screenshot from manual on the shutdown circuit.
I have the whole manual, Raritan emailed it to me.
Sharing it at https://drive.google.com/file/d/0B43EHHTTdLSfbVBCU3ZMODZyVTg/view?usp=sharing

 

I dont know why my mod might not work, it all depends which side of the 400 ohm adjustable resistor the zener is working with?

No. Let's assume the battery voltage is Vbat. Because of the isolation diodes connected to the battery positives, the voltage across the 400Ω pot is Vbat+0.7V. The pot wiper voltage will be an adjustable fraction of that and determines whether or not the right-hand transistor (Q2) turns on. For Q2 to turn on, its emitter must be at least Vbe+Vz (where Vz is the zener voltage) above the wiper voltage, i.e 6.8+0.7=7.5V above the wiper voltage. In your first mod you would be connecting the left end of the zener via the minipot resistance to Vbat+0.7V, so Q2 could never turn on. In your second mod you would be connecting the left end of the zener via the minipot resistance to battery negative (ground), so Q2 would be permanently on.

Your circuit shows more resistance to the zener? 330 ohms plus the minipot pot

That matters not. The zener current is small, so drops very little voltage across the 330Ω or the minipot compared to the 6.8V dropped by the zener. The 330Ω is included to prevent excess current flow if the minipot is inadvertently set to zero resistance.

How does this zener circuit flow current work with the the PNP transistor?

When the battery voltage is low, Q2 emitter is not at a high enough voltage for the zener to conduct. Therefore Q2 is off and the bias resistors on Q1 (the left-hand transistor) base cause Q1 to conduct. This allows gate pulses to be fed to the SCR to turn it on, giving a high charge current for the batteries. When the battery voltage rises sufficiently, the voltage between Q2 emitter and the left end of the zener exceeds 7.5V so Q2 conducts. This pulls the base of Q1 high, Q1 turns off, gate pulses are no longer applied to the SCR, and the only path for charge current is the 100Ω resistor connected to the battery negatives, resulting in just a trickle charge. If either Ign or Shutdown goes high (12V), then similarly Q1 base is pulled high to stop the gate pulses.

 

Much thanks to you on explaining how the circuit works!

I knew somehow the zener reverse breakdown voltage must be controlling the voltage cut off.

Interesting that the charger wants to naturally turn on Q1 and charge, and it is Q2 which is the shut off. I kept thinking Q2 turning on which was turning on Q1 and so was very confused. When really it is Q2 turning on that turns off Q1. This actually starting to make sense to me.

So in a PNP, if the base of a transistor goes high, it turns the PNP transistor off?
And in a NPN, if the base goes high, it turns the NPN transistor on?

Thanks for explaining how this works, I was really wondering and confused, thinking NPN not PNP.
I will need to find some 1/2 watt 330 ohms to try this.
What is the lowest resistor to safely use? Or what range, how high could I go, I may have something here to use.

 

I am thinking, what wattage should these 330 ohm resistors be?
The current flow is returning back through the large power adjustable resistor, which schematic says is 12 watts.
So they should be more hefty than 1/2 watts.

 

So in a PNP, if the base of a transistor goes high, it turns the PNP transistor off?
And in a NPN, if the base goes high, it turns the NPN transistor on?

Yes, if you're considering base 'high' = more positive than emitter.

What is the lowest resistor to safely use? Or what range

You could try anything between, say, 100 and 470 Ohms. Increasing the resistance has the effect that the voltage adjustment range is reduced (so may not cover the actual voltage you want), but the adjustment resolution (fine-ness) is increased.

what wattage should these 330 ohm resistors be?

1/4W will be plenty (they dissipate only ~30mW).

 

1/4W will be plenty (they dissipate only ~30mW).

Interesting. The 12 watt resistor, your sort of paralleling those two 330 ohm resistors with that, so wont a fair amount of current want to flow through them also?
And the minipot is in series with that too.

