# Designing a common emitter audio pre-amplifier

I am trying to design and simulate a common emitter audio pre-amplifier using 2N2219 with characteristics based on industry standards (audio frequency response, etc. Corner frequency must be less than 30Hz, C1, C2 must have an exceptionally low reactance, i.e. be short. The DC bias's Q point must be centered on all cases. where Gain(db)=7, Zout=350, and V1=18V. Edit 1: I tried to calculate it myself but either I don't get sinewave on my oscillator or my Vce is not in the range. please help

Edit 2: After obtaining some insight my zout=R3 value, and modified my circuit below with new values

I am using MATLAB to calculate:

R1= 49.9E3; R2= 11.8E3; R3= 350; R4= 154; R5= 50; RL= 1E6;
%R1= 23.7E3; R2= 6.19E3; R3= 1.33E3; R4= 590; R5= 50; RL= 475;
B= 100;
Zout= ((1/R3)+(1/RL))^-1;  ZoutE= 350;
C= 10E-6;
F= (30:200:12830)';
Vs= 1;
Vcc= 18;
XC1= (1./(2*pi*F*C));
XC2= XC1;

VB= Vcc*(R2/(R1+R2));
VE= VB-0.7;
IE= (VE/R4);
re= (25E-3/IE);
Zin= ((1/R1)+(1/R2)+(1/(B*(R4+re))))^-1;
VCE= 18-(IE*(R3+R4));

Gain= (R3/(R4+re));               GainE= 10^(7/20);
Vout= Vs*(Zin./realsqrt((Zin+R5)^2+(XC1).^2))*Gain;

FC= (1/(2*pi*(R3+RL)*C));


Values of Resistors (except R5) and capacitor changeable.

This is the attempt if zout=R3 This is another attempt if zout=R3||RL • It's pretty clear that R2 is a bit low... ;) Also, what kind of voltage gain are you expecting? And can you help us help you by providing some of your calcs?
– jonk
May 17 at 17:26
• For a good QP point stability factor ... R2 should be 10x-20x R4 ... Always pass by the 3 stages (min) for simulating. 1-DC analysis -QP. 2-AC Analysis. 3-TRAN analysis. 4-Harmonic analysis ... Statistics analysis ... May 17 at 17:29
• @Sabretooth2438 Do you mean that Vce is supposed to be Vcc/2 or do you mean that Vc is? Are you talking about the quiescent collector voltage? Or about the quiescent difference between the collector and the emitter?
– jonk
May 17 at 18:08
• @Sabretooth2438 Thanks. These are significant improvements to your question. It would be helpful if you could somehow include them there. So, it now seems, your real requirement is that whatever two values you use for R3 and RL, their combined impedance must be 350 Ohms? If so, then Zout is not necessarily the same as RL. Just to clarify still more, would it be okay to treat RL as infinity and just make R3 equal to 350 Ohms?
– jonk
May 19 at 16:53
• A Zout of 350 on a 18V supply is quite low for a Common Emitter (CE). I'd cascade it with a Common Collector (CC) a/k/a emitter follower stage. The CC is better for providing a low-output impedance. Then you can use larger-value resistors, save power, use low-noise small signal transistors instead, and get better performance. May 19 at 20:34

## Re-drawing of schematic

For my own clarity, if that of no others, I'm re-drawing your schematic as follows: simulate this circuit – Schematic created using CircuitLab

I'm keeping the schematic generalized in the sense that the BJT stage in the middle has its more negative side tied to $$\V_{_\text{EE}}\$$ instead of ground. There is no harm in this, as we could just set $$\V_{_\text{EE}}=0\:\text{V}\$$. But it's more general to handle it this way and allows for bipolar power supplies to be considered.

Also in the above diagram, the signal source and the load are tied (at their other ends) to ground. That could just as well be $$\V_{_\text{CC}}\$$ or $$\V_{_\text{EE}}\$$ or any other voltage reference point. The main thrust here is that these are single-ended sources and sinks.

