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gm of BJT (Read 5840 times)
Geoffrey_Coram
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gm of BJT
Sep 08th, 2005, 9:05am
 
I'm working with a BJT (doesn't really matter what model), and trying to understand the value of gm.

According to first-order theory, gm = Ic/Vt.  gm is really defined as the partial derivative of Ic with respect to Vb.

Now, if I do a sweep of the base voltage, I can plot Ic/Vt and derivative(Ic) on the same graph, but they don't agree for high base voltage (off by a factor of 2).  Ic/Vt does match the "gm" reported by the device fairly well.  I've tried turning off self-heating, and I've checked that there isn't much base current (Vbe - Ib * Rb hardly differs from Vbe) and there also isn't much avalanche current.

What else needs to be considered?
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Geoffrey_Coram
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Re: gm of BJT
Reply #1 - Sep 8th, 2005, 10:25am
 
OK, Ib is small, but Ie isn't!  So, Vbe - Ib*Rb - Ie*Re is 28mV smaller than Vbe, which is pretty significant.  It's a small-geometry device.
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sheldon
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Re: gm of BJT
Reply #2 - Sep 17th, 2005, 1:42am
 
Geoffrey,

  The gm is in series with the Re of the transistor at high
currents so the effective transcondcutance is limited by the
Re of the transistor. In the earlier days of polysilicon
emitters, this effect was one of the concerns about
migrating to polysilicon emitters along with the stability
of the interface.

                                                     Best Regards,

                                                        Sheldon
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