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I'm a huge neophyte when it comes to electronics, and I'm trying to work out how to create a white noise generator for a project. I've been looking at some to base my examples off of. One of which is below from this site.

enter image description here

I see that Vin starts at the +14V. 1N759 must be the diode that is reverse-biased to produce noise. It is then amplified by two MAX2650 low noise amplifiers and then that amplified noise is outputted. Here are some questions I have about this circuit, pretty much entirely due to my lack of knowledge of electronics (of which I'm looking to better understand here):

  1. What is the purpose of R1 and its connection to ground?
  2. What is the purpose of the 5V output?
  3. What is the purpose of the capacitors here at all? What are they serving to do? Especially C3. It doesn't even look like outputted noise goes near C3.

My complete lack of knowledge is clearly showing. Anything you could recommend that I can read to quickly figure out stuff like this would be greatly appreciated as well.

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  • \$\begingroup\$ your question is not about white noise generators at all ... please change the title of your post to something like basic electronics questions \$\endgroup\$ – jsotola Jan 15 '20 at 0:03
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1) What is the purpose of R1 and its connection to ground?

1N759 is a zener diode (12V). R1 lets current flow through the zener diode (connection in reverse direction is the typical application for zener diods)

2) What is the purpose of the 5V output?

Power supply for the amplifiers

3) What is the purpose of the capacitors here at all? What are they serving to do? Especially C3. It doesn't even look like outputted noise goes near C3.

C1 and C2 are high-pass filters, while C3 stabilizes the power supply for the amplifiers.

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    \$\begingroup\$ C3 is called a bypass capacitor, and should always be installed. Picky circuit designers (of which I am one) would insist on one bypass cap per IC, unless the IC in question has multiple power pins, in which case you use one bypass cap per power pin. \$\endgroup\$ – TimWescott Jan 15 '20 at 3:07
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1) What is the purpose of R1 and its connection to ground?

R1 sets the current through the zener (~50ma in this case).

2) What is the purpose of the 5V output?

It's not an output. That's the conventional way of indicating a connection to (in this case) the 5V power source that powers the circuit- same as the 14V connection.

3) What is the purpose of the capacitors here at all? What are they serving to do? Especially C3. It doesn't even look like outputted noise goes near C3.

C1 isolates the input of the amplifier from the ~1.5 - 2 volts DC that is present at the input. This allows the amplifiers to see only the noise from the diode. C2 does the same thing for the output- the only signal that leaves the circuit is amplified noise- no DC voltage on the output. C3 is called a bypass capacitor. It gets rid of noise that may be on the 5V DC either from the power supply or generated by the amplifier circuit.

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    \$\begingroup\$ R1 won’t keep the input near 0 V. It will keep it near 1 V. The series impedance of the zener is 30 Ohms, which will act as a divider with the 30 Ohms to ground. So the 2 volts left over after the Zener’s 12 V is dropped, will divide to 1 V at the amp’s input. C1 then blocks this 1 VDC. \$\endgroup\$ – Blair Fonville Jan 15 '20 at 4:10
  • \$\begingroup\$ How does R1 keep the input near 1V when it was originally at 14V? For it to drop in voltage, it'd need to pass through the resistor so to speak to my knowledge, but passing through the resistor would cause it to go to ground? And why would C1 charging change the bias on the diode? Because current will flow in the forward bias direction if it is fully charged? \$\endgroup\$ – sangstar Jan 15 '20 at 4:45
  • \$\begingroup\$ If C1 and C2 both eliminate DC from the output, why are two needed? Wouldn't C1 do the job? \$\endgroup\$ – sangstar Jan 15 '20 at 4:51
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    \$\begingroup\$ Because of the presence of C1, the DC level on R1 is irrelevant. No matter what it is, the amps will not see it. Furthermore, it's not clear what "charging and changing the bias on the diode." means. And, since C1 isolates the amp input from the bias, it's not clear how this could matter anyways. \$\endgroup\$ – WhatRoughBeast Jan 15 '20 at 13:49
  • \$\begingroup\$ Why does the presence of C1 cause this? Where can I read more about this? \$\endgroup\$ – sangstar Jan 16 '20 at 3:55

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