Motorcycle charging system - Page 1
Forum section: electronics / beginners

author: Nokia3310
author_rank: Contributor
timestamp: January 29, 2021, 06:13:00 pm
content: HiI'm trying to get a better understanding of how the charging system on my motorbike works.This is a basic schematic of the charging system. It's has a three phase generator and a combined regulator/rectifier. I think the regulator is a shunt regulator. My understanding of how the system works is that as the engine runs the permanent magnets spin past the stator coils which generates AC. That goes to the rectifier bridge which converts the AC to DC and charges the battery and powers the lights etc. As the engine runs and the battery charges the bike's DC voltage level rises. The IC circuit in the regulator measures this DC voltage. When the DC voltage reaches a set point the IC circuit sends a signal to the gate of the SCR and the SCR conduct. When the SCR conduct the phases in the stator are short circuited so the AC output from the stator phase becomes zero. This brings the bike's DC voltage down.Here are the waveforms I took from my bike. The oscilloscope was connected across two of the stator output wires, the yellow wires in the schematic. The oscilloscope had a 20:1 attenuator. The oscilloscope only has one channel so I can only look at one phase.The first waveform was with the stator unplugged so this is the stator's open circuit voltage. It's is showing about 80V p-p. I think this is considered a sine wave?The next waveform is with the probes still on the same yellow wires but now the stator is plugged into the regulator rectifier. The voltage is about 40V p-p. Here I think the bike's DC voltage is still low after start up so the SCR are off and the regulator is allowing current to flow from the stator through the rectifier to charge the battery.Here the DC voltage is rising, the SCR are starting to short some of the positive cycles in the stator phase.In last waveform the battery is now charged and the the lights are turned off so the bike's DC voltage has risen and the regulator has turned on the SCR on all of the negative cycles in the phase.What I'm asking is;Is my understanding of what's happening correct or wrong?Looking at the waveforms can you add any more information?What is inside the "IC Circuit" in the schematic?How does the regulator cut off the peaks of the AC waveform? I can see how the SCR is shorting the phase output to the zero volts line but I don't understand how the AC wave turns from a Sine wave shape to more of a square wave shape?When the stator is shorted by the SCR why does the stator not burn up? I know the regulator/rectifier heats up so some of the current is dissipated as heat but does the full stator voltage get dropped across the SCR? Does all the current go into heating the SCR or is more happening to limit the current? I don't understand how shorting the stator doesn't burn it.If you have any links to good websites or books that explain this type of charging system please post them. Thanks.

author: CaptDon
author_rank: Super Contributor
timestamp: January 29, 2021, 09:17:08 pm
content: I always thought this was a very stupid way to regulate voltage as it wastes power. The stator should be actually inside the engine crankcase (it was on my titan) and on my two stroke bike it was kept cool by the incoming air/fuel charge which was cool because of evaporation/venturi effect of the carburetors which on a two stroke feed into the crankcase through reed valves. It does seem like a strange way to do things. It also seems that when the SCR's conduct there is some kind of core saturation thing happening in the stator and the internal circulating current actually seems to go down??? I put 48,000 miles on a Suzuki Titan 500 with this charging system and never had a failure. In all those miles I only ever replaced tires, brakes, chain and sprocket!!! Never replaced a lamp and because it was a kick start even a crappy battery would last for 5 years!!!

author: Nokia3310
author_rank: Contributor
timestamp: January 29, 2021, 09:56:39 pm
content: A 500cc 2 stroke must have been a lot of fun to ride. I believe the Japanese bikes of that era were a big step forward in terms of performance and reliability.

