Avalanche relaxation oscillator biasing resistor

I've tried the circuit in this article. It works very well and I've obtained 2ns clear pulses at 150 V (the main issue was to find the right avalanche voltage, which turned out to be 150-160V for my 2n3904 transistor).
While the basic principles of operation in this circuit is clear for me, I does not fully understand the role of the resistor R3. In what sense is it supposed to "bias" the transistor?
Also, is it possible to use a TVS or another avalanche type diode in place of the transistor?
 

Tom.G

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R3 provides a DC path for the Base current. If the Base is left floating, the transistor starts conducting as the Collector-Base leakage current starts flowing. This happens at a much lower voltage than the avalanche voltage and the transistor starts operating in the linear region, not in the negative resistance of the avalanche region.

There are two immediate consequences to this:
1) The Collector voltage never gets up to the avalanche voltage
2) The transistor operates as a resistor and just heats up

The end result is you don't have an oscillator any more.

Cheers,
Tom
 
Thank you for your answer. But why not simply connect the base to the ground without any resistor? that's the exact role of the resistor that I don't understand.
 

Tom.G

Science Advisor
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Probably to limit the current thru the base.

You want the current to go thru the Emitter resistor to generate an output signal. Inside the transistor, the Base is between the Collector and Emitter. Once the Collector-Base junction breaks down, the majority of the current will take the lowest resistance path back to the power source.

You could try it and see what happens, assuming you have a spare transistor around to maybe sacrifice.
 

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