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Supplemental Voltage Regulator Ground


Scud

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MartyNZ, you are correct that if the regulator cannot correctly establish the ground reference, then it will output a higher un-regulated output. However, if the ground is not connected, then there is also no return path for the current, so what the positive output goes to is irrelevant, since it is like only connecting one terminal of a battery. So, how does that explain the bright lights on my dash? I think when the ground is lost, the regulator also puts high voltage on the wire for the generator idiot light, which then coupled to all of the dash lights through the generator bulb and over driving the other dash bulbs.

 

Since my new regulator is arriving tomorrow, I'm not going to take time to debug this entire system.  I'll put the new regulator in, make sure it has a clean ground, then go for a nice long night ride to check it out.

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MartyNZ, you are correct that if the regulator cannot correctly establish the ground reference, then it will output a higher un-regulated output. However, if the ground is not connected, then there is also no return path for the current, so what the positive output goes to is irrelevant, since it is like only connecting one terminal of a battery. So, how does that explain the bright lights on my dash? I think when the ground is lost, the regulator also puts high voltage on the wire for the generator idiot light, which then coupled to all of the dash lights through the generator bulb and over driving the other dash bulbs.

 

Since my new regulator is arriving tomorrow, I'm not going to take time to debug this entire system.  I'll put the new regulator in, make sure it has a clean ground, then go for a nice long night ride to check it out.

The regulator sets the Voltage between the black wire and the case to 13.8 Volts.

So just suppose the regulator is not grounded properly and the case is 1 Volt above chassis, then the regulator is set at 13.8 in relation to the case but 14.8 in relation to chassis.

Years ago I took the time to pull a few regulators apart and draw out the internal circuit, I used to have it posted here but I can send you a copy if you PM me with your e-mail addy

IMHO the regulator is a good design but it's let down badly by the Guzzi wiring.

 

The usual fault is one of the diodes overheats and the wire melts off. I think this is a result of the flakey Voltage reference. If you increase the Voltage to a battery the current goes up exponentially.

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....

 

The usual fault is one of the diodes overheats and the wire melts off. I think this is a result of the flakey Voltage reference. If you increase the Voltage to a battery the current goes up exponentially.

Roy, you should go a bit deeper into detail here. For that I think you should for instance consider the fact that the internal resistance of the battery is not constant and that voltage here, as always, is the result of possible power output of the alternator, power consumption of the bike and actual resistance of the whole circuit (bike).

Especially the last part "the current goes up exponentially" went slightly off. At least I do think so. Maybe you could give an easy example based on some resonable numbers/measurements of what you had in mind with 'exponentially'.

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Suppose you have a battery half charged.
If you apply a Voltage and ramp it slowly up the current will remain at zero until the Voltage reaches 13 or so Volts
13.5 might cause 1 amp to flow but 14 Volts will cause a lot more than 2
14.5 will cause several times the current than 13.5, that's what I meant by exponential.
My terminology might not be scientifically correct but then I'm not a scientist I'm a humble journeyman with only 50 years of gathered experience.
Sorry
 

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It is correct that current through a diode grows exponentially with applied tension. However, it is also correct that the current is limited by the power of the source (alternator). So although growth is exponential, there are limits as to what actually can be achieved.

 

Alternator is rated at 350W, which @12V results in ~30A current. The circuit is fused accordingly. Workshop manual talks about DC 27,5A@10k RPM and I'd say we can safely assume that the regulator/diodes in it are also rated accordingly. Based on that, I don't think we have overvoltage or overamperage problem here. If we had, the fuse would burn out.

 

 I'd say the problem is related most likely related to continuos high (but stilll "legal"!!) currents passing through a thermally stressed regulator for prolonged periods, owing to both:

- bad battery or regulator-battery connectivity (high currents)

- placement of the regulator - behind oil cooler and between cylinder heads. (thermal stress)

 

The main current limiter in the charging circuit is battery's state of charge. If that does not go up quickly (within 15-20 minutes) after the engine is started, regulator will be sweating until it dies. 

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As Guzzi3Go says

If the regulator thinks the battery is at 13 Volts when in fact it's much higher it will just keep pumping the Amps to it.

The Voltage reference is what lets the regulator down, too much Voltage across the headlight relay.

