8
\$\begingroup\$

I'm attempting to calculate the base-spreading resistance of BJTs from manufacturer-provided SPICE models, which use the Gummel-Poon model. This Agilent publication (Internet Archive link here, if needed) describes how to calculate the base-spreading resistance from model parameters. I've copied the equation here:

$$ \begin{align} r_{\mathrm{bb}'} &= \mathrm{RBM} + 3(\mathrm{RB}-\mathrm{RBM}) \left(\frac{\tan(z) - z}{z\tan^2(z)}\right),\\ z &= \frac{\sqrt{1 + \left(\frac{12}{\pi}\right)^2 \frac{i_{\mathrm{b}}}{\mathrm{IRB}}} - 1} {\frac{24}{\pi^2} \sqrt{\frac{i_{\mathrm{b}}}{\mathrm{IRB}}}}, \end{align} $$ where \$i_{\mathrm{b}}\$ is base current, RBM is minimum base resistance at high current, RB is zero-bias base resistance, and IRB is current at medium base resistance.

I've applied these equations to several NPNs and compared the results with the measured results from The Art of Electronics (3rd Ed.), and am getting very different results. For instance, OnSemi provides a spice model for the 2N3904, where the relevant parameters are: RB=5.8376 IRB=50.3624 RBM=0.634251. Using these values with the above equations (and base current 100uA) yields an rbb of approximately \$5.8\,\mathrm{\Omega}\$. The Art of Electronics in Table 8.1a, however, gives this as \$110\,\mathrm{\Omega}\$.

That IRB looks high.

What accounts for the difference in rbb? Is this a valid method of estimation?

\$\endgroup\$
13
  • 3
    \$\begingroup\$ Matt, Have you pulled up and read, "The Effects of Distributed Base Potential on Emitter-Current Injection Density and Effective Base Resistance for Stripe Transistor Geometries," by Hauser in 1964, yet? That paper does cover your two equations. Gummel & Poon in their 1970 paper used a simplification. But by the time Spice was implemented, Hauser's earlier 1964 paper was included as a complication, providing a better approximation than Gummel & Poon allowed for. I can dig into the equations and development. But I'm less interested in discussing issues with some Spice program or AofE. \$\endgroup\$
    – jonk
    Commented Sep 4, 2022 at 21:21
  • 2
    \$\begingroup\$ When reading, note that Hauser uses an approach that focuses on dissipation of power to input power, which Spice pretty much doesn't care about. Spice is (and should be) focused on I/V relationships, not power. Getting I/V right helps get power right. But getting power right pretty much does nothing useful in getting the I/V relationship right. Also, Spice cares about having no discontinuities in the first 2 derivatives, plus good behavior at extremes of I and V. That's because such issues are likely during the iterative processes in Spice. Models get massaged to be Spice-friendly. \$\endgroup\$
    – jonk
    Commented Sep 4, 2022 at 21:37
  • 3
    \$\begingroup\$ I just did a search for you and found a paper that appears to cover much of what I has hinting at, above. In fact... I'm surprised how closely, and impressed at how well he covers what was in mind while searching. He does a better job than I could so now I'm not going to bother writing here, at all. Look up a paper in IEEE Microwave Magazine by Steve Maas called "Why I Hate Base Resistance," 2004. I think you will enjoy it! \$\endgroup\$
    – jonk
    Commented Sep 4, 2022 at 21:41
  • 1
    \$\begingroup\$ Again, best wishes, Matt. I like your questioning process! Very detailed. And it is very good to possess that kind of detailed curiosity. You will go far, I imagine!! I'd like to help, if I may. But I suspect you will get it entirely on your own! Your kind of curiosity about nailing down details is exactly the right stuff needed to succeed well and to be able to help others succeed in their own endeavours. I love seeing this kind of question. \$\endgroup\$
    – jonk
    Commented Sep 5, 2022 at 4:32
  • 1
    \$\begingroup\$ Another tidbit here. According to this textbook, the \$z\$ parameter in your equation is actually formed via a MacLaurin series approximation. I doubt it's what is causing such a massive difference, but thought I'd still point it out. \$\endgroup\$
    – Ste Kulov
    Commented Sep 6, 2022 at 18:35

0

Browse other questions tagged or ask your own question.