1
\$\begingroup\$

could you explain what "eoc" and" eos" (edge of conduction and edge of saturation) is?

The problem says "Vbe = 0.7 at edge of conduction" , is that in cutoff mode?

\$\endgroup\$
  • 1
    \$\begingroup\$ Edge of conduction is right on the threshold between cutoff and forward-active (linear) mode. Edge of saturation is right on the threshold between forward-active (linear) and saturation (full-on switch). \$\endgroup\$ – Toor Mar 14 at 21:09
1
\$\begingroup\$

enter image description here The exponential boundary between saturated and linear forward active is called "Threshold of conduction"

BJT Saturation depends on collector voltage or the Δ(Vcb-Vbe)=Vce .

Also hFE drops towards 10% of it's max hFE when Vce=Vce(sat) @ If rated. The actual current ratio varies from device to device.

In datasheets, we define Vce(sat) at some currents Ic ( 1 or more levels) although 10:1 or Ic/Ib=10 is most common, sometimes it defined by OEMs at 20:1 and in super-beta transistors >> 500) it may be defined at 50:1.

As Vbe approaches Vbe(on), the BJT reaches the edge of conduction (EOC), and past that it enters the forward active (FA) region, where it becomes fully conductive. Henceforth, we have Ve = Vb – VBE(on) ≅ Vb – 0.7 V, that is, the emitter will follow the base, albeit with an offset of about –0.7 V ..aka the emitter follower mode.

ref

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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