Hi everyone,

Can you please assist us with the following:

To design a safety circuit that should provide for household

appliances for over-and undervoltage protection. The protective

circuit must immediately switch off upon detection of low-and

high-voltage household appliance and upon detection of normal voltage

switch on again after 3 minutes.

The protective circuit must comply with the following:

If the line voltage is within the normal range (100 to 130V ac), it

will wait for the protective circuit 3 minutes before the output will

be energized. During these 3 minutes there is an amber LED light.

If the line voltage is outside the normal voltage, the output of the

protective circuit will never be under tension.

If the line voltage is less than 100VAC, the protection circuit "low

voltage" must indicate by a red LED that lights up. If the line

voltage is present, the protection circuit must pass a voltage greater

than 105 Vac "normal tension" it will indicate by a green LED that

lights up.

Similarly, the line voltage protective circuit has to be higher than

130V ac "high voltage" will be indicated by a red LED that lights up.

Only when a voltage is less than 125VAC, it must indicate the

protection circuit "normal tension" by a green LED that lights up.

Upon detection of over-and under voltage protection, the circuit

should give a beep of 5 seconds. This should be constructed with an

opamp oscillator circuit in this functionality.

I use a window comparator and a timer for the 3 minute delay.

Please is there someone that could help me to calculate the resistors for the comparator?

Br,

Dennis Tahar

Hi dennis,

For your problem statement the idea of using a windows comparator and timer (555 i guess) is a perfect choice. I assume that you have already found ways to convert the AC input (100V to 200V ) to Analog voltage for the Op-amp. Also, I am not sure on what voltage your op-amp is operating. It might be on 5V or 12V. Either way refer the datasheet for the output saturation voltage of the op-amp. For example an op-amp operating at 5V rail supply (assuming its not rail to rail) will have an output voltage of 3.6V only.

So, now coming to the window comparator, A window comparator is nothing but two voltage comparators combined to maintain a upper and lower limit for the incoming voltage. Hence before you jump into window comparator i would suggest you to learn to use a normal voltage comparator. Once that is done.

http://www.electronics-tutorials.ws/opamp/opamp107.gif?x98918

The above reference will help you in calculating the resistor values of your window comparator. So, basically there will be two reference voltages, your upper reference voltage and your lower reference voltage each of which will be given as individual reference voltage for each op-amp. One op-amp will monitor if the input voltage has fallen below the set limit and the other will monitor if the input voltage has exceeded the set limit. IF it goes beyond the limit the output will be high else low.

Generally the reference voltage (both upper and lower) is set using a potential divider circuit. so, calculate your upper and lower limits of voltage and then decide your resistor for potential divider network.

NOTE: I cannot give you the exact values of the resistors because you have not mentioned how you have converted the AC to DC and how what will be your input voltage to your op-amp. Also I do not know on what voltage your op-amp is operating.

I think i have given you enough insight, if any problem feel free to use the forum.

ALL THE BEST with your project!!

Dear mr Aswinth Raj,

Thank you very much for the explanation.

First there is a transformer and then there is the rectifier. To lower the voltage of 130 V we used a voltage divider for the U equivalent value U equivalent= U(R2/R1 + R2). I choose for R1 = 0.5 K ohm. and for R2 i got 8 K ohm as value. for 130 V (upper trip point).

For the overvoltage i have to calculate the U equivalent values for 130 v and 125 v and for the undervoltage for 105 v and 100 v using the voltage divider:

The U equivalent value for 130V is 7.6 V.

The U equivalent value for 125V is 7.35 V.

After the voltage divider we choose a voltage regulator of12V DC (7812 model).

After that there is the U reference which the comparator will be working at U->U+ , Vo = 0 for the upper trip point:

Ucc = 12 V DC

U reference = Ucc*(R1*(R2//R3)/(R1+(R2//R3)

For the lower trip point

U reference = Ucc*(R2/R1+R3)

i choose 10 K ohm for R1, after calculation i got 15.8 K ohm for R2.

For R3 i got a negative value and that is not a good value for R3 at the positive feedback on the Upper trip point.

I'm still figuring out what mauy be the problem of having negative value.

I am not able to completely understand your circuit.

Provide me the schematics of your project if you need help on anything.

All the best!!

Mr Raj,

I was busy with the schematic, but it doesn't work according the requirements. This is the link of the schematic https://www.flickr.com/photos/147794132@N04/33462041741/in/dateposted-public/

I think that the values of the components or maybe the connections/ type of components or not right.

Best regards