daverclasper Posted November 15, 2021 Report Share Posted November 15, 2021 Hi. Something I've never understood, so would appreciate any advice. From what I understand, running an engine with a dynamo at idle (in my case about 800rpm), with a charged battery (say 12.6v etc) and no load, eg lights, should give a volts reading across the battery of about 13.5v. I don't understand this as a apparently a dynamo wouldn't be charging at these rpm and the ignition would be using some current anyway. Cheers, Dave Link to comment Share on other sites More sharing options...
Andrew Posted November 15, 2021 Report Share Posted November 15, 2021 At idle just about when the red ignition light will flicker the Dynamo will not charge hence the red light. Rev up and if you have a multimeter it will then charge at about 13 to 14 amps. Andrew Link to comment Share on other sites More sharing options...
NonMember Posted November 15, 2021 Report Share Posted November 15, 2021 The "common knowledge" about dynamos "not charging" at idle is... rather an over-simplification at best. They work just as well at idle, relatively, as alternators do. The difference is purely that alternators, being un-commutated, are less sensitive to being "overworked" at very high revs, so they can be geared better for idle. Either way, the output of a generator is vaguely proportional to speed up to the point where it gets limited by some other factor (such as a regulator or winding resistance). You also need to consider that the "output" in this case is more a question of current than voltage (although this all gets complicated!). If the battery is holding 12.6V against the load of the coil, and you add a dynamo into the equation, then the dynamo will pump some current in at any speed above what it would run at as a motor. How much difference that makes to the voltage depends on many factors. It may be that the current available is less than the coil uses but... well... - a coil is about 2A on average (slightly less at high speed if you have fixed dwell angle, quite a lot less at low speed if you have fixed dwell time) - a dynamo can put out 20A at optimum speed - optimum speed is less than the red line, so lets say 4000RPM - if the output is "proportional" to speed, you've got 4A available at idle In reality it's probably not quite that good but you can see that a dynamo should be able to trickle charge your battery even at idle, providing you have no electrical loads other than the coil. And in response to Andrew's comment about the red light... when I had a Vitesse with a dynamo, and a Herald with a dynamo, and a Spitfire with a dynamo, NONE of them used to flicker the ignition light at idle. Link to comment Share on other sites More sharing options...
PeteH Posted November 16, 2021 Report Share Posted November 16, 2021 I have to think about this, So, the water analogy, If you regard the 13V Dynamo output as the Head of Water. The current is the flow rate. If the Basin (Battery) is full, the water will not flow (0amp). If however you drain some water, more water will flow.+ XAmp. You can still measure the head ( Volts, AKA Potential Difference) of water even though no current is flowing. In order to overcome resistance in the Battery (the pipe work) the difference between 12.5 and 13 V creates the means. Simplistic, But is the way I view it. Pete Link to comment Share on other sites More sharing options...
Colin Lindsay Posted November 16, 2021 Report Share Posted November 16, 2021 12 hours ago, daverclasper said: Hi. Something I've never understood, so would appreciate any advice. From what I understand, running an engine with a dynamo at idle (in my case about 800rpm), with a charged battery (say 12.6v etc) and no load, eg lights, should give a volts reading across the battery of about 13.5v. I don't understand this as a apparently a dynamo wouldn't be charging at these rpm and the ignition would be using some current anyway. Cheers, Dave Connect both terminals of the dynamo together (after removing them from the loom) then connect a voltmeter to either terminal ( since they're now linked) and earth the other voltmeter cable. With the engine running at about 2000 rpm it should read about 14v. Being a dynamo the output increases with engine revs but is regulated by the control box to a steadier current. Link to comment Share on other sites More sharing options...
daverclasper Posted November 16, 2021 Author Report Share Posted November 16, 2021 Intresting, thanks. As I thought, electrics aren't necessarily simple!. So, could it be that lower than around 13.5 volts, across the battery at idle is not necessarily a charging issue?. Link to comment Share on other sites More sharing options...
