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I solved the below problem with the think that the emitter current from Q1 is divided to two paths. One is from emitter to 8.5k resistor, and the other is to the base of the Q2. However, from the answer sheet, I found that the emitter current from Q1 does not flow through the 8.5k resistor.

I thought because the base resistance of Q2 is high enough so that we cannot neglect 8.5k resistor. Can anyone explain why am I wrong?


All the answers I found solved problem like the below, ignoring the current to the 8.5k resistor. However, assuming the emitter voltage of Q1, V_E1, (β+1)I_B1 = V_E1 / 8.5k + I_B2

What could be the problem in this approach? Thx in advanceenter image description here

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    \$\begingroup\$ If the voltage exists across 8.5k resistor the current for sure will flow through it. \$\endgroup\$
    – G36
    Commented Oct 10, 2020 at 15:51
  • \$\begingroup\$ That answer is wrong though without thinking through the specifics it might be barely close enough for practical purposes. You might try simulating the circuit as a cross-check for the theory - eg, while you're required to give an algebraic answer, you can use simulation results as a cross check to identify conceptual mistakes. \$\endgroup\$ Commented Oct 10, 2020 at 15:57
  • \$\begingroup\$ @ChrisStratton Thank you. I thought the base resistance r_pi2, is big enough to just ignore the 8.5k resistance. Therefore, significant voltage division would take place at the emitter of Q1. There should be a big difference between the above answer and my thought. Am I thinking right..? \$\endgroup\$
    – Lylac
    Commented Oct 10, 2020 at 16:00
  • \$\begingroup\$ r_pi is a small-signal differential resistance. So you can't include it in the DC situation. The 8.5K resistor will simply increase the Ie1 current from Ib2 to Ie1 = Ib2 + Vbe2+(Ie2*1k) \$\endgroup\$
    – G36
    Commented Oct 10, 2020 at 16:31
  • \$\begingroup\$ @G36 Oh now I got it. I just took r_pi as granted even at DC condition. I really thank you for letting me find out my misunderstanding \$\endgroup\$
    – Lylac
    Commented Oct 12, 2020 at 5:22

2 Answers 2

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Solving this circuit:

enter image description here

Is not an easy task because we have to solve these simultaneous equations

$$V_{CC} = (I_{E1} + I_{C2})R_C + I_{B1}R_B + V_{BE1} + V_{BE2} + I_{E2}R_{E2} $$

$$ V_{BE2} + I_{E2}R_{E2} = (I_{E1} - I_{B2})R_{E1}$$

Using this additional knowledge about "internal" BJT's behavior.

\$I_C = \beta I_B\$

\$I_E = I_B + I_C = I_B + \beta I_B = (\beta+1)I_B\$

\$\frac{I_C}{I_E} = \frac{\beta I_B}{ (\beta+1)I_B} = \frac{\beta}{\beta+1}\$

In fact, if we remove \$R_{E1}\$ from a circuit, this "new circuit" can be solved without any problem

enter image description here

$$V_{CC} = (I_{E1} + I_{C2})R_C + I_{B1}R_B + V_{BE1} + V_{BE2} + I_{E2}R_{E2}$$

and because now:

\$I_{E1} = (\beta_1+1)I_{B1} = I_{B2}\$

and

\$I_{E2} =(\beta_2 + 1)I_{B2}\$

\$I_{C2} =\beta_2 I_{E1}\$

We have

$$I_{B1} = \frac{V_{CC} - 2V_{BE}}{R_B + (\beta_1 + 1)(\beta_2 + 1)(R_C + R_{E2})} = \frac{3.3V - 1.4V}{2k\Omega + (101*101)(750\Omega + 1k\Omega)} = 106.4nA$$

And

\$I_{R_C} = I_{E1} + I_{C2} = I_{E1} + I_{E1}\beta_2 = (\beta_2 + 1)I_{E1} = (\beta_2+1)(\beta_1+1)I_{B1} = 1.086mA\$

And finally:

\$V_{OUT} = V_{CC} - I_{R_C}R_C = 2.486V\$

Now we can bring back the \$R_{E1}\$ and see how \$R_{E1}\$ influence the circuit.
We see that the \$R_{E1}\$ resistor will simply increase the \$I_{E1}\$ current. Because now the \$I_{E1}\$ current is equal to \$I_{E1} = I_{B2} + I_{R_{E1}}\$

And in your example :

\$I_{R_{E1}} \approx \frac{V_{B2}}{R_{E1}} \approx \frac{1.086V}{8.5k\Omega} \approx 210 \mu A\$

And this will increases the \$I_{B1}\$ to \$ 2.08 \mu A\$ and the \$Q_1\$ collector currrent to \$I_{C1} = 208 \mu A\$

Therefore

$$V_{OUT} \approx 3.3V - (210 \mu A + 1mA)750\Omega \approx 2.4V $$

As you can see not a big difference in output voltage value, so we can end-up this iterative process of finding the new operating point.

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Simple analysis of your circuit shows that if Q(1?) is conducting, regardless of Q2, then current must flow in its emitter resistor, which is the 8500 ohms you refer to.

Q2 also conducts, meaning that the voltage across that 8500 ohms which is equal to Vbe(Q2) causes a current to flow in the 8500 ohms equal to Vbe(Q2)/8500.

Vbe(Q2) ~= 0.7V so the current in the 8500 ohms is ~0.7/8500 = 82uA

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  • \$\begingroup\$ Ah, no. There is also this 1k resistor in series with the B-E diode of Q2. \$\endgroup\$ Commented Oct 13, 2020 at 9:31
  • \$\begingroup\$ Bother ! I'd missed that. Will have to calculate again. \$\endgroup\$ Commented Oct 13, 2020 at 9:44

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