How to keep the state of relay
In schematic below when current exceeds the set point of PR1 it toggles the relay and as soon as current drops the relay goes back to normal state:
How can i keep the relay state as triggered even when the current drops and only switch the relay back to normal state when i pressed a push-button?
relay schematics
asked 8 hours ago
newbienewbie
369
add a comment |
In schematic below when current exceeds the set point of PR1 it toggles the relay and as soon as current drops the relay goes back to normal state:
How can i keep the relay state as triggered even when the current drops and only switch the relay back to normal state when i pressed a push-button?
relay schematics
asked 8 hours ago
newbienewbie
369
1
It would be interesting to hear what this circuit will accomplish once it's done. Some background information is what I'm talking about, who knows, maybe, just maybe, there's a better solution to that than what you've proposed.
– Harry Svensson
7 hours ago
1
Can you use a second relay, or a DPDT relay? Or perhaps some digital logic?
– mike65535
7 hours ago
thanks @HarrySvensson found the solution.
– newbie
5 hours ago
thanks @mike65535 found the solution.
– newbie
5 hours ago
add a comment |
In schematic below when current exceeds the set point of PR1 it toggles the relay and as soon as current drops the relay goes back to normal state:
How can i keep the relay state as triggered even when the current drops and only switch the relay back to normal state when i pressed a push-button?
relay schematics
asked 8 hours ago
newbienewbie
369
In schematic below when current exceeds the set point of PR1 it toggles the relay and as soon as current drops the relay goes back to normal state:
How can i keep the relay state as triggered even when the current drops and only switch the relay back to normal state when i pressed a push-button?
relay schematics
relay schematics
asked 8 hours ago
newbienewbie
369
asked 8 hours ago
newbienewbie
369
edited 7 hours ago
newbie
asked 8 hours ago
newbienewbie
369
asked 8 hours ago
newbienewbie
369
asked 8 hours ago
newbienewbie
369
369
1
It would be interesting to hear what this circuit will accomplish once it's done. Some background information is what I'm talking about, who knows, maybe, just maybe, there's a better solution to that than what you've proposed.
– Harry Svensson
7 hours ago
1
Can you use a second relay, or a DPDT relay? Or perhaps some digital logic?
– mike65535
7 hours ago
thanks @HarrySvensson found the solution.
– newbie
5 hours ago
thanks @mike65535 found the solution.
– newbie
5 hours ago
add a comment |
1
It would be interesting to hear what this circuit will accomplish once it's done. Some background information is what I'm talking about, who knows, maybe, just maybe, there's a better solution to that than what you've proposed.
– Harry Svensson
7 hours ago
1
Can you use a second relay, or a DPDT relay? Or perhaps some digital logic?
– mike65535
7 hours ago
thanks @HarrySvensson found the solution.
– newbie
5 hours ago
thanks @mike65535 found the solution.
– newbie
5 hours ago
1
1
It would be interesting to hear what this circuit will accomplish once it's done. Some background information is what I'm talking about, who knows, maybe, just maybe, there's a better solution to that than what you've proposed.
– Harry Svensson
7 hours ago
It would be interesting to hear what this circuit will accomplish once it's done. Some background information is what I'm talking about, who knows, maybe, just maybe, there's a better solution to that than what you've proposed.
– Harry Svensson
7 hours ago
1
1
Can you use a second relay, or a DPDT relay? Or perhaps some digital logic?
– mike65535
7 hours ago
Can you use a second relay, or a DPDT relay? Or perhaps some digital logic?
– mike65535
7 hours ago
thanks @HarrySvensson found the solution.
– newbie
5 hours ago
thanks @HarrySvensson found the solution.
– newbie
5 hours ago
thanks @mike65535 found the solution.
– newbie
5 hours ago
thanks @mike65535 found the solution.
– newbie
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
You can realize the required function using a diode. What needs to be done is to pull one of the inputs of the opamp to such a voltage that the opamp's output stays high (and the relay powered).
Add a diode (1N4148) with the anode to the opamp's output and the kathode to the + input of the opamp. Then when the opamp's output is high (and the relay is powered) the + input of the opamp will be pulled to a higher voltage than R3 and R7 supply (which is about 6 V). The - input of the opamp will not be able to reach that voltage (it will always be at a lower voltage) So the output of the opamp remains high.
