# Amplifier design using BJT [closed]

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.

• neglect Ib and Vce = 16 - (24-16)/Rc * Re Commented Jun 9, 2020 at 12:47
• It is not cecessary to neglect Ib for calculation purposes.. However, you can treat the voltage Vbe as an ideal voltage source and apply superposition rules.This simplifies the calculation (because it is splitted into two parts)
– LvW
Commented Jun 9, 2020 at 12:53
• refer to the following video to replace base biasing with thevenin eqv youtu.be/zTyuzHokWyA Commented Jun 9, 2020 at 12:54
• Please show us all of your work. You said that you tried solving it with KVL and KCL...show us your equations and explain your thinking. Commented Jun 9, 2020 at 13:31
• We do not know what you know, so you have to show us what you know so we can start with what you know. No workings = closed question! Commented Jun 9, 2020 at 14:03

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}}


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..

1. 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.

• it says higher than 600 k ohms, not 600 ohms. Also this is possibly a homework question, so it is fairly common for professors to give shitty specs to make the life a whole lot worse for students. Commented Jun 9, 2020 at 16:42
• @muyustan Ups my bad :S Thanks for pointing that out.. Sure I get that this is homework, and that this is probably why the design specs are weird, but that doesn't change the fact that the first point is for the OP to understand them.
– user173292
Commented Jun 9, 2020 at 16:43