I like your description better than mine. It is clearer and easier to follow. Good job describing it.

I agree that it is a simple matter to follow your procedure.
However, there are two steps which - for my opinion - need some verification (explanation): Steps 5 and step 11.
As mentioned in his first post, Paul wants "to learn how to design a CE amplifier" - not only to blindly follow some listed steps.
May I give some additional comments?

*Step 5: The gain expression for a CE-amplfier is G=-gm*Rx/(1+gm*Re)
with gm=transconductance; Rc=coll. resistor and Rx=Rc||Rload .
(This is the classical feedback formula: "open-loop gain"/(1+"loop gain".)
Without any load (infinite Rload) and Re bypassed with a large C, this expression reduces to G=-Rc/(1/gm + Re).
When Re>>1/g=Vt/Ic it is a good approximation to calculate the resistor ratio with G≈-Rc/Re.
Remark: When Re is bypassed with C the last expression is the theoretical DC gain - and for stabilization purposes it is common practice to stabilize the circuit with a DC gain of app. "10" - with other words: Re≈0.1*Rc.

* Step 11: It is the purpose of the emitter resistor Re to provide negative DC voltage feedback in order to stabilize the designed DC operating point (Ic, Vce) against (a) large tolerances of the ratio B=Ic/Ib and (b) uncertainties of the needed voltage Vbe (often assumed with Vbe=0.65V...0.7V).
The DC operating point is established using a voltage divider which provides the needed DC voltage at the base node (Vb=Ve+0.7V). For exact calculation of the divider resistors we have to consider the base current Ib (which has very large tolerances, while Ic is set at a fixed and desired value).
In order to reduce the influence of these Ib tolerances upon the desired base voltage Vb, the current through the divider should be at least 10 times larger than Ib.
It would be even better to use a larger factor (20 or 50) - that means: The base voltage should be as stiff as possible (remember: The BJT is a voltage-controlled device Ic=f(Vbe)). Negative DC feedback works best for a stiff DC voltage Vb.
However, in this case the input resistance of the whole circuit would decrease.
That means: We have to find a trade-off between DC operating point stability and a sufficiently high input resistance.
The recommended factor of 10 is the result of such a trade-off.
__________________________________________
I hope these additional explanations can help to better understand the design/working principles of a CE stage.

 

Here is the circuitry

Mod: lightened your image.E

How to solve a CE amplifier problem, with almost everything unknown? The only known things are the Vcc which would be in example VCC=6V, Vbe which is stable 0.7V, and the transistor in example 2N3904 with a β at a random value say 200 according to the datasheets. My goal now is to try to find:
1) the required resistors (except Re, which can be found with Ohm's Law),
2) the capacitors,
3) the Vin (this one i need to know in order to find the maximum voltage required to drive the amplifier without signal distortion on acoustic frequencies),
4) Ic, and Ib, (because Ie = Ic + Ib)
5) the voltages Vc, Ve, Vb and Vce, and
6) the Q point of transistor (this one i need to know because not every datasheet provide DC load lines.)


View attachment 250217

Hello,

I havent been able to go through all the posts here there are too many maybe later i can.
But there are some main ideas for designing a CE amplifier i'll list some.

First, you want to add a load to your circuit. Usually a resistive load is ok to start with. The load reduces the voltage gain so if you dont have a load you dont really know the voltage gain. The load is the load you expect to run the amp with in the end application. If you dont want to add a load for some reason then you will have to take the gain as preliminary and maybe modify something later.

You then establish the DC operating point. The DC operating point is the DC output voltage with no input signal. Generally you shoot for 1/2 of the DC supply voltage, so if you have a 10v battery you shoot for 5v DC output. This allows roughly equal positive and negative output signal excursions.
While doing this however you also want to get the collector resistor to correlate to the output load resistance because if you dont have a low enough value the load will not be able to be driven hard enough. For example, with a 10 Ohm output load and a 1k collector resistor there will be a huge voltage drop. Probably something around 10 or 20 Ohms would be required for RC in that case.
At the same time you might think about the distortion as you mentioned. The distortion mainly comes from the transistor non linearity. To for the transistor to be more linear, an emitter resistor is added (as you have shown already). The emitter resistor reduces distortion but also reduces gain. A rough gain calculation is G=RC/RE and that's rough but you can see how raising RE reduces gain. So there is a tradeoff between gain and distortion, and a gain of 10 is probably a good place to start as that gives the intrinsic transistor gain plenty of working overhead (the transistor can reproduce the input better when it has a lot more intrinsic gain than the gain of the circuit).
During this phase you also have to think about the input resistance. You dont want an input impedance too low or the input drive source will get loaded down too much.
You also want to think a little about the quiescent power consumption. CE amps have a high power consumption even with no input signal because of the DC bias which necessitates significant current through the collector resistor. This can run a battery down fast, but if you are stuck with a CE design there is no way around it except to shut off the power.

Now as for the AC gain, the input and output caps have to be sized to pass the frequencies of interest and usually you can go by the model of a single RC circuit to figure that out.
The emitter bypass cap though (across the emitter resistor) is a bit more picky because it increases AC gain but introduces high frequency gain and distortion. So maybe shoot for a DC gain of 10 and AC gain of 20 or even lower on both.

