Avalanche Pulser

Status
Not open for further replies.
G

ghost896

Junior Member level 3
Joined
Jul 3, 2022
Messages
26
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Visit site
Activity points
213
Hello everyone, I am researching on an avalanche based circuit with symmetric output. I have general information about avalanche transistors and how the marx generator works, but there are some parts that I do not understand in the circuit I shared.
The article mentioned that T1 changes the trigger voltage in positive and negative directions. I don't understand how this change happens. In this way, it puts the avalanche transistors in the upper and lower chains into avalanche mode.
Does anyone have a better explanation or understanding of the circuit?
 

Attachments

  • Design_of_Balanced_Output_High-Voltage_Nanosecond_.pdf
    492.1 KB · Views: 250

Sort by date Sort by votes
A variation of the Marx generator has spark gaps instead of transistors. These don't arc until voltage differential is great enough to overcome the gap.

The simulation below is among the menu of circuits in Falstad's animated interactive simulator. The yellow dots represent current bundles (or electrons, depending on how you wish to perceive it).

falstad.com/circuit

All capacitors charge in parallel, until their summed voltage is great enough to overcome all spark gaps, including the final output gap. Then all gaps fire simultaneously. The capacitors rapidly discharge in series.

We can suppose the authors of the document added a symmetrical section to provide a negative polarity to the final output pulse.

 
Last edited:

Upvote 0 Downvote
I don't have a problem with the material, sir, if I understand the logic, I can supply it. What I can't understand here
1) How come the transistor at the input can change the voltage in the bipolar direction like this?
2) How is the marx generator triggered when this change occurs?
 

Upvote 0 Downvote
Thanks for your explanation, actually I know the marx generator you shared and I see that avalanche transistor is used here instead of spark gap. Normally, when the avalanche voltage is exceeded, the transistor will switch to avalanche mode, discharging the capacitors in series. But the part I don't understand is how the trigger starts this job.
1) How come the transistor at the input can change the voltage in the bipolar direction like this?
2) How is the marx generator triggered when this change occurs?
 

Upvote 0 Downvote
Maybe you have problems to see the function of T1 clearly due to the slightly confused drawing arrangement. You should be able to see that T1 Vce is falling from about Vcc to 0 during trigger, respectively it supplies pulses of +/- VCC/2 to both halfcircuits. The polarity is so, that Vce of T2 and T7 is increased by VCC/2. Avalanche breakdown of T2 and T7 triggers the other transistors in each chain.
--- Updated ---

O.K., I see a problem. To operate with 1V input trigger voltage, T1 must be in avalanche breakdown. A large pulse will run back to the trigger generator, potentially destroying it. The trigger pulse would be better applied through a transformer.
 
Last edited:
Upvote 0 Downvote
Another possibility -

Making A Working Jim Williams Based Pulse Generator | Electronics Repair And Technology News Making A Working Jim Williams Based Pulse Generator

#post_contentMaking A Working Jim Williams Based Pulse Generator By Albert From Netherlands
jestineyong.com

Avalanche Pulse Generator

This is a single transistor avalanche relaxation oscillator using a 2N3904. It is capable of producing picosecond rise time pulses for use in measuring oscilloscope bandwidth. It is based on a design from [Kerry Wong] (linked in the LHS column), though with a power supply hacked together from...
hackaday.io

Avalanche Pulse Generator Design

This avalanche pulse generator is a great way to test your mettle as an Electronics Engineer. The challenge is to truly understand how each part of the design works. We certainly got a failing grad…
hackaday.com


Regards, Dana.
 

Upvote 0 Downvote
Thank you for your answer, sorry for my newness, I just really want to understand your logic so I'll ask where I'm confused if you'll excuse me.
I don't know if I understood what you said correctly, I will write my guess for this, if it is wrong, please correct it. As you said, when T1 is triggered, it opens and VCE drops to 0. You said it will generate Vcc/2 pulses. Is it /2 because if we assume R13 and R1 to be equal, it creates a voltage divider for C3 and C3 creates a VCE voltage for transistor T2 as it charges to VCC/2? If this voltage is greater than the avalanche voltage of the transistor, the avalanche chain will start. The upper chain may have a similar structure. However, I couldn't grasp the logic of the voltage being higher and lower than zero in sequence.
You're right, the trigger will encounter a large voltage spike. You suggested a transformer to prevent this. Is this a pulse transformer or a transformer with a N:1 turn ratio to step down the voltage?
 

