# Designing a common emitter amplifier circuit

How I calculated the resistor values: I assumed that Ic = 1mA, and β = 50.5. It is given that VDD = 5V and to design the circuit to get a gain of 10. The transistor is 2N4401

RC = VRE / Ic = (Vdd/2) / Ic = 2.5v / 1mA = 2.5kΩ RE = 10% of VDD / Ic = 0.5v/1mA = 500Ω

IB = Ic/β = 1mA / 50.5 = 19μA VB= VBE + VRE = 0.7 + 0.5 = 1.2v

R1 + R2 = VDD / (Ib * 10) = 5 / (19μA * 10) = 26.3kΩ VB = R2 / (R1+R2) * VDD

R2 = VB(R1+R2) / VDD = 1.2v * 26.3kΩ / 5v = 6.3kΩ R1 = 26.3kΩ - 6.3kΩ = 20kΩ

R1//R2 = 4.8kΩ C = 1 / 2π(4.8k)(1kHz) = 33.2nF

Could someone please explain to me how am I supposed to design this BJT circuit to get the gain of 10? I emailed the Professor's Assistant and he said the figure 2 shows a gain of 3.5 .

• +1 for showing your work, and explaining where you got stuck. That is the correct way to ask questions about academic assignments at this site. Commented Dec 29, 2023 at 16:05

Could someone please explain to me how am I supposed to design this BJT circuit to get the gain of 10?

This was your first error: -

RE = 10% of VDD / Ic = 0.5v/1mA = 500Ω

It's an error for the following reason: you have already calculated Rc to be 2.5 kΩ and, because Rc and Re share (to within a percent or so) the same current, the transistor gain is Rc/Re so, Re would have to be 250 Ω if Rc is 2.5 kΩ.

With Re at 500 Ω the maximum amplification is only 5.

Your other error is not following the guidance in the question about the choice of the input capacitor. You have calculated the capacitor value like this: -

C = 1 / 2π(4.8k)(1kHz) = 33.2nF

But, that will produce a 3 dB reduction in the input voltage at the base so, the net gain of the circuit is 5/1.4142 = 3.54. The 1.4142 represents a 3 dB lowering of the gain in real numbers.

How I calculated the resistor values: I assumed that Ic = 1mA, and β = 50.5. It is given that VDD = 5V and to design the circuit to get a gain of 10. The transistor is 2N4401

You started right out the gate... a little too fast. Return to the specs:

• "Choose $$\R_{_\text{E}}\$$ such that the design is not prone to $$\V_{_\text{BE}}\$$ variations."

$$\V_{_\text{BE}}\$$ variations are due to temperature and due to collector current variations, which in this case vary with the signal. Set the minimum $$\V_{_\text{CE}}=1\:\text{V}\$$ to avoid saturation.

With $$\\mid A_v\!\mid=10\$$ and an input test signal with $$\V_{_\text{IN pk}}=10\:\text{mV}\$$, then $$\V_{_\text{OUT pp}}=200\:\text{mV}\$$. So reserve room for a total of $$\1.2\:\text{V}\$$ for the BJT. This leaves only $$\3.8\:\text{V}\$$ for everything else.

Also, with $$\\mid A_v\!\mid=10\$$, there's no getting around the fact that $$\\frac{3.8\:\text{V}}{1+A_v=10}\approx 345\:\text{mV}\$$ as the quiescent emitter voltage. Signal variations on that are modest here, as $$\\pm 10\:\text{mV}\$$ only represents about $$\\pm 3\%\$$.

This means that what was being requested has to do with temperature and not signal, as over an operating temperature range from $$\-20^\circ\:\text{C}\$$ to $$\+45^\circ\:\text{C}\$$, for example, could mean as much as $$\140\:\text{mV}\$$ change in $$\V_{_\text{BE}}\$$. And that is much more significant with respect to $$\345\:\text{mV}\$$.

Conclusion? Make the quiescent emitter voltage as high as possible. In this case, that means $$\V_{_{\text{E}_\text{q}}}\le 345\:\text{mV}\$$, but as close as possible to that value as can be readily managed.

I would select $$\V_{_{\text{E}_\text{q}}}= 300\:\text{mV}\$$ (because I've a sneaky suspicion about $$\\beta\$$ and my desire to leave a little room) and this means the quiescent voltage drop across $$\R_{_\text{C}}\$$ should be $$\10\times\$$ that or $$\3\:\text{V}\$$. That will leave $$\1.7\:\text{V}\$$ for the BJT.

That should have been your first stop.

From there, $$\R_{_\text{E}}=300\:\Omega\$$ (given $$\I_{_{\text{E}_\text{q}}}\approx 1\:\text{mA}\$$.)

As $$\r_e^{\:'}\approx 26\:\text{m}\Omega\$$ at room temp is negligible, simply figure that $$\R_{_\text{C}}=A_v\cdot R_{_\text{E}}=3\:\text{k}\Omega\$$. But to account for about a 10% loss with the Early Effect and more, I'd increase this to $$\R_{_\text{C}}=3.3\:\text{k}\Omega\$$.