So there are some things to consider when designing a CE amplifier even for just plain audio.
To recap:
DC bias point.
Input impedance.
Output impedance.
DC gain (resistors involved).
AC gain (all three caps involved).

There is a formula for calculating all the resistors in one shot but you need to think about this for yourself first. I am sure you will make out just fine.

 

Hi Here is the way I learned how to design small signal transistor CE amplifiers.
at the hobby level.

I used a 9v voltage source, as my supply.

Than for calculations I used ( 8 V) for V1.

I chose a DC Gain of 10.

I chose a 10K ohm resistor for RC.

than I used the equation (RC / Gain) to give me the value for RE.

Than I use a voltage drop across RC to be around 1/2 the value of V1.

Now I calculate the value of the current through RC using ohms law.

Than I use that same value of current, and using ohms law to calculate the voltage drop across RE.

Now I add 0.7 v to that value, to calculate the value of voltage at the base with respect to ground. This is the voltage drop across R2.
I make the value of R2 to be around 10 to 20 times greater than RE.

Now one more resistor value to solve for, that's R1.

By using ohms law you calculate the value of current through R2, (Vbase / R2)

Than take the value of V1 and subtract the voltage value across R2, and the result will
be the voltage drop across R1.

finally using ohms law, take the value of voltage drop across R1, divided by the current through R2, and the result will be a value for R1.

However this value may be un realized to find a resitor for, so take the next smallest value closest to that calculated value, and use that to build your circuit from.

Simply use 100Uf or greater values for all coupling capacitors.

At this point do not concern yourself with using an emitter bypass cap, as this will only
complicate your learning experiance at this stage of the game.

All this rough calculations should get you close to a working amp.

Now take a voltage reading at the collector of the transistor, and see how close it is to around 1/2 the supply voltage.
if your within a half a volt either direction, you should be fine for testing your amp.

Now check the voltage at the emitter, and the base terminals in that order and see how close you are to your calculated values.

With these rough calculations your voltage readings are going to be off but yet usable to test your amp.

Now if you have acess to an osciloscope, you can learn how to really design an amp for real life situations.

Set both waveform channells to the same voltage readings.

Start by applying a small signal from a audio generator, with a sinewave at around 100mv pk-pk or just plain old pk. and check your collector output for signal clarification.

If you see any distortion, than begin to decrease your generator output until you get non distorted sine wave.

Now check to see if you have any voltage gain from input at the base terminal to output at the collector terminal.

Now experiment with different values of resistors and signal voltages and just learn how the transistor amp behaves.

What you have just done, is not design a working amp, (as that requires a load), how ever you just learned the basics of how to bias a transistor calculating resistor values to get your transistor in it's linear operating range
.
From there you start SLOWLY advancing to working with driving a load, and getting reauired gain, than you continue to advance in designing for specific output voltage swings across loads, sometimes you need to design for a specific input impedance, as you continue to advance you will see what works and what doesn't work for CE amp designs.

Such as if your driving a very low resistanc load, and you need a large power signal across it, yiu will have abandon CE amps, all together, and design push pull and other configs.

But If you can first learn to bias a transistor into it's linear region, by following the above steps, and all the steps others contributed for you to glean from, you than can begin your journey of transistor circuit design, for practical purposes.

 

I found this formula on AAC. I have a question. What f represents here? A frequency? Because I think this is the formula that i miss the most

 

Look at what f is defined as -- it is the ratio of Vth to Vcc.

But look at those equations. Two of them are nonsensical. Near the top left it says that 1/Rth = 1/Rth + 1/Rth. Does this make any sense?

Then the bottom left one claims that R1 = R1*(Vcc/Vth). Does this make any sense?

 

Look at what f is defined as -- it is the ratio of Vth to Vcc.

But look at those equations. Two of them are nonsensical. Near the top left it says that 1/Rth = 1/Rth + 1/Rth. Does this make any sense?

Then the bottom left one claims that R1 = R1*(Vcc/Vth). Does this make any sense?

I.. Don't... Know. It was from a lesson here in AAC. How would I know if it does make sense or not?

 

Look at what f is defined as -- it is the ratio of Vth to Vcc.

But look at those equations. Two of them are nonsensical. Near the top left it says that 1/Rth = 1/Rth + 1/Rth. Does this make any sense?

Then the bottom left one claims that R1 = R1*(Vcc/Vth). Does this make any sense?

https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/biasing-calculations/

Here is the source im talking about

 

I.. Don't... Know. It was from a lesson here in AAC. How would I know if it does make sense or not?

The first one that does not make sense is high school algebra.
The second one only makes sense if Vcc = Vth, which again is highly unlikely.
You are correct that the original text is wrong. The first one should read:

\( R_{th}\;=\;R1\;||\; R2 \)

The above expression means R1 is in parallel with R2. The way to calculate that is:

\( \cfrac{1}{R_{th}}\;=\;\cfrac{1}{R1}\;+\;\cfrac{1}{R2}\;=\;\cfrac{1\cdot R2\;+\;1\cdot R1}{R1\cdot R2} \)

This simplifies to

\( R_{th}\;=\; \cfrac{R1 \cdot R2}{R1+R2} \)

That is DC electronics 101

The second one should read

\( R1\;=\;\cfrac{R_{th}}{\text f}\;=\;R_{th}\cdot\cfrac{Vcc}{V_{th}} \)

Chalk it up to sloppy editing, and nice catch

 

https://www.allaboutcircuits.com/textbook/semiconductors/chpt-4/biasing-calculations/

Here is the source im talking about

Thanks. I'll bring this to the attention of EETech. I don't know when someone will get around to cleaning it up.

