I am preparing for an exam and I am not sure if my solution is correct for this problem:

We have a differential amplifier with 2 NPN transistors (T1, T2). T1's collector is directly connected to the power supply \$V_{CC}=15V\$, while T2's collector is connected to the output and \$R_C\$. T2 base's potential is constant \$V_2=-0.5V\$. After changing \$V_1\$ - T1 base's potential - to \$V_1=0.5\sin(2\pi1000t)-0.5V\$, the peak-peak voltage of the output signal is equal to \$1V\$. Transistors' emitters are connected to \$R_E\$, which is connected to power supply \$V_{EE}=-15V\$. \$I_E=0.5mA\$

  1. Draw the circuit
  2. Calculate values of \$R_E, \ R_C\$
  3. After changing \$V_1\$ to \$V_1=0.01\sin(2\pi10t)-0.5V\$, calculate the maximum and the minimum value of the output voltage and draw the output signal
  4. Calculate T2 collector's potential, if we change \$V_2=-2V\$

1.enter image description here

No. 2

\$V_{diff}=V_1-V_2=0.5\sin(\omega t)-0.5-(-0.5)=0.5\sin(\omega t)\$ We know that \$A_{ in}=0.5V > 50mV\$, hence we knot that the maximum difference \$\Delta U=I_E\cdot R_C=1V\rightarrow R_C=\frac{1V}{0.5mA}=2k\Omega\$.

If we change voltages to \$V_1=V_2=0V\$, then we can easily calculate \$R_E\$:

\$0-0.7-I_E\cdot R_E=V_{EE}\rightarrow R_E=\frac{-V_{EE}-0.7}{I_E}=28.6k\Omega\$

No. 3


\$V_{outMIN}=V_{CC}-I_E\cdot R_E=14V\$

and output voltage will oscillate around \$14.5V\$

Circuit's gain is \$G=-\frac{g_m R_C}{2}=-10V/V\$

So, I think, the output voltage is \$V_{out}=14.5-10\cdot0.01\sin(2\pi10t)V\$

No. 4


I don't know, maybe \$V_{out}=14V\$?


1 Answer 1


i think you mistake, as the DC voltage of Vi1=Vi2=-0.5Volt, your circuit is symmetric, the DC current of each side of the circuit is 0.25mA. so you can calculate RE:


RE=27.6 K ohm

re = 1/gm = VT/IC = 0.1Kohm

iac = (0.5.sinw)/(1/gm + 2.RE)=(0.5 Volt)/(55.2Kohm + 0.1Kohm)= 9 mA

RC.iac = 1 Volt

RC = (1 Volt) / (9mA) = 0.11 Kohm


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