DB26

How to Build a Diode Generator Cut Out Replacement

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Posted (edited)

You do not need to install the diode in the cutout, especially if you need ambient air to cool it.

 

Using a single Diode in lieu of your cutout contact you could install a jumper wire in the cutout, to short the two terminals. and install the diode anywhere along the wire  between the cutout and where it terminates at the other end.  You would need to cut the wire to install the diode unless you were install it at the other end where the wire terminates.

 

Here is a Lucas voltage regulator for a positive ground Triumph TR4 that I converted to solid state last year.  A bit more challenging than a single diode for a cutout.

 

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100_0667.JPG

Edited by Vila (see edit history)
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Impressive work. I’m still very basic at electrical theory, but a I love figuring it out and making things work. 

 

My cutout has another connection here:

 

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it is fused and uses a second wire from the generator. I believe it’s there to blow in case of overcharging. Can that stay connected with the diode modification? 

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Posted (edited)

Do you have a wiring diagram showing the connections?  Was a fuse in it when you opened it?  If yes what was it’s rating?

Edited by TerryB (see edit history)

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17 minutes ago, TerryB said:

Do you have a wiring diagram showing the connections?  Was a fuse in it when you opened it?  If yes what was it’s rating?

Yes I can get one. Haven’t been home in 4 days but will scan a copy tomorrow night.

 

Fuse was in it. Rated 5 amps. The manual states (from my memory) fuse needs to be 6 amps for my 6 volt system. 

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7 minutes ago, DB26 said:

Yes I can get one. Haven’t been home in 4 days but will scan a copy tomorrow night.

 

Fuse was in it. Rated 5 amps. The manual states (from my memory) fuse needs to be 6 amps for my 6 volt system. 

That’s an interesting value for the fuse, it’s not for the generator output at that value, maybe for the field coils in the generator?

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6 minutes ago, TerryB said:

That’s an interesting value for the fuse, it’s not for the generator output at that value, maybe for the field coils in the generator?

That would make a lot of sense. 

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Posted (edited)

Also, I hope you are away from home for entertainment, not work.  I used to be away for days and sometimes weeks when I was working.  Not a fun time.  Fortunately the family understood that’s how the bills got paid!

Edited by TerryB (see edit history)

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23 minutes ago, TerryB said:

Also, I hope you are away from home for entertainment, not work.  I used to be away for days and sometimes weeks when I was working.  Not a fun time.  Fortunately the family understood that’s how the bills got paid!

No, just circumstance unrelated to work thankfully. 

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Most inline diode ( 2 x 10 amps ) in parallel with terminal on both ends can be attached to the output of the cut out to the battery feed will do it. The diode will only allow the current to flow one way. It can be checked with a test light on the battery.. If the current does not flow one way, reverse the terminals. The objective is to prevent accidental flow back from battery to generator.   

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It appears the generator outputs 14 amperes. Here is a snipet from the manual. Including the wiring diagram. 

 

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Talk about a simple wiring circuit!  That helps make life simple for sure.  The field fuse blew when your battery cable came off from what I gather in the Dodge discussion area.  Good that it did or the generator could have been damaged.  My 1937 Dodge pickup had a third brush regulator but no fuse.  When I opened up the generator to inspect it the generator was shot due to some problems it had before I bought it.  Solder was all over the inside of the case so it got HOT!

 

You are on the right track with the diode modification yet I never had any problems with the cutout relay the 12 years I had my truck.  You might need to increase the generator output when using the diode as a diode needs at least 0.7 volts of bias for it to turn on.  What that means in real world is if the battery is 6.3v, the generator will have to be putting out a minimum 7.0v for the diode to allow current to flow to the battery.  7.5v output is probably even better.   I think keeping the field fuse is a good idea too if you can.

Terry

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Good to hear I’m heading in the right direction. And yes, I’m very glad that fuse saved my butt. I should be trying to make the diode circuit in the coming week. I have this thread for help and another fellow member helping me through PMs and one more through email. This community is amazing. 

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Having a hard time figuring this out...

 

The fused field coil protection has one leg of its wire connected to one coil / relay in the cutout. How will I recreate this connection when I have removed the coils / relays?

 

The thin wire in the photo connects to one side of the fuse. 

 

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Posted (edited)

Where does the other end of the thin wire terminate?  Does it go to ground?  Can you measure the resistance of the coil with the thin wire?

Edited by TerryB (see edit history)

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Ok, the fuse is just a special type of current sensitive wire that connects the field coils to ground.  It has zero resistance for all intents and purposes and is there to provide an overload protected path to ground for the generator field coils.  The cut out thin wire coil wire is getting a path to ground where it is connected to the fuse. If the thin wire is connected at the same point the field wire is connected, it means the thin wire coil will have no path to ground if the fuse blows.  If it’s connected where the fuse is grounded then the fuse has no impact on the small coil operation, it’s just a convenient place to make the connection.

 

The small coil, from what I can find on cut out operation theory, is the part of the relay that starts the cut out to attempt to close the contacts as the generator output is increasing.  When the generator output is up to snuff and able to charge the battery it’s the coil with the thick wire that keeps the cutout pulled in to charge the battery.  Theory says the battery charge current is going through the thick wire coil which makes for a strong contact closure.

 

So with that information it seems there is no need to worry about the thin coil connection to the fuse.  If the voltage at the diode is now the value that makes the whole show work the small coil and it’s connections should not matter.  The field coil connection to the fuse must be maintained and one side of the fuse connected to ground for the generator to operate.  The fuse can be replaced with just plain wire but then you risk a generator meltdown if an open circuit develops in the charging circuit.

 

Note:  all this is based on my background in electricity and what I have read on cutout operation on the internet.  I have not actually done what you are proposing.  