## List out what you know will be specified as inputs to the design

The first thing I try to achieve well is listing out the details I think I know. In your case, I think you know these things:

• $$\A_v=7\:\text{dB}\approx 2.24\times\$$
• $$\\beta_{_\text{MIN}}=100\$$
• $$\V_{_\text{CC}}=18\:\text{V}\$$
• $$\V_{_\text{EE}}=0\:\text{V}\$$ (ground-referenced emitter side)
• $$\V_{_\text{CE}}=\frac{V_{_\text{CC}}-V_{_\text{EE}}}2=9\:\text{V}\$$
• $$\R_{_\text{C}}\mid\mid R_{_\text{L}}=350\:\Omega\$$ (you've called this $$\Z_{_\text{OUT}}\$$ in your writing)
• $$\R_{_\text{S}}=50\:\Omega\$$

While I've provided values above, it's not necessary that the values be these particular values. We can keep the analysis general.

Again, please note that I'm going to retain $$\V_{_\text{EE}}\$$ in what follows because it is more generalized to do so. Also, for these purposes $$\A_v\$$ is the absolute magnitude of the gain and is always positive.

## List out what else you may need to specify

There are a couple of others things which come to mind:

• Base biasing stiffness equal to, say, 10% of $$\I_{_{\text{C}_\text{Q}}}\$$: $$\k=0.1\$$
• Base-emitter forward-biased voltage at the quiescent operating point: $$\V_{_\text{BE}}\approx 700\:\text{mV}\$$ (TBD)
• Thermal voltage at junction's quiescent operating point: $$\V_T\approx 26\:\text{mV}\$$

Those are commonly used rules applied. But you need them, or concepts relatable to them, in order to complete the design.

## Specify other relevant relationships

You may need only some of these for a solution, but it is worth listing those that come to mind quickly:

• $$\r_e^{\:'}=\frac{V_T}{I_{_{\text{E}_\text{Q}}}}\$$
• $$\A_v=\frac{R_{_\text{C}}}{R_{_\text{E}}+r_e^{\:'}}\$$
• $$\V_{_{\text{E}_\text{Q}}}=V_{_\text{EE}}+I_{_{\text{E}_\text{Q}}}R_{_\text{E}}\$$
• $$\V_{_{\text{C}_\text{Q}}}=V_{_\text{CC}}-I_{_{\text{C}_\text{Q}}}R_{_\text{C}}\$$
• $$\V_{_{\text{B}_\text{Q}}}=V_{_{\text{E}_\text{Q}}}+V_{_\text{BE}}\$$
• $$\V_{_\text{CE}}=V_{_{\text{C}_\text{Q}}}-V_{_{\text{E}_\text{Q}}}\$$
• $$\I_{_{\text{E}_\text{Q}}}=\frac{\beta_{_\text{MIN}}+1}{\beta_{_\text{MIN}}}I_{_{\text{C}_\text{Q}}}\$$
• $$\I_{_{\text{B}_\text{Q}}}=\frac{1}{\beta_{_\text{MIN}}}I_{_{\text{C}_\text{Q}}}\$$
• $$\R_2=\frac{V_{_{\text{B}_\text{Q}}}-V_{_\text{EE}}}{k\: I_{_{\text{C}_\text{Q}}}}\$$
• $$\R_1=\frac{V_{_\text{CC}}-V_{_{\text{B}_\text{Q}}}}{k\: I_{_{\text{C}_\text{Q}}}+I_{_{\text{B}_\text{Q}}}}\$$ (to include enough extra for base recombination)

In addition, we need to keep in mind that $$\R_{_\text{C}}\mid\mid R_{_\text{L}}=350\:\Omega\$$. But here I'm going to take the easier path by setting $$\R_{_\text{L}}=\infty\:\Omega\$$ so that I can simplify the remaining analysis by setting $$\R_{_\text{C}}=350\:\Omega\$$.

## Identify and then resolve the unknowns

From the above, I can see that there are two key pieces needed in order to resolve a design: $$\I_{_{\text{C}_\text{Q}}}\$$ and $$\R_{_\text{E}}\$$. This suggests that I should be able to find two equations to help me out.

