# Electrical – quiescent point of two transistors BJTs

bjttransistors

Assuming that:

\$\beta=200\$

\$|V_{BE ON}|=0,6 \,V\$

\$|V_{CE SAT}|=0,2\,V\$

\$r_0=50\,k\Omega\$

\$c_{\pi}=4\,pF\$

\$c_{\mu}=4\,pF\$

How can I calculate the quiescent point of each of the transistors?
I need to know and DC voltages on all nodes and \$I_{CQ}\$ and \$V_{CEQ}\$ of both transistors. Can you help me, please?

Note: The two floors must be analyzed simultaneously, they can not be analyzed separately.

Without capacitors:

Small-signal equivalent circuit for mid frequency:

For the first transistor (Q1):

\$g_m=\frac{I_{C1}}{V_T}=\frac{1,446\times 10^{-3}}{25\times 10^{-3}}=57,84\,mS\$

\$r_{\pi}=\frac{\beta V_T}{I_C}=\frac{200\times 25\times 10^{-3}}{1,446\times 10^{-3}}=3457,81\,\Omega\$

\$r_0=50\,k\Omega\$ (teacher gives this number to us)

For the second transistor (Q2):

\$g_m=\frac{I_{C1}}{V_T}=\frac{5,50\times 10^{-4}}{25\times 10^{-3}}=22\,mS\$

\$r_{\pi}=\frac{\beta V_T}{I_C}=\frac{200\times 25\times 10^{-3}}{5,50\times 10^{-4}}=9090,9\,\Omega\$

\$r_0=50\,k\Omega\$ (teacher gives this number to us)

Calculation of \$R_{out}\$, \$R_{in}\$ and \$A_V\$:

For the first transistor:

\$R_{out}=r_0//R_1=\Big(\frac{1}{50\times 10^3}+\frac{1}{6,8\times 10^3}\Big)^{-1}=5985,92\,\Omega\$

\$R_{in}=R_g+R_4//r_{\pi}=1\times 10^3+\Big(\frac{1}{270\times 10^3}+\frac{1}{3457,81}\Big)^{-1}=4414,09\,\Omega\$

\$A_V=\frac{v_0}{v_i}=\frac{-g_m v_{\pi}R_{out}}{v_{\pi}}=-57,84\times 10^{-3}\times 5985,92=-346,23\$

For the second transistor:

\$R_{out}=r_0//R_2//R_L=\Big(\frac{1}{50\times 10^3}+\frac{1}{4,7\times 10^3}+\frac{1}{10\times 10^3}\Big)^{-1}=3005,12\,\Omega\$

\$R_{in}=R_1//r_{\pi}=\Big(\frac{1}{6,8\times 10^3}+\frac{1}{9090,9}\Big)^{-1}=3890,16\,\Omega\$

\$A_V=\frac{v_0}{v_i}=\frac{-g_m v_{\pi}R_{out}}{v_{\pi}}=-22\times 10^{-3}\times 3005,12=-66,11\$

For this circuit

simulate this circuit – Schematic created using CircuitLab

You can write these KVL equations:

$$V_E = V_{BE1}+I_{B1}*R_F$$

$$V_E = (I_{E2}-I_{B1})*R_E$$

And this give us

$$(I_{E2}-I_{B1})*R_E = V_{BE1}+I_{B1}*R_F$$

$$I_{B1}=\frac{I_{C1}}{\beta1}$$

we end up with this:

$$(I_{E2}-\frac{I_{C1}}{\beta1})*R_E = V_{BE1}+\frac{I_{C1}}{\beta1}*R_F \;\;(1)$$

And another KVL equation is:

$$V_{CC} - I_{RC1}*R_{C1} - V_{BE2} = V_E$$

where :

$$I_{RC1} = I_{C1}+I_{B2} = I_{C1}+\frac{I_{C2}}{\beta2}$$

Hence :

$$V_{CC} - \left ( I_{C1}+\frac{I_{C2}}{\beta2} \right)*R_{C1} - V_{BE2} = \left (I_{E2}-\frac{I_{C1}}{\beta1}\right)*R_E \;\;(2)$$

So, we have two equations and two unknowns. Because Q2 emitter current is:

$$I_{E2} = I_{C2}*\frac{\beta2 +1}{\beta2}$$

And if we assume \$\beta = \infty \$

We will end up the this:

$$I_{C1} = \frac{V_{CC} - 2V_{BE}}{R_{C1}}$$ $$I_{C2} = \frac{V_{BE}}{R_E}$$