Motorcycle Forum banner

1 - 20 of 42 Posts

·
Premium Member
Joined
·
5,384 Posts
That system, as described, would draw lots of power when the fans are not on.
My S50 uses the first circuit in the linked drawing, but, if you already have a thermostat as described in that text, you could use the second, which would draw lots less current while the fans are off. BTW, that's an error in the second description; it should say 'While closed, relay is on, keeping fan OFF'
 

·
Registered
Joined
·
1,133 Posts
Discussion Starter #3
resisted circuit

I think that if a powered circuit is resisted until it no longer has current, it no longer uses any current(power either).
so when a switch(thermostat or any other switch) is installed in the circuit, and the switch is on, the devices in the load do not operate. It would seem that if the switch was turned off, that the devices would operate, but without some kind of a shunt in the resistors, the devices would not get a ground(or the negative side). I think there has to be some kind of seperator between the resisted circiut and the grounded circuit for them to function in the switch. So if the fans ground after the resistors and the switch, I think the circuit would cancel itself out before it reached the ground(no current-no power to drive the fans or the current to the ground), but when the switch is off(no resistance to stop the fans) they would blow plenty of air...and allow the current to flow through the circuit.
 

·
Premium Member
Joined
·
5,384 Posts
Not sure what you are trying to say, in terms of circuitry. The diagram you posted makes no sense, but, to stop a fan when closed, the thermostat would have to be parallel to it, electrically, thus shorting the current away from the fan. To do this, the resistor would have to be in series with both, or you short the battery.
 

·
Pale Rider
Joined
·
528 Posts
Your fans draw 20 Amps!!?!?!?!? That is a huge amount of current. Check the maximum output of your alternator, then add up the amperage of all circuits. If this number is higher than the maximum output of your alternator, the fans will drain your battery, and your bike will die in the middle of nowhere. You will also likely destroy your alternator, as well as your Rectifier/Regulator.

In all honesty, your description seems excessive. Let's start from the beginning: what type of bike, and why the need for the cooling fans?

Is an oil cooler an option? My air-cooled '79 Honda 750 ran its oil at 250 F coming out of the engine, which killed the oil quickly. I added an oil cooler, with an automatic, mechanical thermostat, which cooled the oil to 210 F, at highway speeds, with a heavy load. That basically doubled the life of the engine, and it took 0 Amps of power to do it. I did this modification in 2010, not in the 70's. Cheers!
:coffee:
 

·
Registered
Joined
·
1,133 Posts
Discussion Starter #6 (Edited)
yahmeesbike

the bike is an XS1100 yamaha, the 81 took the 1/4 mile from honda in 81 along with alot of other races... what I am trying to do is turn on an ignition coil cooling fan/regulator cooler fan, with a thermostat out of a heater....
exp.
There is an electronic circuit that shorts out a device using resistors that = the amperage of the device.... then when the switch turns off(like the thermostat on a heater where the thermostat turns the element off but its on all the rest of the time)...the device turns on because the resistor circuit is turned off and the device goes straight to ground, its called a ground circuit or something like that...
this is what I want to build for the fans in my coil cooler, when the heat rises, the resistance circuit will shut off, allowing the fans to blow, and the thermostat to cool off. kind of like this diagram, but Im not sure exactly...
 

·
Registered
Joined
·
1,133 Posts
Discussion Starter #7 (Edited)
A maybe1

now the theory is right, cause you can stop a fan with a resistor...I did it.

So how about this?

assuming that the temperature doesn't effect the transistor that handles the npn, and that the resistor doesn't bleed off if the fans turn everso slightly from wind currents while Im riding, and the powersource doesn't spike when the regulator turns on and off causing the switch to turn them on....

this maybe worky!
 

·
Pale Rider
Joined
·
528 Posts
Believe it or not, I did component-level repair, and reverse-engineering as an electronic technician for 3.5 years. I really don't follow your circuits very well.

Set that aside. An ignition coil is relatively small. It may get hot, but I doubt it gets hot enough to need a 20 Amp fan to cool it! If it does, then you are running something which belongs on a drag strip, not a roadway. Heavier duty coils may be called for: they would not generate as much heat, so external cooling would not be needed.

Just how hot do the ignition coils get? Have you used a thermometer to get some numbers? A 20 Amp fan will push a lot of air, but if the coils get that hot (above 120 F), I suspect they would burn up before your 20 Amp fan could move enough air to properly cool them. The varnish coating around the copper wire windings within the coils can't take heat much above 120 F, before the varnish burns up, ruining the coils. I'd really need to see some temperature readings on the outside of the ignition coils to get an idea of what level of cooling is needed. Cheers!
:coffee:
 

·
Premium Member
Joined
·
5,384 Posts
Neither circuit will work. The transistor holds ground on the fans, whether or not the resistor is connected through the thermostat. The only thing the thermostat does is increase the power dissipated by the transistor. If it is increased enough, yes, the fans will stop running - forever, because the transistor melts. I'm ignoring the fact that you show no current limiting resistor on the base of the transistor.

Trust me, the circuits I showed are far simpler, and efficient, with no excess heat generated in either.
 

