A few questions on JW pulse generator based on 2N3904 (and other transistors)

[JacquesBBB]

 I have made a minimal pulse generator based on
Jim Williams application note 47
http://cds.linear.com/docs/en/application-note/an47fa.pdf
in  the variant using a 2N3904 transistor, as mentioned several time on the Web, for example in
http://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 the
power supply, and just made the pulse generator part.
with the following setting (AN47, FIg. D1)

Q1 : 2N3904
C1 : 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 from
the 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 it
stabilised 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 changes
a 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).

[T3sl4co1l]

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

[JacquesBBB]

Thanks Tim for the feedback.

I have continued my experiments. First by adding another BNC connector to hook a coax,
after reading the  discussion of
https://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 overhead
due 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 DS1054z

rise 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.

[T3sl4co1l]

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 ?

When 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.

The thing which surprises me is the very good  computed bandwidth of the   Rigol DS1054z

rise time rt=1.2 ns  so BW = 0.35/rt= 292 Mhz.  Way over the announced  100Mhz.

Nope:
https://www.eevblog.com/forum/beginners/bandwidth-calculation-caveats!/msg807413/#msg807413
This 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

[JacquesBBB]

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.

I 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 ?

[T3sl4co1l]

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

[JacquesBBB]

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 transistor
is practically null, so the remaining impedance is the one of  the output  BNC connector, given by the associated resistor.

Is this correct ?

[T3sl4co1l]

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

[macboy]

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.

I 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 ?

Your 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 are clearly visible on your screen capture.

[JacquesBBB]

Your 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 are clearly visible on your screen capture.

I 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 on
another thread
https://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.

[JacquesBBB]

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, following
closely JW AN94 note 
http://cds.linear.com/docs/en/application-note/an94f.pdf



I 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 attenuator


and the second one with a home made minimal 20 db 16 db attenuator


They both give similar results. In all cases, there was a 50R plug (with a T) at the scope.

[HighVoltage]

Nice findings and thanks for the explanations.

[JacquesBBB]

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 pulse



and for R=200R, the damping is probably excessive, as in fig 11 of JW AN94.


Fig11


I choose finally R=75R which looked similar to fig. 13 of AN94

[JacquesBBB]

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 have

rt= 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.