In this post I have explained a simple yet very accurate temperature controller circuit using IC LM35, featuring an automatic cut off with a push button latch.
LM35 is a precision temperature sensor IC, which can be effectively used for detecting temperature differences accurately. For more details regarding this IC you can refer to this article.
Circuit Diagram
How the Circuit Works
The IC1 741 is configured as a comparator circuit.
Its pin#3 which is its non-inverting input pin, is attached with the output of the heat sensor IC2 LM35.
The inverting input pin#2 of IC1 is configured with a preset P1, which is used for setting the reference voltage at pin#2.
Whenever pin#3 voltage exceeds pin#2 voltage, the IC1 output at pin#6 turn high.
Conversely, as long as pin#3 voltage remains below pin#2 voltage, the output pin#6 of IC1 remains at 0V, or logic low.
At normal room temperature the LM35 output will be low which will keep the pin#3 potential of the 741 lower than its pin#2 reference.
Therefore, when the push button is pressed, the IC741 momentarily activates sending a 0V (low) to the base of BC557 which now conducts so that the whole circuit now gets latched even if the push button is released, and stays powdered through the BC557.
In this situation the relay switches ON and its contacts shift to the N/O points. Since the heater is wired with the supply through the N/O contacts of the relay, it is also switched ON.
The heater temperature now starts rising. Since the IC2 LM35 is attached with the heater, the LM35 also starts heating up which causes its output voltage at pin#3 of IC 741 to rise proportionately.
When the LM35 output voltage at pin#3 of IC1 741 reaches a point where it exceeds its pin#2 potential, the output of IC 741 instantly reverts and goes high.
As soon as the output of IC 741 goes high, it inhibits the negative or 0V supply to the base of BC557. Due to this the BC557 turns off.
As BC557 turns OFF, it breaks the latch so that the 12V supply to the circuit is cut off, the whole circuit along with the relay is turned off.
This also turns OFF the heater and the circuit returns to its previous original condition.
How to Setup and Test
Assemble the circuit as shown in the above diagram. Initially, do not connect any heater with the relay contacts.
Keep the P1 slider fully at the positive level or fully towards R2.
Switch ON the 12V DC to the circuit.
Push and release the SW1 switch momentarily. This should instantly cause the circuit and the relay to switch ON and get latched. Both the LEDs must illuminate.
Next, using a soldering iron heat the LM35 to the desired level. Monitor the temperature with an accurate thermometer, make sure it does not exceed 100° Celsius.
Once the desired temperature is reached, slowly adjust the P1 preset until the IC1 output turns high causing the latch to break and turn OFF the relay. This is indicated by the LEDs which are now turned OFF.
Repeat the above procedure a few times to confirm the results.
The circuit is now set and ready to work with an actual heater attached with the relay contacts.
Troubleshooting
If somehow the circuit does not work, please implement the following improvements in the circuit.
Please connect a 1uF/25V capacitor across the base/emitter of the BC557 transistor. This will ensure that the circuit does not start by itself during power switch ON, rather initiates only when the push-button is pressed.
Please remove R2 (10k resistor) and replace it with a jumper. This will ensure that when initially the slider of the preset is held at the positive side, the op amp output delivers a low logic without fail.
Also, connect a 1uF/25V across pin#3 of the IC and ground, this is optional.
After finishing the above operations, please follow the steps explained under "How to Setup and Test."
Parts List
- All Resistors are 1/4 watt 5% CFR
- 1k, 4.7k = 1 each
- 10k = 2
- 4.7k preset = 1
- Semiconductors
- IC 741, IC LM35 = 1 each
- Transistor BC557 = 1
- LEDs 20mA, 5mm = 2
- Relay 12V 10 amp = 1
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