Bipolar Transistor 1/gm

Defined as 1/gm=I/VT , Where I is current an VT is about 25mv. What hapens when I (current) is large - say 8ma? 1/gm should get big, but it must have an upper limit.

K
Reply to
HardySpicer
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"HardySpicer" schrieb im Newsbeitrag news: snipped-for-privacy@b9g2000prd.googlegroups.com...

Hello K,

Gm is limited by the emitter's series resistance Re, the internal base resistance Rb and the external generator's source resistance.. Re and Rb mostly depend on the size of the transistor. RE and RB are SPICE-parameters of transistors.

Maximum Gain of the Common Emitter Circuit

Gm = 1/(Vt/Ic+Re+(Rb+Rg)/BF)

Vt=26mV, RE, RB, BF see SPICE parameters BF is the DC current gain beta=Ic/Ib. It depends on Ic. Rg is the generator's series resistance

Best regards, Helmut

.model 2N3906 PNP(IS=1E-14 VAF=100 BF=200 IKF=0.4 XTB=1.5 BR=4

  • CJC=4.5E-12 CJE=10E-12 RB=20 RC=0.1 RE=0.1 TR=250E-9 TF=350E-12
  • ITF=1 VTF=2 XTF=3)
Reply to
Helmut Sennewald

A transistor's transconductance is not _defined_ as gm = Vt/Ie. One can _derive_ the _approximation_ that gm = Vt/Ie if one ignores various aspects of the transistor's behavior.

Depending on the transistor, 8mA may be "omigod" large, it may be "where was that signal again?" small, or it may be "yawn" perfectly 'normal'.

The transconductance gain of a transistor is _defined_ as the ratio of change in collector current to the (small) change of base voltage when a bunch of other assumptions are met. A good large part of it, in normal operation, is the Vt/Ie component. A normally-smaller part of it is the emitter's ohmic resistance, although at high currents that may start to dominate, or at least play a part.

gm = Vt/Ie is a pretty good approximation for back-of-the-envelope calculations, particularly when you take into account the fact that Vt varies with temperature (the 't' in Vt is 'temperature') and that Ie can never be pinned down with precision unless you're using some exceedingly fancy bias schemes.

If you want the whole story, get a good book on transistor circuit design. I think that AoE goes into this pretty thoroughly -- certainly when I skimmed the relevant chapters recently looking for some refresher numbers it all seemed to jibe with what I already knew.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
Reply to
Tim

can

e

Got loads of books, none of them tell you what happens under limiting conditions.

Reply to
HardySpicer

Sorry, that should read gm=3DI/VT

Reply to
HardySpicer

It'll be in the fine print.

At the high-current end you are (I think) mostly limited by the emitter ohmic resistance (and, possibly, by the thing going up in smoke). At the low-current end there's some gain-destroying process that happens, but I'm not sure what it is!!

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
Reply to
Tim
[snip]

Also current crowding at the edge of the emitter, aka IKF in your model card.

Carrier recombination due to contaminents in the semiconductor, aka ISE in your model card.

In the "good ol' days" I had to write my own models for BJT's. I explained the math to my son Aaron and he wrote me an executable (back in DOS days ;-) that took the raw lab data and spit out the model card parameters.

Modeling CMOS, particularly below threshold, is nasty and complex. For discretes I often cheat and take a full ASIC model card and scale L and W until I get a reasonable match. ...Jim Thompson

[On the Road, in New York]
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Reply to
Jim Thompson

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