# Why does a Common Base amplifier gives non inverting output?

I am unable to understand why a Common base amplifier gives a non-inverting output. 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?

• 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! – Tony Stewart EE75 May 18 at 11:20

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.