Again, thank you so much for what you are doing. It's not in vane; now I can see where I began to get lost.

It doesn't matter much if it won't work in real life; it was just an example chosen at random (something at hand). The main point was learning how you decide and do the calculations for a common-emitter amp.

I've also been all night trying to find a single formula that would give me Rc, and I was unable to find it. Every single place I looked literally said "you chose Ic" and then calculated Rc accordingly; and it puzzled me how Ic could be just a matter of "choice". Anyway, the closest I got to finding a formula were the ones you posted "Rc = 0.1*RL" and another one in a page that I was unable to find again (I even think it was a thread on this forum), on which I could get Rc through this formula "Ic = (Vcc - Vce) / RL".

In any case, this is as far as I got yesterday with the experiment. It works!... sort of. I'll explain what I did later on (in case anyone is interested). And I'll get right away studying all the calculations and considerations on your post.

 

And I'm again stuck with "Then choose an emitter resistor for a desired collector current (since Ic ≈ Ie)." What's the Ic; the maximum current through the load (the 200mA)?
I always get stuck because of the quiescent Ic.

You choose Ic based on multiple considerations. A transistor's datasheet shows many characteristics that change depending on Ic, such as β, collector-emitter saturation voltage, base-emitter on voltage, etc. Emitter resistance re goes down with increasing Ic. Input resistance increases with increasing β and increasing re (but an emitter resistor Re will dominate). Collector output resistance increases with decreasing Ic (although the amplifier's output resistance will usually be dominated by Rc). And of course you have to keep Ic within the transistor's operating range.

If you're making a voltage amplifier you generally want high input resistance, so high β is good. Check your transistor's datasheet to find out a good Ic value for high β. Then add an emitter follower to the output to buffer the high output resistance of the CE amp and provide an output with low resistance.

 

Here you have a example of a amplifier you should never build.
You want 0.3W of power at 8Ω speaker. So we need 200mA of a RMS current or 283mA peak current. This correlates with a voltage drop across the speaker 2.27V. So we need Vcc larger then 4.55V. But in real life we need much more than this 5V. It's impassible to achieve such a a large voltage swing without huge distortions.

I see that I began making a very basic mistake in my calculations; I got V from the RMS current needed for the load, and not from the peak current.

So we normally select quiescent Ic current much larger than desired load current.
Icq >> ILoad. I pick Icq = 0.5A

This is the step that most confuses me; and for 2 reasons:

1 - You seem to just "choose" Icq, instead of calculating it. Of course, you guys have the knowledge and experience to do that; a beginner like me can't even dream of doing such things.

2 - I would imagine that Icq would be at least half of the 283mA. Why? because if I were to use the characteristics curve graph to find the quiescent current, I would put the maximum current to the load at the top of the Y axis, and then trace a line down to Vcc in the X axis; which by force would give me Icq around half the maximum load current (283mA) -or even lower than half.

I'm guess that the approach I'm describing is for a common-emitter configuration in which the load goes in series with the collector-emitter (Load in place of Rc) and not in parallel (like we are doing now).

Rc = 2.5V/0.5A = 4.7Ω The power dissipation in Rc resistor is 1.2W so we need 2W resistor.
For this Icq value the maximum positive current at load is equal to:
IL_max = Icq * Rc/(Rc + RL) = 185mA peak.
Next we need BJT type we want to use I pick BD243
http://hvt.bme.hu/~beged/elakszim/bd243.pdf
The power dissipation in BJT will also be equal to 1.2W (Icq*Vceq), so we don't need any heat sink. Transistor case will be very hot.
Transistor case will reaches 76°C (Tc = (P * Rthja) + Ta) degrees higher temperature then the ambient temperature.
Of course we need a capacitor in series with the speaker.
C = 0.16/(F*R) = 0.16/(20Hz*8Ω) = 1000μF
Now we need to choose RB resistor. This type of a biasing network you choose (single RB resistor) is very poor and should never be used in real life. Why? Because of the transistor beta variation. Every transistor will have a slightly different current gain. Also we don't know exact value for beta the transistor we want to use. Of course we can use a typical value from data sheet. For example for BD234 the beta can vary from typical value hfe = 100 for Ic = 0.5A to hfe = 20. So what we can do is to choose
Rb = (5V - 0.7V)/(Icq/hfe) = 4.3V/(0.5A/100) = 4.3V/5mA = 860Ω I peak 820Ω or 910Ω from the standard E24 series.
And next we are force to use "Edison method" and select Rb resistor value on the bench to get Vce = 2.5V.
Also we need a input cap
Cin = 0.16/(F * rin).
So we have this poorman amplifier we should never build.

