I have an old (1968) Kohler 2500W (120VAC) generator that starts automatically (12VDC) on demand and shuts off when the load is removed. I would like to replicate the switching to turn my inverter (120V) on and off automatically as well. It draws 1-2 amp when on. The problem is I don't have the schematic for the generator control box nor a part list. Any help in design and parts needed would be great.

 

Is it possible that Kohler uses a special receptical, with something simple like a switch connected to rod that runs through the center of the ground pin. Don't know why but I invision the ground pin on the plug pushing a rod that will close the switch. I also could be a mile off target.

 

The set up is: the generator is in a boathouse and 300 feet away is the fuse panel in a cottage. All normal wiringand recepticles. When a load of 40W or more is demanded the generator starts. When the load is removed the generator stops.
My understanding is there is a 12V open circuit thru the line. When a load is applied (say a 60W light bulb) is turned on, it closes that circuit. Using relays it closes the circuit to the starter selanoid and closes the 110V circuit into the line and opens the 12V. There is a centrifically operated switch on the engine so once it reaches a certain RPM the starter is shut off. There is also a circuit to run a battery charger.

What I want to do is the automatic switching without the starter etc.

I've found a schematic that I think is close to what I have. http://www.oldengine.org/members/diesel/Kohler/Kohler24.htm What do you think?

 

I think that this will imitate what your generator circuit does. Please note that my Spice didn't have a 3PDT relay so I had to draw it with separate relays though it's really one relay RL1. Keep in mind that the load must provide a DC path for this circuit to work.


Edit: I removed the schematic because I neglected to provide a latch action. I'm working on it...

 

OK, I changed the design to a 4PDT relay with a set of contacts being used to latch the coil. Problem with this scheme is that if you unplug the device from the inverter the inverter will still be on because the relay has latched. I decided to post the schematic anyway as it should stimulate discussion and the dendrites. In order to add automatic reset to this circuit either input current or output current is going to have to be sensed and used as a control level to unlatch the relay.

 

That looks great! What I should do now is pick up a relay so I can physically see how it would work.

Could a second relay be used on the AC load to latch RL1? Then when the AC load is removed RL1 would reset?

 

That looks great! What I should do now is pick up a relay so I can physically see how it would work.

Whoa, not so fast! This circuit still needs to sense when the load is unplugged or shut off. Let's wait to hear from others while I think this out. This is also Shore Power that should always receive an extra measure of attention to safety.

 

Oh, I wasn't planning on running it live. I have a healthy respect for electricity.
I won't hook anything up until I know it will work and how it will work. I have a couple of weeks before I'm on site again and this is not a manditory modification. I was thinking of building a proto-type and use batteries (low power) so I can test the function and make modifications without curling my hair.

 

Please note that the schematic in post 5 was modified because R1 was wired to the wrong relay contact.

 

I don't know if this helps or not but I just found on a Kohler forum this schematic of a typical automatic control unit.

 

I don't know if this helps or not but I just found on a Kohler forum this schematic of a typical automatic control unit.
View attachment 10112

Not really, because you're needing to control an Inverter, not start a Gen. See the attached schematic where I added a comparator chip.
In your first post you stated that the Inverter pulls about 1.5 Amps. I presume this is with no load. The theory of operation below is based on the fact that the Inverter input current will increase proportionately as the load on the output increases.

Theory of operation: K1 is normally de-energized. When a load is plugged in a DC path to GND is provided for the K1 coil and it energizes. This action switches the contacts of K1 to power the input of the Inverter and to also switch the load to the AC output of the Inverter. When this happens R_Sense develops enough voltage drop to trigger the comparator and turn its output (High) which latches K1 through the collector of Q2. When the load is unplugged or turned off the comparator will turn Q2 off and de-energize K1. VR1 is set to not trigger U1 on until more than ~1.5 A is being drawn by the Inverter.
Note: K1 is a 6V relay used with the dropping resistor (R7) or the load resistance..

I have not been able to fully spice this so I welcome critiques and thoughts from the membership. I'm also falling behind on my work, so anyone who wants to help out here feel free to.