Author Topic: A BJT avalanche question  (Read 369 times)

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Offline 6SN7WGTBTopic starter

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A BJT avalanche question
« on: September 06, 2024, 06:00:57 pm »
On the back of a BNC connector I have knocked together a simple 2N3904 avalanche pulse generator.

HV in to C of 2N3904 via 100k resistor, 7.5pF to gnd, then B of Tr to gnd via 10k and E output via a 68/100/68 Pi pad.

It's producing a decent pulse of around 900ps rise time and 10V pp. I'm driving it from a stabilised 0-160V DC PSU.

My puzzlement is that it only 'avalanches' with an HV between 129 and 132V quite repeatably.

I'd have expected it to need 100+V but not to have such a 'notch' window of operation. I was going to make the usual 1.5V cell to HV boost circuit, but it's not likely to work.

Why is this?  (Specific Tr I am using?).

 

Offline moffy

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Re: A BJT avalanche question
« Reply #1 on: September 06, 2024, 07:17:08 pm »
Are you talking about it pulsing then recovering, then pulsing again when the cap recharges i.e. a pulse train? If so then the recovery current could be quite sensitive, like an SCR that has just fired needs its current to drop below a threshold level before it can block again.
 

Offline 6SN7WGTBTopic starter

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Re: A BJT avalanche question
« Reply #2 on: September 06, 2024, 07:24:24 pm »
Yes, that's correct. So am I understanding correctly that this may be due to 'carrier multiplication'? i.e. the BJT becomes suddenly very sensitive to voltage?

But why the sudden cessation at a only slightly higher voltage?
 

Offline moffy

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Re: A BJT avalanche question
« Reply #3 on: September 06, 2024, 07:43:28 pm »
Yes, that's correct. So am I understanding correctly that this may be due to 'carrier multiplication'? i.e. the BJT becomes suddenly very sensitive to voltage?

But why the sudden cessation at a only slightly higher voltage?
At a slightly higher voltage the current through the 100k resistor exceeds the minimum recovery current. The low voltage is the zero current breakdown voltage the second is the maximum recovery current for your configuration. Might find that these voltages are temperature dependent also.
 

Offline T3sl4co1l

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Re: A BJT avalanche question
« Reply #4 on: September 06, 2024, 09:03:49 pm »
Try varying the base resistor.  Exact thresholds vary between parts.  I found 2N3904 to work more reliably with 4.7k.

For Rbe --> infty, breakdown is stable at Vceo.  For Rbe --> 0, breakdown is stable at Vcbo.  Somewhere inbetween, breakdown changes, and it is on that slope between where pulse avalanche breakdown occurs.

And yeah, it's pretty particular about voltage and current.  I think mine do it around 120V (10k supply resistor?).

Apparently the breakdown proceeds as a current filament (impact ionization, avalanche cascade), which injects carriers into the base, multiplying at a much faster rate than normal.  The excess conductivity discharges junction capacitance into it, causing intense local heating; this turns the filament site intrinsic, causing punch-through, i.e. the NPN junctions look like NIN, so basically, a short circuit.  This maintains while the external circuit discharges, and takes about 10µs to cool down and deionize.

At low currents (just below total breakdown), the waveform looks like this:



left is a full discharge, the rest is little bits and pieces simmering near breakdown but not fully discharging.

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 
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Online PCB.Wiz

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Re: A BJT avalanche question
« Reply #5 on: September 06, 2024, 11:01:50 pm »
On the back of a BNC connector I have knocked together a simple 2N3904 avalanche pulse generator.
I think I recall seeing xx918, xx2369 and BF570 devices also used for avalanche pulse generation. 
These have vceo = 15V and vcbo = 40V
 


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