femtosecond pulser! - Page 1
Forum section: electronics / projects

author: electrolust
author_rank: Unknown
timestamp: April 01, 2017, 05:49:59 pm
content: april fool's!!No but really, even commercially pulsers seem to be limited to 10-11(tens of picoseconds) rise times. And this seems to have been the case for 50 years. Is this some physical limit that's being approached? Or no practical application? Or the limits of measurement?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 01, 2017, 06:50:20 pm
content: Yes.As they say, "RF doesn't like to stay in wires". Metals are too lossy in the low THz to optical range, and radiation is better anyway. Dielectric waveguides and free waves are used instead. Hence femtosecond lasers.Tim

author: ap
author_rank: Frequent Contributor
timestamp: April 03, 2017, 12:17:50 pm
content: Nonlinear transmission lines e.g. are used as pulse shapers to generate pulses with <5ps rise/fall time. So <10ps is available commercially (PPL). One problem certainly, besides generating it, is also to measure this. Sampling scopes with quite a bandwidth are needed to measure it:BW (GHz) = 0.35/RT(ns) can be used as a rule of thumb to assess bandwith required, which means a bandwidth of about 70GHz is needed at 5ps RT. That also means special PCB material, 1mm connectors...Below that, custom built research stuff essentially.

author: tszaboo
author_rank: Super Contributor
timestamp: April 03, 2017, 12:29:24 pm
content: Just calculate, how much current is required to charge any small capacitance at those frequencies. Or how much current per second change. The thing about current, it takes "big" devices to generate large currents, and big devices have large capacitance.

author: rstofer
author_rank: Super Contributor
timestamp: April 03, 2017, 03:28:44 pm
content: It's all Maxwell's fault - he told us that everything has mutual capacitance. So, we are stuck with some amount of capacitance that we want to charge in zero time.i = C dv/dt where dv/dt approaches infinity (because dt goes to zero) takes an infinite amount of current even for very tiny amounts of C.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 03, 2017, 05:21:10 pm
content: Quote from: rstofer on April 03, 2017, 03:28:44 pmWell, no. It just means things get smaller.There is no such thing as pure capacitance, or pure inductance, in reality. Only impedance and distance. (L and C are low frequency asymptotic approximations.) Even impedance is rather contrived: you can only measure impedance at a point (i.e., the current in and out of the conductors of a zero-width transmission line, or \$\vec{E}\$ and \$\vec{B}\$ at a point in space), which isn't very helpful when you have waves smaller than a given boundary condition (i.e., your coax turns into waveguide at high enough frequencies).So you make things smaller, to preserve the TE0 transmission line mode. But smaller things tend to have proportional losses, i.e., the same loss per scaled distance. So a meter of RG-59 is just fine at 100MHz, and 0.1m of microcoax is just fine at 1GHz, and so on.So that's a short reason why you don't see circuits (i.e., systems built with TE0 transmission lines as wires) with risetimes below about 1ps: even on a monolithic structure (like InP with aluminum metallization, or whatever exactly it is they're doing), the losses, or dispersion, is so great that the wave gets sloppy after some mm of distance. (It doesn't help that monolithic structures aren't as conductive as properly formed cables in the first place, but it also doesn't help that metal itself is increasingly lossy at higher and higher frequencies.)Tim