Yes, I know: they have lower noise levels. What I mean is how is this achieved? And I guess there must be a trade-off as well (otherwise all transistors would be made as low-noise?)

  • \$\begingroup\$ I know one thing that could be helpful. If you reverse bias an NPN, you're unlikely to damage it, even a low noise one, but you are very likely to increase the noise level permanently. \$\endgroup\$
    – Thomas O
    Jun 13, 2011 at 18:26
  • 1
    \$\begingroup\$ @ThomasO That sounds like damage to me \$\endgroup\$
    – endolith
    Aug 25, 2015 at 19:17

3 Answers 3


These devices address burst, avalanche, flicker, and thermal noise.

Burst noise is the result of inconsistent ion deposition in the semiconductor fabrication process. It is reduced by escalating the selection/rejection criteria stringency, selling chips at different grades (eg: fast, slow); by changes in layout to better account for process variations; and by changes in the fab process itself to improve deposition homogeneity.

I think of Avalanche noise as amplified shot noise. Under reverse bias some electrons collide with the lattice in the PN junction depletion region with enough energy to form an electron-hole pair. Depending on the reverse bias voltage and junction characteristics, an avalanche breakdown may propagate, registering as a current spike. It is reduced by manufacturers by design and process changes to both increase the length of the depletion region (reduced field) and increase the energy required to free nearby electron-hole pairs.

Flicker noise, also called 1/f and pink noise, comes from "slow fluctuations of properties of ... materials"[1] during operation. As it is a sum of other sources of low frequency noise, it is addressed as these sources are identified.

Thermal noise is directly proportional to temperature, so any change that lowers local temperatures improves this figure. For example, changing die package for better dissipation; or layout changes to spread out local current hotspots.


Now that I've read more about this, properties of transistors that affect noise:


  • Low base spreading resistance rbb (and emitter bulk resistance Ree?) reduces (Gaussian, white) thermal noise.
  • High transconductance reduces base current, which reduces both (Gaussian, white) shot noise and (pink) flicker noise caused by modulation of base current.


  • Large transconductance reduces thermal noise caused by the channel resistance
  • Also, you can reduce thermal noise by connecting multiple JFETs in parallel, but when the parallel input capacitance exceeds the source capacitance, it will start degrading again by attenuation. So lower input capacitance helps here.
  • Gate leakage current causes shot noise. This is normally extremely small, but can become noticeable if JFET is hot.
  • Burst or popcorn noise caused by manufacturing defects, but should be minimal in modern devices

The other causes of noise are properties of the circuit, not the device.


  • Self, Small Signal Audio Design, Ch. 1
  • Horowitz & Hill, The Art of Electronics, 7.1 3 Transistor amplifier voltage and current noise
  • Hood, The Art of Linear Electronics, Ch. 16 Noise and Hum

These are discouraged for other purposes:

Avoid using “low noise” JFETs advertised for audio applications. These devices achieve their low noise performance at the expense of larger input capacitance and leakage current.

An experimenters approach to detecting the Schumann Resonances


Noise is variation due to external conditions. There are many different types of noise. Some can be reduced, some cannot. e.g. Thermal noise cannot be removed but electrostatic noise can be removed/reduced by using different types of material for packaging. This is a very common technique used. Also, by changing materials and doping, we can bring about reduction in noise. Yes, that will definitely affect the performance like reduce the amplification factor if used in an amplifier, etc but it is a compromise worth taking if you can compromise on search factors.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.