Given the following circuit with \$\beta=80\$, \$V_{BE(on)}=0.7V\$, \$V_A=\infty\$, for an input signal \$V_s=18cos(\omega t) mV\$, I would like to calculate the power dissipated in the transistor.


I began by calculating the thevenin equivalent voltage and resistance in order to solve for the base current as follows:

\$R_{thev} = \frac{R_1 R_2}{R_1+R_2} = \frac{30k\Omega \times 125k\Omega}{30k\Omega + 125k\Omega} = 24.194k\Omega\$

\$V_{thev} = V_{cc} \frac{R_2}{R_1+R_2} = 2.32V\$

The base current can be solved for by writing a KVL equation around the base-emitter loop and substituting \$I_e=(\beta+1)I_b\$:

\$-V_{thev} + R_{thev}I_b + V_{BE(on)} + R_eIe = 0\$

\$-V_{thev} + R_{thev}I_b + V_{BE(on)} + R_eIb(1+\beta) = 0\$

and solving for \$I_b\$:

\$I_b = \frac{V_{thev}-V_{BE(on)}}{R_{thev} + (1+\beta)R_e} = 25.1\mu A\$

Then \$I_c = \beta I_b = (80)(2.006 \mu A) = 2.01mA \$

The collector-emitter voltage can be solved for by writing the equation around the collector emitter loop:

\$V_{CE} = V_{cc} - R_c I_c - R_e I_e = 6.973 V\$

The power dissipated in the transistor is given by

\$PQ = I_{CQ}V_{CEQ}-\frac{1}{2}I^2_cR_c\$

And presumably, the power dissipated in the load is given by

\$PL = \frac{1}{2}I^2_cR_L\$

First I will need to solve \$I_c\$

I write te expression as:

\$I_c = \beta(\frac{R_{thev}}{R_{thev} + R_{ib}})(\frac{V_s}{R_{is}})\$

In the expression above \$R_{ib}\$ is the internal resistance of the transistor in the small signal model given by:

\$R_{ib} = r_\pi + (1+\beta)R_e\$ with \$r_\pi = \frac{V_T}{I_b}\$

\$R_{is} = Rs + R_i\$ (in this case no source resistance was indicated so it is just \$R_i\$)

And \$R_i = R_{thev}||R_{ib}\$

Therefore substituting the above expressions into \$I_c\$

\$I_c = \beta(\frac{R_{thev}}{R_{thev} + r_\pi + (1+\beta)R_e })(\frac{V_s}{R_{thev}||R_{ib}})\$

\$ = (80)(\frac{24194}{24194+40513})(\frac{V_s}{15148}) = 1.974\times 10^{-3} V_s\$

From this I calculate the power dissipated in the transistor to be

\$PQ = I_{CQ}V_{CEQ}-\frac{1}{2}I^2_cR_c = (2.01mA)(6.973V) - 0.5(1.974\times 10^{-3} \times 0.018)^2(2000) = 13.98mW\$

However the textbook answer gives \$13.0mW\$ rather. Where am I going wrong?

  • \$\begingroup\$ In your last formula, did you mean for the units to come out as mA or mW? \$\endgroup\$
    – The Photon
    Commented May 8, 2020 at 14:43
  • \$\begingroup\$ @ThePhoton my mistake, I have fixed it . \$\endgroup\$
    – Blargian
    Commented May 8, 2020 at 15:26
  • \$\begingroup\$ Isn’t the power dissipation for a transistor simply P=Ic*Vce? \$\endgroup\$
    – Leoman12
    Commented May 8, 2020 at 15:47
  • 1
    \$\begingroup\$ FWIW, simulation gives 13.037mW \$\endgroup\$ Commented May 8, 2020 at 16:19
  • 1
    \$\begingroup\$ @Blargian but you can ignore Rth also because we know that Rs = 0Ω thus vin = vpi \$\endgroup\$
    – G36
    Commented May 8, 2020 at 17:50

1 Answer 1


In short.

$$P_Q = I_{CQ} \times V_{CEQ} \approx 2\text{mA} \times 6.97\text{V} \approx 13.94 \text{mW}$$

And the AC component of a collector current is equal to

$$i_c = g_m v_{\pi} = 1/13\Omega \times 18 \text{mV} \approx 1.38 \text{mA}$$

Therefore power dissipated in the transistor is:

$$P_Q = I_{CQ}V_{CEQ}-\frac{i^2_c \left(R_C||R_L\right)}{2} = 13.94\text{mW} - \frac{1.38\text{mA}^2 \times 1\text{k}\Omega}{2} \approx 13\text{mW}$$

  • \$\begingroup\$ Thank you for your answer. One thing I'm still not understanding is the distribution of power between the transistor and the load. I read that as the AC input signal increases the distribution shifts from the transistor to the load. How is the AC component of collector current distributed between the collector resistor and the load itself? \$\endgroup\$
    – Blargian
    Commented May 8, 2020 at 17:19
  • \$\begingroup\$ @Blargian In this case, a current divider rule will do the job. $$i_{RL} = i_c \times \frac{R_C}{R_C + R_L} $$ or $$i_{R_C} = i_c \times \frac{R_L}{R_C + R_L} $$ \$\endgroup\$
    – G36
    Commented May 8, 2020 at 17:48

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