Additional I add Re resistor to reduce the voltage gain. I also include the LTspice file.

The rest I understand it; perhaps with the exception of this formula:

"For this Icq value the maximum positive current at load is equal to: IL_max = Icq * Rc/(Rc + RL) = 185mA peak."
​

And why you chose 2.5V for the mid point of the voltage swing. Is 0.5*Vcc a rule of thumb, or could I have chosen a voltage that is at least the 2.27V that we needed for the voltage swing? (disregarding distortions and other factors)

And next we are force to use "Edison method" and select Rb resistor value on the bench to get Vce = 2.5V.

One last clarification: I assume you meant Re -since you already calculated Rb-; which I read should be Re = 0.1 * Rc.

 

You choose Ic based on multiple considerations. A transistor's datasheet shows many characteristics that change depending on Ic, such as β, collector-emitter saturation voltage, base-emitter on voltage, etc. Emitter resistance re goes down with increasing Ic. Input resistance increases with increasing β and increasing re (but an emitter resistor Re will dominate). Collector output resistance increases with decreasing Ic (although the amplifier's output resistance will usually be dominated by Rc). And of course you have to keep Ic within the transistor's operating range.

If you're making a voltage amplifier you generally want high input resistance, so high β is good. Check your transistor's datasheet to find out a good Ic value for high β. Then add an emitter follower to the output to buffer the high output resistance of the CE amp and provide an output with low resistance.

All those considerations for choosing Ic are way beyond my current knowledge. Is not that I don't know what those concepts are; I just wouldn't be able to consider how they would influence my choice of Ic unless I have a formula (or various formulas) taking them all into account.

 

Why you try to find a formula for Rc? I know that is is very common for the beginners that they look "formula" for everything. But you don't find "formula" for every component in the circuit. Sometimes we as a designer need to choose some component values and use trial and error to see if we meet our designing goals.
But sometimes we can create our formulas simply be using a math.
For example if our main goal is to achieve P = 0.3W for a given RL resistance at Vce = 0.5Vcc.
We can write this

IL_max = Icq*Rc/(Rc + RL) (1)

Ic = 0.5Vcc/(Rc) (2)

And after we solve this we get formula for Rc:

\(\LARGE R_c = \frac{0.5Vcc}{IL_{max}} - R_L\)

 

Probably because the simple explanations, while useful, are technically incomplete. Books tend to be more in depth.

No,the books assume you already know the basis of all types of amplifying devices.

Transistors in Common Emitter,FETs in Common Source,Vacuum tubes in Common Cathode,are all basically Voltage Dividers,with one section of the divider variable by the application of a control signal.

Representation of these devices as combinations of Voltage or Current sources may be very valuable in circuit analysis,but tend to obscure the basic facts of life for beginners.

 

Why you try to find a formula for Rc? I know that is is very common for the beginners that they look "formula" for everything. But you don't find "formula" for every component in the circuit. Sometimes we as a designer need to choose some component values and use trial and error to see if we meet our designing goals.

Well, the main reason is that I imagine there would be a formula for calculating Rc. I already use the trial and error all the time, I just thought that wasn't the proper way of doing things; since I assume that the function of all components in a circuit can be calculated mathematically -therefore, the must be a formula for the most common.

But sometimes we can create our formulas simply be using a math.
For example if our main goal is to achieve P = 0.3W for a given RL resistance at Vce = 0.5Vcc.
We can write this

IL_max = Icq*Rc/(Rc + RL) (1)

Ic = 0.5Vcc/(Rc) (2)

And after we solve this we get formula for Rc:

\(\LARGE R_c = \frac{0.5Vcc}{IL_{max}} - R_L\)

I keep on seeing a paradox (for lack of a better word) in that formula: you cannot calculate Icq until you find Rc, or calculate Rc until you find Icq. And since Icq ≠ the current needed for the load (283mA), you need experimental knowledge or practice for choosing a value for Icq. For example: if I were to chose Icq I would go with half 283mA, while you know by experience it should be much higher than 283mA -you chose around twice that value-.

 

1 - You seem to just "choose" Icq, instead of calculating it. Of course, you guys have the knowledge and experience to do that; a beginner like me can't even dream of doing such things.

How can you design the circuit without knowledge ?
Simple for CE amp Icq >> IL_max

2 - I would imagine that Icq would be at least half of the 283mA. Why? because if I were to use the characteristics curve graph to find the quiescent current, I would put the maximum current to the load at the top of the Y axis, and then trace a line down to Vcc in the X axis; which by force would give me Icq around half the maximum load current (283mA) -or even lower than half.