·
Registered
Joined
·
1,133 Posts
Discussion Starter #10
diagram question

Is the dotted line in your simpler circuit a shunt? if it is, what kind of shunt, and is the resistor value = fan value?
 

·
Registered
Joined
·
1,133 Posts
Discussion Starter #11 (Edited)
anothre idea?

is this better?

according to this brain strain, the cap won't fire until there is enough current in the line to fire it, so I guess the question is how many uf is the cap going to need, to keep from firing when the grounded out circuit is on, how much it will require to fire when the circuit is freed, and how much charge will there be, when the switch shuts off, that could burn out the cap.

now I know that if the cap is too high, it won't fire at all, and I know that a grounded electronic circuit is not always considered trustworthy.

so those 20 amps might blow a fuse when the circuit spikes huh!
 

·
Registered
Joined
·
1,133 Posts
Discussion Starter #12
found this on Wiki....

Capacitance article in the wiki said this...
"
In electric power distribution, capacitors are used for power factor correction. Such capacitors often come as three capacitors connected as a three phase load. Usually, the values of these capacitors are given not in farads but rather as a reactive power in volt-amperes reactive (var). The purpose is to counteract inductive loading from devices like electric motors and transmission lines to make the load appear to be mostly resistive. Individual motor or lamp loads may have capacitors for power factor correction, or larger sets of capacitors (usually with automatic switching devices) may be installed at a load center within a building or in a large utility substation.
"
this is what I am talking about.
 

·
Premium Member
Joined
·
5,384 Posts
Is the dotted line in your simpler circuit a shunt? if it is, what kind of shunt, and is the resistor value = fan value?
Standard notation for a coil is to show which switch it moves by connecting it with a dashed line, in case there is more than one switch on the diagram.
 

·
Premium Member
Joined
·
5,384 Posts
I don't know what you mean by the cap 'firing'. A capacitor is like a jar; when you connect it to a current source, it fills at a rate determined by the impedance between it and the source. When it approaches the voltage of the source, the rate slows down and stops. In that circuit, when you close the switch, the cap will discharge through one resistor (or both) to a voltage somewhat above ground. When the switch is opened, it will charge again, through the fans it is connected to, and reach battery voltage. The current through the fans will be at a maximum when the switch is closed, as determined by the resistors, and drop as the cap charges. The lowest voltage on the cap will be controlled by the ratio of resistance of the fans and the resistors, the highest will be the battery voltage.

Say a fan is 10 Ohms, and the resistor connected to it is 8 Ohms also; with the switch closed, the voltage will be 8/(10+8), or 0.44 times the battery voltage, or about 5.3V. If your drawing implies the cap, both fans, and resistors are connected at the same point, the voltage on the cap with the switch closed will be the ratio of the parallel resistance of the fans and the parallel resistance of the resistors. For example, if one resistor is 10 Ohms and the other is 8 Ohms, the parallel resistance is (10*8)/(10+8), or 4.44 Ohms. If the fans are also 10 and 8 Ohms, the voltage will, be 1/2 battery voltage. Ohm's laws, BTW.
 

·
Registered
Joined
·
31 Posts
Why is everybody trying to make this so complicated? Just run a positive wire to the positive side of the fan. Run a wire from the negative side to the thermostat. Add a ground wire from the other side. Just remember the KISS method.
 

·
Premium Member
Joined
·
5,384 Posts
That's one of the ways I drew, above; if you have a thermostat that opens when it gets hot, you need a relay to invert the signal - the other way I drew. Doesn't get much simpler, either way.
 

·
Registered
Joined
·
1,133 Posts
Discussion Starter #17
finally some kind of idea...

I think I got it this time....

So either way, if my switch is backwards, I need two circuits to make this work

circuit 1 = thermo to ground with resistors in it. (when its on, the fans are off because of resistors in the load)
circuit 2 = relay from thermo to ground fired when thermo is off(resistors out of the circuit, fans run at high speed... because the resistors are removed by the relay switch.)

Man that was worse than the first time through!!!!
 

·
Premium Member
Joined
·
5,384 Posts
Which circuits are you referencing, the ones I posted, or the ones you came up with?

Also, you say 'at high speed'; are you trying to run the fans at two speeds, low speed all the time, and high speed when hot? If so, you will need high-power resistors in series with the fans, to drop the voltage through them, and the thermostat would bypass the resistors when hot.

If, however, you are trying to stop the fans by connecting resistors in parallel to them, it will NOT work; the only way to stop the fans is to open the connection between power and ground.
 

·
Registered
Joined
·
1,133 Posts
Discussion Starter #19
single pole double throw

if a SPDT switch was installed in such a way, as to have the thermostat release a "usually on" relay, and that relay has a ground on the second throw, when the thermo cut-out, the relay would start the fans....


+-fan-thermo_/_ relaythrow1>-resistor-ground(closed thermo)
_relaythrow2>-ground(open thermo)
negative relay ground

wow!
ps can't be a king so be a really great peasant
they were my circuits....
 
1 - 20 of 42 Posts
Top