From all of the above, I see these coming to mind:

• $$\V_{_\text{CE}}=\frac{V_{_\text{CC}}-V_{_\text{EE}}}2=V_{_\text{CC}}-V_{_\text{EE}}-I_{_{\text{C}_\text{Q}}}\left(\frac{1+\beta_{_\text{MIN}}}{\beta_{_\text{MIN}}}R_{_\text{E}}+R_{_\text{C}}\right)\$$
• $$\A_v=\frac{R_{_\text{C}}}{R_{_\text{E}}+\frac{V_T}{\left[\frac{1+\beta_{_\text{MIN}}}{\beta_{_\text{MIN}}}\right]I_{_{\text{C}_\text{Q}}}}}\$$

So I would set out to solve these, simultaneously, in order to find $$\I_{_{\text{C}_\text{Q}}}\$$ and $$\R_{_\text{E}}\$$.

I believe that, with those in hand, the rest can be worked out using the listed information.

## Solution

Compute these two terms:

• $$\A_v^{\:'}=\frac{1+\beta_{_\text{MIN}}}{\beta_{_\text{MIN}}}A_v\$$
• $$\V_{_\text{CE}}^{\,'}=V_{_\text{CE}}+V_T=\frac{V_{_\text{CC}}-V_{_\text{EE}}}2+V_T\$$

And these two factors:

• $$\m_r=1-\left(1+A_v^{\:'}\right)\frac{V_T}{V_{_\text{CE}}^{\,'}},\quad\quad V_{_\text{CE}}^{\,'}\gt \left(1+A_v^{\:'}\right)V_T\$$
• $$\m_i=\frac{A_v^{\:'}}{1+A_v^{\:'}}\$$

Noting that as $$\A_v^{\:'}\$$ grows larger in magnitude, $$\m_r\$$ tends toward 0 and $$\m_i\$$ tends toward 1.

Then find:

• $$\R_{_\text{E}}=m_r\cdot\left[\frac{R_{_\text{C}}}{A_v}\right]\$$
• $$\I_{_{\text{C}_\text{Q}}}=m_i\cdot\left[\frac{V_{_\text{CE}}^{\,'}}{R_{_\text{C}}}\right]\$$

Take note of what these say. One thing is that $$\R_{_\text{E}}\$$ is approximately $$\\frac{1}{A_v}R_{_\text{C}}\$$, but not quite. It will be smaller than this estimate by the factor $$\m_r\$$. Another is that $$\I_{_{\text{C}_\text{Q}}}\$$ is approximately $$\\frac{V_{_\text{CE}}}{R_{_\text{C}}}\$$, but again not quite. (That fact alone is kind of interesting in its own right.) But it will also be smaller, adjusted by another factor. This time: $$\m_i\$$.

Here is what I'll set up in LTspice:

* Input specifications:
.param VCC = {18V}
.param VEE = {0V}
.param BETAMIN = {100}
.param AVdB = {7}
.param RC = {350V/1A}
.param VCE = {(VCC - VEE) / 2}
.param VBE = {700mV}
.param STIFFNESS = {0.1}

* Input-derived values:
.param AV = {10**(AVdB/20)}
.param AVprime = {AV * (1 + BETAMIN) / BETAMIN}
.param VCEprime = {VCE + VT}
.param mr = {1 - (1 + AVprime) * VT/VCEprime}
.param mi = {AVprime / (1 + AVprime)}

* NPN model eliminating Early effect, current crowding, & bulk impedences:
.model MyNPN ako:2N2222 NPN(Bf={BETAMIN} VA=1E7 IKF=100 RE=1n RC=1n)

* Simultaneous solution provides:
.param RE = {mr * RC / AV}
.param ICQ = {mi * VCEprime / RC}

* Develop RB1 and RB2:
.param VE = {VEE + (BETAMIN+1)/BETAMIN * ICQ * RE}
.param VB = {VE + VBE}
.param RB2 = {(VB - VEE) / (ICQ * STIFFNESS)}
.param RB1 = {(VCC - VB) / (ICQ * STIFFNESS + ICQ/BETAMIN)}


Given that model of the NPN BJT which follows the Level I Ebers-Moll DC model, how close will the calculations be when they are compared to what a sophisticated simulator produces?

(There are at least 3 Ebers-Moll models, plus a great many improved models that have come still later, but the first one from Ebers-Moll is both simple and good enough for most DC operating point desires. Other models include the Gummel-Poon (GP) and its various modifications, VBIC and its evolution, and today the MEXTRAM is probably already in its 6th or 7th major revision, by now.)