Colin Lindsay Posted November 16, 2021 Report Share Posted November 16, 2021 Some cars with a perfectly good output at speed show the red charging light at idle; it's too slow for the dynamo to have any real effect on the system. I've just changed the brushes in one that wasn't running very well, and that along with a good clean of the commutator seems to have gotten a few more sparks out of it even at low revs. Link to comment Share on other sites More sharing options...
Andrew Posted November 16, 2021 Report Share Posted November 16, 2021 My point is as a old triumph mechanic once told me that if your car does not have a Rev counter to set the idle speed by the best guide is set the idle to when the red light just flickers on. When this speed is set and the red light is flickering the Dynamo is not charging for obvious reasons. This is quite normal and no harm done. Andrew 1 Link to comment Share on other sites More sharing options...
Chris A Posted November 17, 2021 Report Share Posted November 17, 2021 10 hours ago, Andrew said: set the idle speed by the best guide is set the idle to when the red light just flickers on. That's interesting. I'll try and remember that and try it with my 13/60 Link to comment Share on other sites More sharing options...
johny Posted November 17, 2021 Report Share Posted November 17, 2021 I agree with that idea but it should be remembered that the ignition light going out just means that the difference in battery volts and dynamo volts is insufficient to illuminate the bulb. The dynamo could be giving 3v less than the battery so the battery isnt charging and current will still be flowing from the battery to the dynamo but the 3v across a 12v bulb isnt enough to light up its filament... Link to comment Share on other sites More sharing options...
Andrew Posted November 17, 2021 Report Share Posted November 17, 2021 Bloody hell let’s complicate it. Simple if red light on no charge to battery. So left long enough flat battery end of Link to comment Share on other sites More sharing options...
johny Posted November 18, 2021 Report Share Posted November 18, 2021 well its just that light out doesnt necessarily mean system healthy - you have to take some measurements to confirm that👍 Link to comment Share on other sites More sharing options...
NonMember Posted November 18, 2021 Report Share Posted November 18, 2021 18 hours ago, johny said: I agree with that idea but it should be remembered that the ignition light going out just means that the difference in battery volts and dynamo volts is insufficient to illuminate the bulb. The dynamo could be giving 3v less than the battery so the battery isnt charging and current will still be flowing from the battery to the dynamo but the 3v across a 12v bulb isnt enough to light up its filament... Does it? Because that would imply the light is wired between battery and "dynamo output", but the dynamo output is effectively connected to the battery and the warning light is driven by the control box. There are no diodes in there to allow the interpretation you've put on it. The warning light means the control box has judged that the dynamo output is insufficient but that's not the same thing at all. Link to comment Share on other sites More sharing options...
johny Posted November 18, 2021 Report Share Posted November 18, 2021 In my diagram the warning light is connected between the battery and dynamo output but at the input to the regulator not the output. Then with engine stopped, ignition on, the current flows through the bulb and to earth through the dynamo. When started the dynamo output at the regulator increases and opposes the ignition light current flow so as I say the bulb can go out because the current isnt high enough to illuminate the filament but that doesnt mean the dynamo output is a higher voltage than the battery. In fact if adjusted correctly the regulator shouldnt even connect the dynamo output to the battery until its producing 12.7 - 13.3v... Link to comment Share on other sites More sharing options...
NonMember Posted November 18, 2021 Report Share Posted November 18, 2021 59 minutes ago, johny said: In my diagram the warning light is connected between the battery and dynamo output Ah, that's not what the diagram I looked at (in the proper WSM) shows. Perhaps yours is an early 1200 and I was looking at the 13/60 one. Link to comment Share on other sites More sharing options...
johny Posted November 18, 2021 Report Share Posted November 18, 2021 6 minutes ago, NonMember said: Ah, that's not what the diagram I looked at (in the proper WSM) shows. Perhaps yours is an early 1200 and I was looking at the 13/60 one. mines from the Vitesse/GT6 WSM and I would have thought all cars with a regulator box would be basically the same? When running it makes sense for the battery to supply the cars systems while the regulator powers up the dynamo (dependant on revs as well) until its output volts are sufficient and then the cut off contact closes to let it take over from the battery while also recharging it... Link to comment Share on other sites More sharing options...
