# How would this circuit work as a white noise generator?

Got it from an electronics magazine from 1979.

From what I gather, the first transistor's collector and emitter are wired together so it acts as a diode, and the thermal noise from the junction is supposed to be amplified by the two stage common emmiter. I don't really get why the noise would flow through the first capacitor and how would the transistors amplify it, as they seemingly draw all current in the base through the bias resistor and not from the previous stage.

• Your going to have to name the transistors. It would be great if you could draw this with the circuit tool that is provided with the site. Nov 6, 2017 at 23:47
• I can't find what you're reffering to exactly, can you link it? Nov 6, 2017 at 23:53
• Edit your question and hit Ctrl-m, or edit and hit the circuit editor button Nov 6, 2017 at 23:57
• @RalphSnow or while you're in that falstad editor (which I love), just hit "t" on your keyboard and click with your mouse button and a text box will appear. Also, in the future, don't forget to add ground. It sets your reference point and tells us what your 0 voltage potential is. Nov 7, 2017 at 0:19
• I just copied the circuit exactly like it was in the magazine. I always thought in circuits like this ground was everything connected to the battery's negative output and the ground symbol only really used in schematics as a shortcut -- when it's impractical to draw a wire from the component to ground. Nov 7, 2017 at 0:27

The left-hand transistor is being used as a diode - not only that, as a reverse-biased diode operated in breakdown. This process is notably noisy, and the 100 nF capacitor blocks the DC level from the subsequent amplification. A lot of NPN signal transistors have base-emitter breakdown voltage in the range of 6 to 9 volts, so a supply voltage of at least 12 volts is a good idea for this approach to work, although with proper transistor selection you might get away with a 9 volt battery. Note that, for instance, a 2N3904 has a minimum breakdown voltage of 6 volts, but that is a minimum value and cannot be relied on.

• The article suggests BC548/542 transistors and a 9V battery. So if it's in breakdown, that means some current is flowing through it to ground, right? Why doesn't the noise signal flow the same way? Nov 7, 2017 at 0:16
• @RalphSnow - Much of it does. However, the 100 nF/1k/base-emitter/ground path also gets some of the current, which is how the signal is amplified. At most frequencies of interest, the impedance of this path is comparable to that of the diode connection, so some current flows through it. Nov 7, 2017 at 3:10
• That really cleared up things for me, thanks a lot! Nov 7, 2017 at 16:49

This noise generator uses the reverse base emitter breakdown of the transistor. Typically this is in the region of 5-7v for a typical BC169 or 2N3904 device.

The battery voltage must be at least a bit larger than the breakdown voltage of the junction.

The collector does not usually play a part in the operation so it could be disconnected.

The noise voltage may be only a few millivolts and so is amplified by the succeeding stages.

The capacitors couple the changes in voltage to the next stage and so change the bias current slightly. This change is amplified to appear as a larger signal at each collector.

Here is a more modern version from http://holdenc.altervista.org/avalanche/:

• Thanks, could you point me a resource to learn more about operational amplifiers, specially how they're wired? Could I use a 4558 as a dual stage amplifier, replacing the two transistors in the original circuit I posted? I just happen to have one lying around. Nov 7, 2017 at 0:21
• @RalphSnow TI doc SLOD006B is very good. Nov 7, 2017 at 3:31