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I am unable to understand why a Common base amplifier gives a non-inverting output. enter image description here My reasoning so far: Let's take a look at the Common Collector (Emitter Follower) Amplifier. Here \$V_e = V_b -0.7\$. So on differentiating we get \$ d{V_e} = d{V_b}\$ which clearly implies that output is non-inverting. Similarly in the case of Common Emitter amplifier \$ V_{out} = V_{cc} - I_cR_c \$. which implies that \$d{V_{out}} = -d{{I_c}R_c}= \beta d{I_b}R_c \$. The negative sign implying that output is inverted. But in the case of a common base amplifier, I can't figure out with similar reasoning. When the emitter voltage increases then \$I_e\$ increases which in turn increases the \$I_c\$ and the situation again becomes like that of a CE amplifier. Please provide an explanation of why this is the case and where am I wrong?

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  • \$\begingroup\$ Because Ie = Ic+Ib must go in the same direction! Whereas when base Vbe rises then a rise in Ic means a drop in Vce! \$\endgroup\$ – Tony Stewart EE75 May 18 at 11:20
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No. When \$V_E\$ increases, \$I_E\$ decreases, because \$V_{BE}\$ decreases. Any extra current that flows through \$R_E\$ is supplied by the signal source, not the transistor.

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  • \$\begingroup\$ I_e decreases because its an exponential function of V_be. Am i right? \$\endgroup\$ – shahrOZe May 18 at 11:33
  • \$\begingroup\$ Yes, that's correct. \$\endgroup\$ – Dave Tweed May 18 at 11:35
  • \$\begingroup\$ Thanks for the answer. Now i have understood. \$\endgroup\$ – shahrOZe May 18 at 11:36
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    \$\begingroup\$ The emitter must be driven with a voltage source as low as re for max efficiency, unity current gain and then impedance ratio yields voltage gain. \$\endgroup\$ – Tony Stewart EE75 May 18 at 11:46

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