Hi Just to say as you confirmed via your test and many people are confused over this that a 6 v dynamo’s output is upto 20volts DC not 6 - 8 as people assume 😊 This is only achieved once the regulator is set up correctly, when the machine is idling then there is no output at around 1k revs or less , Once the engine revs up to around 1.5 thousand revs then the output of upto 20volts will be achieved at the regulator on the “D” terminal input this then activates both shunt coils “inner coils “ not outer thick coils The cut off coil the R/H coil will operate and close once set up to an input voltage of 6.3 - 6.7 volts which then goes to the voltage regulator then down to the ammeter / headlight terminal “A” which then charges the battery . The voltage input once achieved from the dynamo to the backplate of the regulator which operates and vibrates at around 50 times a second this needs setting to 7.7 - 8.1 volts max This will stabilise the voltage that then goes down to the field coil on the dynamo marked “F” Once the battery is fully charged the voltage then has to go somewhere and travels down the resistor and reduces the voltage which then reduces the current flow other wise the battery would explode Once the battery is fully charged then the cut out coil contact to the right opens until the battery is asking for current It also opens when on tick over 😊😊 The cut out tho doesn’t open at 3000 revs it opens when fully charged or at rest, it should always be open at rest or at low revs Hope this helps your subscribers I’m a qualified electrician and have studied these myself I need to do a new video myself It’s not the easiest as you know to get across to people without knowledge All the best Phil Green
Hi Phil. Thanks for that excellent explanation. I'm sure that will be very helpful. A video with diagrams would be the best way to fully explain it I reckon, but clearly you are much more knowledgeable on the subject than I am!