The Voltage drop in the headlight circuit should normally be ~0.5 Volts but I have seen mine over 1 Volt after sitting all winter, just caused by resistance in the relay contacts.

I have pulled several regulators apart and found the leads melted off the diodes with signs of arcing.

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The main current limiter in the charging circuit is battery's state of charge. If that does not go up quickly (within 15-20 minutes) after the engine is started, regulator will be sweating until it dies. 

I've been preaching this ever since luhbo showed me that my arcane charging protocols and ancient chargers were flawed.

 

caused by resistance in the relay contacts.

I have pulled several regulators apart and found the leads melted off the diodes with signs of arcing.

Huh . . . yeah . . . those relays . . .

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Suppose you have a battery half charged.

If you apply a Voltage and ramp it slowly up the current will remain at zero until the Voltage reaches 13 or so Volts

13.5 might cause 1 amp to flow but 14 Volts will cause a lot more than 2

14.5 will cause several times the current than 13.5, that's what I meant by exponential.

My terminology might not be scientifically correct but then I'm not a scientist I'm a humble journeyman with only 50 years of gathered experience.

Sorry

No Roy, I still cannot see it. What if you just take Ohm's law here: I = U/R ? That's as linear as it could be. Why do you call that 'Exponentially'? What physical laws do you apply for your 13-13.5-14V current estimation?

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It is correct that current through a diode grows exponentially with applied tension. ...

Totaly out of context, I'd say. He's talking of current through a battery, not about diodes. Besides that, regulator or whatever diodes you have in mind would still see exactly the current that goes through the regulator. A bike is not a laboratory or diode test stand.

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Totaly out of context, I'd say. He's talking of current through a battery, not about diodes. Besides that, regulator or whatever diodes you have in mind would still see exactly the current that goes through the regulator. A bike is not a laboratory or diode test stand.

True. Diode's UI characteristics is of little or no relevance in this discussion.

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As Guzzi3Go says

If the regulator thinks the battery is at 13 Volts when in fact it's much higher it will just keep pumping the Amps to it....

Regulators don't pump Amps. Nothing pumps Amps. The alternator creates a certain potential difference, then according to the resistance behind it a certain current will start to flow.

Now I assume a more or less fixed resistance for the harness,for the battery a floating one (rising remarkably with higher voltage), consider that I'm looking at a voltage divider (the battery is parallel to the harness, not in series), then I can really not see any reason for any "exponential Amps pumping", especially not with this rather small 350W alternator.

 

BTW: had one dead regulator after 10 years or so. The main problem with these units is the shrinking green stuff used for sealing the electronics. Moisture gets in, corrosion starts, the green stuff stresses the connections due to incompatible thermal expansion coefficients, things like that. Electrically they're ok, as well as the whole system including reference pick up and so on. The culprit is that most of the components are just too cheap (fuse holder, connectors and such). But then, how many of us have bought their bike new, and of those who did, how many have thought 2 or 3 times about buying it because of its price, and how many of all those buyers put more than 50.000 km on their bikes?

A bike's a leisure product, like a MP3 player, or a leg shaver for the ladies, its calculated life span goes against Zero when you compare it with an average car.

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Well, if we want to go nitpicking, mentioning battery's internal resistance is also out of context. It is very small, in the mOhm range, so even if it would be changing drastically (factor 2x, 3x or 10x, which it does not), it would still have practically no influence on the current flowing through the battery. Internal resistance plays a role only in short circuit scenarios, where it is comparable with applied load.

 

What acts as a main current limiter in the charging-battery circuit is the battery's tension. Once battery reaches regulator's cutt-off point, no current flows into the battery.

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Call it nitpicking, it's ok. Anyhow, watch the charging curve of a good Hawker and you'll think twice about the internal resistance. I saw them taking 8A at 12.5V, what later went down to 0.3A at 15.2V (no positive iones left). Figure the factor.

I drove the V11 with a flat battery and also for some days at probably 18 or 19V because of a dead regulator - and nothing got damaged besides the battery and two headlamps (that's how I noticed it).

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That is not due to increased internal resistance, but state of charge (U=Q/C, Charge/Capacity). The reason why the current went down from 8A@12V to 300mA@15 is because the battery was charged and its tension went up. Consequently, less volts remained available for pushing current through the circuit. Ic = (Ur-Ub)/(Rb+Rc). Lowercase R,B,C being regulator, battery, circuit.

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