A few questions on  JW pulse generator based on 2N3904 (and other transistors) - Page 1
Forum section: electronics / beginners

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: November 30, 2015, 12:25:26 am
content: I have made a minimal pulse generator based onJim Williams application note 47http://cds.linear.com/docs/en/application-note/an47fa.pdfin the variant using a 2N3904 transistor, as mentioned several time on the Web, for example inhttp://www.kerrywong.com/2013/05/18/avalanche-pulse-generator-build-using-2n3904/As I do not intend to use this pulse generator outside of my lab, I did not bother to make thepower supply, and just made the pulse generator part.with the following setting (AN47, FIg. D1)Q1 : 2N3904C1 : 5pf (with some margin, this is a 45 years old component, but was the only ceramic cap I had in this range with high voltage (500V) ).resistors : 1M, 12k, 51 ohm.I had thus the pleasure to use my Fluke 415B HV power supplied which I salvaged some time ago fromthe dumpster, but had no use of it yet.I increased the voltage until I got the avalanche behaviour at 100V, but with some flickering, and itstabilised at 102V.But as time goes on, I had to increase the voltage to 106 V to keep a stable pulse.I verified that the change did not come from the power supply.Question 1 : Does the avalanche behaviour change as the transistor heats ?In his note, JW mention that all transistors do not present avalanche behaviour, and that in a 50 samples set, only 82%presented this behaviour.I tested only one which worked, so my sample is not sufficient to make statistics.Question 2 : Is the selection just to have the avalanche effect ? Doe the rapidity of the avalanche changesa lot with the various transistors among the same denomination (2N3904).Will the avalanche threshold voltage change a lot ?My design is not as compact as Jim Williams' (Fig. D3 in AN47).Question 3 : Should I gain a lot in rising time with a more compact design ?In any case, I do not have the proper gear to test a fast pulse.Below is what I got with my rigol DS1054z (100 Mhz).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 30, 2015, 02:05:59 am
content: Avalanche breakdown voltage increases with temperature, although I'm not sure about this mechanism specifically (as it depends somewhat on base leakage current, which increases more quickly with temperature, and might be expected to have the opposite effect due to amplification of base current).2N3904s can be expected to be slower than smaller switching or RF transistors (like the recommended 2N2369), but should still be 1ns or below.I think part of the selection process arises from differences in leakage current. If you try different base resistors for each transistor, you will find a sweet spot, where avalanche is most consistent over the widest range of supply voltages. The value of base resistor, and range of supply voltages, both vary with type.I haven't tested nearly enough to be able to tell if avalanche is a universal phenomenon, that all BJTs can exhibit (given the right combination of terminal voltages, currents, resistances), or if it's characteristic of a more limited set (due to device geometry or doping or what).The other part of selection, is simply finding transistors that switch faster than all the others: I think JW found a range from ~300ps to 800ps or something like that, for the 2N2369s he tested.Tim