To add the reset function, add a 1 kohm resistor in series with the diode and add a switch to ground from the + input to force it low so that the output of the opamp becomes low as well:
simulate this circuit – Schematic created using CircuitLab
answered 7 hours ago
BimpelrekkieBimpelrekkie
47k240104
A downside of this solution is that it will disarm the relay even in the active area. If it's an overvoltage protection this might be not desirable. Maybe a slight modification could help, using a normal closed push-button in series with D1 instead?
– Dorian
6 hours ago
Oh, already sugested by Sphero
– Dorian
6 hours ago
@Dorian A downside of this solution is that it will disarm the relay even in the active area I fail to see how that would happen. Can you explain.
– Bimpelrekkie
6 hours ago
When you push the switch the relay will be off no matter the input.
– Dorian
6 hours ago
1
@newbie Why do you think you need a 2 A diode? If you do not have the 1N4148 you can use almost any other diode including the 1N4007 that's already in your schematic. Also "tolerance" relates to how accurate something is. Your sentence should have been: ... to tolerate more current up to 2A?. Note that there is a resistor in series with the diode, in this circuit the highest current through the diode is 12V/1kohm = 12 mA which is much less than 2 A.
– Bimpelrekkie
4 hours ago
|
show 3 more comments
You can add some positive feedback, with a N.C. pushbutton in series.
For example, a diode from the op-amp output to non-inverting input. With said switch in series.
Edit: This is similar to Bimpelrekkie's solution except it requires a N.C. pushbutton, which is less common, and does not force the relay state when the button is pushed. The diode can be any common silicon signal diode such as 1N4148.
Note: If you require a particular guaranteed state at power-up you might have to add some circuitry.
answered 7 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
i'm new to electronics, can you please explain with schematic?
– newbie
5 hours ago
2
Indeed I did not suggest this solution as push-to-open push buttons are hard(er) to find than push-to-close push buttons.
– Bimpelrekkie
4 hours ago
add a comment |
If you have a chance to change your relay, you may try to use latching relays.
I'm adding a latching relay below.BTW I couldn't check current and voltage ratings of your circuit. This component is just an example.
https://www.digikey.com/product-detail/en/kemet/EE2-3SNUH-L/399-11008-1-ND/4506460
there is 2 control input in this relay. once you triggered first input and the other one is logic0, relay switches inputs. Until you apply trigger to the second input this will stay same.
answered 6 hours ago
KorayKoray
403
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can realize the required function using a diode. What needs to be done is to pull one of the inputs of the opamp to such a voltage that the opamp's output stays high (and the relay powered).
Add a diode (1N4148) with the anode to the opamp's output and the kathode to the + input of the opamp. Then when the opamp's output is high (and the relay is powered) the + input of the opamp will be pulled to a higher voltage than R3 and R7 supply (which is about 6 V). The - input of the opamp will not be able to reach that voltage (it will always be at a lower voltage) So the output of the opamp remains high.
To add the reset function, add a 1 kohm resistor in series with the diode and add a switch to ground from the + input to force it low so that the output of the opamp becomes low as well:
simulate this circuit – Schematic created using CircuitLab
answered 7 hours ago
BimpelrekkieBimpelrekkie
47k240104
A downside of this solution is that it will disarm the relay even in the active area. If it's an overvoltage protection this might be not desirable. Maybe a slight modification could help, using a normal closed push-button in series with D1 instead?
– Dorian
6 hours ago
Oh, already sugested by Sphero
– Dorian
6 hours ago
@Dorian A downside of this solution is that it will disarm the relay even in the active area I fail to see how that would happen. Can you explain.
– Bimpelrekkie
6 hours ago
When you push the switch the relay will be off no matter the input.
– Dorian
6 hours ago
1
@newbie Why do you think you need a 2 A diode? If you do not have the 1N4148 you can use almost any other diode including the 1N4007 that's already in your schematic. Also "tolerance" relates to how accurate something is. Your sentence should have been: ... to tolerate more current up to 2A?. Note that there is a resistor in series with the diode, in this circuit the highest current through the diode is 12V/1kohm = 12 mA which is much less than 2 A.
– Bimpelrekkie
4 hours ago
|
show 3 more comments
You can realize the required function using a diode. What needs to be done is to pull one of the inputs of the opamp to such a voltage that the opamp's output stays high (and the relay powered).