Let's see what happens with the above, starting with the assumption of $$\V_{_\text{BE}}=700\:\text{mV}\$$. (This is only important for the base biasing pair of resistors. It doesn't impact the emitter resistor calculations.)

This won't be right and the simulation will deviate a bit from calculations. But that can be adjusted for a 2nd run to see how well the calculations perform when provided with an improved DC operating point estimate.

The first approximation yields this result: The above shows the two base resistors and the calculated emitter resistor shown in blue, sitting just above the schematic. The value for the emitter resistor is quite close to what you show in your schematic, today. Good!

Also note the values it computed for the base biasing pair. (These will change if we adjust the estimate for $$\V_{_\text{BE}}\$$.)

And above you can also find the computed and actual quiescent current (they are present, one above the other, about in the middle of the long operating point list of values over on the lower right.) Some of the values for comparison are circled to help make it easier to see computed vs simulated values. Note that they are within 1% of the predicted values.

Now, computing $$\V_{_\text{BE}}=3.49973\:\text{V}- 2.76852\:\text{V}= 731.21\:\text{mV}\$$ and substituting back in to get a second approximation that will slightly adjust the base biasing pair: The calculated vs the simulated values are now quite close: within about 100ppm of the predicted values, at worst. And $$\V_{_\text{BE}}\$$ doesn't shift enough to worry about for a 3rd shot at it. So we can stop here.

Probably could have just stopped at the first attempt, to be honest. It was already close enough.

## Source impedance

I set $$\R_{_\text{S}}\approx 0\:\Omega\$$ in the above. For the DC operating point that doesn't matter. There will be a small unaccounted (by me, above) impact on the resulting voltage gain, though, given $$\R_{_\text{S}}= 50\:\Omega\$$.

Speaking of which, let's see the gain with $$\R_{_\text{S}}\approx 0\:\Omega\$$: That looks pretty close to plan. About $$\6.91\:\text{dB}\$$. This difference is due to other model elements coming into play that I didn't remove.

What about with $$\R_{_\text{S}}= 50\:\Omega\$$? Less. Now it is about $$\6.61\:\text{dB}\$$. A drop representing $$\-0.3\:\text{dB}\$$ (about -3.5%.)

Given that the biasing pair alone represents a load of slightly less than $$\1.6\:\text{k}\Omega\$$. With the base loading included, it drops to around $$\1.4\:\text{k}\Omega\$$. So we might have predicted this drop as $$\\frac{1.4\:\text{k}\Omega}{50\:\Omega+1.4\:\text{k}\Omega}\approx 0.966\$$.

Pretty close to $$\10^{^\frac{-0.3}{20}}=0.966\$$.

One approach to factoring this back in would be to simply increase the targeted voltage gain by this amount and then re-running the simulation to get new values.

But the difference is small enough in this case that it doesn't seem worth the trouble. Besides, a real BJT won't exactly fit the assumptions used by theory or simulations, anyway. It's close enough.

We know why it is happening and we can say by just how much. And that is what's important.

## Notes

Once you have a better idea about the value for $$\I_{_{\text{C}_\text{Q}}}\$$, you can refer to a BJT datasheet and its typical curves in order to make a better estimate of $$\V_{_\text{BE}}\$$. Use that as a double-check on the earlier guess or else take a 2nd hack at refining the numbers, if comparing results suggest a need to do that.

A caution, as well: While theory and simulation can and should reach similar results as shown here, it's also true that in real life transistor parts are not so nice. (You can get reliable values for resistors.) The BJTs will vary, one amongst another. And they will vary a lot over temperature. And they will drift over time, as well. The values used for $$\\beta\$$ may be different, one part to another, by at least $$\\pm 50\%\$$ around the average, for example. And the base-emitter voltage will also differ, somewhat. Both will vary with temperature and operating currents. Etc.

So while a design may seem good, there is another step to perform. And that is a sensitivity analysis to find the boundaries of how the circuit behaves over the range of its expected applications.

While quantitative sensitivity analysis isn't covered here, I can add a few thoughts to consider:

• Changes in $$\V_{_\text{BE}}\$$, whether it's over temperature or over BJT variations, will tend to narrow or widen the base-emitter voltage and this will impact the quiescent collector current and therefore also the base current. A stiffer base-biasing pair will reduce the impact of these changes.
• Changes in operating $$\\beta\$$, whether it's over temperature or over BJT variations, will similarly affect the base current. So, again, a stiffer base-biasing pair will reduce the impact of these kinds of changes.