Do you suppose that 12 watt rating is superfluous, unneeded? How much current is really going to flow there? Do you think just the minipot would be sufficient? , use instead of the power resistor?

330 + 330 = 660 ohms plus minipot 500 ohms = 1100 ohms, paralleled next to 400 ohms, so wont the current flow be split ?
Like two channels in the road back? I suppose the 1/2 watt rating on the minipot, it wont matter if the resistors are also 1/2 watt rated.

Also , this is dropping the 400 ohms down to maybe 300 ohms and that is ok? How much current flow can that 2n525 handle?

 

I cant find any info except this on 2N525 (wrong on NPN)
http://alltransistors.com/transistor.php?transistor=4674
which says the collector current max is 0.5 amps
So if 15 volts is on collector, trans max is 0.5 amps the wattage is 7.5?
so too much for just the minipot of 0.5 watts

 

Thinking more on how this works I found this that says a PNP transistor works as a switch when the base goes to ground, it turns on.
http://www.rason.org/Projects/transwit/transwit.htm

All the zener is doing is it's break down voltage exceeds that 6.8v, (grounding the base, there is a current flow from emitter to base charging the zener) and it conducts in reverse bias which turns on Q1 (off transistor), and that turns off Q2 (on transistor) which turns off the charging.
Seeing there is an oscillating rectified power , DC voltage rises and falls coming from the rectifiers, it just keeps turning on and off and on and off forever.
So why not just run that zener wire through minipot and maybe extra resistors right to ground? Which is what the original circuit does with those two 300 ohm resistors already in the circuit.

Positive volts in at top left , negative at bottom

 

In theory the bridge works as replacement for the stud diode and may continue to work for some time however, I have had some bad experience using this particular type and construction bridge rectifiers. Personally, I shall rather replace that kind of bridge with stud diodes. If the bridge is not Chinese, I might consider using it the way you did (something like IXYS ( http://static1.tme.eu/katalog_pics/a/7/3/a73a91b9e119a665b1cdb5b691f80964/dsei2x31-06c.jpg ) is real good quality, but expensive).

I have found that the plug-in terminals on the bridges tend to overheat whether you plug in wires or whether you solder them. All the same.
That being said, I wish you success with your mod and agree, free is free!

 

The 12 watt resistor, your sort of paralleling those two 330 ohm resistors with that, so wont a fair amount of current want to flow through them also?

Their peak current is only ~12mA and RMS current is only 9.5mA.

Do you suppose that 12 watt rating is superfluous, unneeded? How much current is really going to flow there?

Peak current through the 400Ω pot is ~36mA; RMS current is ~28mA. With 15V across it it would be dissipating an average 420mW. So the 12W rating is indeed excessive.

Do you think just the minipot would be sufficient? , use instead of the power resistor?

The minipot would then have to dissipate ~270mW. While that is within its rating, I think it is uncomfortably close to 500mW for reliable use in a marine/automotive environment. However, if the minipot were used in conjunction with the two 330Ω (or even 100Ω) resistors then the power dissipation would be significantly reduced and the 400Ω pot would not be needed.

the collector current max is 0.5 amps

Provided the battery voltage doesn't drop below 10V, Q1 current will be below 0.5A.

So why not just run that zener wire through minipot and maybe extra resistors right to ground?

Q2 would then turn on with the battery only partly charged.

Which is what the original circuit does with those two 300 ohm resistors already in the circuit.

It doesn't. Ground is battery negative, which is wheret Q1 emitter connects.

 

Thanks for bearing with me on this. I appreciate all of it and have lots of things to try when I get the parts.

Their peak current is only ~12mA and RMS current is only 9.5mA.

Peak current through the 400Ω pot is ~36mA; RMS current is ~28mA. With 15V across it it would be dissipating an average 420mW. So the 12W rating is indeed excessive.The minipot would then have to dissipate ~270mW. While that is within its rating, I think it is uncomfortably close to 500mW for reliable use in a marine/automotive environment. However, if the minipot were used in conjunction with the two 330Ω (or even 100Ω) resistors then the power dissipation would be significantly reduced and the 400Ω pot would not be needed.

