Voltage for transistors in saturation

Question

I need to find the voltage Vi for Q3 saturated with B = 100 for transistors.

I did the calculations and then simulated the circuit and i had similar results. the Question is: the analysis is right ?

My calculations:

If Q3 is in saturation then vce3 = 0.2v and with: 20 - 0.2 = 5kIc3 Ic3 = 0.00396 A, also Ib3 = 39.6 uA

If i suppose Vce1 aprox 0.2v and vcb3 = -0.5v: 19.8 - 0.2 - 0.5 - 1.98 = 5kIe1

then:

5kIe1 = 17.12V and Ie1 = 0.003424 A, Ib1 = 34.24 uA

but: Vcb1 = 1.712V and Q1 is not in saturation. Because: Vcb(sat) = -0.5v

Using:

5kIe1 + Vce1 = 17.32v , Vcb1 = 500Ie1 , Vce1 = Vcb1 + 0.7

solving:

      5KIe1 + Vcb1 = 16.62  
      500Ie1 - Vcb1 = 0

Ie1 = 0.003A, Ib1 = 30 uA, Vce1 = 2.2 v, Vcb1 = 1.5v 

For Q2 i can say that Vce2 = 2.68v and Vcb2 = 1.98v from 50kIb3 + 0.7 = Vce2 I3(5k) = Vce2/5k = 0.000536 A

With Ib1 + Ic1 = I3 + Ie2 + Ib3 Ie2 = Ib1 + Ic1 - I3 - Ib3

Ie2 = 2.4544 mA, Ib2 = 24.544 uA

Ending with:

     Vi = 1.98 - 100kIb2,   

     Vi = - 0.4744 V 

are the calculations right ?. I doubt about the ecuations:

        5kIe1 + Vce1 = 17.32V
        Vce1 = Vcb1 + 0.7 
        Vcb1 = 500Ie1

Seems right, but i didnt take in consideration the rest of the circuit.

Thanks a lot for your help and time.

Answer 1

Your circuit doesn't really make any sense, in my opinion. Chances are, it was created almost at random and without any "intelligent actor" behind it. You'll see why I have that opinion in a moment.

Let's redraw that crazily laid out schematic, to start.

^{simulate this circuit – Schematic created using CircuitLab}

Let's analyze the schematic first by assuming that \$Q_2\$ is completely off (\$V_I=20\:\textrm{V}\$) and see where that leaves things.

Here's the KCL nodal equation for \$V_X\$:

$$\begin{align*} \frac{V_X}{R_3} + \frac{V_X}{R_6} &= I_E + \frac{700\:\textrm{mV}}{R_3}\tag{Eq. 1} \end{align*}$$

Where we know this from KVL:

$$\begin{align*} V_X&=20\:\textrm{V} -I_E\cdot R_2-.7-\frac{I_E}{\beta_1+1}\cdot R_5\label{vx}\tag{Eq. 2} \end{align*}$$

Let's see what assuming \$\beta_1=100\$ gets us:

$$\begin{align*} I_E &\approx 1.916\:\textrm{mA}\\\\ V_X &\approx 8.772\:\textrm{V} \end{align*}$$

From here, we can work out that \$I_{R_3}=\frac{V_X-700\:\textrm{mV}}{R_3}= 161.44\:\mu\textrm{A}\$. Given your own computation that \$I_{C_3}=3.96\:\textrm{mA}\$, this already works out to \$\beta_3\approx 24.5\$ and that is already well below your nominal value of \$100\$.

So it's clear, now, that without any contribution from \$Q_2\$ the circuit already puts \$Q_3\$ in at least a shallow saturation situation.

Okay. So now what? Well, \$Q_2\$ won't contribute anything until its base voltage is at least \$700\:\textrm{mV}\$ below the above-computed value for \$V_X\$. So \$V_I\le 8\:\textrm{V}\$, roughly speaking. As you imagine the value of \$V_I\$ declining towards (and perhaps below ground), \$Q_2\$ will start pulling away current and thereby gradually moving \$Q_3\$ out of saturation.

That will happen about when \$\beta_3=100\$, or when \$V_X\approx 2.7\:\textrm{V}\$. From \$\ref{vx}\$ above, we can easily work out that this happens when the emitter current is \$I_{E_1}\approx 3\:\textrm{mA}\$. Subtracting \$I_{B_3}\approx 40\:\mu\textrm{A}\$ and \$I_{R_3}\approx 540\:\mu\textrm{A}\$, this leaves about \$I_{C_2}\approx 2.42\:\textrm{mA}\$. As \$V_{CE_2} \approx 2.7\:\textrm{V}\$, \$Q_2\$ is not in saturation at this time so the full \$\beta_2=100\$ can be used and we figure \$I_{B_2}\approx 24.2\:\mu\textrm{A}\$. We know that \$V_{B_2}\approx 2\:\textrm{V}\$, so this means that \$V_I\le 2\:\textrm{V} - 100\:\textrm{k}\Omega\cdot 24.2\:\mu\textrm{A}\approx -420\:\textrm{mV}\$.

So now we know approximately where \$Q_3\$ leaves saturation and enters into active mode: \$V_I\le -420\:\textrm{mV}\$.

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\$Q_3\$ is in saturation for all positive values of \$V_I\$ and computing the point where it leaves saturation is a bit nuanced. This is why I think the whole question was made up without much attention given to it. (It might be a good question if you are somewhat advanced in DC analysis.)