Analyzing a simple circuit with transistor
Well, we have the following circuit:
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 \$\beta\to\infty\$:
{{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}\$ (\$\beta\to \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:
$$\begin{align*}I+\frac{V_\text{B}}{R_{\text{B}_1}}+\frac{V_\text{B}}{R_{\text{B}_2}}+I_\text{B}&=\frac{V_\text{D}}{R_{\text{B}_1}}+\frac{V_\text{CC}}{R_{\text{B}_2}}\\\\1\:\text{mA}+\frac{V_\text{B}}{1\:\text{k}\Omega}+\frac{V_\text{B}}{1\:\text{k}\Omega}+0\:\text{A}&=\frac{5.6\:\text{V}}{1\:\text{k}\Omega}+\frac{600\:\text{mV}}{1\:\text{k}\Omega}\\\\&\therefore\\\\V_\text{B}&=2.6\:\text{V}\\\\V_\text{E}&=2.0\:\text{V}=V_\text{B}-V_\text{BE}\\\\V_\text{C}&=4.6\:\text{V}=V_\text{CC}-I_2\cdot R_\text{C}\\\\V_\text{CE}&=2.6\:\text{V}=V_\text{C}-V_\text{E}\\\\I_\text{D}&=2\:\text{mA}=\frac{V_\text{B}-V_\text{D}}{R_{\text{B}_1}}\end{align*}$$
Second Circuit
Since you only want guidance, for now, I'd suggest the following steps in the indicated order:
- Label the top node as the unknown value \$V_\text{CC}\$.
- Write down the nodal equation for \$V_\text{CC}\$.
- Write down the nodal equation for \$V_\text{B}\$.
- Write down the nodal equation for \$V_\text{E}\$.
- Write down the equation relating \$V_\text{B}\$ to \$V_\text{E}\$.
- Write down the equation relating \$V_\text{CC}\$ to \$V_\text{C}\$.
- 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.