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: November 30, 2015, 10:58:21 pm
content: Thanks Tim for the feedback.I have continued my experiments. First by adding another BNC connector to hook a coax,after reading the discussion ofhttps://www.eevblog.com/forum/blog/eevblog-306-jim-williams-pulse-generator/.I tried several coax before finding one that behave well.question 4 : Why choosing a 50R coax here. There is nothing that relates this impedance to the board. I would think that any impedance would fit. Do I miss something ?Then I look in my supplies which transistor I could try. I had some KSC1730, with very low capacitance (1.5pF)and low CB voltage max (30 V).So I installed a socket instead of the 2N309 and put the KSC1730. Beware, the KSC1730 is ECB !!The avalanche occurs at 70 V and the performances seems better than the 2N3904, despite all the overheaddue to the socket.I should try later with various base resistors.The thing which surprises me is the very good computed bandwidth of the Rigol DS1054zrise time rt=1.2 ns so BW = 0.35/rt= 292 Mhz. Way over the announced 100Mhz.On the other hand, on my TDS460 I have rt = 1ns so BW = 350 Mhz which is comparable to the announced 400 Mhz.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 01, 2015, 03:58:25 am
content: Quote from: JacquesBBB on November 30, 2015, 10:58:21 pmWhen the transistor "fires", it goes C-E short (or actually more like 10 ohms). The pulse line is dumped into the output transmission line.If the output line is 50 ohms and terminated, then you get zero reflection when the pulse line is also 50 ohms. Well, give or take switch resistance. To adjust for that, you may need a little resistance in series or parallel with the output connector.Mind you're also putting half the supply voltage into the oscilloscope, which it may not appreciate. The pulse line charges up to 70 or 106V or whatever, the transistor closes, half the voltage goes one way (53V down the output line) and half goes the other (-53V down the pulse line -- superimposed on the static 106V, so it also sees 53V with respect to ground). These wavefronts propagate out, until one is absorbed by the scope (maybe -- don't expect it to be linear with that much voltage going into it), and the other reflects off the open end of the pulse line.Off an open circuit, the reflection is in phase, so the -53V wavefront hits the end of the pulse line, becomes -106V (thus fully draining the line), and propagates back. By the time this reaches the scope, a full positive pulse has been seen: the time from rising to falling edge is equal to twice the delay length of the pulse line. (Evidently, your pulse line is about 7ns long, or 1.4 meters if it's typical coax.)You really need an attenuator in there. A typical 50 ohm 20dB attenuator has 61.1 ohms to GND (at the transistor emitter), 247.5 ohms in series, then another 61.1 ohms to GND (across the output connector).Place the resistors as low against the ground plane as possible, and keep some distance away from the transistor, so the switching edge doesn't couple in by proximity alone.QuoteNope:https://www.eevblog.com/forum/beginners/bandwidth-calculation-caveats!/msg807413/#msg807413This isn't entirely fair, because you do look to have a step sort of waveform. But it's not apparent how much overshoot or ringing you have. Given the loose construction, maybe quite a bit?The other issue is that you're probably clipping the front end, causing it to read much faster than it can. There is also internal filtering and interpolation done on the data, before it gets displayed. The ringing just before and after the edge makes me think it's interpolating an "instantaneous" edge. Try again, with sinc (sin x/x) interpolation off, or in dot (not line/vectors) mode. These settings should be under Display or Horiz. or Config or something like that.Tim

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 01, 2015, 10:17:20 pm
content: Quote from: T3sl4co1l on December 01, 2015, 03:58:25 amI understand that the line should be open. Otherwise you are shorting a 106 V with a 50 R resistor.But my question is wether it is important that it is a 50 R impedance coax.Why not a 75 R coax (with open termination of course) ?It will have the same effect ? Where will be any mismatch impedance ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 02, 2015, 04:21:15 am
content: If the lines are mismatched, then when the reflected wavefront returns to the transistor, you want all that energy to proceed into the output transmission line, without reflecting again. If the two lines are mismatched, a fraction of that wave will propagate back down and up the pulse line again, leading to additional echoes. This is apparent in your waveform.In fact, since your waveform shows an overshoot, it must be that your transmission line has a lower impedance than the pulse line, which is reflecting some energy back out of phase (which again reflects off the far end of the pulse line in-phase, then comes back, and so on). It would seem adding a series resistor to the output transmission line would help (maybe 22 ohms?).Reflections between transmission lines work exactly like reflections off transparent objects, like light entering clear glass. The glass is denser (in optics, it is said to have a higher index of refraction; in electronics, it has a lower characteristic impedance), causing some light to reflect off the surface (out of phase). Light still continues on through. The pulse echoes you can think of as the faint secondary images you see in a back-silvered mirror, or in a double pane window.Tim

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 02, 2015, 10:20:42 am
content: OK,now I understand. The problem I had was to evaluate the impedance of the circuit at the point of connection of the delay line coax.I was troubled by the presence of the transistor. But I I understand well, when the avalanche occurs, the resistor of the transistoris practically null, so the remaining impedance is the one of the output BNC connector, given by the associated resistor.Is this correct ?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 02, 2015, 01:08:36 pm
content: In fact, if you know the impedances, you can use the mismatch to measure the transistor's "on" resistance. As I recall, it's around 10 ohms.So over perhaps 300ps, the transistor goes from a fairly open circuit (leaking <1mA at 100V, i.e., >100kohms), to ~10 ohms. What happens afterwards can be modeled simply by assuming the transistor acts as a resistor.The transistor also remains conducting for a relatively long time, something like 10us. This is similar to the storage time (when using BJTs normally), but supercharged due to the extreme current density (about 10x the maximum rating of the device!). If you view the waveform on the pulse line, you'll see it's held low for some time, before recharging and eventually repeating.Tim

