Author Topic: simple controller problem  (Read 1988 times)

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Offline kafor1Topic starter

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simple controller problem
« on: December 16, 2018, 04:17:25 pm »
so i was ask to make a controller for the pressure washer, and i made it. but my problem is when i looked at the ratings of the pressure washer which is run by a single phase induction motor the current rating is 10.5A, which is my relay rating is only 10A.. i wonder if its possible, if not what should i do

- i used TW14 wire for the connector of my controller
- i used 2 spdt relay since my local store does not sell dpdt
- 1 relay is for light indicator and the other is for controlling
- i have no protection diode for my relay.

im really a beginner at electronics and im trying my best to learn,
and also can someone explain to me the term load because the term is always used in here. it would be very appreciated if it was added with example.. thank you so much
 

Offline jmelson

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Re: simple controller problem
« Reply #1 on: December 16, 2018, 09:42:55 pm »
You have no diodes to absorb the inductive spike when the relays are turned off.  This will likely cause the 555 to be damaged.

A 10 A induction motor likely draws close to 100 A for a few milliseconds when starting.  This will likely damage the relay contacts.

Jon
 

Offline tpowell1830

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Re: simple controller problem
« Reply #2 on: December 16, 2018, 10:40:13 pm »
Circuit looks okay, however, I don't understand the purpose of D1 (I would use reference designator LED1), why not just tie to V+ to show you have power on? Also, you could have less devices and more universal relay driver if you use a NPN transistor to pull your external relay coils to ground, as shown in this similar circuit: https://electronicsarea.com/on-off-switch-circuit-using-555-timer/ . Don't forget to use D1 on relay coil, as shown (D2 on second relay coil). Also, there are plenty of DPDT relays available, you just need to look around.

This looks suspiciously like a 9 volt battery driving your circuit and will last 10 minutes on your circuit driving the relays. I would find a small wall wart power supply
that delivers the current that you need at 9 to 15 volts.

I would also add a power on toggle switch for your circuit, otherwise the battery will run down completely after one use. Strike that, just because the battery is not a good supply and you need a way to power on and off your circuit (see wall wart above).

EDIT: I would label the power switch that I mentioned "POWER", or "PWR" if you have limited space, with reference designator "S1" and label your SW1 "RUN" with reference designator "PB1" and label SW2 "STOP" with reference designator "PB2". But that is my style...
« Last Edit: December 16, 2018, 10:46:14 pm by tpowell1830 »
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Offline kafor1Topic starter

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Re: simple controller problem
« Reply #3 on: December 17, 2018, 02:07:17 am »
You have no diodes to absorb the inductive spike when the relays are turned off.  This will likely cause the 555 to be damaged.

A 10 A induction motor likely draws close to 100 A for a few milliseconds when starting.  This will likely damage the relay contacts.

Jon

so ill i would just  confirm if the circuit would be ok if ill add 1n4001 in both relay. aaand im really a beginner at this so pardon me for my question.. im just using a 15A extension wire is it possible to draw 100A in it? and so my 10A relay can handle 10.5A of current?

and plz explain me the term LOAD in electronics, and provide example, if its not a hassle.. thank you so much
 

Offline kafor1Topic starter

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Re: simple controller problem
« Reply #4 on: December 17, 2018, 02:15:26 am »
Circuit looks okay, however, I don't understand the purpose of D1 (I would use reference designator LED1), why not just tie to V+ to show you have power on? Also, you could have less devices and more universal relay driver if you use a NPN transistor to pull your external relay coils to ground, as shown in this similar circuit: https://electronicsarea.com/on-off-switch-circuit-using-555-timer/ . Don't forget to use D1 on relay coil, as shown (D2 on second relay coil). Also, there are plenty of DPDT relays available, you just need to look around.

This looks suspiciously like a 9 volt battery driving your circuit and will last 10 minutes on your circuit driving the relays. I would find a small wall wart power supply
that delivers the current that you need at 9 to 15 volts.

I would also add a power on toggle switch for your circuit, otherwise the battery will run down completely after one use. Strike that, just because the battery is not a good supply and you need a way to power on and off your circuit (see wall wart above).

EDIT: I would label the power switch that I mentioned "POWER", or "PWR" if you have limited space, with reference designator "S1" and label your SW1 "RUN" with reference designator "PB1" and label SW2 "STOP" with reference designator "PB2". But that is my style...


aah yes the purpose of D1 is to show that it is OFF since it is red, D2 is green it would indicate that its ON, and your opinion is verry goood i didnt think of that, i guess if i didnt add that i would put my hole in my picket since ill be buying lots of 9v bat hahah. ill add that to my circuit. and so can my relay handle 10.5A ? since the rating is just 10A

edit.. and also if its not a hassle can u explain me the term load, and provide example is its ok with you  :) :)
 

Offline tpowell1830

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Re: simple controller problem
« Reply #5 on: December 17, 2018, 02:56:39 am »


aah yes the purpose of D1 is to show that it is OFF since it is red, D2 is green it would indicate that its ON, and your opinion is verry goood i didnt think of that, i guess if i didnt add that i would put my hole in my picket since ill be buying lots of 9v bat hahah. ill add that to my circuit. and so can my relay handle 10.5A ? since the rating is just 10A

edit.. and also if its not a hassle can u explain me the term load, and provide example is its ok with you  :) :)

Your relay is rated at 10 amps, but could handle 10.5 amps for a short period of time. My rule of thumb on relay contact parameters is to make them 20% higher current carrying than rated load. So, in this case, 20% of 10.5 amps is approximately 12 amps, so I would pick the closest or higher current rating to 12 amps, which the standard is probably 15 amps for most relays.

