5
\$\begingroup\$

I have been reading a couple of papers (1,2) that use a Step-Recovery Diode to generate sub-nanosecond electrical pulses, for the purposes of pulsing an LED. I am simply trying to understand the purpose of each of the components in the pulse generation circuit, which is shown in the attached figure, any help would be much appreciated.

SRD pulse generator circuit, from (1). Pulse generator shown on left of figure.

I understand that the SRD and short-circuited coax cable (RG-316) create the short electrical pulse as follows: The circuit is triggered by an input square wave. When the SRD switches from forward bias ("on") to reverse cut-off ("off") a reverse current flows briefly as the charge stored in the diode is discharged. This pulse splits in to two paths - one to the output and one along the shorted coax. The pulse reflected by the short circuit reaches the output with a delay proportional to the length of the coax, where it interferes destructively with the other pulse...the short output pulse is equal to this delay as this is the part of the waves that is not "cancelled out" by the destructive interference.

The other components in the SRD generator I'm not so clear of their purpose. I believe the 470nf cap and 33µH inductor are a bias-tee that allows a DC bias to be added to the output pulse. The 100nF cap and 100R resistor I think are a high-pass filter to AC-couple the trigger input. What I'm least clear on is the purpose of the Schottky Diode (SD) and the adjacent 470nF cap and 56R resistor, any thoughts on those would be most welcome.

References: (1) Binh et al., "A simple sub-nanosecond ultraviolet light pulse generator with high repetition rate and peak power", Review of Scientific Instruments, 84 (2013) (2) Lee and Nguyen, "Uniplanar picosecond pulse generator using step-recovery diode", Electronics Letters, 37 (2001).

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Simulation would be a good route here if you can obtain part numbers for the diodes. \$\endgroup\$
    – Andy aka
    Commented Dec 23, 2021 at 15:25
  • \$\begingroup\$ I don't know how that circuit works but you tend to need to reverse bias it to sweep carriers from the depletion region to shut off an LED rapidly. Similarly, you need to momentarily forward bias it harder than normal to turn it on rapidly so maybe examine it with those in mind. \$\endgroup\$
    – DKNguyen
    Commented Dec 23, 2021 at 15:26
  • \$\begingroup\$ examples ? cdn.macom.com/datasheets/MMDx_SMMDx%20Series.pdf arxiv.org/pdf/1610.07115.pdf \$\endgroup\$
    – Antonio51
    Commented Dec 23, 2021 at 17:07

1 Answer 1

Reset to default
3
\$\begingroup\$

The description you gave about the circuit is correct.

Regarding your question, the Schottky diode is added in series to clip the output, to prevent the edges of the input waveform to be transmitted to the output. If you remove the Schottky, apart from the main pulse generated by the SRD+stub, you will see two small pulses corresponding to the transmission of the rising and falling edges of the input signal.

Those resistors to ground are most of the times cited as "impedance matching networks". I have been studying this type of generators and I never found them useful, from a theoretical nor practical standpoint.

\$\endgroup\$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.