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schematic

simulate this circuit – Schematic created using CircuitLab

I'm trying to draw the large transfer characteristics for this common base BJT below. I redrew figure 1 to look like figure 2 so it can help better visualize what's going. I want to plot Vo Vs Vi

I used KVL to come up with this equation -Vcc + RLIc + Vo + Vi = 0

So then Vo = Vcc -RLIc - Vi

Ic = Is x e(Vbe/Vt)

Vo = Vcc - [Is x e(Vbe/Vt)RL) - Vi

My confusion is that the base is grounded. Does this now mean that Vbe = -Vi? Or is it that Vbe = 0 since the base is grounded?

If Vbe = 0 then Vo = Vcc - (IsRL) - Vi

Or Vbe = 0 -Vi then Vo = Vcc - [Is x e(-Vi/Vt)RL] -Vi. Can you please tell me what's right here? Or maybe I get it all wrong then I will need some guidance. Thanks!

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  • \$\begingroup\$ Please use Mathjax to make your question more readable. Note on EE we start and end inline Mathjax with \$ instead of just $. \$\endgroup\$ – The Photon Sep 19 '16 at 2:11
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Does this now mean that Vbe = -Vi? Or is it that Vbe = 0 since the base is grounded?

\$V_{be} = V_b - V_e\$

So yes, in your diagram, \$V_{be}\$ is \$-V_i\$.

\$V_{be}\$ is only equal to 0 if the base and emitter are at the same potential.

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  • \$\begingroup\$ Yes, I thought so but I wasn't quite sure. Thanks! \$\endgroup\$ – Patrick Sep 19 '16 at 2:16

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