TC-33 electrical nightmare ENDS ! HAPPILY !

[Superduper]

Hey guys, all that is fine and dandy but if we go back to the first post, how again does this all relate to that post where the user surmised that a loose/corroded power terminal is causing blown fuses in lighting circuit every few months. NO other symtoms were offered including starting related issues, etc.

This discussion has gotten kind of long so lets all take a break here and go back to review the original question.

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BLOWS FUSE IN LAMP CIRCUIT EVERY FEW MONTHS:
found loose terminal:
repaired terminal, fixed??
(A) Yes, bad connections causes blown fuses in lamp circuit
(B) No, bad connections will not likely cause overcurrent in lamp circuit, suggest looking further for intermittent short in lamp circuit.
(C) Whoa, I know of someone who had bad ground and it blew fuses....

Through lots of twists and turns, we are now here talking about starting motors, solenoids, relays and windings, impendance, inductance, capacitance, ohms law, arc potential, etc. etc.
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(1) George, Shouldn't the engineers in sizing the lamp circuit take these voltage fluctuations into consideration and employ enough leeway to ensure the proper operation of the circuit? Note that this is NOT a common problem that I have seen. These fluctuations are common and fairly normal and I haven't seen a huge group of people complaining of blowing fuses (in lamp circuit only) nor does it take like six months to blow. Tractors often sit for extended periods of time and a low battery condition, especially when tractors used seasonally is NOT a fluke circumstance, it happens all the time.
(2) Guy, If the extra resistance caused by bad pos/neg connection increases the heatup duration of lamp filament, further resulting in an extended low-load-resistance condition; doesn't that require the voltage to be low (poss very low) in order to have this condition and if so, doesn't the low voltage condition also result in lower current consumption too? You can't have it both ways here: lowered voltage = lowered current consumption. I'm going to continue to assume that engineers designed the lamp circuit correctly with satisfactory margin as again, I do not see widespread complaints about this particular circuit?

So after all these techno-electrical-conspiracy-theories, I am still under the impression that: An intermittent short to ground in the lamp circuit is the most likely culprit, that any problems introduced by the loose/corroded terminal sufficient enough to cause problems in the lighting circuit would most likely be noticed elsewhere, especially in the very high current consumption area of the diesel starting circuit and glow plug area, that the duration between blown fuses (months) is consistent with intemittent shorts.

Remember that in automotive applications, loose/bad/corroded terminals are quite common. In practice, such conditions normally results in lamps appearing to operate normally but hitting the starting circuit causes the system (all) to cut out completely, if temporarily. This is because the high current draw of the starting circuit causes heat to build up in the area of the bad contact and this additional heat increases the resistance to the point where voltage/current is reduced to a useless trickle. After a short waiting period, power will then appear to be restored sufficiently for normal lamp circuit operation. These are classic examples of loose or corroded terminal connections and I have seen this dozens of times. I would expect to see some problems here way before I see any problems in the lamp circuit.

 

[Keoke]

Quote from Superduper follows:

(1) George, Shouldn't the engineers in sizing the lamp circuit take these voltage fluctuations into consideration and employ enough leeway to ensure the proper operation of the circuit? Note that this is NOT a common problem that I have seen. These fluctuations are common and fairly normal and I haven't seen a huge group of people complaining of blowing fuses (in lamp circuit only) nor does it take like six months to blow.

Superduper, generally automotive lighting circuits are fused at 130% of normal load to accommodate cold filament surge and a minor increase in source voltage. The key words here are "normal load" very seldom is trailer lighting (eg. post #12) taken into consideration by vehicle owner when adding additional lighting. Also, often when large sized vans are converted into campers the additional lighting is usually tied into the vehicles OEM lighting circuits. This additional current use brings on excessive voltage drop and circuit protection problems not considered in original circuit design.
If done correctly additional camper lighting (tail circuit) should be used only to trigger a relay that could power up all the additional lighting from the main bat/alt circuit.

cheers,

 

[John_Mc]

It takes hsp (746 watts=1hsp) to turn a starter. A slightly weak battery (low voltage) will cause starter to possibly draw a little more amps then a strong one because the missing voltage will be made up with more amps. eg a 6 volt starter will draw more amps then a 12 volt starter on the same engine (IXE= P) so if the same hsp is needed and the voltage is less then amps must be used to make up the needed watts. We are talking motors here not restive loads (light bulbs). This is why 6 volt battery cables are fatter then 12 volt cables.

You are assuming that power stays constant in this situation. It doesn't. While it may be true that it takes a given amount of power to turn a starter (under a specified load) at a certain speed. That's not what happens when the battery is low. The starter turns more slowly, if at all. What stays constant is the resistance. The correct formula in this case is E=IR or rephrased: I=E/R (Current = voltage/resistance). Since R is constant, if you drop the volts by say 10%, then current will drop by 10% as well.

If you hooked up a 6 volt battery to the circuit, in place of the 12 volt it was designed to take, the starter will turn very slowly, if at all, and pull 1/2 the amps. The reason a system designed for 6 volts takes more amps to turn the starter (assuming the same load), is that the starter is designed differently, with lower resistance to pull more amps. This is where your formula (IxE=P) comes in. If it takes a given amount of power to spin the starter at a given speed and load, and it's a 6 volt system, they will design it to pull more current (lower resistance) to get the same power output. If you hooked a 12 volt battery up to this system, it would draw twice the current, assuming you didn't just burn something out.