author: Slh
author_rank: Regular Contributor
timestamp: January 29, 2021, 11:03:14 pm
content: Not a complete explanation but hopefully a few answers.The generator has a maximum short circuit current (one of the design parameters). I think it's a magnetic saturation effect but not really sure. The stator is sized to take the short circuit current continuously. The main issue is I^2R losses. It just means that you need thick wire in the stator and lots of cooling for the thyristors. Actual wasted power in the regulator is surprisingly small (say 15A short circuit current, 2V drop in the diode and thyristor is 30W losses in the regulator). Not nothing but not much on an 11kW+ engine.Your first trace is kind of sinesoidal but has a lot of distortion. That's just a feature of the arrangement of the stator and rotor poles.The square wave clipping is nothing to do with the control. If you look at the current from the stator it will look kind of sinesoidal but with deadline around the zero crossing. This is because the battery is a fairly stiff 12-14V and will take whatever current you give it without increasing much. As the battery approaches fully charge, the thyristor starts to clip the end of the cycle to reduce the average current. When it's fully charged then it just turns on 100% of the time to prevent overcharging.More modern systems have MOSFETs that can be turned off to stop charging. More like a brushless motor controller. The only issue is that they have to take the full voltage (80V+) and control is a tiny bit harder but it's easy enough these days and it means that you don't always have full short-circuit current in the stator.

author: Nokia3310
author_rank: Contributor
timestamp: January 30, 2021, 12:53:55 am
content: That's great information thanks. I'll measure the diameter of the stator wire and look up what current that diameter is rated for. The stator is rated at 12V 18A at 5,000 r/min. It would be interesting to see what safety margin they allowed for in the selected wire.I would like to look at the current output but I don't have a suitable ammeter or current clamp but if the stator is designed to be shorted without damaging it would it be possible to short it with a resistor and then measure voltage drop across the resistor and use I= V/R to calculate current? I could calculate the current at a certain RPM for each phase and compare them. If I connected the oscilloscope across the resistor would the voltage drop waveform show the current waveform? It would need a fairly high Watt resistor though.The stator specs areStator output 12V 18A at 5000 RPMStator coil resistance 0.36 to 0.44 Ω at 20 degrees C

author: Slh
author_rank: Regular Contributor
timestamp: January 30, 2021, 08:07:58 am
content: The current rating depends on the wire insulation as well as the wire size. It's going to be enamelled magnet wire and you won't be able to find the temperature rating. Also, the stator is attached to a big piece of metal that's going to be very hot most of the time. I expect that the copper could easily be >120C in normal operation.A resistor in series with a phase would work. A 50W 0.05R resistor should do the job.Short circuit current should be fairly constant over say around 1000rpm. It's all down to how the generator is designed.

author: Nokia3310
author_rank: Contributor
timestamp: January 30, 2021, 01:05:19 pm
content: Good point about the insulation. I'll buy a resistor and check the current from each phase.This will sound funny but I was looking around to see if if I have anything with a suitable resistor in it. I couldn't find a resistor but I did find a heating element from a washing machineCould I use that? I think it's basically just a big resistor? I think with the heating element I could get a comparison waveform for each phase but the problem could be if the resistance of the heating element changes as it heats up I wouldn't be able to calculate the current using I=V/R.It says on the heating element 230V 2000W Bleckmann UF70 SHD SA14Or maybe the idea wouldn't work at all because what I would see on the oscilloscope is the waveform of a washing machine heating element instead of a stator

author: Nokia3310
author_rank: Contributor
timestamp: February 01, 2021, 09:45:49 pm
content: Hi SlhOne more question. I'd like to do the two tests shown in the attachment. Before I do them I'd like to work out the power rating for the resistors.Test 1 - I think this is adding resistors in series? A resistor in series would allow me to measure voltage drop across the resistor and with that I could calculate the current coming from each phase when the regulator plugged in and the bike is running.Test 2 - I think this is adding a resistor in parallel? I would disconnect the stator and connect a resistor between two phases. Doing this would short the stator. The voltage drop across the resistor would allow me to calculate the maximum current output from that phase. I could then compare the maximum output from the three phases.The question is how do I calculate what power rating should be for resistor R1 and R2?The stator spec isStator output 12 V 18 A at 5,000 r/minStator coil resistance 0.36 - 0.44 Ω at 20 degreesCan I just say the stator output is 18A, so 18A/3 = 6A per phase @ 12V 6A x 12V = 72WSay 100W to be on the safe side. Does that seem correct?Would there be any drawbacks using a 200W resistor? I think the highest Amp motorcycle stators go to is about 50A so 50A/3=16.67A @ 12V =200W.

