High-speed widerange voltage to current converter

Hello.

I have differential voltage signal, which represents pulses. Form of this pulses can be rectangular, gauss, exponential and parabolic. Duration of this pulse can be between 50 and 1000 ns. Rise time and fall time of rectangular pulse 2-5 ns. Amplitude of this pulses can be between 10 mV and 2.5 V. I need to convert this voltage pulses to current pulses with amplitude (current signal also should be differential) from 10 nA to 1 mA. Furthermore, output of this converter should be isolated from external circuit, because there is high voltage (about 300 V). So, i need high-speed wide-range voltage to current converter. Do you have any suggestions?

One question, you are I assume using them to either measure and/or control some process, what are the input characteristics of the signal processing system,
cheers,
Richard

It is for measure process.

Input impedance of processing system is between 50 Om and 10 kOm. There is high voltage in someone systems.

The input impedance of your input system is low enough that you could possible use simple series resistors of around 10Mohm, which would certainly look like a current source. The resistor(or resistors) would need a 300V rating.
Isolation is not trivial if you have to preserve the DC response, not knowing what your circuit generating the pulses looks like means I can't comment on that aspect of it. With the rise and fall times, LED type isolation is out.
Isolation using fibre optics may be possible, only thing likely to work with your band width, but a lot of design work needed.
Cheers,
Richard

Pondering this problem more, even using the usual tricks of low inductance resistors, in a series string, to reduce the capacitance, at 10Mohms you will end up with excessive strays, which will destroy the frequency response, and probably cause excessive ringing as well. The input capacitance of your measuring system will also no doubt be high.
Active voltage to current converters are unlikely to have the frequency response without an extraordinary effort, and capacitance at the receiving end will not help either.
I get the feeling that you may need to re think what you are doing, could be wrong, but!
cheers,
Richard