 

I.. Don't... Know. It was from a lesson here in AAC. How would I know if it does make sense or not?

You need to look at them and ask if they make sense. You need to get in the habit of doing this habitually, both with your own work (because you WILL make mistakes) and when you are drawing upon the work of others (because others WILL make mistakes).

Your ability to answer the question about whether or not something makes sense will vary and evolve. But when basic arithmetic/algebra mistakes are made, you should be able to spot these. The next type of "make sense" evaluation you should get in the habit of making are extreme-case estimates. Often times the end equation for a system is so complicated that it's nearly impossible to tell if the full thing makes sense. But look at the limiting cases. What if one of resistors is shorted (set to zero resistance)? What if one is opened (set to infinite resistance)? What if all the resistances are set to the same value? Often times the response for special situation can be determined by inspection. When you then apply those constraints to the complex full-solution equation, does it reduce to what it needs to for the special cases? If it doesn't, then something doesn't make sense. Identifying WHAT doesn't make sense is a critical step if figuring out WHY it doesn't make sense, which will usually provide strong clues as to what needs to be done to fix it so that it does make sense.

 

You need to look at them as ask if they make sense. You need to get in the habit of doing this habitually, both with your own work (because you WILL make mistakes) and when you are drawing upon the work of others (because others WILL make mistakes).

You ability to answer the question about whether or not something makes sense will vary. But when basic arithmetic/algebra mistakes are made, you should be able to spot these. The next type of "make sense" evaluation you should get in the habit of making are extreme-case estimates. Often times the end equation for a system is so complicated that it's nearly impossible to tell if the full thing makes sense. But look at the limiting cases. What if one of resistors is shorted (set to zero resistance)? What if one is opened (set to infinite resistance)? What if all the resistances are set to the same value? Often times the response for special situation can be determined by inspection. When you then apply those constraints to the complex full-solution equation, does it reduce to what it needs to for the special cases? If it doesn't, then something doesn't make sense. Identifying WHAT doesn't make sense is a critical step if figuring out WHY it doesn't make sense, which will usually provide strong clues as to what needs to be done to fix it so that it does make sense.

Thank you for the correction. I will try to study these mistakes too

 

The first one that does not make sense is high school algebra.
The second one only makes sense if Vcc = Vth, which again is highly unlikely.
You are correct that the original text is wrong. The first one should read:

\( R_{th}\;=\;R1\;||\; R2 \)

The above expression means R1 is in parallel with R2. The way to calculate that is:

\( \cfrac{1}{R_{th}}\;=\;\cfrac{1}{R1}\;+\;\cfrac{1}{R2} \)

This simplifies to

\( R_{th}\;=\; \cfrac{R1 \cdot R2}{R1+R2} \)

That is DC electronics 101

Oh, i see. Thank you. It's just im starting with electronics, and the only formula i know for parallel resistor circuitry is Rtotal = (R1×R2)/(R1+R2). The formula 1/Rtotal = (1/R1)+(1/R2)+(1/Rn) is very uncommon for me to use.

 

Oh, i see. Thank you. It's just im starting with electronics, and the only formula i know for parallel resistor circuitry is Rtotal = (R1×R2)/(R1+R2). The formula 1/Rtotal = (1/R1)+(1/R2)+(1/Rn) is very uncommon for me to use.

The second one is the "correct" one -- or more properly the more foundational one. For a parallel circuit of resistors, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistors. A better way to put it is that, for resistors in parallel, the equivalent conductance is the sum of the individual conductances. If you have encountered conductance, it is merely the reciprocal of the resistance. The resistance tells you how much the component resists current flowing through it, while the conductance tells you how much the component allows current to flow though it.

The formula you are familiar with is merely the more general one under the constraint that there are exactly two resistors.

 

Oh, i see. Thank you. It's just im starting with electronics, and the only formula i know for parallel resistor circuitry is Rtotal = (R1×R2)/(R1+R2). The formula 1/Rtotal = (1/R1)+(1/R2)+(1/Rn) is very uncommon for me to use.

Hello there,

You can learn a lot about a transistor by considering that it is just a current controlled current source.
That is, as you increase base current the collector current also increases by the factor called "Beta" in the linear mode. That makes the first equation simple:
iC=iB*Beta
Beta is just the current gain we take to be constant for some time.
So if you know iB, you know iC, and since iE is the sum, we have:
iE=iB+iC

Those simple equations will get you quite far in understanding the operation of a transistor.
You can consider the base emitter voltage to be a constant 0.7v or you can start off with zero. Either way you will gain a lot of understanding about this.

If you know the base current you multiply that by Beta and that gives you the collector current, and if you have a collector resistor then that resistor drops more voltage as you increase the base current. That means the collector voltage falls (NPN). Thus, for some small base current you get a larger collector current and thus a large change in output voltage.
Now how did we get that base current.
We get it by applying a voltage though an impedance (could be just a resistor for now) connected in series with the base emitter.. When we increase the input voltage that gives us an increase in base current and thus a larger change in collector current. Because we only need a small base current we can apply a small voltage and see the collector voltage change by a lot more. That's voltage gain. As you thought, the gain is Vout/Vin.

