Simple alternator control by zener diode 1

[Oakhouse]

Hi,
British motorbikes of the 1960s were fitted with a simple two-wire stator alternator,consisting of six poles wired in series,with a permanent magnet rotor inside.
Voltage control was a zener diode in parallel with the battery and bridge rectifier output.
I wish to know what happens to the alternator (EMF,magnetic fields,output,whatever)when the zener conducts through itself,ie across the rectifier output terminals.
Thanks.

 

[melone]

It clamps the voltage to the zener voltage.

 

[Oakhouse]

Hi Melone,
what I need to know,is when the zener diode conducts it effectively shorts the alternator output leads together.What then happens internally to the alternator with regard to magnetic field strength,EMF and output?

 

[BrianG]

Hi, I owned several Triumph motorcycles in my youth with that kind of charging control. The permanent magnet type alternator is like a constant current source so the voltage can vary quite significantly and is therefore unsuitable for direct connection to the battery. The zener clamps the voltage to a safe "float" voltage for the battery when it is fully charged.

The thing to remember is that the zener is NOT a direct short circuit, but that it has a definite slope resistance, so its effect on the alternator is much the same as a normal load. The excess power is simply dissipated as heat by the zener internal resistance.

The zener therefore has to be mounted on a large heatsink - on Triumphs early models used a flat heatsink under a side panel, later ones had a bullet-shaped finned version in the front forks right in airflow under the headlight. If I remember correctly the zeners were rated at 150W.

 

[Oakhouse]

Hello BrianG,
thanks for the reply.
I know zeners produce (unwanted) heat in operation,but their main task is to pass current,or conduct,at a given voltage.
When they conduct, they pass current effectively across the alternator output leads.
The Lucas workshop manual of the time details a bench test of the zener,where a current of two amps should be seen to flow through the zener,if zener is OK.
So, what effect does this "return" current have on the alternator operation?

 

[BrianG]

Hi Oakhouse, I am not sure what you are trying to say here. Of course a zener will start to pass current at a certain voltage and in that respect it is no different to any other kind of load, say turning the headlight on, so any reasonable load will reduce the output voltage from the alternator. Under these conditions, like any other circuit, there will be a "return" current equal to the load current.

If I understand your question about testing the zener in the workshop, I have not seen the Lucas test but would imagine it uses a separate (variable) voltage source to give the 2A test current you mention. I would be very surprised if this test were supposed to be carried out with the zener still connected to the bike's wiring as this would make it impossible to test unless the battery is also disconnnected. Provided the alternator is NOT running the question of return current during the test is therefore irrelevent.

Remember that the alternator produces d.c. to charge the battery through the bridge rectifier, so, when the alternator is not running, apart from a tiny leakage current in the rectifier, all return current into the alternator is blocked by the rectifier. (Otherwise your battery would get discharged by the alternator)

Note also that the zener voltage is chosen to start loading up the alternator only when its output voltage starts to approach the maximum battery "on charge" voltage. This current increases through the zener progressively as the the alternator output tries to rise, e.g. with increased r.p.m. so that an approximate "equilibrium" is achieved.

When the alternator stops, and charging stops, the battery terminal voltage drops below the zener voltage. This prevents the zener from discharging the battery (barring a faulty component).

Hopefully the above explains how the whole system works.

 

[uppili11]

I have some more issues to add about the general system.The modern alternators do not have permanent magnet but a field system whose supply is controlled by an electronic regulator which consists of a zener diode and a darlington switching transistor which switches on and off to control the output voltage.but if the loading of the alternator is full the field excitation will be continuous to cater to the load.(ofcourse there is an additional auxillary bridge rectifier or the field which is also used for the warning lamp mechanism)
The output voltage of alternator is rectified by diodes which are independent of the regulator circuit.
But in some alternators even today the normal diodes are replaced by diodes called avalanche diode which clamps the ouput voltage but the diodes must be rated for the full output of the alternator but I forgot the advantage of this avalanche diode.
Also the most important feature of an alternator compared to a dynamo or dc generator the output of an alternator beyond a specific speed say around 6000rpm is constant and any amount of acceleration will not yield extra ouput because of the industive reactance saturation of the stator winding because of high frequency.
So coming to the basic question the zenor diodes acts as a rectifier as well clamp the voltage and is rated For the full load capacity of the alternator whether the system draws the power or is wasted across which is highly inefficient form of control because of the permanent magnet rotor which is almost non existant today in commercial automobiles.sorry for the very long discussion

 

[Oakhouse]

Hi Brian G,
I don't think I'm explaining myself very well here.
Forget zeners,bikes and batteries for now.
What is the theoretical relationship between EMF,back EMF, magnetic flux strength,current etc etc in a simple alternator (such as these) under conditions of no load,light load,full load and short circuit?
Disregard any outside factors such as regulators,fuses etc for the purpose of this example.

 

[uppili11]

Normally the generated voltage is proportional to

frequecy,flux per pole and turns per phase.

the generated voltage = terminal voltage plus current*reactance

so the output depends on the frequency(the speed of rotation)-higher the speed -higher the output
number of turns-more turns-more output

more flux means more output.

again less reactance means less loss but as the frequency increases in alternator the reactance also increases and beyond certain limit you will not get any output because of saturation due to very high reactance.


