Can rx be calculated from Vn?

In transistor specifications such as these:

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There is at several frequencies an Rn/50 specification. My understanding is Rn is a resistance that would give the same thermal noise density as a transistor?s input noise:

Vn = SQRT( 4 k T Rn B ) Where B is the bandwidth (area under the frequency response curve).

Equation 43 on page 11 of:

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Gives Vn as a function of rx (base spreading resistance), and collector current. Assuming given value for Rn is accurate (I am aware it might not be), is it valid to calculate an rx (base spreading resistance) value by setting the above two equations for Vn equal to each other and solve for rx?

After doing this and squaring both sides this would be:

4 k T Rn B = 4 k T rx B + 2 k T (VT/IC) B Where: VT is the thermal voltage k T / q . IC is the collector current Rn is specified at.

The solution for rx would be: rx = Rn - VT/(2 IC)

I would use the value of rx given for the frequency that is closest to the frequencies the transistor would be used at.

Another question: Why is it that in transistor spec sheets the given value for Rn has been divided by 50?

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Artist
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  • Rbb' or Rbb prime is the designation for base spreading resistance.
  • With Vn measures at different collector currents, one _might_ be able to calculate Rbb'. Depends on the data collected.
Reply to
Robert Baer

In that article I referenced:

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on page 9, in the 3rd paragraph under section "XVI The BJT Noise Model" rx is identified as the base spreading resistance. In articles and texts elsewhere I have seen both rx and rbb' used for base spreading resistance. rx is an alternative to rbb' for this.

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Artist

Good question. Setting the two Vn equations equal to each other seems valid to me. In regards to your second question, the normalized resistance rn is used to plot the noise circles on Fig 12 of the datasheet. rn is derived from Rn.

rn = Rn / Zo = Rn / 50

Thank you,

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Don Kuenz

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