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I want to build a frequency multiplier with high multiplication factor. I read some application notes of Hewlett Packard on Step Recovery Diodes. I think a SRD can do what I want. However, I have never used one before. Since they are difficult to obtain, I searched for a replacement and came up with standard PIN diodes.

I searched for PIN diodes with low capacitance and short carrier lifetime which should in my opinion give a large number of harmonics. My preferred type of PIN diode is HSMP-3822, but I don't have one here, so I used HSMP-3832 for some first experiments.

Indeed, I got harmonics up to 100 MHz with a 5 MHz drive signal, which is pretty good in my opinion. However, I don't have an idea what the exact design procedure is for a SRD comb generator.

How do I find the correct drive level and what other components do I exactly need for the SRD?

Ricardo
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T. Pluess
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  • Some people use the step recovery effect of RF transistors: http://www.adv-radio-sci.net/2/7/2004/ars-2-7-2004.pdf – starblue Jul 28 '16 at 14:34

1 Answers1

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A PIN diode is not a good substitute for a step recovery diode. If you want to bodge (use something cheaper) and I have, then a switching diode like the BA682 (smd) or BA482 (leaded) works much better. This has worked well at 130MHz input to 1500MHz output, but misbehaved at 500MHz in to 3GHz out.

In outline, you need to engineer a low pass filter to the drive side, and a high pass filter to the output side. The first element of the output HPF should be an inductor to ground. It is the current flowing through this inductor that gets interrupted when the SRD 'snaps off' that generates the output voltage. The drive LPF components and the output inductor need to be matched to your input drive voltage and frequency for best efficiency. This part is amenable to straightforward modelling, have a play with SPICE. Note that the diode will draw a unidirectional current, so you need to provide a DC bias path.

It's very easy to accidentally get parametric division of the input signal, unless you can make its environment look lossy at those low frequencies. Unfortunately, parametric misbehaviour is very difficult to model well, especially in the very non-linear environment, and this is where the design can degenerate into black art.

At 5MHz input, you could well expect good behaviour, and good prediction from your simulator.

Unfortunately, if you don't want to afford the 10s or 100s of $ for a specified Step Recovery Diode, other diodes types are not going to spell out in their data sheets whether they are any good or not. It is up to you to do the experiments, and the research.

You have started well, by exposing your assumptions to the views of experts, who have told you that this specific part number of band switching diode works better as a SRD than other types of diode, and especially better than PIN diodes.

Of course if you do not like this advice, there is absolutely no need to take it. What you do need to do is build a model, either SPICE, real hardware, or preperably both, and test some diodes in there.

I draw your attention to the fact that 1N400x and 1N540x diodes are quite snappy, which is why they are often shunted with capacitors in high quality audio amplifier power supplies. Without the capacitor (which slows the rate of change of capacitance versus voltage at end of conduction, the type of behaviour a high doping density will get you in a good rectifier, band switch or step diode, but not in PINs), they often generate a slew of high harmonics of mains which is difficult to eliminate from the audio circuits.

Neil_UK
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  • How to find a good substitute for a SRD? I thought the trick of the SRD is the long carrier life time. And this is what PIN diodes also have, don't they? or, asked differently: which parameter in the BA682 datasheet tells me that this diode could be used as a SRD substitute? – T. Pluess Jul 30 '16 at 08:37
  • No, the most important thing is a fast snap-off, which means high levels of doping. PINs have low levels which leads to a wide intrinsic region (the I in PIN), which gives very low off capacitance. The band switch diode is optimised for low on resistance, which requires heavy doping. However, at 5MHz, you could achieve some sort of comb with any diode, even a 1N4148. If the question is one of efficiency, amount of power in the harmonics versus drive power, when pushing the frequency up, that demands fast snap-off. – Neil_UK Jul 30 '16 at 08:44
  • I think in the usual datasheets of diodes, the snap-off is not shown. Is it possible to determine it from other diode parameters? I read the band switching diode's datasheet. For me, it looks like some sort of PIN diode, however, it is intended only for switching. Is this true? – T. Pluess Aug 03 '16 at 06:19
  • @T.Pluess updated question with more info – Neil_UK Aug 03 '16 at 09:32
  • I am absolutely not in doubt about your advice concerning the bandswitch diodes. I just wonder what the difference between a PIN and a bandswitch diode is. – T. Pluess Aug 05 '16 at 06:18
  • The doping density => the rate of change of capacitance with time at charge depletion, PINs have a purposely created I region (the clue is in the name) whereas bandswitch diodes minimise it. – Neil_UK Aug 05 '16 at 06:46