John Mc

 

[Keoke]

You are assuming that power stays constant in this situation. It doesn't. While it may be true that it takes a given amount of power to turn a starter (under a specified load) at a certain speed. That's not what happens when the battery is low. The starter turns more slowly, if at all. What stays constant is the resistance. The correct formula in this case is E=IR or rephrased: I=E/R (Current = voltage/resistance). Since R is constant, if you drop the volts by say 10%, then current will drop by 10% as well.

If you hooked up a 6 volt battery to the circuit, in place of the 12 volt it was designed to take, the starter will turn very slowly, if at all, and pull 1/2 the amps. The reason a system designed for 6 volts takes more amps to turn the starter (assuming the same load), is that the starter is designed differently, with lower resistance to pull more amps. This is where your formula (IxE=P) comes in. If it takes a given amount of power to spin the starter at a given speed and load, and it's a 6 volt system, they will design it to pull more current (lower resistance) to get the same power output. If you hooked a 12 volt battery up to this system, it would draw twice the current, assuming you didn't just burn something out.

John Mc

John, all your ohms law calcs are correct as far as fixed starter resistive loads are concerned.
But you are leaving out one significant fact.
As the starter armature turns it generates an opposing voltage that flows in the opposite direction from the voltage coming from the battery(cemf). Accordingly, the total voltage drop (operating voltage) across the starter drops off as the starter speed picks up.
A fully charge 12v battery will spin the starter faster thus generating slightly more cemf. This in turn will cause the use of slightly less current draw then a battery with a slightly less then full charge. Key word here is slightly.
A 6 volt starter running on 12 volts will not use TWICE as many amps as one running on 6 volts. Because of it's lower resistance value, a six volt starter wired to 12v will have a lot more hsp thus will spin both the engine and it's own armature a lot faster. This in turn will generate considerable cemf (opposing) voltage and the consequent lowering of the voltage drop and amps across the starter. Bottom line, due to cemf the operating voltage for this starter, while more then 6v, will not be a full 12 volts either.
Motor/generators have been used as starters/battery chargers fof some time on small engines and new applications of these units are now being used on electric cars. The cars electric motor can be converted to a generator/alternator on deceleration or braking.
PS, in a generator the armature is spun while in a alternator the field coil is spun and the armature (stator) is fixed.

take care,

 

[John_Mc]

As the starter armature turns it generates an opposing voltage that flows in the opposite direction from the voltage coming from the battery(cemf). Accordingly, the total voltage drop (operating voltage) across the starter drops off as the starter speed picks up.
A fully charge 12v battery will spin the starter faster thus generating slightly more cemf. This in turn will cause the use of slightly less current draw then a battery with a slightly less then full charge. Key word here is slightly.
A 6 volt starter running on 12 volts will not use TWICE as many amps as one running on 6 volts. Because of it's lower resistance value, a six volt starter wired to 12v will have a lot more hsp thus will spin both the engine and it's own armature a lot faster. This in turn will generate considerable cemf (opposing) voltage and the consequent lowering of the voltage drop and amps across the starter. Bottom line, due to cemf the operating voltage for this starter, while more then 6v, will not be a full 12 volts either.

George -

If I'm following this, it sounds as though you are saying the effect of the cemf on the current flowing is greater than the effect of the increasing voltage. That's the only way I could see that you'd get a decrease in current by increasing the voltage from say, 11 volts to 12 volts. I can see that the cemf would cause the current to decrease in a steady state spinning starter as compared to one that was just starting to turn and ramping up to speed. I have a hard time grasping how it could decrease the current in a starter turning at "full RPM" by more than the effect of the increase in current you get from an increased voltage. Sounds like free energy to me.

On the other hand, it's been a long time since I had to deal with any of this... my college physics classes are long behind me, and I had a long time keeping up with "Supersonic Surber" the professor in the one electrical engineering class I took. (WOW, I can't believe that name even came back to me! It's been 25 years.) I'm probably living proof that "if you don't use it, you lose it".

John Mc
(not an engineer, nor do I play one on TV... nor did I stay at a Holiday Inn Express last night)

 

[Keoke]

George -

If I'm following this, it sounds as though you are saying the effect of the cemf on the current flowing is greater than the effect of the increasing voltage. That's the only way I could see that you'd get a decrease in current by increasing the voltage from say, 11 volts to 12 volts. I can see that the cemf would cause the current to decrease in a steady state spinning starter as compared to one that was just starting to turn and ramping up to speed. I have a hard time grasping how it could decrease the current in a starter turning at "full RPM" by more than the effect of the increase in current you get from an increased voltage. Sounds like free energy to me.

John, see if this makes sense. When starter first comes on the voltage drop across starter is 12v. On the hot side 12v B+ and on the gnd side 0v for an operating voltage of 12. This high current shot is referred to as "strike amps" and because it momentarily includes the solenoid pull in winding draw it is very high.
Now, as the starter motor ramps up and increases it's speed the armature generates it's own voltage that tries to counter the voltage coming from the battery. So the internal voltage drop across the starter is no longer 12v.
EG, if the armature voltage (cemf) is 2v then the true internal operating voltage of the starter is 10v. Accordingly, a current drop would follow the internal operating voltage drop.
Any armature spinning withing a magnetic field (starter pole shoes) will generate a voltage of it's own.

John, hopes this makes sense.

take care,
keoke