Add a diode (1N4148) with the anode to the opamp's output and the kathode to the + input of the opamp. Then when the opamp's output is high (and the relay is powered) the + input of the opamp will be pulled to a higher voltage than R3 and R7 supply (which is about 6 V). The - input of the opamp will not be able to reach that voltage (it will always be at a lower voltage) So the output of the opamp remains high.
To add the reset function, add a 1 kohm resistor in series with the diode and add a switch to ground from the + input to force it low so that the output of the opamp becomes low as well:
simulate this circuit – Schematic created using CircuitLab
answered 7 hours ago
BimpelrekkieBimpelrekkie
47k240104
A downside of this solution is that it will disarm the relay even in the active area. If it's an overvoltage protection this might be not desirable. Maybe a slight modification could help, using a normal closed push-button in series with D1 instead?
– Dorian
6 hours ago
Oh, already sugested by Sphero
– Dorian
6 hours ago
@Dorian A downside of this solution is that it will disarm the relay even in the active area I fail to see how that would happen. Can you explain.
– Bimpelrekkie
6 hours ago
When you push the switch the relay will be off no matter the input.
– Dorian
6 hours ago
1
@newbie Why do you think you need a 2 A diode? If you do not have the 1N4148 you can use almost any other diode including the 1N4007 that's already in your schematic. Also "tolerance" relates to how accurate something is. Your sentence should have been: ... to tolerate more current up to 2A?. Note that there is a resistor in series with the diode, in this circuit the highest current through the diode is 12V/1kohm = 12 mA which is much less than 2 A.
– Bimpelrekkie
4 hours ago
|
show 3 more comments
You can realize the required function using a diode. What needs to be done is to pull one of the inputs of the opamp to such a voltage that the opamp's output stays high (and the relay powered).
Add a diode (1N4148) with the anode to the opamp's output and the kathode to the + input of the opamp. Then when the opamp's output is high (and the relay is powered) the + input of the opamp will be pulled to a higher voltage than R3 and R7 supply (which is about 6 V). The - input of the opamp will not be able to reach that voltage (it will always be at a lower voltage) So the output of the opamp remains high.
To add the reset function, add a 1 kohm resistor in series with the diode and add a switch to ground from the + input to force it low so that the output of the opamp becomes low as well:
simulate this circuit – Schematic created using CircuitLab
answered 7 hours ago
BimpelrekkieBimpelrekkie
47k240104
You can realize the required function using a diode. What needs to be done is to pull one of the inputs of the opamp to such a voltage that the opamp's output stays high (and the relay powered).
Add a diode (1N4148) with the anode to the opamp's output and the kathode to the + input of the opamp. Then when the opamp's output is high (and the relay is powered) the + input of the opamp will be pulled to a higher voltage than R3 and R7 supply (which is about 6 V). The - input of the opamp will not be able to reach that voltage (it will always be at a lower voltage) So the output of the opamp remains high.
To add the reset function, add a 1 kohm resistor in series with the diode and add a switch to ground from the + input to force it low so that the output of the opamp becomes low as well:
simulate this circuit – Schematic created using CircuitLab
answered 7 hours ago
BimpelrekkieBimpelrekkie
47k240104
answered 7 hours ago
BimpelrekkieBimpelrekkie
47k240104
answered 7 hours ago
BimpelrekkieBimpelrekkie
47k240104
answered 7 hours ago
BimpelrekkieBimpelrekkie
47k240104
47k240104
A downside of this solution is that it will disarm the relay even in the active area. If it's an overvoltage protection this might be not desirable. Maybe a slight modification could help, using a normal closed push-button in series with D1 instead?
– Dorian
6 hours ago
Oh, already sugested by Sphero
– Dorian
6 hours ago
@Dorian A downside of this solution is that it will disarm the relay even in the active area I fail to see how that would happen. Can you explain.
– Bimpelrekkie
6 hours ago
When you push the switch the relay will be off no matter the input.
– Dorian
6 hours ago
1
@newbie Why do you think you need a 2 A diode? If you do not have the 1N4148 you can use almost any other diode including the 1N4007 that's already in your schematic. Also "tolerance" relates to how accurate something is. Your sentence should have been: ... to tolerate more current up to 2A?. Note that there is a resistor in series with the diode, in this circuit the highest current through the diode is 12V/1kohm = 12 mA which is much less than 2 A.
– Bimpelrekkie
4 hours ago
|
show 3 more comments
A downside of this solution is that it will disarm the relay even in the active area. If it's an overvoltage protection this might be not desirable. Maybe a slight modification could help, using a normal closed push-button in series with D1 instead?