Nominally, this means we have $$\5\:\text{V}-3.3\:\text{V}-300\:\text{mV}=1.4\:\text{V}\$$ for the BJT. We know we needed at least $$\1.2\:\text{V}\$$. So that's okay.

We do have a problem on the base biasing side. I was right to worry.

From what I can see of the problem statement, you cannot just assume $$\\beta=50.5\$$. The datasheet provides a minimum $$\\beta=40\$$ for the device at the indicated collector current, but Figure 15 in this OnSemi datasheet suggests that over temperature, from $$\-20^\circ\:\text{C}\$$ to $$\+45^\circ\:\text{C}\$$, I might see a minimum of anything between $$\\beta=20\$$ to $$\\beta=50\$$, worst case.

Yet, higher values will likely be found in practice. Probably more around $$\\beta=100\$$. Maybe still higher yet! So we are talking about a base current recombination current variation of 5:1, with $$\10\:\mu\text{A} \le I_{_{\text{B}_\text{q}}} \le 50\:\mu\text{A}\$$!

We need to either create some margin for that kind of variation on the emitter collector side (there's some there but not a lot to be honest) or else radically stiffen up the base biasing pair to handle the worst case BJT situation at the worst case operating temperature.

Since this is for education and since I want to move more quickly to working out the base biasing pair and avoid discussions about how to balance trade-offs at the emitter-collector side, as well, I'll simply state that the base biasing pair should be very stiff.

How stiff? Well, the base biasing pair should carry about $$\10\times\$$ the base current. In this case, the worst case base current is $$\\frac{1\:\text{mA}}{\beta=20}=50\:\mu\text{A}\$$. So this means $$\500\:\mu\text{A}\$$ in the biasing pair!! For practical circuits, this is crazy. And would suggest performing some trade-offs over on the emitter-collector side to accommodate this. But I'm cutting straight to the chase. So I will use $$\500\:\mu\text{A}\$$.

From Figure 17 in the same OnSemi datasheet and at $$\I_{_{\text{C}_\text{q}}}=1\:\text{mA}\$$ I think I would estimate $$\V_{_{\text{BE}_\text{q}}}\approx 650\:\text{mV}\$$. So I'd target $$\V_{_{\text{B}_\text{q}}}=950\:\text{mV}\$$.

This means $$\R_2=\frac{1.0\:\text{V}}{500\:\mu\text{A}}=1.9\:\text{k}\Omega\$$. Make it $$\R_2=1.8\:\text{k}\Omega\$$. So $$\R_1=\frac{5.0\:\text{V}-950\:\text{mV}}{\frac{950\:\text{mV}}{1.8\:\text{k}\Omega}+50\:\mu\text{A}}\approx 7.0\:\text{k}\Omega\$$. I'd round this up to $$\R_2=7.5\:\text{k}\$$ since we are using aggressively bad $$\\beta\$$ for the design and to make sure that the quiescent emitter voltage stays low enough.

So I'd use $$\R_{_\text{E}}=300\:\Omega\$$, $$\R_{_\text{C}}=3.3\:\text{k}\Omega\$$, $$\R_1=7.5\:\text{k}\Omega\$$, and $$\R_2=1.8\:\text{k}\Omega\$$.

The input impedance will be around $$\1.2\:\text{k}\Omega\$$. (The parallel of $$\R_1\$$, $$\R_2\$$ and the $$\\beta\$$ multiplied value of $$\R_{_\text{E}}\$$, approximately.) Capacitor values will be in quadrature to this so one can get away with a lot more than a simplistic calculation would suggest. Also, it should hold so that at least 90% or more of the signal is available to the final gain at the output. Luckily, the instructions say you may make adjustments. So I'll leave that to you to perform. But my calculations say that $$\C\gt 274\:\text{nF}\$$ to achieve at least 90%. But I'd make that $$\\ge 1\:\mu\text{F}\$$ to start.

(Unfortunately, the worst case assumptions about $$\\beta\$$ are affecting this choice.)

Here's an LTspice run given the above design. (Only just now performed it, as I was curious how it would come out.)

It gives $$\\mid A_v\!\mid =10.003\$$ and that seems pretty close to intents. (Okay. I admit being surprised about how close. I expected a worse fit.)

Here's the results over the entire temperature range:

These were with $$\\beta\approx 117\$$ for that device in that circumstance. I tweaked things to so that the model produced $$\\beta=20\$$ and then at room temp I get $$\\mid A_v\!\mid =9.469\$$ for the worst case circumstance possible.

I would consider that acceptable.

Final note: This is with very stiff biasing. There are much better ways to handle BJT variations while also getting even higher precision $$\A_v\$$, using global NFB. But this is the topology you are stuck with.