Over & Under Voltage Protection Circuit
A OV-UV Protection circuit/ Circuit breaker is an automatically operated electrical switch designed to protect an electrical circuit from damage caused by excess current, typically resulting from an overload or short circuit. Its basic function is to interrupt current flow after a fault is detected. Unlike a fuse, which operates once and then must be replaced, a circuit breaker can be reset (either manually or automatically) to resume normal operation. Circuit breakers are made in varying sizes, from small devices that protect low-current circuits or individual household appliance, up to large switchgear designed to protect high voltage circuits feeding an entire city. The generic function of a circuit breaker, RCD or a fuse, as an automatic means of removing power from a faulty system is often abbreviated to ADS (Automatic Disconnection of Supply). < Wikipedia >
Terminology: You Must Know
1. Voltage Divider
A voltage divider (also known as a potential divider) is a passive linear circuit that produces an output voltage (Vout) that is a fraction of its input voltage (Vin). In other words if you have a voltage supply of 10 Volt but you want only 6 Volt as excitation of your circuit, then the simplest solution is to use a voltage divider circuit(only two resistors).
How to do? => Choose a random value for R1 let’s say 1kΩ then to find R2’s value use this formula
Vout = Vin(R2/(R1+R2))
Hence R2 is 1.5kΩ and your circuit should look like
A comparator circuit compares two voltages and outputs either a VDD (the voltage at the plus side) or a 0 (the voltage at the negative side) to indicate which is larger. Comparators are used to check whether an input has reached some reference value. In most cases some comparator IC’s like LM339 is used for this purpose, but op-amps may be used as an alternative. In this exercise we are using a comparator IC LM339 which works as explained in diagram below.
A typical digital logic output pin can only supply tens of mA of current. Larger devices like Light Bulbs, Motors, Fan, TV, Refrigerators might even need several amps. But a Relay requires only a few mA to operate and can be used as a switch for such high current seeking devices. To switch ON and OFF relay a Transistor based driving circuit is needed as shown in figure below. When Base current is not present Transistor is in cut-off region (switch off) but when a sufficient amount of current is provided on the Base terminal Transistor comes under saturation region (switch ON)
A Zener diode allows current to flow from its anode to its cathode like a normal semiconductor diode, but it also permits current to flow in the reverse direction when its "Zener voltage" is reached.
Zener diodes are widely used in electronic equipment of all kinds and are one of the basic building blocks of electronic circuits. They are used to generate low power stabilized supply rails from a higher voltage and to provide reference voltages for circuits, especially stabilized power supplies. They are also used to protect circuits from over-voltage and provides a constant value across it as shown in figure.
What are we designing?
We are going to design a protection circuit for electrical appliances say a Refrigerator. This circuit will protect (cut-off) your refrigerator if the Line Voltage (AC Voltage coming to your electric plug) exceeds 250V or reduce below 180V. The block diagram below showing different units of a circuit breaker.
We will design this circuit in reverse order (Last block will be designed first) to maintain the accuracy.
1. Designing Relay Driving Circuit:
Designing of Relay Driving Circuit is discussed is previous exercise which involves following steps
è Measurement of Relay Coil Resistance (Rcoil)
è Find Current required by coil/ collector current for transistor (Ic)
è Selection of transistor and current gain (β)
è Calculate Base Current(IB)
è Choose Base voltage as 3.3V (TTL VOH(min)) and find value of Base Resistance (RB)
Values for our design are
Rcoil=71.9Ω, Ic=69.541mA, βmin=35 , IB=1.986mA , RB=1.359 KΩ
NOTE: Value of RB for your circuit may be different because of different Coil resistance
Now test this circuit in all respects and be sure that it is working properly.
2. Design of Comparing Circuit:
For this design we are using LM339 Quad-Comparator IC. Now to set two reference voltages for comparator inputs we will use Zener diodes as voltage regulator. This manual is using a 4.7V Zener (Corresponding to 180V AC) and a 5.6V Zener (corresponding to 250V AC) to set reference voltages as shown in figure. (You can use any value of Zener diode but make sure both should be different)
Selection of Rs
The current through the resistor Rs is given by (KVL and Ohm's law):
I= (Vin - Vz)/Rs
This is the maximum current through the Zener diode and so, is chosen to limit the current through the diode to some value such that the maximum power dissipated by the diode is below the maximum power rating.
