking about the equation you had
Variation of the resistance with T: R(T)=R(T0)exp[B(1/T)-(1/T0)]
The to convert Kelvin to Celsius, you subtract 273.15.
298.15 Kelvin is 25 ºC.
 

Thread Starter

MrP90

Joined May 19, 2013
12
I have a Kelvin to Celsius degree in my circuit. This is realized by the amplifier A2, but I don´t understand how that happens... I want the temperature shown in the LCD monitor in ºC, so I have to convert it.
 

LDC3

Joined Apr 27, 2013
924
An increase in 1 ºC is an increase in 1 Kelvin. It all depends on what you want to call zero. After all, the freezing point of water varies as to the pressure. Celcius picked the freezing point of water as zero. Scientists needed a point where vibrational energy was zero; this is how the Kelvin scale is determined.

To monitor the temperature, you need 2 things. At what voltage setting do you get a certain temperature (like 68 ºF or 20 ºC). What is the voltage change for every degree change in temperature (either 1 ºF or 1 ºC). With these 2 bits of information, you can use any scale you want since it is using the units you want in the display.
 

Thread Starter

MrP90

Joined May 19, 2013
12
When the RA resistor is 70Celsius, my voltage is about 0.7V. So the voltage is proporcional to the temperature, how/why does this happen?
 

LDC3

Joined Apr 27, 2013
924
When the RA resistor is 70Celsius, my voltage is about 0.7V. So the voltage is proporcional to the temperature, how/why does this happen?
All I can say is that the diode has a temperature dependency on the zener voltage. How much is dependent on the design of the diode.
 
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