So that's your basic analysis. It's actually very rewarding to calculate the different things about a CE amplifier either for audio work, RF work, or even to act as a switch. The operation of the switch application is a little different and it's actually even simpler because we just apply two different voltages to the input such as 0v and 3v. We then look at the output voltage and see it switch from maybe 10v down to 0.2 to 0.5v usually. That's how the switch application works.

If you compile all these things together you can create equations that solve for every single resistor in the circuit and even the capacitors given the specifications you need for the amplifier.

 

Hello there,

You can learn a lot about a transistor by considering that it is just a current controlled current source.
That is, as you increase base current the collector current also increases by the factor called "Beta" in the linear mode. That makes the first equation simple:
iC=iB*Beta
Beta is just the current gain we take to be constant for some time.
So if you know iB, you know iC, and since iE is the sum, we have:
iE=iB+iC

Those simple equations will get you quite far in understanding the operation of a transistor.
You can consider the base emitter voltage to be a constant 0.7v or you can start off with zero. Either way you will gain a lot of understanding about this.

If you know the base current you multiply that by Beta and that gives you the collector current, and if you have a collector resistor then that resistor drops more voltage as you increase the base current. That means the collector voltage falls (NPN). Thus, for some small base current you get a larger collector current and thus a large change in output voltage.
Now how did we get that base current.
We get it by applying a voltage though an impedance (could be just a resistor for now) connected in series with the base emitter.. When we increase the input voltage that gives us an increase in base current and thus a larger change in collector current. Because we only need a small base current we can apply a small voltage and see the collector voltage change by a lot more. That's voltage gain. As you thought, the gain is Vout/Vin.

So that's your basic analysis. It's actually very rewarding to calculate the different things about a CE amplifier either for audio work, RF work, or even to act as a switch. The operation of the switch application is a little different and it's actually even simpler because we just apply two different voltages to the input such as 0v and 3v. We then look at the output voltage and see it switch from maybe 10v down to 0.2 to 0.5v usually. That's how the switch application works.

If you compile all these things together you can create equations that solve for every single resistor in the circuit and even the capacitors given the specifications you need for the amplifier.

If it would be easier done than said. In example, my first post that i sent in this thread, on how to calculate CE amplifier with everything unknown. And im sure that lots of people who are interested in electronics and entering there, would also ask for this question. To calculate a CE amplifier there are at least 7 or 10 formulas, if not more. Here are some of them:
Ic=β×Ib. Consider β to be 100. The Ib? The Ib is equal to Ib=(Vbb-0.7)/(Rb+Re(101)). The Rb=(R1×R2)/(R1+R2). These values R1 and R2 are unknown, so you can't use Rb formula, and then you can't use Ib formula. But then you can't use the Ic formula. Vce=Vcc-Ieq(Rc+Re). This formula whole can't be used because there are too many unknown values. You don't know what is Ieq equal to, neither Rc or Re, because Re depends from Ic current and because Ic almost equal to Ie, with ohms law, you can find Re. Vbb = Vcc(R1/(R1+R2)). Can't be used too. If you have only Vcc as known, say 6V, but the others you don't even know how to calculate.. Then what do i do? Ic(sat) = Vcc / (Rc+Re). Can't be used because of unknown values. Ieq = (1+β)*Ib. Can't be used. Ve = Vbb - Vbe. Can't be used. Vbb is unknown Ve is too.. So what to do? I have read many, many books about these unknown values.. But I can't find any information. For example let's suppose for now that Rb is 47K. There are too many variations you can do to create a 47K voltage divider. So how to know which value best fits for R1 and R2? Is the Vbb equal to the voltage drop of R2? If yes then what purpose does R1 serves? How to calculate the voltage drop for R1. How to calculate the current flowing through R1 and R2? Is R1 an important resistor? 2 more questions. Does C1 and C2 coupling capacitors serve as frequency filters (from 20-20Khz)? Or they used to just couple the next stages? If yes, then why we don't use 10nF or 1000uF as any capacitor blocks the DC and allow AC to pass? If no, then is it the Cb bypass capacitor the one that acts as frequency filter designed to allow acoustic frequency to pass? If yes, then how a single capacitor manage to control such a large difference of different frequencies such as 20-20khz? So many questions and these questions i can't solve on any book i read. The books i read have a brief idea of how a CE works. There are no detailed explanations on how to design a CE amplifier. But even if there is, there are always already known values. I don't need that. With this way i can't learn properly. If these books doesn't give information, then at least provide the formulas so i could calculate by myself. I want to learn, but this is really frustrating. At the beginning of each chapter, there always say "in this chapter you will learn how to calculate CE amplifier, define voltage divider" and so on. But when i read the whole chapter, there is practically nothing from what the beginning of the chapter is promising.. And after that i have to search hundreds of books, and still don't get the answer. Sometimes I just ask myself, why are these books even has been written when they are not able to teach what they promise you to teach. So after that i have no other choice but pay electronics tutorial a loads of money just to learn something that books don't learn. And this is the exact reason why I hate reading and try to teach myself through experiments.