Load, when mentioned in this context, is current in amps. Your inductive motor has a load of 10.5 amps for the contacts to "carry" (like a truck carries a load).

Hope this helps...
« Last Edit: December 17, 2018, 02:59:32 am by tpowell1830 »
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Offline wraper

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Re: simple controller problem
« Reply #6 on: December 17, 2018, 03:02:22 am »


aah yes the purpose of D1 is to show that it is OFF since it is red, D2 is green it would indicate that its ON, and your opinion is verry goood i didnt think of that, i guess if i didnt add that i would put my hole in my picket since ill be buying lots of 9v bat hahah. ill add that to my circuit. and so can my relay handle 10.5A ? since the rating is just 10A

edit.. and also if its not a hassle can u explain me the term load, and provide example is its ok with you  :) :)

Your relay is rated at 10 amps, but could handle 10.5 amps for a short period of time. My rule of thumb on relay contact parameters is to make them 20% higher current carrying than rated load. So, in this case, 20% of 10.5 amps is approximately 12 amps, so I would pick the closest or higher current rating to 12 amps, which the standard is probably 15 amps for most relays.

Load, when mentioned in this context, is current in amps. Your inductive motor has a load of 10.5 amps for the contacts.
With inductive load you should derate relay like 2-3 times. Often you can find about this in datasheets. Snubber should be used as well. Also don't forget about inrush current.
 
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Offline tpowell1830

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Re: simple controller problem
« Reply #7 on: December 17, 2018, 03:10:46 am »


aah yes the purpose of D1 is to show that it is OFF since it is red, D2 is green it would indicate that its ON, and your opinion is verry goood i didnt think of that, i guess if i didnt add that i would put my hole in my picket since ill be buying lots of 9v bat hahah. ill add that to my circuit. and so can my relay handle 10.5A ? since the rating is just 10A

edit.. and also if its not a hassle can u explain me the term load, and provide example is its ok with you  :) :)

Your relay is rated at 10 amps, but could handle 10.5 amps for a short period of time. My rule of thumb on relay contact parameters is to make them 20% higher current carrying than rated load. So, in this case, 20% of 10.5 amps is approximately 12 amps, so I would pick the closest or higher current rating to 12 amps, which the standard is probably 15 amps for most relays.

Load, when mentioned in this context, is current in amps. Your inductive motor has a load of 10.5 amps for the contacts.
With inductive load you should derate relay like 2-3 times. Often you can find about this in datasheets. Snubber should be used as well. Also don't forget about inrush current.

Thanks Wraper, I agree completely, my mind was derailed when I mentioned this. But it is often derailed these days....

Inductive motor load switches should use a much higher current rating and, I would say (normally, if my head was screwed on right), get a proper motor starter relay with built in thermal overload protection.
PEACE===>T
 
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Offline kafor1Topic starter

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Re: simple controller problem
« Reply #8 on: December 17, 2018, 04:26:53 am »


aah yes the purpose of D1 is to show that it is OFF since it is red, D2 is green it would indicate that its ON, and your opinion is verry goood i didnt think of that, i guess if i didnt add that i would put my hole in my picket since ill be buying lots of 9v bat hahah. ill add that to my circuit. and so can my relay handle 10.5A ? since the rating is just 10A

edit.. and also if its not a hassle can u explain me the term load, and provide example is its ok with you  :) :)

Your relay is rated at 10 amps, but could handle 10.5 amps for a short period of time. My rule of thumb on relay contact parameters is to make them 20% higher current carrying than rated load. So, in this case, 20% of 10.5 amps is approximately 12 amps, so I would pick the closest or higher current rating to 12 amps, which the standard is probably 15 amps for most relays.

Load, when mentioned in this context, is current in amps. Your inductive motor has a load of 10.5 amps for the contacts.
With inductive load you should derate relay like 2-3 times. Often you can find about this in datasheets. Snubber should be used as well. Also don't forget about inrush current.

Thanks Wraper, I agree completely, my mind was derailed when I mentioned this. But it is often derailed these days....

Inductive motor load switches should use a much higher current rating and, I would say (normally, if my head was screwed on right), get a proper motor starter relay with built in thermal overload protection.

so am i doing something dangerous, not using any motor load switches, ,motor starter relay,, just plugging the cord to 15A extension wire to energize and unplugging the cord to deenergize the motor? since i dont have any switches

Edit: also i dont know what is the rating of snubber should i buy,or where i should connect it in my circuit
« Last Edit: December 17, 2018, 04:29:12 am by kafor1 »
 

Offline spec

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Re: simple controller problem
« Reply #9 on: December 17, 2018, 08:42:58 am »
Hi kafor1

Bad news. The current rating for a relay is defined with a resistive load. The relay current rating is much reduced for an inductive load like a motor. Without seeing the data sheet for your relay or motor, I would guess that you would need a relay with a resistive load rating of around 25A at least for switching the motor. You would also need snubbing/catching circuits as already described.

But in any case, there is little danger, all that would happen is that the relay contacts would burn out or weld together and keep the motor permanently on. So you can carry on using the relay, but don't be surprised if it does not last too long. :)
 
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