Upvote 0 Downvote
I can understand that the circuit works in the logic of the marx generator, but I can't understand the triggering part.
Regards
 

Upvote 0 Downvote
When we short-circuit the base and emitter pins of the avalanche transistor, it actually turns into an upstream diode and this diode has a breakdown voltage. When this voltage is exceeded, the current starts to flow in the opposite direction and remains in avalanche mode until the potential difference is close to 0. Normally, I can see this in simulation, but when T1 is involved, I can't understand the trigger, sir. What exactly does the trigger here do? It opens transistor T1 and discharges capacitor C2. I'm not sure if the T1 is running in avalanche mode here either.
--- Updated ---

When I trace the circuit in the simulation it works fine. When I check the base current of T1 I see about - + 200 mA current. But how did this current come about? Since reverse current occurs, I think there may be a current flow from the collector to the base pin.
Easy peasy said:
Do you know how avalanche transistors work ?
Click to expand...
 
Last edited:

Upvote 0 Downvote
Easy peasy said:
ckt
Click to expand...
Sorry, I don't understand
--- Updated ---

Actually I have 3 questions with your permission sir
1) How is it that the small trig voltage we give to the input can reach larger voltages in the + and then - direction?
2) How does this + and - voltage start the avalanche chain?
3) In the simulation, even if I give the source voltage more than the avalanche voltage of the transistors (350V), the capacitor C2 does not exceed 350V.
Easy peasy said:
T1 is turned on then off to trigger the ckt - quite quickly - per the diagram supplied above ...
Click to expand...
--- Updated ---

FvM said:
The trigger pulse would be better applied through a transformer.
Click to expand...
4) You suggested a transformer so as not to disturb the trigger source. Is this a different transformer than 1:1? So if the trig voltage rises to 30 volts (it looks like it in the Simulation) is it to keep the trigger pin within the withstand voltage range?

Although I studied the circuit, I could not fully understand it. I would be very happy if you can help.
 
Last edited:

Upvote 0 Downvote
The circuit should not be viewed as having two outputs, one positive and one negative. It has one differential output, and the discharge should always be done between the two output connections, never to GND.
Othwerwise, strange things will happen around T1.

Going to the right in the circuit diagram the resistors will se a higher and higher voltage during the pulse. Resistors around T6 and T11 must handle the full output voltage (where do you find 10 kV resistors?), and that is not necessary if the capacitors are charged with resistors from the previous stage instead of having all resistors connected to VCC or GND.
 
Upvote 0 Downvote
Hello,
Actually I am trying to generate 2.5 kV, not 10kV. The breakdown voltage of each avalanche transistor is 350 (fmmtd417). Output voltage 350 * 2 * N
N : cascade number
*2 : because symmetrical
The voltage graph I get at the output:


Trig voltage:


But I do not fully understand the working logic of the circuit. What are your ideas?
 

Upvote 0 Downvote
First, do you need the differential output? If you want a single-ended high voltage to ground, I think you should use a more "standard" Marx generator. The generator in this thread is only OK if you do a differential discharge between the output terminals.

Before the trigger all capacitors and all transistors will have the VCC voltage across them.
This voltage should be slightly below the avalanche voltage of the transistors.
When the trigger pulse comes, the potential of C3 (not the voltage across it) will increase and the potential of C2 will decrease. This will increase the voltage across T2 and T7 to trigger the avalanche reaction, and this will lead to a chain reaction which causes the avalanche reaction in all transistors. At least in your simulation, the avalanche doesn't happen exactly at the same time in all transistors. This casuses the strange jumps you see at T1. Every avalanche in the upper row of transistors will cause a positive jump in the potential of T1 (all 3 connections) and every avalanche in the lower row will cause a negative jump.
Count the number of positive and negative jumps in your graph!
The circuit is not very kind to T1. The emitter and collector will jump up and down, and the input capacitor will try to hold the base at a somewhat fixed potential. This will probably cause the avalanche to always happen first in the upper row of transistors, both in simulation and in a real circuit.
When I look at the pictures in the article referenced in post #1, it seems to be two transistors instead of only T1, one for each half of the circuit. That is probably a better solution.
 

Upvote 0 Downvote
Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…