1
\$\begingroup\$

In the datasheet of the family of step-recovery diodes (they are essentially PIN diodes with the I layer "very thin") MMDx of MACOM (link here), following figure is given

enter image description here

I understand that transition time is the time it takes for the diode to go from the low to high impedance state, once all the charge has been removed (which is a large amount of charge for an SRD). I have observed that this time depends on the rise time of the driving waveform. However, I'm failing to understand what the quantity "drive" of the x axis represents.

I refered to the literature and found nothing. Moll and Hamilton 1969 paper which is usually cited for an explanation of the SRD workings says nothing about dependences of the transition time. In fact, it says it can be estimated as 10 ps per micron width of the I layer, but nothing about influence of other factors.

\$\endgroup\$
1
  • \$\begingroup\$ Not writing answer as I don't know, but I can guess that drive in Coulombs is the amount of charge that you 'charge up' the diode with in the conducting state, before you reverse bias it and suck the charge out while waiting for tau to elapse. It could therefore be related to the drive current, times a factor for that diode. It could also be the reverse current times tau, which has units of charge. The two should be related, less some recombination loss. \$\endgroup\$
    – Neil_UK
    Commented Dec 29, 2021 at 11:08

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.