Could just take it out of the circuit then using two 330 ohm resistors and the minipot as the new 'power resistor'?
How do you know what the current flow is through that resistor? The OEM put a odd circular heat sink 'thermalloy' on Q1
I thought that would indicate it takes more current? But it is not too involved in the power resistor circuit? Maybe they chose that component as it was sourceable and robust compared to what was available 40 years ago?

Q2 would then turn on with the battery only partly charged.

Even with adding enough resistance to keep the zener from firing (reverse breakdown)?

It doesn't. Ground is battery negative, which is where Q1 emitter connects.

Ok, I am thinking ground for the circuit is not battery negative but the centertap wire going back to the transformer

 

In theory the bridge works as replacement for the stud diode and may continue to work for some time however, I have had some bad experience using this particular type and construction bridge rectifiers. Personally, I shall rather replace that kind of bridge with stud diodes. If the bridge is not Chinese, I might consider using it the way you did (something like IXYS ( http://static1.tme.eu/katalog_pics/a/7/3/a73a91b9e119a665b1cdb5b691f80964/dsei2x31-06c.jpg ) is real good quality, but expensive).

I have found that the plug-in terminals on the bridges tend to overheat whether you plug in wires or whether you solder them. All the same.
That being said, I wish you success with your mod and agree, free is free!

It was sort of an ASAP fix. Dont want to leave the boat without a charger, bad things regarding bilge pumps draining batteries and sinking could happen.
So far been ok, the aluminum plates are not getting hot and the voltages on all 3 banks are pretty close.

What I find interesting is their is some independence on all 3 banks. One bank is house battery, second bank is starter battery, third bank is gen battery.
If I turn on a 12v device connected to the house bank, the charger responds by upping the voltage on the house bank to 14.3v while keeping both starter and gen banks at 13.6

Why is that? I thought all the banks would be treated equally? Something to do with sensing power demand from a bank, how does it adapt the volt levels? IMO, then it is doing better, I though it would jump all 3 banks up equally.

The charger you can also hear a tiny click like noise, it is a definite tiny pulse observed on the ammeter when just charging 3 batteries. Put a load on and it revs up the charger, ammeter jumps up and tiny click noises are more rapid sounding. I wonder how it makes that noise. It also has a slight hum that increases under a load. None of it objectionably load. Whatever it is is normal, been like that since I owned it from 1998.

 

Could just take it out of the circuit then using two 330 ohm resistors and the minipot as the new 'power resistor'?

Simulation says it should be ok; but it might be handy to leave it there to provide physical anchor points for the added components.

How do you know what the current flow is through that resistor?

Simulation with LTspice.

The OEM put a odd circular heat sink 'thermalloy' on Q1

Because the Q1 (left-most transistor) collector current peaks around the 500mA maximum rated current when a battery is charging. As you say, "it is not too involved in the power resistor circuit". In fact the current through the power pot (and minipot if used) goes to ground (battery neg), not through Q1.

Even with adding enough resistance to keep the zener from firing (reverse breakdown)?

Added resistance merely reduces the zener breakdown current; it still allows enough base current for Q2 to turn on when the battery voltage rises far enough.

there is some independence on all 3 banks

Because of the three isolating diodes.

I thought it would jump all 3 banks up equally

In the long term it should, but in the short term the battery with the lowest voltage (e.g. the one you're pulling juice from) will hog most of the charging current.

you can also hear a tiny click like noise

Anything with a coil in it, passing high current pulses, is liable to make a noise because of magnetic effects. You meter is probably of the moving-coil type. The transformer in the power supply also has coils.

 

I got another circuit idea from a friend. To wit, string together 4 100 ohm 5 watt resistors.
On the lowest junction attach the minipot, and centertap minipot to the zener diode.
That would eliminate the 400 ohm pot from the circuit.

Think that would work?

