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For this circuit i need to find the Vi voltage to get Q2 in active mode.

The result i got is not close to the values in the simulator, probably theres something wrong with the circuit analisys.

Where´s the mistake ?.

Easy to see that:

Vce4 = 0.7v
from 20 - 1kI - Vce4 = 0 and vce4 - Vcb4 - 0.7 = 0 if Vcb4 = 0

Then:

20 - 0.7 = 1kI having 19.3v in R4 with 0.0193 A.

Also:

      The Vce3 = 19.3  from 20 - Vce3 - 0.7 = 0.

Thats the easy part, now i know the Q3/Q4 is a current mirror. So, the I current I i just calculated is aprox to Ic3.

If i suppose Q1 OFF:

     Ie2 = 0.0193A; that means Ib2 + Ic2 = 0.0193A. 

If Q2 is in active mode:

     Ie2 = Ib2(B + 1) , Ib2 = Ie2/(B+1)  if  B=100   

then Ib2 = Ie2/101 and Ic2 = Ie2/(1+ 1/B):

     Ic2 = 0.0191 A. 

That sounds good but when Q2 is saturated. Vce2 = 0.2v and Vcb2 = -0.5v:

       20 + 10 - 19.3 - 1kIc2 - 0.2 = 0

       Ic2 = 0.0105 A.

Q2 is in saturation mode.

From Ie2 = Ib2 + Ic2, Ib2 = Ie2 - Ic2 = 0.0088 A.

Now:

If i suppose Q1 is in active mode having in mind that -0.7v + Vce1 = 0 Vce1 = 0.7v

With

       Vi - 100kIb1 - 0.7 - 19.3 + 20 = 0
       Vi = 100kIb1. This sounds weird. 

Is Q2 is in active mode then: Ic2 = Ib2; 10.7 = Vce2 + 1kIc2; Ie2 = Ic2(1+1/B)

From Ic2(sat) = 0.0105 A that would mean: Ib2 = 0.000105A

Using:

  Ie2 + Ie1 = 0.0193A --> Ie2 = 0.0193 - Ie1  and Ie2 = Ic2 + Ib2
  Ie2 = 0.010605 then  Ie2 < 0.010605  for 0.010605 < 0.0193 - Ie1
  Ie1 > 0.008695 A.

Ib1 = Ie1/(B+1) and Ib1 (B+1) > 0.008695

  Ib1 = 8.6x10^-5 A

Then:

      Vi/100k  = Ib1 
      Vi > 8.6 V  for Q2 in active mode.

I simulated the circuit and for Vi = 8v the Vce2 = 0.4V. Also the current Ice3 is always 24.2mA, and The Vce3 is not 19.3V, its aprox to 18.4V for low Vi voltages. But that would mean the Vbe2 is not 0.7V.

Thanks for your help and time.

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    \$\begingroup\$ That's a nasty looking schematic. \$\endgroup\$ Nov 1, 2017 at 14:15
  • \$\begingroup\$ 1) Q3 and Q4 are not the same type, since this is a current mirror, they should be. 2) Ic_Q3 should be roughly around 20 mA so that's OK. 3) Vi = 8V, yeah, then Q1 will take all the current so there will be no current left for Q2. Try again but make Vi = 0 V and then Vi = 10 mV. Conclusion you're overdriving the diff. pair Q1, Q2. \$\endgroup\$ Nov 1, 2017 at 14:19
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    \$\begingroup\$ You have too much current through the current source - it is saturating Q2. Increase R4 or reduce the value of V4. \$\endgroup\$ Nov 1, 2017 at 14:20
  • \$\begingroup\$ Kevin is right, even in balanced state this circuit is wrongly dimensioned. Ic_Q3 = 20 mA, so in balanced state Ic_Q2 = 10 mA that would mean 10 V drop across R3. The supply is +10 V so that means 0 V at the collector of Q2. That is on the edge, I would halve R3 in value (so 500 ohms) or double R4 in value so the current lowers. \$\endgroup\$ Nov 1, 2017 at 14:24
  • \$\begingroup\$ I think this is homework. And I don't think the course is all that good. The mis-numbering of the transistors is probably the OP's fault in layout here. I don't think the OP gets to change the topology or values of anything but \$V_i\$. \$\endgroup\$
    – jonk
    Nov 1, 2017 at 16:47

1 Answer 1

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I agree with your work regarding estimating the current in \$R_4\$ as \$I_{R_4}\approx 19.3\:\textrm{mA}\$. So your analysis there is fine. Yes, \$Q_3\$ and \$Q_4\$ make up a current mirror, so you'd expect a similar sinking current in the collector of \$Q_3\$. (I'm also assuming here that we do not need to worry about the Early Effect, which in this case would otherwise probably be an issue.) \$Q_3\$'s collector should have a compliance voltage down to perhaps \$-19.3\:\textrm{V}\$. (That won't really matter because the emitter of \$Q_2\$ will keep it from getting anywhere below \$-1\:\textrm{V}\$.)

Now, you need to assume that \$Q_2\$ is active. This means \$V_{B_2}=V_{C_2}=0\:\textrm{V}\$. (Easily seen in the circuit.) Until this point, \$Q_2\$ will be saturated and a lot of the current mirror's sinking current will also come from the base of \$Q_2\$. But right at the point where \$Q_2\$ becomes active we can assume \$\beta=100\$ and its base current will be "normal." This also means that the current in \$R_3\$ must be \$I_{R_3}=\frac{10\:\textrm{V}-0\:\textrm{V}}{R_3=1\:\textrm{k}\Omega}=10\:\textrm{mA}\$. So the emitter current will be the collector current plus 1% for the base current, or \$I_{E_2}=10.1\:\textrm{mA}\$.

\$Q_1\$'s collector has the remaining difference of \$I_{C_1}=19.3\:\textrm{mA}-10.1\:\textrm{mA}=9.2\:\textrm{mA}\$. With \$\beta=100\$, the base current for \$Q_1\$ must be \$I_{B_1}=\frac{9.2\:\textrm{mA}}{100}=92\:\mu\textrm{A}\$. That has to be supplied through \$R_1\$. So the voltage across it must be \$\mid\:V_{R_1}\mid\:=92\:\mu\textrm{A}\cdot 100\:\textrm{k}\Omega=9.2\:\textrm{V}\$.

The only remaining question is, "what is the voltage at the base of \$Q_1\$?" Well, we don't actually know exactly. But it must be very close to \$0\:\textrm{V}\$. But I think your problem assumes all the \$V_{BE}=700\:\textrm{mV}\$; and in any case we know that \$Q_1\$ itself would just be barely active if its base were at \$0\:\textrm{V}\$. So I think it's reasonable here to argue that the problem wants you to take \$V_{B_1}=0\:\textrm{V}\$. And there isn't anything much here to argue strongly against that assumption. So there it is.


So this means the answer is, \$Q_2\$ becomes active when \$V_i\ge 9.2\:\textrm{V}\$.

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  • \$\begingroup\$ I made a big mistake with the Q2 saturation and Vc2 = 0, also i think my analysis should be more clear and organized. Thanks for your help and time jonk, have a nice day. \$\endgroup\$
    – Samu R
    Nov 1, 2017 at 23:32
  • \$\begingroup\$ @SamuR I'm really glad to see you working hard on these things. You laid out your work, too. Very nice. Thanks. I'm glad it helped and thanks for letting me know. \$\endgroup\$
    – jonk
    Nov 1, 2017 at 23:48

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