Electronic – How to lengthen a short pulse

Your requirements are conflicting. You say 1 A at 3 kV for no longer than 10 µs, but then say you need about 30 mJ, which is 1 A at 3 kV for 10 µs. It seems there is very little wiggle room.

My first reaction is to use a car ignition coil. They can certainly put out 3 kV for short periods of time, and the total energy is nicely limited by what is stored in the magnetic field before the primary is shut off. If you really care most about delivered energy, then this will work well.

However, getting a square pulse at a controlled voltage and time will be more difficult. If you are willing to waste most of the energy, then a 3 kV clamp would work. You adjust the energy so that the time at the 3 kV clamp limit is the 10 µs you want. Possibly multiple ignition coils will need to be in parallel.

Update based on comments

So you have a solenoid with 9 Ohm resistance and need to push 500 mA through it. At this peak moment you'll have 4.5 Volts (9 * 0.5) on this solenoid, so 3.6 do not look enough (if you are sure it is enough, then probably the required current is less).

I would take some step-up convertor chip (based on what I can get ready from the stores around - in my case it is MC33063 which is available in convenient DIP or SOIC packages for far less than $1. Then I google for "MC33063 datasheet pdf" and browse this document to find out the typical schematics. I see I'll need few more components - resistors, capacitors, inductance. They say the chip is able to provide up to 1.5 A so the rest you need is to choose proper values according to their instruction, for, say, 5V output voltage. Your batteries will be serving almost 3A at this moment so it is important they are fresh.

I suspect here is some misunderstanding there :)

You usually can control either voltage or current, not both. I.e. you can apply given voltage to solenoid and see what current will flow (it will not rise immediately).

The core idea of step-up conversion, uses inductance itself. Imagine that your controller apply the voltage to the coil. Coil starts rising the current and when the current becomes large enough, you switch it to capacitor. Coil could not stop the current instantly so it charges capacitor until the current falls to zero. At this moment capacitor has the maximum charge and maximum voltage. Often diode prevents back flow from capacitor to the coil.

Look at this picture: http://paginas.fe.up.pt/~ee07229/images/classicDCDCconverter400px.gif

Imagine there is a MCU-controlled transistor (perhaps, MOSFET) instead of the key.

Now we only need to understand better what type of load / solenoid you use. perhaps it can get more than 500mA current with less than 3.6 Volts. This depends on its resistance...

Anyway it may happen that cheapest / simplest way is to use the third battery.