 

If it would be easier done than said. In example, my first post that i sent in this thread, on how to calculate CE amplifier with everything unknown. And im sure that lots of people who are interested in electronics and entering there, would also ask for this question. To calculate a CE amplifier there are at least 7 or 10 formulas, if not more. Here are some of them:
Ic=β×Ib. Consider β to be 100. The Ib? The Ib is equal to Ib=(Vbb-0.7)/(Rb+Re(101)). The Rb=(R1×R2)/(R1+R2). These values R1 and R2 are unknown, so you can't use Rb formula, and then you can't use Ib formula. But then you can't use the Ic formula. Vce=Vcc-Ieq(Rc+Re). This formula whole can't be used because there are too many unknown values. You don't know what is Ieq equal to, neither Rc or Re, because Re depends from Ic current and because Ic almost equal to Ie, with ohms law, you can find Re. Vbb = Vcc(R1/(R1+R2)). Can't be used too. If you have only Vcc as known, say 6V, but the others you don't even know how to calculate.. Then what do i do? Ic(sat) = Vcc / (Rc+Re). Can't be used because of unknown values. Ieq = (1+β)*Ib. Can't be used. Ve = Vbb - Vbe. Can't be used. Vbb is unknown Ve is too.. So what to do? I have read many, many books about these unknown values.. But I can't find any information. For example let's suppose for now that Rb is 47K. There are too many variations you can do to create a 47K voltage divider. So how to know which value best fits for R1 and R2? Is the Vbb equal to the voltage drop of R2? If yes then what purpose does R1 serves? How to calculate the voltage drop for R1. How to calculate the current flowing through R1 and R2? Is R1 an important resistor? 2 more questions. Does C1 and C2 coupling capacitors serve as frequency filters (from 20-20Khz)? Or they used to just couple the next stages? If yes, then why we don't use 10nF or 1000uF as any capacitor blocks the DC and allow AC to pass? If no, then is it the Cb bypass capacitor the one that acts as frequency filter designed to allow acoustic frequency to pass? If yes, then how a single capacitor manage to control such a large difference of different frequencies such as 20-20khz? So many questions and these questions i can't solve on any book i read. The books i read have a brief idea of how a CE works. There are no detailed explanations on how to design a CE amplifier. But even if there is, there are always already known values. I don't need that. With this way i can't learn properly. If these books doesn't give information, then at least provide the formulas so i could calculate by myself. I want to learn, but this is really frustrating. At the beginning of each chapter, there always say "in this chapter you will learn how to calculate CE amplifier, define voltage divider" and so on. But when i read the whole chapter, there is practically nothing from what the beginning of the chapter is promising.. And after that i have to search hundreds of books, and still don't get the answer. Sometimes I just ask myself, why are these books even has been written when they are not able to teach what they promise you to teach. So after that i have no other choice but pay electronics tutorial a loads of money just to learn something that books don't learn. And this is the exact reason why I hate reading and try to teach myself through experiments.

There exists a straightforward design procedure that starts with things you can specify or choose and proceeds in a systematic way to a finished design. Using this procedure does require some insight into the why and the how of defining the requirements and making the initial choices. I would suggest that before you can design you need to be comfortable with the analysis. I don't think you can get by without that step.

 

If it would be easier done than said. In example, my first post that i sent in this thread, on how to calculate CE amplifier with everything unknown. And im sure that lots of people who are interested in electronics and entering there, would also ask for this question. To calculate a CE amplifier there are at least 7 or 10 formulas, if not more. Here are some of them:
Ic=β×Ib. Consider β to be 100. The Ib? The Ib is equal to Ib=(Vbb-0.7)/(Rb+Re(101)). The Rb=(R1×R2)/(R1+R2). These values R1 and R2 are unknown, so you can't use Rb formula, and then you can't use Ib formula. But then you can't use the Ic formula. Vce=Vcc-Ieq(Rc+Re). This formula whole can't be used because there are too many unknown values. You don't know what is Ieq equal to, neither Rc or Re, because Re depends from Ic current and because Ic almost equal to Ie, with ohms law, you can find Re. Vbb = Vcc(R1/(R1+R2)). Can't be used too. If you have only Vcc as known, say 6V, but the others you don't even know how to calculate.. Then what do i do? Ic(sat) = Vcc / (Rc+Re). Can't be used because of unknown values. Ieq = (1+β)*Ib. Can't be used. Ve = Vbb - Vbe. Can't be used. Vbb is unknown Ve is too.. So what to do? I have read many, many books about these unknown values.. But I can't find any information. For example let's suppose for now that Rb is 47K. There are too many variations you can do to create a 47K voltage divider. So how to know which value best fits for R1 and R2? Is the Vbb equal to the voltage drop of R2? If yes then what purpose does R1 serves? How to calculate the voltage drop for R1. How to calculate the current flowing through R1 and R2? Is R1 an important resistor? 2 more questions. Does C1 and C2 coupling capacitors serve as frequency filters (from 20-20Khz)? Or they used to just couple the next stages? If yes, then why we don't use 10nF or 1000uF as any capacitor blocks the DC and allow AC to pass? If no, then is it the Cb bypass capacitor the one that acts as frequency filter designed to allow acoustic frequency to pass? If yes, then how a single capacitor manage to control such a large difference of different frequencies such as 20-20khz? So many questions and these questions i can't solve on any book i read. The books i read have a brief idea of how a CE works. There are no detailed explanations on how to design a CE amplifier. But even if there is, there are always already known values. I don't need that. With this way i can't learn properly. If these books doesn't give information, then at least provide the formulas so i could calculate by myself. I want to learn, but this is really frustrating. At the beginning of each chapter, there always say "in this chapter you will learn how to calculate CE amplifier, define voltage divider" and so on. But when i read the whole chapter, there is practically nothing from what the beginning of the chapter is promising.. And after that i have to search hundreds of books, and still don't get the answer. Sometimes I just ask myself, why are these books even has been written when they are not able to teach what they promise you to teach. So after that i have no other choice but pay electronics tutorial a loads of money just to learn something that books don't learn. And this is the exact reason why I hate reading and try to teach myself through experiments.

