Direct ignition coil as a "safe" HV generator?

Hello,
I would like to know some things about direct ignition coils as a means to safely generate HV for some HV projects experiments.

1. These have 4 pins, what are the pins connections and voltages that must apply to them?
2. I think they generate AC, how can I get DC out of them?
3. I believe they generate more than 20KV, is there any way to limit this voltage, so I can make a variable HV PSU?
4. Any idea about the current they can supply?

1. The four pins are the high and low voltage terminals of the step-up transformer. (Often there are only 3 pins, one being common)
2. To reduce the output voltage, use an external shunt capacitor. (For a rough idea, the internal capacitor is about a micro farad)
3. To obtain DC, short the internal capacitor externally and use the device in the fly-back mode. (Current limiting resistor will be required in series with the secondary)

1. "1. These have 4 pins, " , The standard ignition coil has two terminals and the EHT connector, the terminals are labelled + and -, the can must be earthed. Any other transformer is a manufacturers specific, and might work at any voltage, i.e. capacitor discharge.
2. You can not get DC out of them, you must use a EHT diode to rectify the output (steal one from a scrapped CRT television - they are good for 8KV).
3. To reduce the output , you must reduce the input voltage. A lot of ignition coils have 6V input. This is so when the car is started on a cold day, there is more then enough voltage for a good spark, when the ignition key is released from the "start" position to the run position, a resistor is put in the 12V feed so the coil and points only get 6V. These coils take about 4A.
4. What current?, I would have though that they can deliver a steady 20 Watts, so that will be several milliamps. Its hard to say as they could deliver 20 KV, but typically the spark would occur at 7 KV, so after this voltage they are operating into a short circuit (or 22K if the spark plugs have suppressors in their leads). The current would be limited by the inductance of the secondary and primary.
Frank