– Dorian
6 hours ago
Oh, already sugested by Sphero
– Dorian
6 hours ago
@Dorian A downside of this solution is that it will disarm the relay even in the active area I fail to see how that would happen. Can you explain.
– Bimpelrekkie
6 hours ago
When you push the switch the relay will be off no matter the input.
– Dorian
6 hours ago
1
@newbie Why do you think you need a 2 A diode? If you do not have the 1N4148 you can use almost any other diode including the 1N4007 that's already in your schematic. Also "tolerance" relates to how accurate something is. Your sentence should have been: ... to tolerate more current up to 2A?. Note that there is a resistor in series with the diode, in this circuit the highest current through the diode is 12V/1kohm = 12 mA which is much less than 2 A.
– Bimpelrekkie
4 hours ago
A downside of this solution is that it will disarm the relay even in the active area. If it's an overvoltage protection this might be not desirable. Maybe a slight modification could help, using a normal closed push-button in series with D1 instead?
– Dorian
6 hours ago
A downside of this solution is that it will disarm the relay even in the active area. If it's an overvoltage protection this might be not desirable. Maybe a slight modification could help, using a normal closed push-button in series with D1 instead?
– Dorian
6 hours ago
Oh, already sugested by Sphero
– Dorian
6 hours ago
Oh, already sugested by Sphero
– Dorian
6 hours ago
@Dorian A downside of this solution is that it will disarm the relay even in the active area I fail to see how that would happen. Can you explain.
– Bimpelrekkie
6 hours ago
@Dorian A downside of this solution is that it will disarm the relay even in the active area I fail to see how that would happen. Can you explain.
– Bimpelrekkie
6 hours ago
When you push the switch the relay will be off no matter the input.
– Dorian
6 hours ago
When you push the switch the relay will be off no matter the input.
– Dorian
6 hours ago
1
1
@newbie Why do you think you need a 2 A diode? If you do not have the 1N4148 you can use almost any other diode including the 1N4007 that's already in your schematic. Also "tolerance" relates to how accurate something is. Your sentence should have been: ... to tolerate more current up to 2A?. Note that there is a resistor in series with the diode, in this circuit the highest current through the diode is 12V/1kohm = 12 mA which is much less than 2 A.
– Bimpelrekkie
4 hours ago
@newbie Why do you think you need a 2 A diode? If you do not have the 1N4148 you can use almost any other diode including the 1N4007 that's already in your schematic. Also "tolerance" relates to how accurate something is. Your sentence should have been: ... to tolerate more current up to 2A?. Note that there is a resistor in series with the diode, in this circuit the highest current through the diode is 12V/1kohm = 12 mA which is much less than 2 A.
– Bimpelrekkie
4 hours ago
|
show 3 more comments
You can add some positive feedback, with a N.C. pushbutton in series.
For example, a diode from the op-amp output to non-inverting input. With said switch in series.
Edit: This is similar to Bimpelrekkie's solution except it requires a N.C. pushbutton, which is less common, and does not force the relay state when the button is pushed. The diode can be any common silicon signal diode such as 1N4148.
Note: If you require a particular guaranteed state at power-up you might have to add some circuitry.
answered 7 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
i'm new to electronics, can you please explain with schematic?
– newbie
5 hours ago
2
Indeed I did not suggest this solution as push-to-open push buttons are hard(er) to find than push-to-close push buttons.
– Bimpelrekkie
4 hours ago
add a comment |
You can add some positive feedback, with a N.C. pushbutton in series.
For example, a diode from the op-amp output to non-inverting input. With said switch in series.
Edit: This is similar to Bimpelrekkie's solution except it requires a N.C. pushbutton, which is less common, and does not force the relay state when the button is pushed. The diode can be any common silicon signal diode such as 1N4148.
Note: If you require a particular guaranteed state at power-up you might have to add some circuitry.
answered 7 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
i'm new to electronics, can you please explain with schematic?
– newbie
5 hours ago
2
Indeed I did not suggest this solution as push-to-open push buttons are hard(er) to find than push-to-close push buttons.
– Bimpelrekkie
4 hours ago
add a comment |
You can add some positive feedback, with a N.C. pushbutton in series.
For example, a diode from the op-amp output to non-inverting input. With said switch in series.