It seems like the approach you are trying to use is to throw equations you've seen somewhere at the problem hoping that somewhere in that process you'll stumble across one that might be useful.

There's a reason that analysis comes before design. If you want to design a CE amplifier (or whatever) you need to first come up with your desired outcomes. Otherwise it would be like asking someone how to build a wall. It's an unanswerable question. A wall isn't built to be a wall (well, some artists might beg to differ, but...). A wall is built to serve a purpose and you can't design the wall unless you start with the purpose. Is the purpose to finish a basement? Is the purpose to enclose a flower bed? Is the purpose to support a driveway? Is the purpose to hold back a river to create a reservoir? Very different purposes, very different walls.

The same with your CE amplifier. What is it's purpose? To amplify music and drive a set of headphones? To amplify the 100 MHz signal from a radio antenna? What is the input signal coming from? How big is the input signal? What is the output driving? How big does the output signal need to be? Answering these questions will allow you to start figuring out a set of performance specifications such the gain needed, the input impedance, the output impedance, the input voltage swing, the output voltage swing, the output current drive capability. Those are some of the most basic ones. Then you have others such as the bandwidth (and potentially several other frequency-related ones) and the linearity and others that get at how good does the quality of the amplification need to be in order to be considered "good enough".

Once you have a set of working specs, at least the main ones, you are ready to start designing. But this requires that you have knowledge of how the different components in your amplifier affect the performance of that amplifier against the target specs. That knowledge comes from analyzing CE amplifiers and playing with the results in order to get a feel for what parameters have the greatest impact on which performance results.

To really get a feel for that, you need to not only be able to analyze the DC behavior of the circuit, but also the small-signal (aka, the AC) behavior as well. That's were the very concept of the transconductance parameter that is so central to the design of these kinds of amplifiers comes from. Unless you spend the time to study these topics and do the analysis, you will be stuck trying to throw equations that someone else came up with and that you don't understand at a problem hoping that something, somehow, happens to work at some point.

 

So what to do? I have read many, many books about these unknown values.. But I can't find any information.
:::::::::::::::::::::::::::::::::::::
So many questions and these questions i can't solve on any book i read. The books i read have a brief idea of how a CE works. There are no detailed explanations on how to design a CE amplifier. But even if there is, there are always already known values. I don't need that. With this way i can't learn properly. If these books doesn't give information, then at least provide the formulas so i could calculate by myself. I want to learn, but this is really frustrating.
::::::::::::::::::::::::::::::::::::
And after that i have to search hundreds of books, and still don't get the answer. Sometimes I just ask myself, why are these books even has been written when they are not able to teach what they promise you to teach. So after that i have no other choice but pay electronics tutorial a loads of money just to learn something that books don't learn. And this is the exact reason why I hate reading and try to teach myself through experiments.

Paul, let me first say that I do understand your frustation. But I have a question to you:
From your text, I have derived that you have consulted many many books. Did you notice that in some books the BJT is described and explained as a CURRENT-CONTROLLED device and in some other books as a VOLTAGE_CONTROLLED device?

To me, THIS fact is really frustrating. It is really surprising (and frustrating) that even 70 years after the BJT was introduced two different opinions still exist about the basic working principles of this device.
You can be sure - only ONE explanation is correct: Undoubtly, the BJT is VOLTAGE-CONTROLLED.
The basic relation is: Ic=Is[exp(Vbe/Vt)-1].
And all classical steps for designing BJT based gain stages reflect this fact!
This "Ebers-Moll equation" (derived from Shockleys famous formula desribing the voltage-current relation for any pn junction) is, therefore, implementred in all electronic simulation programs.
There are many examples, explanations, formulas, observations and physical constants which clearly proove that Ic does depend only on the voltage Vbe - and the base current Ib=Ic/B is nothing else than an unwanted side effect. I cannot understand that still there are people who think that a very small current should be able to control a current that is 200 times larger. Even from the energy aspect this is impossible!

Of course, the base current Ib does exist and can be considered during design of the biasing network - but it has no controlling function at all.
All people and books stating that the BJT would be current-controlled cannot present any evidence - it is nothing else than a claim! A formula Ic=B*Ib (derived from Ib=Ic/B) cannot say anything about cause and effect. Of course, this also applies to BJT switching applications. It is simply false to say that a large base current wouls drive the BJT into saturation. A large base current is the RESULT of saturation but never its cause.

