Electronics > Beginners
simple controller problem
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kafor1:
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
jmelson:
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
tpowell1830:
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...
kafor1:

--- Quote from: jmelson 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

--- End quote ---

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
kafor1:

--- Quote from: tpowell1830 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...

--- End quote ---


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  :) :)
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