Edit: This is similar to Bimpelrekkie's solution except it requires a N.C. pushbutton, which is less common, and does not force the relay state when the button is pushed. The diode can be any common silicon signal diode such as 1N4148.
Note: If you require a particular guaranteed state at power-up you might have to add some circuitry.
answered 7 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
You can add some positive feedback, with a N.C. pushbutton in series.
For example, a diode from the op-amp output to non-inverting input. With said switch in series.
Edit: This is similar to Bimpelrekkie's solution except it requires a N.C. pushbutton, which is less common, and does not force the relay state when the button is pushed. The diode can be any common silicon signal diode such as 1N4148.
Note: If you require a particular guaranteed state at power-up you might have to add some circuitry.
answered 7 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
edited 4 hours ago
answered 7 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
answered 7 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
answered 7 hours ago
Spehro PefhanySpehro Pefhany
204k4150408
204k4150408
i'm new to electronics, can you please explain with schematic?
– newbie
5 hours ago
2
Indeed I did not suggest this solution as push-to-open push buttons are hard(er) to find than push-to-close push buttons.
– Bimpelrekkie
4 hours ago
add a comment |
i'm new to electronics, can you please explain with schematic?
– newbie
5 hours ago
2
Indeed I did not suggest this solution as push-to-open push buttons are hard(er) to find than push-to-close push buttons.
– Bimpelrekkie
4 hours ago
i'm new to electronics, can you please explain with schematic?
– newbie
5 hours ago
i'm new to electronics, can you please explain with schematic?
– newbie
5 hours ago
2
2
Indeed I did not suggest this solution as push-to-open push buttons are hard(er) to find than push-to-close push buttons.
– Bimpelrekkie
4 hours ago
Indeed I did not suggest this solution as push-to-open push buttons are hard(er) to find than push-to-close push buttons.
– Bimpelrekkie
4 hours ago
add a comment |
If you have a chance to change your relay, you may try to use latching relays.
I'm adding a latching relay below.BTW I couldn't check current and voltage ratings of your circuit. This component is just an example.
https://www.digikey.com/product-detail/en/kemet/EE2-3SNUH-L/399-11008-1-ND/4506460
there is 2 control input in this relay. once you triggered first input and the other one is logic0, relay switches inputs. Until you apply trigger to the second input this will stay same.
answered 6 hours ago
KorayKoray
403
add a comment |
If you have a chance to change your relay, you may try to use latching relays.
I'm adding a latching relay below.BTW I couldn't check current and voltage ratings of your circuit. This component is just an example.
https://www.digikey.com/product-detail/en/kemet/EE2-3SNUH-L/399-11008-1-ND/4506460
there is 2 control input in this relay. once you triggered first input and the other one is logic0, relay switches inputs. Until you apply trigger to the second input this will stay same.
answered 6 hours ago
KorayKoray
403
add a comment |
If you have a chance to change your relay, you may try to use latching relays.
I'm adding a latching relay below.BTW I couldn't check current and voltage ratings of your circuit. This component is just an example.
https://www.digikey.com/product-detail/en/kemet/EE2-3SNUH-L/399-11008-1-ND/4506460
there is 2 control input in this relay. once you triggered first input and the other one is logic0, relay switches inputs. Until you apply trigger to the second input this will stay same.
answered 6 hours ago
KorayKoray
403
If you have a chance to change your relay, you may try to use latching relays.
I'm adding a latching relay below.BTW I couldn't check current and voltage ratings of your circuit. This component is just an example.
https://www.digikey.com/product-detail/en/kemet/EE2-3SNUH-L/399-11008-1-ND/4506460
there is 2 control input in this relay. once you triggered first input and the other one is logic0, relay switches inputs. Until you apply trigger to the second input this will stay same.
answered 6 hours ago
KorayKoray
403
answered 6 hours ago
KorayKoray
403
answered 6 hours ago
KorayKoray
403
answered 6 hours ago
KorayKoray
403
403
add a comment |
add a comment |
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1
It would be interesting to hear what this circuit will accomplish once it's done. Some background information is what I'm talking about, who knows, maybe, just maybe, there's a better solution to that than what you've proposed.
– Harry Svensson
7 hours ago
1
Can you use a second relay, or a DPDT relay? Or perhaps some digital logic?
– mike65535
7 hours ago
thanks @HarrySvensson found the solution.
– newbie
5 hours ago
thanks @mike65535 found the solution.
– newbie
5 hours ago