# Differential amplifier exercise problem

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 $V_2=-2V$

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

$V_{outMAX}=V_{CC}=15V$

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

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

$V_{in}=V_1-V_2=0.01\sin(2\pi10t)+1.5V$

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

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=0.5-VBE-RE*(0.5mA)=-15V

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