Knowing this, it is not a problem to design a CE amplifying stage. And the "secret" consists of "negative feedback" - in most cases an emitter resistor RE is used for this purpose. Negative feedback has the big advantage that the circuit performance (DC bias point, voltage gain,...) has a remarkably reduced sensitivity to the BJT parameters. In particular, it is not too important if your calculation assumes Vbe=0.7 volts or Vbe=0.75 volts ....
(Sensitivity of Ic against this voltage is one of the arguments of the "current-control-party" against voltage control - not really an argument!).
If you want I can list the main steps for designing a CE stage ...it is really not a problem.....

Short summary: It is really not a big problem to design a well-working gain stage....and it is pretty logical to see how and why the relevant formulas are to be applied. HOWEVER - the most important step is to UNDERSTAND HOW THE TANSISTOR REALLY WORKS! Without such an understanding it must be "frustrating" to follow blindly some formulas/relationships which can be found in the literature.

PS: Did you consult also the electronic engineers "bible" ("The Art of Electronics" , Horowitz/Hill) ?
Quote: "But to understand differential amplifiers,......and other important applications you must think of the transistor as a transconductance device - collector current is determined by base-to-emiter voltage" (End of quote).

 

If it would be easier done than said. In example, my first post that i sent in this thread, on how to calculate CE amplifier with everything unknown. And im sure that lots of people who are interested in electronics and entering there, would also ask for this question. To calculate a CE amplifier there are at least 7 or 10 formulas, if not more. Here are some of them:
Ic=β×Ib. Consider β to be 100. The Ib? The Ib is equal to Ib=(Vbb-0.7)/(Rb+Re(101)). The Rb=(R1×R2)/(R1+R2). These values R1 and R2 are unknown, so you can't use Rb formula, and then you can't use Ib formula. But then you can't use the Ic formula. Vce=Vcc-Ieq(Rc+Re). This formula whole can't be used because there are too many unknown values. You don't know what is Ieq equal to, neither Rc or Re, because Re depends from Ic current and because Ic almost equal to Ie, with ohms law, you can find Re. Vbb = Vcc(R1/(R1+R2)). Can't be used too. If you have only Vcc as known, say 6V, but the others you don't even know how to calculate.. Then what do i do? Ic(sat) = Vcc / (Rc+Re). Can't be used because of unknown values. Ieq = (1+β)*Ib. Can't be used. Ve = Vbb - Vbe. Can't be used. Vbb is unknown Ve is too.. So what to do? I have read many, many books about these unknown values.. But I can't find any information. For example let's suppose for now that Rb is 47K. There are too many variations you can do to create a 47K voltage divider. So how to know which value best fits for R1 and R2? Is the Vbb equal to the voltage drop of R2? If yes then what purpose does R1 serves? How to calculate the voltage drop for R1. How to calculate the current flowing through R1 and R2? Is R1 an important resistor? 2 more questions. Does C1 and C2 coupling capacitors serve as frequency filters (from 20-20Khz)? Or they used to just couple the next stages? If yes, then why we don't use 10nF or 1000uF as any capacitor blocks the DC and allow AC to pass? If no, then is it the Cb bypass capacitor the one that acts as frequency filter designed to allow acoustic frequency to pass? If yes, then how a single capacitor manage to control such a large difference of different frequencies such as 20-20khz? So many questions and these questions i can't solve on any book i read. The books i read have a brief idea of how a CE works. There are no detailed explanations on how to design a CE amplifier. But even if there is, there are always already known values. I don't need that. With this way i can't learn properly. If these books doesn't give information, then at least provide the formulas so i could calculate by myself. I want to learn, but this is really frustrating. At the beginning of each chapter, there always say "in this chapter you will learn how to calculate CE amplifier, define voltage divider" and so on. But when i read the whole chapter, there is practically nothing from what the beginning of the chapter is promising.. And after that i have to search hundreds of books, and still don't get the answer. Sometimes I just ask myself, why are these books even has been written when they are not able to teach what they promise you to teach. So after that i have no other choice but pay electronics tutorial a loads of money just to learn something that books don't learn. And this is the exact reason why I hate reading and try to teach myself through experiments.

Hello again,

I can see you are a deep thinker and that's good because if you keep at it you will eventually find what you are looking for, if not right now. I too have been bugged by the lack of good, general ways to handle some simple electronic problems and so i went after some of those problems and came up with solutions. The solution to the CE amplifier comes from considering everything at the same time. That means relating one thing to another until you can single out each resistor or other part. Circuit synthesis is the reverse of analysis. Once you know how to analyze you can begin to find was to calculate the different components. You then end up with a process as well as a group of formulas. For a really simple example:
R1=R2+R3
and you dont yet know what R2 and R3 are so how do you calculate R1.
Well, in some cases you dont calculate R1, yet, you just use that R2+R3 in the next equation where you see R1 and then you may end up with a formula for R2 and R3, which is now void of R1, and then you can go back and calculate R1 with no problem.
This is what we can do with the CE amplifier, and it starts with the DC bias point.

