In a common emitter amplifier with \$h_{re}\$ and \$h_{oe}\$ unimportant, and a load $$R_{CL} = R_C//R_L$$ I'm studying the gain voltage $$ A_v = \frac{v_{ce}}{v_g} = \frac{v_{ce}}{v_{be}}$$ So $$i_c = h_{fe}i_b = h_{fe}\frac{v_{be}}{h_{ie}}$$$$ v_{ce} = -R_{CL}i_c$$ Finally $$A_v = -R_{CL}\frac{h_{fe}}{h_{ie}} $$ My doubt is about \$ v_{ce} = -R_{CL}i_c\$. Why is there a minus?

  • \$\begingroup\$ Can you think of if the output voltage goes positive or negative, when input voltage goes positive? \$\endgroup\$
    – Justme
    Oct 8, 2021 at 13:04
  • \$\begingroup\$ I do not understand \$\endgroup\$
    – Simone
    Oct 8, 2021 at 13:35
  • \$\begingroup\$ When Ib increases, how does it affect Ic, and how does Ic affect Vbe? \$\endgroup\$
    – Justme
    Oct 8, 2021 at 13:47
  • \$\begingroup\$ With the current arrows in the directions shown, \$v_g\$ is positive with respect to ground, and \$v_{ce}\$ is negative with respect to ground. \$\endgroup\$
    – Chu
    Oct 8, 2021 at 14:40
  • \$\begingroup\$ The negative symbol indicates that the output signal is inverted. \$\endgroup\$
    – Audioguru
    Oct 8, 2021 at 16:12

1 Answer 1


Notice that small letters are used for voltages and currents in your equations. This convention (using small letters vs capital letters) is indicative of the method used in your analysis. Namely, this fact indicates: this analysis uses small signal approximation for solving your circuit (CE amplifier).

Now, small signal approximation is not necessarily a trick with the signal represented as a sum of the large DC component and the small AC (harmonic) component. At its source, the small signal approximation is used to analyze the non-linear circuit behavior when currents and voltages are close to those of the circuit biased to its quiescent point.

In this approximation, the voltage across the transistor is written as \$V_{ce} = V_{Q\_ce} + v_{ce}\$ and the base-emitter voltage is \$V_{be} = V_{Q\_be} + v_{be}\$, where \$V_{Q\_ce} \,, V_{Q\_be}\$ are the collector and base voltages at the Q point and \$v_{ce} \,, v_{be}\$ are small deviations from these quiescent values. The small signal component \$v_{be}\$ of the base voltage is \$v_g\$ in your drawing, but forget for a moment the \$v_g\$ designation, because it labels the AC voltage source and can be interpreted as the amplitude of a sine wave. The analysis is more straightforward when one considers voltage/current small changes and not AC waves.

The circuit is drawn in such a way that \$i_b\$ and \$i_c\$, small deviations of the base current and the collector current, are positive. But \$V_{ce}\$ decreases with a \$V_{be}\$ increase for the CE amplifier: a higher base voltage increases the current through the load. The upper terminal of the load (not shown in this drawing) is tied to the supply voltage source and its potential is constant. The higher voltage drop across the load (due to a greater current thru the load) lowers the potential at the lower terminal of the load in order to cause the difference between the upper and lower load terminals to increase. As a result, \$v_{ce}\$, the small deviation of the collector voltage (not shown in this drawing), is negative. Remember the collector is connected to the load at its lower terminal. This explains a minus sign in the formula \$v_{ce} = -R_{CL}i_c\$. No contradiction with the Ohm law: the formula is written for value changes, not for the values themselves.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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