# quiescent point of two transistors BJTs

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:

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

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

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

• No work shown, at all? How would you see approaching this problem? Are you able to write out nodal equations and solve them? What other simplifications may you apply to help? Are you allowed to use iteration? And what is a "floor"? (I'm unfamiliar with its use here.)
– jonk
Commented May 12, 2017 at 20:46
• @jonk I do not know if I can do this, but I thought I'd take all the capacitors off the circuit. But I can not find Kirchoff's equations. I'll put the circuit image without the capacitors. Could you please help me? Commented May 12, 2017 at 21:45
• @jonk I make the equivalent resistance between 4,7k (R2) and 6k8 (R1) in serie. Commented May 12, 2017 at 21:50

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:

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

• For $\beta = 200$ and $V_{BE} = 0.6V ; I_{E2}=I_{C2}$ My math software compute this result Ic1 = 1.44557mA and Ic2 = 0.547367mA
– G36
Commented May 15, 2017 at 17:09
• @CarmenGonzález As for emitter current for every transistor we can write this: $$I_E = I_B + I_C = I_B+I_B*\beta = (\beta +1)I_B$$ and $$I_C = I_B*\beta$$ so we have $$\frac{I_E}{I_C} = \frac{(\beta + 1)I_B}{\beta I_B} = \frac{\beta + 1}{\beta }$$ and this why we get this $$I_E = I_C * \frac{\beta + 1}{\beta}$$
– G36
Commented May 15, 2017 at 18:09
• Your small signal model looks good, but notice that RL and transistor emitter are connected to ground also. And you don't know haw to calculate the Rin, Rin and the gain?
– G36
Commented May 16, 2017 at 17:07
• Yes, RL should be in the same position as R2
– G36
Commented May 16, 2017 at 17:42
• Looks good, but signal source resistance (Rg) don't belong to amplifier Rin. Also, the unit of a voltage gain is not omega. And to be able to find Rin, Rout and Av. Simply connect this two models together and solve for Av, because Rin_T1 and Rout_T2 will stay the same and Rin of the hole amplifier is Rin_T1.
– G36
Commented May 16, 2017 at 19:38

How can I calculate the quiescent point of each of the transistors?

generally it is very difficult to calculate the dc operating points of a dc-coupled multi-stage circuit.

in this particular case, assuming sufficiently high beta, the voltage drop over R3 is Vbe of Q1. From that, you can calculate Q2's Ie/Ic.

and Q1's collector sits at 2 Vbe. from that, you can calculate its Ic and then Ib....

from there, go back and double check to make sure that there isn't lots of voltage drop on R4....

• I did not realize what equations you want me to solve. Could you put it in mathematical format please? The beta they give in the statement, I suppose we'll have to use it for something. My question is: - Can I remove the capacitors and everything in series with them to do the polarization? -And how do I now write Kirchoff's Laws and what are the equations I have to solve in order to discover the potentials at each node, the quiescent points and the currents? Commented May 12, 2017 at 22:29

The easiest approach to hand analysis on a problem like this is to appreciate how the circuit is going to behave, so you can make reasonable approximations. Transistor Q2 is operating as an emitter-follower, and thus has an input resistance of $\beta R_E$ which in this case in negligible compared to the 6k8 resistor. Hence Q2 appears as a high-impedance buffer.

Redrawing the schematic as the following,

simulate this circuit – Schematic created using CircuitLab

You can write out the equations for $I_C$ and $I_B$ then equate them as $I_C =\beta I_B$.

Edit:

$$I_C = \dfrac{V_{CC} - V_x}{R_c}$$

$$I_B = \dfrac{V_x - V_{BE} - V_{BE}}{R_f}$$

$$I_C = \beta I_B$$

Solve for $V_x$ then you have the DC-operating point (quiescent point).

• My questions are: - Can I remove the capacitors and everything in series with them from the polarization? -And how do I now write Kirchoff's Laws and what are the equations I have to solve in order to discover the potentials at each node, the quiescent points and the currents? Commented May 13, 2017 at 0:34
• @Carmen González yes, at DC the capacitors are open-circuit. Added equations as an edit. Commented May 13, 2017 at 0:47
• My teacher does not want the equations made with a OPAMP. It wants the equations of Kirchhoff's Laws, for the transistors. (Without the capacitors supposedly for polarization). I have doubts about what equations I have to write to determine the potential on all nodes, currents, and the quiescent point. Could you help me and write the equations in your answer but with the figure of the transistors, please? And you could write the formulas to calculate the quiescent points and the currents, please? Commented May 13, 2017 at 1:01
• @Carmen González The opamp dosen't show up in the equations (its symbolic of how Q2 is behaving, as a buffer) the base current isn't loading the collector node. Rf = R4, Rc = R1, Vx is the voltage at the collector of Q1. Commented May 13, 2017 at 1:10
• But I really really appreciated if you could do the equations with the figure of the transistors, because with OPAMP I do not understand much. Besides that I will have to calculate the other voltages at the transistor nodes. Please, I need your help, it's very precious. Commented May 13, 2017 at 1:29

Your second circuit is not equivalent to the first. Vcc should not be connected directly to Q2 base.

Yes, you can open circuit the capacitors for DC analysis.

Just use the fist circuit and start by assuming it is biased in the linear region.

since IC=200*IB

Vcc-(200*IB1)*6.8K-0.6V-IB1*270k-.6V=0

This is a loop equation for VCC through the collector resistor of Q1, the base emitter voltage of Q2, the base resistor of Q1, and the base emitter of Q1. From this you should be able to calculate all the rest.

I almost forgot, this assumes the base current of Q2 is negligible compared to IC1. It is the standard way to do this hand analysis. If you are not allowed to make this assumption, you must put in a large signal model for each transistor and write loop or node equations. The simplest large signal model is a 0.6V battery form B-E and a controlled current source with gain=beta from C-E.

Good luck to you.

• I realized the equation you made. The teacher gave permission to make approximations as long as they are justified. Why do we consider that the current from collector 1 enters the base of Q2? Should not the current of the base always be zero? And what are the approximations we can make in your equation? How do I discover the other currents and the other potentials? Commented May 13, 2017 at 12:12
• No, we are considering the base current of Q2 to be 0 and the collector current going out the emitter of Q1. We can do this so long as IC1 is big with respect to IB2. Solve the equation I gave you for IB1. Multiply this by 200 to get IC1. Use IC1, Vcc, and R1 to caluclate the voltage at C1. Subtract 0.6V from the voltage at C1 to get Q2 emitter voltage. Use the Q2 emitter voltage and R3 to get IE2. Assume IE2 = IC2. Use IC2, RC2, and Vcc t calculate the voltage at C2. Commented May 13, 2017 at 16:01