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This is a general question. Suppose I have a circuit like below from CircuitDigest:

enter image description here

My supervisor said that about the circuit that..

[The] circuit does actually involve an amplifier, even though it’s not super obvious: the noisy current in the Zener diode is amplified by the transistor, and R1 effectively converts current fluctuations into voltage fluctuations.

I had no idea it involves an amplifier - I don’t see one here. I see a transistor only. He also says the resistors effectively converts current fluctuations and voltage fluctuations.

What can I read to be able to make deductions such as this? Any recommendations (textbooks and the like) would be appreciated, as well as any useful links.

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  • \$\begingroup\$ I don't know how much you know. But I'll suggest something and see if it works for you. The reverse-biased zener's breakdown voltage has a very sharp knee, so the voltage won't vary much even when the current varies a lot. So the shot noise (and flicker noise, separately?) may yield significant current variations. This leaves current fluctuations which are amplified by the BJT \$\beta\$ to yield voltage changes across R1. These changes yield current changes in R2, but they are absorbed by C1 (hopefully) which stabilizes the bias point of the BJT. \$\endgroup\$ – jonk Feb 4 at 7:29
  • \$\begingroup\$ You see the transistor ... you should be able to see it's configured as a common emitter amplifier with the output voltage developed across R1. \$\endgroup\$ – Brian Drummond Feb 4 at 14:05
  • \$\begingroup\$ @jonk What do you mean by a “sharp knee”? Do the current variations (brought on by diode breakdown) cause voltage variations simply by Ohm’s Law? Why does R1 effectively convert current fluctuations to voltage fluctuations? What would happen if it were not there? What is the purpose of R2? \$\endgroup\$ – sangstar Feb 4 at 22:37
  • \$\begingroup\$ @sangstar Sorry, I could have written better. By "sharp knee" I actually meant to point out that it's been designed to hold a fairly flat (changing only very slowly) voltage difference between its leads when contrasted (plotted) against the current through it. It's a zener, after all. And if R1 weren't there the circuit wouldn't function. I don't understand why you don't understand that detail. That's flummoxing, to me. I guess there are a lot of reading and understanding steps to get you from where you are to where you need to be? \$\endgroup\$ – jonk Feb 4 at 22:48
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I think any basic electronics book will get you to where you want to be, things like the art of electronics by Horowitz, Razavi or Sedra/Smith

to be honest though I think you can figure this out with a circuit simulator to get an intuition of what is going on.

what does a resistor do? remember ohm's law? if there is a current through a resistor, the shape of the voltage will be identical since V=IR where V is voltage, I is current and R is the resistance value.

the amplifier way of thinking... If you think of the transistor as an ideal switch controlled by voltage, any small current that the diode drives into the transistor is going to make the 26V from DC line be fully applied(or partly depending on saturation) onto the resistor.

I would not call it an amplifier myself since I don't understand what is the input to have a ratio, this is just me not understanding this particular circuit, not a concept issue.

To understand what your instructor meant, you need to understand Ohm's law, transistor operation, diode operation, Kirchhoff's Voltage and current laws I would say.

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