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I’ll have to probe around again and see if I get ground on that connection

 

if thats the case, it makes things much simpler, I hope. 

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5 hours ago, TerryB said:

Ok, the fuse is just a special type of current sensitive wire that connects the field coils to ground.  It has zero resistance for all intents and purposes and is there to provide an overload protected path to ground for the generator field coils.  The cut out thin wire coil wire is getting a path to ground where it is connected to the fuse. If the thin wire is connected at the same point the field wire is connected, it means the thin wire coil will have no path to ground if the fuse blows.  If it’s connected where the fuse is grounded then the fuse has no impact on the small coil operation, it’s just a convenient place to make the connection.

 

The small coil, from what I can find on cut out operation theory, is the part of the relay that starts the cut out to attempt to close the contacts as the generator output is increasing.  When the generator output is up to snuff and able to charge the battery it’s the coil with the thick wire that keeps the cutout pulled in to charge the battery.  Theory says the battery charge current is going through the thick wire coil which makes for a strong contact closure.

 

So with that information it seems there is no need to worry about the thin coil connection to the fuse.  If the voltage at the diode is now the value that makes the whole show work the small coil and it’s connections should not matter.  The field coil connection to the fuse must be maintained and one side of the fuse connected to ground for the generator to operate.  The fuse can be replaced with just plain wire but then you risk a generator meltdown if an open circuit develops in the charging circuit.

 

Note:  all this is based on my background in electricity and what I have read on cutout operation on the internet.  I have not actually done what you are proposing.  

You are RIGHT, it is a common ground. The fuse and relay just happen to use the same location. I poked around with a meter and it has continuity with ground. So, problem solved. Thank you. You gave me the clarity I needed. 

 

Onto making the diode circuit. 

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I have made the diode replacement and reassembled my generator. Preliminary tests seem to show a success. I will come back at you with photos and results soon. 

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Here are a few photos of the new cutout. Uses a 25 amp diode. Installing back on car in a few days. 

 

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Okay, it’s installed and I have some results. I have NOT done any adjustments yet. 

 

I bench tested the generator with a drill and got it to output around 7 volt at a medium drill speed. 

 

After I installed the generator back onto the car I did a baseline reading of the battery which was at about 6.15 volts. 

 

After starting the vehicle, at idle, there was no change in the voltage. 

 

When I increased RPMs, the most I saw at the battery was 6.35 volts. 

 

I did a reading at the diode cutout and saw only .5 (half a volt) coming from the generator at high RPMs. 

 

I assume I need to move the 3rd brush now.

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Posted (edited)

Before adjusting, do you have an amp meter installed in the car, if so check what it reads. it should be on the charge side of the gauge.

 

The voltage should increase slowly and reach up to high 6's (volts) as charge goes into the battery from the  generator. Do note, this would only occur well above idle speeds.

 

Are you measuring across the diode? If so then your reading of 0.5v is about right.

Edited by maok (see edit history)
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1 hour ago, maok said:

Before adjusting, do you have an amp meter installed in the car, if so check what it reads. it should be on the charge side of the gauge.

 

The voltage should increase slowly and reach up to high 6's (volts) as charge goes into the battery from the  generator. Do note, this would only occur well above idle speeds.

 

Are you measuring across the diode? If so then your reading of 0.5v is about right.

 

I just re polarized my generator, and got a different set of readings for you. 

 

Generator output before diode, at idle:

 

Ranging from 3.80 to 4.40 volts, and after a few minutes dropped to 1 volt. I think the idle settled down. 

 

Generator output before diode, at higher RPM:

 

About 7.00 volts

 

Reading after diode at idle:

 

6.05 volts

 

Reading after diode at higher RPM:

 

About 7.00 volts

 

Reading at battery negative post at idle:

 

6.15 volts

 

Reading at battery negative post at higher RPM:

 

6.60 volts. 

 

The ammerter is at a slight discharge at idle. And jumps into the charge portion when it’s at the higher rpms. 

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They looked to be ideal readings.

 

While driving, monitor your charge rate on the ammeter and adjust the third brush if you need more or less charge. Turning it with the rotation gives you more current output and anticlockwise gives you less.

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If your generator runs off the fan belt then you could change the pulley on it for smaller diameter one which will give you increased output at idle. If driven from the timing chain then ignore the above and increase your idle speed on the carb for more output at idle.

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3 minutes ago, maok said:

They looked to be ideal readings.

 

While driving, monitor your charge rate on the ammeter and adjust the third brush if you need more or less charge. Turning it with the rotation gives you more current output and anticlockwise gives you less.

 

+1 to this. Another thing to do is periodically check the battery water. If the water level is low then some has "boiled off" (not really boiling but rather splitting the water molecules into separate hydrogen and oxygen) indicating overcharging. And check the density to see if it is undercharged. In the third brush systems the battery was used to even things out, you just want to make sure that on average the battery ends up fully charged when you park the car but not over charging on the road.

 

Back in the old days when these cars were driven year round on a daily basis the third brush needed to be adjusted fairly often to account for season and driving style. For example it typically takes more power to start an engine in freezing weather and in winter the days are shorter so your commute has the lights on more. The reverse for summer.

 

1 minute ago, maok said:

If your generator runs off the fan belt then you could change the pulley on it for smaller diameter one which will give you increased output at idle. If driven from the timing chain then ignore the above and increase your idle speed on the carb for more output at idle.

 

Not necessarily a good idea. The third brush mechanism "kind of sort of" turns the generator into a constant current device with the peak output, based on pulley sizes, final drive ratios, wheel size, etc., usually between 25 and 35 MPH. So you may be trading better charging at idle for worse charging while actually moving.

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