author: CaptDon
author_rank: Super Contributor
timestamp: February 02, 2021, 03:42:55 pm
content: The heating element may be a bad idea as I think its resistance will change drastically with current flow and temperature. Also the current waveform is so distorted it is hard to know what the true rms value of the current is with so much distortion. Best wishes in your efforts!! The Suzuki 500TT Titan was a real fun bike to ride. It had something like 44 horsepower at 450lbs of total weight. It used a controlled oil injection pump and the crankshaft was drilled similar to a 4 stroke crankshaft with forced lubrication to the rod bearings and crank bearings. The tach went to 12,000rpm and I missed a gear (you could find a weird neutral in between 2nd and 3rd if you didn't shift forcefully) and the tach needle was over the neutral indicator lamp at an estimated 14,000 rpm. I waited for the rpms to die off before I tried to put it back into 3rd gear for fear of shearing off the dogs in the transmission. I always ran the synthetic 'Yamalube' from Yamaha as it was the best 2 stroke oil available although the Suzuki CCI oil was also very very good! From a dead standing start a Harley Davidson could pull a few feet a head of me off the line with their big low end torque, but once I hit about 5,000 rpm that Suzuki would drive past those Harley idiots like they were standing still!!! Harley guys hated 2 stroke Jap bikes!! I ended up working at Harley in York back in 1992 and rode a GS750 Suzuki to work there every week day!!! Hehehehehe, they made me park in the contractor lot!!!!!

author: Nokia3310
author_rank: Contributor
timestamp: February 02, 2021, 08:38:06 pm
content: Quote from: CaptDon on February 02, 2021, 03:42:55 pmHaha brilliant. I can see it now. You pulling up for work at Harley, dressed head to toe in Suzuki gearThe forced lubrication is probably a big part of what got you to 48,000 miles without a rebuild! I wonder why more 2 strokes didn't have that?I think you're probably right about the heating element.

author: mikerj
author_rank: Super Contributor
timestamp: February 02, 2021, 09:15:29 pm
content: Quote from: CaptDon on January 29, 2021, 09:17:08 pmOnly a small amount of power in the scheme of things, a permanent magnet alternator behaves more like a current source than a voltage source i.e. you don't get huge currents flowing when the SCRs are conducting. It's a simple, inexpensive and functional way to regulate these alternators, all the things that are music to a manufacturers ears.A scheme that opens the stator rather than shunts it has to survive fairly high voltages when the engine is running at high RPM, as would all the insulation on the stator and wiring.

author: CaptDon
author_rank: Super Contributor
timestamp: February 03, 2021, 03:32:53 pm
content: Good point about the high voltage at high RPM with a no load condition on the stator!!! That would also mandate the use of diodes that could stand the voltage and potential inductive kick!! Many years back they used to sell a little switch box for cars that jumped out the alternator regulator and disconnected the battery and you could get huge 100+vdc from the alternator to run electric drills or almost any brush type 120vac motor as well as incandescent 120vac lighting. I imagine there were a lot of fried alternators at the local rebuild shops back then!!! The other downside was having to rev the car engine to about 3000 rpm!!!

author: Renate
author_rank: Super Contributor
timestamp: February 03, 2021, 06:26:18 pm
content: Yeah, I'm certainly familiar with the permanent magnet/3 phase stator/shunt regulator.Years ago I had a Honda CL350.The battery was giving me trouble.Back in ye olde day, big honking electrolytics were a thing on the surplus market.Everybody had one, the kind of thing that came out of a mainframe linear power supply.I had a 100,000 µF cap. I put it where the battery was.With the ignition off, you kick started a few times to charge the cap.Then you kicked if over one more time, switching on the ignition just as you did that.It worked like a charm.

