Analyzing a simple circuit with transistor

Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When analyzing a transistor we need to use the following relations:

  • $$text{I}_text{E}=text{I}_text{B}+text{I}_text{C}tag1$$
  • Transistor gain $beta$:
    $$beta=frac{text{I}_text{C}}{text{I}_text{B}}tag2$$

Now, using KCL, we can write:

$$
begin{cases}
text{I}_text{x}=text{I}_3+text{I}_4\
\
text{I}_3=text{I}_1+text{I}_text{t}+text{I}_text{d}\
\
text{I}_2=text{I}_text{x}+text{I}_5\
\
0=text{I}_text{d}+text{I}_1+text{I}_5\
\
text{I}_2=text{I}_text{t}+text{I}_4\
\
beta=frac{text{I}_4}{text{I}_text{t}}
end{cases}tag3
$$

Using KVL, we get:

$$
begin{cases}
text{I}_3=frac{text{V}_text{x}-text{V}_1}{text{R}_3}\
\
text{I}_4=frac{text{V}_text{x}-text{V}_5}{text{R}_4}\
\
text{I}_2=frac{text{V}_3-text{V}_4}{text{R}_2}\
\
text{I}_text{d}=frac{text{V}_2}{text{R}_1}\
\
text{V}_1-text{V}_3=alpha\
\
text{V}_1-text{V}_2=eta
end{cases}tag4
$$

Solving the circuit using your given values, and using $beta=10^{20}$ and $alpha=eta=frac{7}{10}$ leads to:

FullSimplify[
 Solve[{Ix == I3 + I4, 
   I3 == (1*10^(-3)) + It + Id, (1*10^(-3)) == Ix + I5, 
   0 == Id + (1*10^(-3)) + I5, (1*10^(-3)) == It + I4, 10^(20) == I4/It, 
   I3 == ((56/10) - V1)/1000, 
   I4 == ((56/10) - V5)/1000, (1*10^(-3)) == (V3 - V4)/1000, 
   Id == V2/1000, V1 - V3 == 7/10, V1 - V2 == 7/10}, {Id, It, Ix, I3, 
   I4, I5, V1, V2, V3, V4, V5}]]

So:

{{Id -> 3900000000000000000029/2000000000000000000020000, 
  It -> 1/100000000000000000001000, 
  Ix -> 7900000000000000000069/2000000000000000000020000, 
  I3 -> 5900000000000000000069/2000000000000000000020000, 
  I4 -> 100000000000000000/100000000000000000001, 
  I5 -> -(5900000000000000000049/2000000000000000000020000), 
  V1 -> 5300000000000000000043/2000000000000000000020, 
  V2 -> 3900000000000000000029/2000000000000000000020, 
  V3 -> 3900000000000000000029/2000000000000000000020, 
  V4 -> 1900000000000000000009/2000000000000000000020, 
  V5 -> 2300000000000000000028/500000000000000000005}}

And:

{{Id -> 0.00195, It -> 1.*10^-23, Ix -> 0.00395, I3 -> 0.00295, 
  I4 -> 0.001, I5 -> -0.00295, V1 -> 2.65, V2 -> 1.95, V3 -> 1.95, 
  V4 -> 0.95, V5 -> 4.6}}

EDIT 1

For $betatoinfty$:

{{Id -> 39/20000, It -> 0, Ix -> 79/20000, I3 -> 59/20000, 
  I4 -> 1/1000, I5 -> -(59/20000), V1 -> 53/20, V2 -> 39/20, 
  V3 -> 39/20, V4 -> 19/20, V5 -> 23/5}}

Wich is:

{{Id -> 0.00195, It -> 0., Ix -> 0.00395, I3 -> 0.00295, I4 -> 0.001, 
  I5 -> -0.00295, V1 -> 2.65, V2 -> 1.95, V3 -> 1.95, V4 -> 0.95, 
  V5 -> 4.6}}

EDIT 2

FullSimplify[
 Solve[{Ix == I3 + I4, I3 == I1 + It + Id, I2 == Ix + I5, 
   0 == Id + I1 + I5, I2 == It + I4, β == I4/It, 
   I3 == (Vx - V1)/R3, I4 == (Vx - V5)/R4, I2 == (V3 - V4)/R2, 
   Id == V2/R1, V1 - V3 == α, V1 - V2 == η}, {Id, It, Ix, 
   I3, I4, I5, V1, V2, V3, V4, V5}]]

Which gives:

