Amplifier design using BJT [closed]

Question

For the circuit above $$Vc=16V, β=200, VT =25mV \quad and\quad |VBE|=0.6V $$ are given. I want to design a circuit with following conditions

1. The current on \$R_1\$ will be 0.1*Ic1. Pmax (DC maximum power consumption) =10mW
2. there will be no resistance value higher than 600kΩ

I tried to solve it using KVL and KCL but each time I got stucked with voltage drop across BJT's itself because I could not find \$I_{B}\$. Are there any suggestions for how could I find it?

P.S: I do not ask for anybody to solve whole problem. I just want suggestions to solve by myself. So please just guidance, no complete solution.

Answer 1

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\$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$
• Emitter voltage: $$\text{V}_\text{BE}=\text{V}_2-\text{V}_3\tag3$$

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_\text{x}=\text{I}_\text{C}+\text{I}_3\\ \\ \text{I}_3=\text{I}_\text{B}+\text{I}_4\\ \\ \text{I}_\text{x}=\text{I}_\text{E}+\text{I}_4 \end{cases}\tag4 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_\text{C}=\frac{\text{V}_\text{x}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_\text{E}=\frac{\text{V}_3}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_\text{x}-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag5 $$

Now, I use Mathematica to solve your problem using \$\$:

In[1]:=VBE = -6/10;
\[Beta] = 200;
Vx = 24;
V1 = 16;
V2 = 25*10^(-3);
FullSimplify[
 Solve[{IE == IB + IC, \[Beta] == IC/IB, VBE == V2 - V3, 
   Ix == IC + I3, I3 == IB + I4, Ix == IE + I4, IC == (Vx - V1)/R1, 
   IE == V3/R2, I3 == (Vx - V2)/R3, I4 == V2/R4, 
   I3 == (1/10)*IC, (Vx - V2)*I3 == 10*10^(-3)}, {IE, IB, IC, I3, I4, 
   V3, R1, R2, R3, R4, Ix}]]

Out[1]={{IE -> 201/47950, IB -> 1/47950, IC -> 4/959, I3 -> 2/4795, 
  I4 -> 19/47950, V3 -> 5/8, R1 -> 1918, R2 -> 119875/804, 
  R3 -> 919681/16, R4 -> 4795/76, Ix -> 22/4795}}

In[2]:=N[%1]

Out[2]={{IE -> 0.00419187, IB -> 0.0000208551, IC -> 0.00417101, 
  I3 -> 0.000417101, I4 -> 0.000396246, V3 -> 0.625, R1 -> 1918., 
  R2 -> 149.098, R3 -> 57480.1, R4 -> 63.0921, Ix -> 0.00458811}}

Answer 2

First step is to understand the task.

1. The current on R1 will be \$0.1\cdot I_{c1}\$. \$ P_{max}\$ (DC maximum power consumption) \$=10mW\$

Why?

Those don't seem like sensible design specs to work from, perhaps they are derived from other design specs?.

• The current on R1 will be \$0.1\cdot I_{c1}\$.

  Don't you have any load? It doesn't really make sense to design an amplifier to not drive any load, and if you do have a load then \$I_{c1}\$ will depend on this load.

• \$ P_{max}\$ (DC maximum power consumption) \$=10mW\$

  Again; \$ P_{max}\$ would depend on the load, which in you circuit is non-existing. Do you mean that you want the unloaded power consumption to be no more than \$=10mW\$?? In which case; what size of load do you need to drive? that will determine some of the component values and hence affect the minimum unloaded power consumption. You cannot just design a circuit to not have unloaded power consumption above \$10mW\$ without knowing what size of load it need to drive..

2. there will be no resistance value higher than 600kΩ

Again; Why???

The value \$600\Omega\$ is very often used as the characteristic impedance of audio circuits, such as audio amplifiers, but this doesn't mean that "no resistance value is higher than \$600\Omega\$", I don't see why you would ever want that to be a design parameter.??

EDIT!: I misread the question, I didn't see that it said 600kOhm, but I still don't understand why this would be a specific design spec.

Once you have those things sorted you can solve it using Jan's approach.