author: Slh
author_rank: Regular Contributor
timestamp: February 05, 2021, 07:08:21 am
content: The current rating for the stator is per phase. Your resistor will be conducting 18Arms all the time (as it's either going straight into the battery or shorted by the thyristors). You need to choose a resistor that will give you a big enough signal while also limiting your losses (I^2*R) to a sensible level. 0.05R to 0.1R rated for 100W would probably be ok.Incidentally, you'll need to be careful when you connect your oscilloscope. I don't recognise your scope but if it's got two channels then make sure you connect the grounds to the same point. This is the sort of test where one could easily think it's safe to measure voltage and current at the same and blow up the scope by shoving 18A through the ground leads.

author: Nokia3310
author_rank: Contributor
timestamp: February 05, 2021, 08:02:52 pm
content: Quote from: Slh on February 05, 2021, 07:08:21 amThanks for that. I was visualising all the phases contributing at the same time for a total of 18Amps but now that you say it, it makes sense that each phase is designed to deliver the full 18Amps.The phases are 120 degrees out so at any one time in the cycle one phase will have the most positive voltage. The diodes on that phase will be forward bias. That will cause the voltage at the cathodes of the other diodes to be higher than the voltage at their anodes so they will be reverse bias. That means only one phase will ever be conducting and will have to deliver the full 18 Amp. The other two phases don't contribute at this time because their diodes are reverse bias. Then as the voltage in the phase that's conducting drops, the next phase in the cycle becomes the most positive and it will start conducting.For the testing I can see now that when a phase conducts it gives it's maximum output for that RPM or it gives no output if the SCR is conducting, so the only test that needs to be done to see the maximum output of each phase is the series test at specified RPM with a discharged battery.I am thinking about getting this resistorhttps://uk.rs-online.com/web/p/panel-mount-fixed-resistors/0188043/?relevancy-data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searchHistory=%7B%22enabled%22%3Atrue%7DThe oscilloscope I'm using is a DSO150https://jyetech.com/wp-content/uploads/2019/01/dso150-shell-user-manual.pdf

author: riichi
author_rank: Newbie
timestamp: November 13, 2022, 07:08:48 am
content: You may have already solved it.I tried a simulation with an IC circuit that I imagined.When the battery voltage is low, the thyristor does not turn on, and as the battery voltage increases, the turn-on time increases.It looks like a good match with the oscilloscope waveform.As mikerj said, the AC waveform peaks are cut because the armature acts as a current source.The inductance (L1, L2, L3) is the cause. It seems to be equivalent to the armature reaction.In the following figure, the horizontal axis is the number of revolutions. Compare when the thyristor is always on and when it is always off.From the second graph, the current is the same whether the thyristor is on or off, except for idling. The generator is acting as a current source.From the graph at the bottom, if the generator is short-circuited, it will not generate power.(I'm not good at English. It's a machine translation, so please forgive my strange English.)

author: brucehoult
author_rank: Super Contributor
timestamp: November 13, 2022, 09:22:17 am
content: Quote from: CaptDon on February 02, 2021, 03:42:55 pmSounds fun. I just got back from a quick 80 km run into town and back to get nasi goreng on my 1995 BMW R1100RT. That's 628 lb fully fuelled and 90 HP, so quite a bit more HP/lb, especially once you add in the rider. Being a boxer twin it's got the low end torque like Harleys, but it actually goes around corners. When I came out with my food some Maori teenagers were gathered around it. "Cool bike, mister". Not bad for a 27 year old machine.Recent pic:https://pbs.twimg.com/media/FbY9JObaIAAghJw?format=jpg&name=largeOh, relevant to the thread: it is rated to power 800 W of accessories! Typically heated riding suits, though I've never had one. The standard heated grips and flipping the vanes to direct hot air from the oil cooler into the cockpit (vents just below the mirror/turn signal) instead of out the slots above the cylinders (for summer) has always done the job for me.

author: Nokia3310
author_rank: Contributor
timestamp: November 13, 2022, 04:26:45 pm
content: Quote from: riichi on November 13, 2022, 07:08:48 amWow that's fantastic work. Well done. I will study it closely. Thanks for posting.