NonMember Posted November 18, 2021 Report Share Posted November 18, 2021 OK, I wasn't looking at the internal wiring of the regulator. The light is connected to the "WL" terminal but I see that's internally connected to "D". Link to comment Share on other sites More sharing options...
johny Posted November 18, 2021 Report Share Posted November 18, 2021 One thing Im not sure about is the route of the current flow from the ignition bulb before the dynamo output voltage exceeds the battery - whether it goes through the reg to the field winding and earth or back through the commutator and winding to earth or maybe both.... Link to comment Share on other sites More sharing options...
daverclasper Posted November 18, 2021 Author Report Share Posted November 18, 2021 Wow. Not so simple. Now an easy question, hopefully?. It appears the Control Box black earth wire, is part of the loom that comes into the C.B. Where is that wire eventually earthed (as wiring diagrams don't appear to show this, as such). Ta Link to comment Share on other sites More sharing options...
daverclasper Posted November 21, 2021 Author Report Share Posted November 21, 2021 On 18/11/2021 at 18:53, daverclasper said: Now an easy question, hopefully?. It appears the Control Box black earth wire, is part of the loom that comes into the C.B. Where is that wire eventually earthed (as wiring diagrams, don't appear to show this, as such). Any help with with this please Link to comment Share on other sites More sharing options...
johny Posted November 21, 2021 Report Share Posted November 21, 2021 Its a difficult one this and if you have any concerns with the earth to the control box the best is probably just to run a new separate earth cable... Link to comment Share on other sites More sharing options...
Wagger Posted November 22, 2021 Report Share Posted November 22, 2021 Some of you, like me, are Electrical Engineers and the water analogy is very good. At about 800rpm the dynamo will just be at the point where it is generating a voltage equal to that of the battery. The ignition light will flicker at this point. That could be anything between 12 and 13.5 volts. The ony load with nothing else switched on is the coil. This is intermittent because the points actually keep disconnecting it. There is a capacitor involved too and, at some point, this is almost a series tuned circuit. A series tuned circuit draws minimum current at resonance. As little as 0.5 amps. The intermittent load is the cause of the ignition lamp flickering. When my dynamo did not work. My Vitesse got me to work and back in daylight for five days dong 30 miles per day. I carried a spare battery in those days, just in case. 25 amp dynamo's were not common at my local scrapyards, so I had to wait a while. Two cars ago, I mounted a second battery in the spare wheel well (as it had run-flats) and connected it so that it was only in parallel with the main one whilst the car was running. The car had an intermittent current drain fault. A flying jump lead was a real asset. Link to comment Share on other sites More sharing options...
Wagger Posted November 22, 2021 Report Share Posted November 22, 2021 The three bobbin regulator has four coils. (one of them has two separate windings). There is a voltage regulator, a current regulator and a change-over relay. As Colin Lindsay describes, the dynamo would produce a voltage directly proportional to its rpm between the the two terminals and chassis if the two spades were permanently connected. It would burn out due to over current if used connected this way. This device is a clever piece of electro magnetic apparatus. No electronics involved. It keeps the voltage constant for charging the battery and powering the whole car's loads. It also limits the full load current so that the dynamo and wiring do not burn out. The change over relay switches from battery to dynamo as rpm increases. If you set tick-over so that the ignition light is dimming, then that will prevent the battery discharging too fast when you are idling in traffic. In certain circumstances, if the fan belt breaks, or jumps off, you will hear a whirring under the bonnet. Open it and you will dee the dynamo running as a motor because the change-over relay cannot switch off. Disconnecting the battery will stop it, and it will not re-start if you re connect the battery. Alternators will not run as motors unless they are supplied with a.c. and would need a 'Start' winding or conversion to 2 or 3 phase. Strangely enough, many are multi phase with full wave rectification on each phase to produce a smooth D.C. All of this is from memory, so expect some variations as devices evolved. Look up 'How does the Lucas RB340 work' and you will find full descriptions. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now