The formulas you have presented here are probably all correct. The only missing thing is you have to just think a little bit more and see how to simplify things so much that you end up with an equation for one resistor, or something close to that. Sometimes solutions come out as ratios though, because more than one value for two or more parts may work the same or nearly the same. So then you have to choose one (say resistor) and then calculate the other. So you may choose 1k, 2.2k , etc., then calculate the other resistor.

I can show you a procedure for calculating the resistors for a CE amplifier if you like, it's not all that difficult, but it will be a set of formulas and a procedure how to use them It takes maybe 5 minutes. The only thing is though that i dont think this will answer your more general question of how to go about solving circuits LIKE this as well as this one. That would require you to learn a little more about how to analyze a circuit. Not super difficult either though, but you need algebra and that's all for now.

As to the capacitor values you talked about, the coupling capacitors are chosen for not just one but two reasons. The first is to remove the DC component so you just get the AC signal and that means it does not alter the DC bias point of either circuit being coupled. The second is it has to be of large enough value to be able to pass the LOWEST frequency you need to amplify without too much attenuation. This is actually a simple problem too.
The choice of just using a capacitor solves the first problem: the coupling. That's done.
The choice of value is based on frequency, and so you have to do another little equation for the impedance of the capacitor. The impedance is Z=1/(j*w*C) where 'j' is the imaginary operator. In may cases though you can estimate this as Z=R=1/(w*C) (w=2*pi*frequency). that gives you some idea how it will act with different frequencies.
For example, if we approximate 'pi' as simply '3', we can write:
R=1/(2*3*f*C) which is R=1/(6*f*C)
Now for a frequency 'f' of 100 Hertz we get:
R=1/(600*C)
and solving for C we get:
C=1/(600*R)
and so if we want the capacitor to have low impedance 'R' we make R a low value like maybe 100. That gives us:
C=1/(60000)
and that comes out to about 17uf so we might use a 22uf capacitor.

So there is a logic to it, it just takes a little more effort and some analysis.

If you'd like to see the procedure for biasing a CE amplifier, i can show you that, but it will just be a formula or two, somewhat longer than you've seen so far, that takes into account more of the other components too. You may not understand where this formula came from though, unless you are willing to do a little more analysis, but that just means using a lot of symbolic algebra which modern software does pretty good with ease.
In fact, why dont i show you how to develop a simpler aspect of this process so you can get a better idea how it all comes together. I think you may have actually done some of this already.

 

Paul, let me first say that I do understand your frustation. But I have a question to you:
From your text, I have derived that you have consulted many many books. Did you notice that in some books the BJT is described and explained as a CURRENT-CONTROLLED device and in some other books as a VOLTAGE_CONTROLLED device?

To me, THIS fact is really frustrating. It is really surprising (and frustrating) that even 70 years after the BJT was introduced two different opinions still exist about the basic working principles of this device.
You can be sure - only ONE explanation is correct: Undoubtly, the BJT is VOLTAGE-CONTROLLED.
The basic relation is: Ic=Is[exp(Vbe/Vt)-1].
And all classical steps for designing BJT based gain stages reflect this fact!
This "Ebers-Moll equation" (derived from Shockleys famous formula desribing the voltage-current relation for any pn junction) is, therefore, implementred in all electronic simulation programs.
There are many examples, explanations, formulas, observations and physical constants which clearly proove that Ic does depend only on the voltage Vbe - and the base current Ib=Ic/B is nothing else than an unwanted side effect. I cannot understand that still there are people who think that a very small current should be able to control a current that is 200 times larger. Even from the energy aspect this is impossible!

Of course, the base current Ib does exist and can be considered during design of the biasing network - but it has no controlling function at all.
All people and books stating that the BJT would be current-controlled cannot present any evidence - it is nothing else than a claim! A formula Ic=B*Ib (derived from Ib=Ic/B) cannot say anything about cause and effect. Of course, this also applies to BJT switching applications. It is simply false to say that a large base current wouls drive the BJT into saturation. A large base current is the RESULT of saturation but never its cause.

Knowing this, it is not a problem to design a CE amplifying stage. And the "secret" consists of "negative feedback" - in most cases an emitter resistor RE is used for this purpose. Negative feedback has the big advantage that the circuit performance (DC bias point, voltage gain,...) has a remarkably reduced sensitivity to the BJT parameters. In particular, it is not too important if your calculation assumes Vbe=0.7 volts or Vbe=0.75 volts ....
(Sensitivity of Ic against this voltage is one of the arguments of the "current-control-party" against voltage control - not really an argument!).
If you want I can list the main steps for designing a CE stage ...it is really not a problem.....

Short summary: It is really not a big problem to design a well-working gain stage....and it is pretty logical to see how and why the relevant formulas are to be applied. HOWEVER - the most important step is to UNDERSTAND HOW THE TANSISTOR REALLY WORKS! Without such an understanding it must be "frustrating" to follow blindly some formulas/relationships which can be found in the literature.

PS: Did you consult also the electronic engineers "bible" ("The Art of Electronics" , Horowitz/Hill) ?
Quote: "But to understand differential amplifiers,......and other important applications you must think of the transistor as a transconductance device - collector current is determined by base-to-emiter voltage" (End of quote).

I think i have this book too, and if I remember well, i found no useful information about what i search. But I will try to search again. Maybe I missed something