In the absense of a regulator you can load the alternator
to any extent but the terminal voltage drops depending on the load but if you have regulator the switching may be continuous or in pulses which regulates the field current.and hence mantains a constant voltage.

 

[Oakhouse]

Hi Uppili11,
OK,I got all that.
Now what happens if you short together the two output leads when-
A) the rotor is stationary,then begins to rotate to full speed,and
B) the rotor is spinning at full speed, and the alternator is at max output?
Thanks

 

[uppili11]

A. THE ALTERNATOR GENERATES OUTPUT IF THE CONDITIONS LIKE THE SPEED,FLUX AND THE CONDUCTORS ARE IN PLACE.IF YOU MEAN SHORT CIRCUIT -IN INDICATES VERY LOW RESISTANCE AND IN THE ABSENSE OF A RECTIFIER OR REGULATOR THE ENTIRE CURRENT WILL CIRCULATE WITH IN THE SHORTED LEAD LEADING TO HIGH DISSIPATION OF POWER.IN SOME CASES THE SHORTED LEAD MAY FUSE IF IT IS NOT THICK ENOUGH TO CARRY THE CURRENT.PLEASE REMEMBER THE ALTERNATOR WILL NOT GENERATE MORE OUTPUT THAN SPECIFIED AT THE SPECIFIC SPEED /FIXED FLUX.THIS CONDITION IS LIKE A HEATER COIL WHERIN THE RESISTANCE IS SO LOW AND A HIGH VOLTAGE IS APPLIED AND DRAWS POWER FROM THE SUPPLY.

B.IF YOU SHORT CIRCUIT WHILE RUNNING-THE BASIC THEORY IS THE CURRENT TAKES THE LEAST RESISTANCE PATH AND INSTEAD OF POWER DRAWN BY OTHER CIRCUITS IT WILL CIRCULTE BETWEEN THE ALTERNATOR AND SHORTED LEAD.PLEASE NOTE ALL THE ABOVE CONDITIONS ARE ASSUMING THE FIELD IS EXCITED SEPARATELY AND THERE ARE NO RECTIFIERS OR BATTERY CONNECTED ACROSS.

IF YOU TRY THIS WITH BATTERY CONNECTED YOU WILL ENJOY A BLAST!!!!(I REMEMBER ONE OF THE INTERN MEASURED THE DISTANCE BETWEEN TWO LEADS OF THE BATTERY USING A VERNIER!!!!)

 

[uppili11]

ALSO ANOTHER ISSUE:THE TERMINAL VOLTAGE WILL DROP
IF THE CURRENT DEMAND BY THE SHORTED LEAD IS SO MUCH
AND AT ONE SPECIFIC POINT THE ALTERNATOR WILL NOT BE ABLE TO DELIVER THE LOAD AT SPECIFIC VOLTAGE .E.G ALTERNATOR OUTPUT IS 40AMPS.ASSUMING THE SHORTED LEAD IS 0.1 OHM.IF THE TERMIANL VOLTAGE IS MAINTAINED AT 12 VOLTS THE CURRENT DRAWN WILL BE 120AMPS WHICH IS NOT POSSIBLE BECAUSE THE ALTERNATOR CAN NOT DELIVER SUCH OUTPUT. THE TERMINAL VOLTAGE WILL DROP AND COMES TO A STABILISED POINT AS PER KIRCHOFF LAW WHEREIN THE SUM OF VOLTAGE DROP IN A CLOSED CIRCUIT WILL BE ZERO WHEREIN THE DROP ACORSS THE LEAD AND THE INTERNAL DROP OF THE ALTERNATOR WILL BECOME EQUAL .
MAY BE WILL TRY THIS CONDITION IN A LAB AND GIVE YOU THE RESULT

 

[Oakhouse]

Hi Uppili11,
when the leads are shorted together,a high current will flow in the windings.Do you agree that this current will cause a magnetic field in opposition to the existing permanent magnet magnetic field (Lenzs Law) meaning that generated EMF value will suddenly drop?

 

[melone]

Off-Topic:

PLEASE STOP TYPING IN ALL CAPITALS!!!!! That is interpretted as shouting.

 

[uppili11]

will continue this in the week end.Thanks Mr melone!!.

 

[uppili11]

Yes again uppili11.i fully agree that there will be demagnetization but there are three cases:

1.0 unity power factor:
In this case the armature flux is cross magnetizing.The result is that the flux at the leading tip of the poles is reduced while it increased at the trailing tips.However these two effects nearly offset other leaving the average field strength constant.In other words ,armature reaction for unity p.f is distortional

2.0 zer p.f lagging:

Here the armature flux is in direct opposition to the main flux.Hence the main flux is decreased.Therfore it is found that armature reaction is wholly demagnetizing with the result that due to weakening of main flux ,less emf is generated

3.0 zero p.f leading:

In this case the armature flux has moved forward by 90 deg so that it is in phase with the main flux .this results in added main flux.Hence in this case armature reaction is wholly magnetizing which results in greater induced e.m.f..iN ORDER TO KEEP THE GENERATED VALUE OF E.M.F SAME ,FIELD EXCITATION WILL HAVE TO BE REDUCED.

considering the short circuit -in my opinion is Upf(let some more experts join here to clarify)and hence will have cross magnetizing effect.So under shorted condition the chances of winding burn out is a high possibility.