{{Id -> -((I1 R3 - Vx + (I2 R3)/(1 + β) + η)/(R1 + R3)), 
  It -> I2/(1 + β), 
  Ix -> I2 - (I2 R3)/((R1 + R3) (1 + β)) + (
    I1 R1 + Vx - η)/(R1 + R3), 
  I3 -> (I2 R1 + 
    I1 R1 (1 + β) + (1 + β) (Vx - η))/((R1 + 
      R3) (1 + β)), I4 -> (I2 β)/(1 + β), 
  I5 -> (-I1 R1 - Vx + (I2 R3)/(1 + β) + η)/(R1 + R3), 
  V1 -> (-I2 R1 R3 - 
    I1 R1 R3 (1 + β) + (1 + β) (R1 Vx + R3 η))/((R1 +
       R3) (1 + β)), 
  V2 -> (R1 (-I1 R3 + Vx - (I2 R3)/(1 + β) - η))/(R1 + R3),
   V3 -> -α - (I2 R1 R3)/((R1 + R3) (1 + β)) + η - (
    R1 (I1 R3 - Vx + η))/(R1 + R3), 
  V4 -> -I2 R2 - α - (
    I2 R1 R3)/((R1 + R3) (1 + β)) + η - (
    R1 (I1 R3 - Vx + η))/(R1 + R3), 
  V5 -> Vx - (I2 R4 β)/(1 + β)}}

First Circuit

Let’s look at the base voltage of the first circuit using what I believe to be your stated facts:

  • $I_text{B}=0:text{A}$ ($betato infty$)
  • $V_text{BE}=600:text{mV}$
  • $V_text{D}=600:text{mV}$
  • $V_text{CC}=5.6:text{V}$

Swapping the diode near the base with its series resistor, without damage to the analysis, we can now write using KCL:

$$begin{align*}I+frac{V_text{B}}{R_{text{B}_1}}+frac{V_text{B}}{R_{text{B}_2}}+I_text{B}&=frac{V_text{D}}{R_{text{B}_1}}+frac{V_text{CC}}{R_{text{B}_2}}\\1:text{mA}+frac{V_text{B}}{1:text{k}Omega}+frac{V_text{B}}{1:text{k}Omega}+0:text{A}&=frac{5.6:text{V}}{1:text{k}Omega}+frac{600:text{mV}}{1:text{k}Omega}\\&therefore\\V_text{B}&=2.6:text{V}\\V_text{E}&=2.0:text{V}=V_text{B}-V_text{BE}\\V_text{C}&=4.6:text{V}=V_text{CC}-I_2cdot R_text{C}\\V_text{CE}&=2.6:text{V}=V_text{C}-V_text{E}\\I_text{D}&=2:text{mA}=frac{V_text{B}-V_text{D}}{R_{text{B}_1}}end{align*}$$

Second Circuit

Since you only want guidance, for now, I’d suggest the following steps in the indicated order:

  1. Label the top node as the unknown value $V_text{CC}$.
  2. Write down the nodal equation for $V_text{CC}$.
  3. Write down the nodal equation for $V_text{B}$.
  4. Write down the nodal equation for $V_text{E}$.
  5. Write down the equation relating $V_text{B}$ to $V_text{E}$.
  6. Write down the equation relating $V_text{CC}$ to $V_text{C}$.
  7. Solve the above 5 equations simultaneously for these five unknowns: $V_text{CC}$, $V_text{B}$, $V_text{E}$, $V_text{C}$, and $I_text{D}$.

(Keep in mind that you know the value for $V_text{D}$, $V_text{BE}$, and $I_text{CC}$.)

As a hint you may use to verify things, the resulting values are as easy to write down as the answers found for the first circuit.

😉

Since someone decided to mark this question down (I don’t mind, because it gives me fun excuses), I think I’ll just write out the complete solution for you. You got lucky!

Using sympy:

var('vcc vc ve vb rb1 rb2 rc re id icc vbe vd')
eq1=Eq(vb/rb1+vb/rb2,vcc/rb2)
eq2=Eq(vcc/rb2+vcc/rc+id,icc+vc/rc+vb/rb2)
eq3=Eq(ve/re,(vcc-vc)/rc+id)
eq4=Eq(vb,ve+vbe)
eq5=Eq(vc,vcc-vd)
ans=solve([eq1,eq2,eq3,eq4,eq5],[vcc,vb,ve,vc,id])
for x in ans:x,ans[x].subs({vd:.6,vbe:.6,rb1:800,rb2:2000,re:1000,rc:1200,icc:3e-3})
(vb, 1.60000000000000)
(id, 0.000500000000000000)
(vc, 5.00000000000000)
(vcc, 5.60000000000000)
(ve, 1.00000000000000)

It solves out.

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