author: macboy
author_rank: Super Contributor
timestamp: December 02, 2015, 03:17:50 pm
content: Quote from: JacquesBBB on December 01, 2015, 10:17:20 pmYour scope must be terminated at 50 ohms. Your DS1054z does not have this feature, so you need to use a 50 ohm pass-through terminator, or at least a BNC tee with a 50 Ohm terminator on the other side. You must use a 50 ohm BNC cable to connect the pulser to the scope. Finally, you must use a 50 ohm cable as the charge line. When all these things come together, then the result is a good pulse that will not suffer from the reflections that areclearlyvisible on your screen capture.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 02, 2015, 06:25:14 pm
content: Quote from: macboy on December 02, 2015, 03:17:50 pmI have realized that. I have put a T terminated by a 50 ohm plug. This gets rid of the main reflections, but other remains.@Tim : Thanks for the clarification.I have continued the discussion of the measure of the Rigol DS1054z bandwidth onanother threadhttps://www.eevblog.com/forum/testgear/pulse-generator-rise-time-and-rigol-ds1054z-bandwidth/and will continue the present one on the construction of various pulse generators.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 06, 2015, 10:34:46 am
content: I have continued my experiments on pulse generator.The first thing I can say is that Tim (T3sl4co1l) was right. It looks like the sin(x)/x, combined with some overshoot on the pulse generation is artificially increasing the slope of the step function on the Rigol DS1054z.To track the possible origine of the ringing that occurred in the pulse, I added a pots for control or the output impedance, and of the base resistor.Nothing really improved from that, apart from realizing that 10 k was a good value for the base resistor and lower values did not alloy as easily the avalanche (for the KSC1730).Then I added the very small coil L4 and a 200R (damping in JW note ) pot at the 10pF cap, followingclosely JW AN94 notehttp://cds.linear.com/docs/en/application-note/an94f.pdfI had neglected these before, and this was the origine of the overshoot. With JW setting, you can tune the overshoot with the potentiometer. I must say I am impressed to see the effect of such a ridiculously small coil.After all these modifications my device began too look very cluttered and probably not too much RF efficient.So I did it again from new, trying to keep all the output wires as short as possible. I changed also the input resistor from 1M to 560k to allow a little bit more current( this is still a max of about 0.2 mA). By the way, does anybody have any clue on what should be the best for this max current value ?The resulting device is much nicer. See the very little coil on the middle BNC plug. This plug is for easily hooking a BNC cable of various length.and after tuning the pot, I got a very clean pulse with a rising time of about 1.9 ns.The first plot is without any attenuatorand the second one with a home made minimal20 db16 db attenuatorThey both give similar results. In all cases, there was a 50R plug (with a T) at the scope.

author: HighVoltage
author_rank: Super Contributor
timestamp: December 06, 2015, 10:52:48 am
content: Nice findings and thanks for the explanations.

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 07, 2015, 08:51:34 pm
content: This is an update on the effect of the damping pot.To tune it, I used the old Tek TDS460A that was given to me some time ago. Its 400Mhz, but with only 100Ms/s. Not as responsive as the Rigol, but with a cleaner signal when it triggers properly.For R=0R, there is a lot of overshoot in the pulseand for R=200R, the damping is probably excessive, as in fig 11 of JW AN94.Fig11I choose finally R=75R which looked similar to fig. 13 of AN94

author: JacquesBBB
author_rank: Frequent Contributor
timestamp: December 07, 2015, 09:05:53 pm
content: I have no access to a very fast scope ( 3Ghz) where I could properly evaluate the rising time rt of this pulse generator, but I can make an estimate, just to see if it makes sense.The observed rising time ro should be ro = sqrt(rt^2 + rt'^2) where rt' is the rising time of the oscillo. In the TDS doc, we find rt' = 0.875 ns. So we should havert= sqrt (ro^2 - rt'^2),that is, for ro=1.176 ns, rt = 0.785 ns.which is quite possible, but additional measure with more precise scope should still be done.

author: rolfdegen
author_rank: Regular Contributor
timestamp: April 24, 2025, 04:45:57 am
content: I tried 19x 2N3904 transistors in my Avalanche Pulse Generator. None worked. I use power supply of +124 volts. I don't have any more voltage.I need a little bit help. ThanksMy circuit is like this.

author: rolfdegen
author_rank: Regular Contributor
timestamp: April 27, 2025, 08:19:58 pm
content: Continue here of my Avalanche pulser..https://www.eevblog.com/forum/blog/eevblog-306-jim-williams-pulse-generator/625/