So we can conclude that using stiffer biasing for the BJT tends to make the circuit more predictable. But the price is the current wasted through the divider pair and therefore also the power expended. There is always a balancing act made here. But that's what engineering is about. That said, it is better to put quantities to the question. But doing so involves more equations and I think it is time to put this to bed.

• What mr and mi refer to? And is AV prime is the one I use to get Vout? May 19 at 7:03
• @Sabretooth2438 They, mr and mi, will be used when I write the final answer. I haven't done that, yet. I was hoping you'd solve that simultaneous equation, first. And Av prime is used to calculate both RE and ICQ, as are mr and mi. Finally, you've introduced a serious problem when you mentioned RL. You need to clarify that issue, as well. Is it part of the specification? Or not? In fact, RL is such a problem in my mind that I'm not sure I can properly complete an answer until I know what you meant when you meantioned RL.
– jonk
May 19 at 7:06
• @Sabretooth2438 Tell you what. Clarify what you mean by RL, sufficient that I can understand it well enough to move forward, and if you can add anything of note about the source impedance (RS) and whether or not I should care about it, and I'll finish the answer. I won't even wait for your simultaneous solution work. Deal?
– jonk
May 19 at 7:18
• @TonyStewartEE75 To a fairly close approximation, THD drives this equation for this kind of stage: $I_{_{\text{C}_Q}}=\frac12\cdot A_{v}\cdot\frac{V_{T_{_\text{MAX}}}}{R_{_\text{C}}}\left[1+\sqrt{1+\frac43\frac{v_{_\text{PP}}}{V_{T_{_\text{MIN}}}}\frac1{{T\! \small HD}}}\right]$. I've discussed this elsewhere here. It's not for me to ram this into a poor student who is just struggling to bias the stage in the first place.
– jonk
May 20 at 0:02
• @TonyStewartEE75 The equation I wrote out is developed out of carefully crafted theoretical results. So that's the approach I'd take as it is thorough, complete and will "just work right" for the right reasons. I dislike/distrust anything that smells of "rules of thumb," unsupported thoroughly by theoretical infrastructure that proves their approach But in no way does that mean other approaches are necessarily wrongly applied. Just that if they aren't underpinned well (and completely) by theoretical analysis, I (personally) won't trust them much. I have to see the approach to believe.
– jonk
May 20 at 5:22

Why are you overloading the circuit with a 350 ohms load? An audio preamp usually drives an amplifier with an input resistance that is 20 times or more the output impedance of the preamp. Then your 10uF output capacitor will pass 30Hz with much less loss (10uF into 350 ohms= 46Hz.

Reduce the resistance of R3 so that the gain is 7dB and re-bias the base so that the output can swing symmetrically.

I simulated your overloaded circuit: • I used the corner frequency formula Fc=1/(2*pi*(RL+RC)*C) which I got less values, but my I am still not sure is my zout=Rc or Zout=Rc||RL May 18 at 15:52
• Usually the output resistance of an amplifier is much less than the load resistance then the capacitor is calculated for the load resistance. I increased the output capacitor value so that it produces almost no loss at 30Hz. Then the gain of the circuit is a loss of 0.5 times. simulation May 18 at 20:04
• @Sabretooth2438 As you say correctly, output fc = 1/(2*pi*(RL+RC)*C). There is also a highpass filter caused by C1 acting with the input resistance. The two high pass filters have an accumulative effect to determine the overall -3dB frequency. The output resistance is equal to Rc, which is R3 (350R) in the circuit diagram in your question. RL is the load resistance which is separate to the output resistance. Ignoring any signal source resistance, the gain = (Rc//RL)/(RE + re) where re is the emitter's intrinsic resistance which is equal to approx 25mV/ IE at 25 degC. (IE = emitter current). May 18 at 20:40
• What is the load that has a resistance as low as 350 ohms? An amplifier or a recorder has an input resistance of 10 thousand ohms or higher. To have a voltage gain of 6dB then the load resistance must be much higher than the collector resistor value and the emitter resistor value must be a little less than half the collector resistor value. May 18 at 21:22