If max Power Rating =500mW(datasheet), Vz=5.6V then Iz=89.28mA and IIO=50nA(max) then from above equation Rs=(12-5.6)/(89.28m-50n). Hence Rs is nearly 72Ω but for safety purpose we are using 1KΩ instead.
Working of Comparator Blockà Negative input of C1 and positive input of C2 are referenced as 5.6V and 4.7V respectively. Hence if voltage at positive input of C1 is greater than 5.6V than voltage at Vout will be higher. It will also higher if voltage at negative input of C2 will be lesser than 4.7V and LOW otherwise.
Coupling 1st and 2nd Block
Working : As mentioned above whenever the voltage at comparator side reach beyond 5.6V or below 4.7V, Vout becomes high and provide a sufficient amount of current to base of transistor which energize the relay and circuit connected in its output start functioning. But when the input of comparator is in the range of 4.7V to 5.6V Vout remains 0V, no current flows through Base and hence relay will remain off.
Need of D1 & D2
The working of our circuit is such that output of either one of the two comparator will high or both zero. So in case C2’s output is high D1 will not allow the current IB to enter into C1. D2 also perform the similar task in case output of C1 is high.
3. Voltage Step down from 230V to 12V
Now we need some samples from voltage supply of our house (which is also the input for your Refrigerator) and we have to convert this voltage to a small value Vc such that
Vc = 5.61V when VAC(supply) =250V
Vc = 4.69V when VAC(supply) =180V
So that when the supply goes beyond 250V, output of Block 1 will be 5.61V which will be the input of Comparators and output of C1 will high. In case input is 4.69V (for AC 180V), output of C2 will high and turn on the relay which will switch off your Refrigerator.
So for this purpose we need two voltage divider circuit as shown in figure. To calculate the values of R1, R2, R3 and R4 follow these steps
i. Assemble this circuit without connecting R1, R2, R3 and R4 in the circuit.
ii. Choose an appropriate value of Capacitor for filter circuit.
iii. Increase the input AC voltage to 250V and measure the output voltage (Vo1) across capacitor. Now to divide this Vo to get VB = 5.61V
a. Select R2’s value as 5.6kΩ.
b. Calculate R1 as
VB = Vo1(R2/(R1+R2))
c. For this circuit it is calculated as R1=19kΩ
iv. Decrease the input voltage to 180V and measure the voltage (Vo2) across capacitor. And follow the same procedure to calculate the value of R3 and R4.
Coupling 1st, 2nd and 3rd Block
· --> Connect VA to ‘–‘pin of comparator 2 (C2).
· --> Connect VB to ‘+’ pin of comparator 1 (C1).
Still there are two additional DC power supplies of 6V (for Relay) and 12V (for comparator IC) are needed to operate this circuit. But the final product should not require such additional supplies, right?? So now we will work for removing these.
· Connect a 7812 Voltage Regulator IC across the capacitor to provide 12V DC to Comparator IC (as shown in figure).
· To provide Voltage to relay (6V) again we need a voltage regulator circuit
i. We make use of 6V Zener diode for this purpose.
ii. Connect a 100 Ω resistor in series with Zener to provide sufficient amount of current to relay coil (needs at least 70mA which is Ic).
Now test this circuit by varying the line voltage from 100V-260V and observe the results.
How My Circuit should function?
· If your Line voltage is in the range of 180V-250V, output of both comparator C1 and C2 will be 0V, no Base current will flow in the transistor so collector current will flow. Because Ic = 0A, relay will not energized and hence light bulb connected in output circuit will remain ON.
· As soon as line voltage decreases below 180V, VA (Output of Block 3) becomes nearly 4.69V (or <4.7V) and hence the output of comparator C2 goes high and a sufficient amount of current flow in the base of transistor which drive collector current to energize relay. Now relay disables output circuit and the electric appliance connected in output goes OFF.
· When line voltage reaches beyond 250V, VB (Output of Block 3) reaches 5.61V (or >5.6V) because of which output of comparator C1 goes high and a Base current IB flows to energize the relay and disable the output circuit to protect electric appliance (Bulb/Refrigerator) connected in that circuit.