 

And got some more suggestions like this one. I have 10, 100 ohm 5 watt resistors to try out in one of these type circuit designs.
I will setup several of the suggested ones and see how they work or dont work.
It will be a little while to report back due to waiting on parts.

Which of all these will work or won't work?

This one keep the 400 ohm pot and lets the minipot fine tune circuit

or this, uses 4 100 ohm resistors and all 3 leads on the minipot

 

I predict the post #54 circuit won't work. The charge shut-off point will be essentially fixed regardless of the minipot setting. You need to use all 3 terminals of the minipot, i.e use it as a potentiometer not as a rheostat.

Regarding post #55, suppose the voltage at the top rail is 15V, Q1 collector is 0V.
The minipot wiper can be set from 5.9V-9.1V in the first circuit, 6.1V-8.8V in the second, 0V-15V in the third , 3.9V-7.2V in the fourth cicrcuit. The first and second options should give enough adjustment range to enable the charge cut-off point to be set at ~13.7V. The fourth option probably won't. The third option, while providing maximum adjustment, carries the risk that if the minipot is set near minimum resistance then the 15Ω resistor is not enough to prevent excess current damaging the minipot, the zener and possibly Q2.
None of the options justifies the use of 5W resistors; ½W will be perfectly adequate.

 

I haven't been following this all that closely, so may be wrong on what your trying to do. But to keep three seperate batteries charged, as needed, while at dockside, wouldn't it make more sense to have three window comparators? One for each battery? The charger would be on at all times(at dockside) but the comparators would only charge a battery when it goes below it's 'needs charged voltage'. Each comparator would turn on a mosfet that feeds that battery. All three mosfets would be connected to the charger, but only turn on as needed.

Sorry if what I suggested is not what your trying to do.

 

I predict the post #54 circuit won't work. The charge shut-off point will be essentially fixed regardless of the minipot setting. You need to use all 3 terminals of the minipot, i.e use it as a potentiometer not as a rheostat.

Regarding post #55, suppose the voltage at the top rail is 15V, Q1 collector is 0V.
The minipot wiper can be set from 5.9V-9.1V in the first circuit, 6.1V-8.8V in the second, 0V-15V in the third , 3.9V-7.2V in the fourth cicrcuit. The first and second options should give enough adjustment range to enable the charge cut-off point to be set at ~13.7V. The fourth option probably won't. The third option, while providing maximum adjustment, carries the risk that if the minipot is set near minimum resistance then the 15Ω resistor is not enough to prevent excess current damaging the minipot, the zener and possibly Q2.
None of the options justifies the use of 5W resistors; ½W will be perfectly adequate.

You are right about the 5 watt resistors, but the price is only pennies difference, so I thought why not?
How about the volts range on this one. I changed bottom resistor to 50 ohms with total column at 350 ohms and upper leg at 150 ohms (from Q1).
Also how would increasing the spread between the minipot legs affect the volts? As in the upper leg at 200 ohms ( from the bottom at Q1 ) and lower leg at 50 ohms?

 

You are right about the 5 watt resistors, but the price is only pennies difference, so I thought why not?

You might find the 5W type more difficult to solder reliably, because of their greater thermal capacity.

How about the volts range on this one.

Why did you reduce the resistor value to 50Ω? Your aim is a range centred on ~7.5V (assuming a 6.8V zener). As shown you'd only get a volts range of 2.25V-6V (too low). Moving the minipot top leg up by one resistor (so that the minipot bridges two 100Ω resistors) would give a volts range of 2.6V-9.8V.

 

Ok thanks , I am trying to understand the pattern that when resistors are in certain positions what the volts will be and your a big help!
Interesting that a greater increase of resistance in the center will give a greater adjustable range.

I was also thinking how the ignition shutdown works. ?
How do the two diodes determine the voltage cut off as in when the engine cranks, volts drop and charger gives a starting boost.
Then when volts are normal, the charger turns off. Are those two diodes doing a reverse breakdown like a zener?