Dropping 12v dc to 6v dc

There seems to be a lot of mis-understanding here. As Andrew has said, the coil is not simply resistive and you can't treat it as such.

A trembler coil is an early type of transformer. It has a low voltage primary winding and a high voltage secondary. Ohms Law simply does not apply. You need to use calculus to determine the current at any given moment in time but can't do so unless you know the inductance value, the resistance, and the applied voltage. Adding a series resistor (or lamp) effectively adds to the coil resistance.

The operation is roughly as follows; A voltage is applied to the primary and the current in the primary starts to rise from zero at a rate determined by the inductance. It then follows an exponential function involving the resistance as well. The current results in energy being stored in an increasing magnetic field. When the magnetic field reaches a set level a contact opens and the current can no longer flow in the primary. The energy in the magnetic field has to go somewhere and that somewhere is the secondary winding. The secondary voltage rises rapidly until the spark occurs and the energy is dumped in the spark. The magnetic field collapses and the contact closes again starting the cycle over.

The current measured by an ammeter will be roughly half the current at which the contact opens. If a resistor is used to reduce the measured current to that value the contact will never open and the system will not work. What is needed is a constant voltage to drive the trembler coil.

Russell

No, a resistor will work fine, though you may need to experiment to get the right value.

The danger to the coil is overheating and as the thermal time constant of the coil will be vastly higher than the rate at which the unit 'trembles' the fact the voltage fluctuates over the cycle will be irrelevant.

In the early days 6 volts for a trembler coil was often got from 4 Everready no 6 cells, sometimes known as telephone batteries, I'v seen a power supply made up to resemble a box of no 6 cells. There is a site some where that you can down load a copy of the wrapper for a no 6, I think you can just squeeze it onto a sheet of A4 paper.

Ian S C

An interesting video here about testing Ford Model T Trembler Coils.

I picked up:

• A Model T coil should be adjusted to draw about 1.3A at 6V. (The amount of current is determined by the size of the gap on the primary contact, which means it is frequency dependent.)
• The Model T has 4 trembler coils, one for each cylinder, so a trembler box would need 5 or 6A. (This might be where Vintage's lots of amps comes from.)
• Judging by the tone of the buzz, I guess the trembler is running at about 200Hz. As I have a poor ear, could a musical member please confirm.
• Adjusting the trembler requires a test rig fitted with an ammeter

Someone said the coil should be run from a low-impedance current source. That's dead easy if you have a 6V accumulator. The problem with a resistor is Neil's "you may need to experiment to get the right value". I'm not comfortable with that because the 'right value' isn't a constant - it depends on the internal resistance of the battery, the size of the contact gap, the value of the condenser, and the AC resistance of the coil, which also depends on the tremble frequency, and engine rpm. No doubt these errors tend to balance out, but you now have an awkward problem if you wire the thing up and find the engine doesn't work properly. Perhaps the spark is too weak, or a double spark is firing the mixture early. It could be the trembler is out of adjustment, or that the resistor is throwing the system out over part of its working range.

John Fletcher said 'why don't you just measure it with an ammeter'. Very good point! I'm of an age where I assume decent ammeters to be expensive and hard to obtain. I expect to spend a month's salary plus on an Avo. Wrong again! For less than £10 you can have an accurate digital ammeter. Have a look at ebay. At those prices it's not worth messing with light bulbs.

If a resistor is used, then a lamp might be the best way of providing one. Although filaments are sensitive to vibration, in this application their resistance changes helpfully with temperature.

Dave

Edited By SillyOldDuffer on 30/05/2018 11:28:42

Yes, a resistor will allow it to work but the value will need to be somewhat lower than that calculated from the measured current.

Not much danger of overheating the coil unless the contact is faulty and remains closed. In operation the current is a sawtooth waveform and the average current will be determined by the magnetic flux required to operate the contact not by the resistance. A fuse would be advisable to protect the coil in the event of a stuck contact.

Russell

A trembler coil will draw a very spiky current, it's not certain that a cheapo digital ammeter will accurately measure it. An old moving coil would be more suitable.