But at the top of the Y axis transistor enters saturation region and this corresponds to the negative half at the load.
But in this amplifier we have more problems with positive half, when BJT enters cut-off region. And the transistor is no longer able to provide any current to the load. And this is why we need Icq >> IL.

The rest I understand it; perhaps with the exception of this formula:

"For this Icq value the maximum positive current at load is equal to: IL_max = Icq * Rc/(Rc + RL) = 185mA peak."​

For the AC signal Rc is connect in parallel with RL.
So I simple use current divider rule
http://en.wikipedia.org/wiki/Current_divider

And why you chose 2.5V for the mid point of the voltage swing.
Is 0.5*Vcc a rule of thumb, or could I have chosen a voltage that is at least the 2.27V that we needed for the voltage swing? (disregarding distortions and other factors)

Yes it is a rule of thumb. Normally we choose Vceq > V_load_max to prevent clipping.

One last clarification: I assume you meant Re -since you already calculated Rb-; which I read should be Re = 0.1 * Rc.

No, I meant RB not Re. You must understand that this simple bias network (RB only) is very sensitive on beta variation. Yes, I calculated Rb, but I assume Hfe = 100. But it is certain that the transistor I gonna to use will have completely different hfe value. The hfe of any particular transistor isn't set in stone, it varies from one transistor to another of the same type. And this is why i need to tweak RB value after I build the circuit.

 

I keep on seeing a paradox (for lack of a better word) in that formula: you cannot calculate Icq until you find Rc, or calculate Rc until you find Icq. And since Icq ≠ the current needed for the load (283mA), you need experimental knowledge or practice for choosing a value for Icq. For example: if I were to chose Icq I would go with half 283mA, while you know by experience it should be much higher than 283mA -you chose around twice that value-.

I don't see any paradox here.
You simply solve this simultaneous linear equations for Rc and Ic

IL_max = Icq*Rc/(Rc + RL) (1)

Ic = 0.5Vcc/Rc (2)

\(\LARGE I_c = \frac{Vcc*IL_{max}}{Vcc - 2*IL_{max}*R_L}\)

\(\LARGE R_c = \frac{0.5Vcc}{IL_{max}} - R_L\)

so all you need to know is RL and IL_max

 

How can you design the circuit without knowledge ?
Simple for CE amp Icq >> IL_max

Yes, I'm starting to realize that the reason the books I read don't get into explaining all this things along the basics of CE amps is because they would completely overtake the content and way surpass the scope of the lesson. That's why they just always give you the value of Rc, instead of teaching you how to obtain it when designing a CE amp.

But at the top of the Y axis transistor enters saturation region and this corresponds to the negative half at the load.
But in this amplifier we have more problems with positive half, when BJT enters cut-off region. And the transistor is no longer able to provide any current to the load. And this is why we need Icq >> IL.

I just gave it a try, and I actually have no problem analyzing the circuit once I have all the values; the problems come when I have to evaluate and consider all these things while designing it -which is what I want to do: design circuits, not analyze them. I'm probably asking for something impossible, akin to running before I learned how to walk.

For the AC signal Rc is connect in parallel with RL.
So I simple use current divider rule
http://en.wikipedia.org/wiki/Current_divider

Yes it is a rule of thumb. Normally we choose Vceq > V_load_max to prevent clipping.

No, I meant RB not Re. You must understand that this simple bias network (RB only) is very sensitive on beta variation. Yes, I calculated Rb, but I assume Hfe = 100. But it is certain that the transistor I gonna to use will have completely different hfe value. The hfe of any particular transistor isn't set in stone, it varies from one transistor to another of the same type. And this is why i need to tweak RB value after I build the circuit.

I see what you mean now in all those points.

I don't see any paradox here.
You simply solve this simultaneous linear equations for Rc and Ic

IL_max = Icq*Rc/(Rc + RL) (1)

Ic = 0.5Vcc/Rc (2)

\(\LARGE I_c = \frac{Vcc*IL_{max}}{Vcc - 2*IL_{max}*R_L}\)

\(\LARGE R_c = \frac{0.5Vcc}{IL_{max}} - R_L\)

so all you need to know is RL and IL_max

Yes, I see how it's done now that I know I have to chose a value for Icq.

I was just puzzled about something, you calculated the current needed for the speaker to be 283mA peak, but never used it in any calculation; instead, you calculated again IL_max; when I assumed it would be the 238mA.

 

I've also been all night trying to find a single formula that would give me Rc, and I was unable to find it. Every single place I looked literally said "you chose Ic" and then calculated Rc accordingly; and it puzzled me how Ic could be just a matter of "choice". ......

View attachment 61126

Hi Adam, in this context, just a general remark:

It would be a very simple task to design an amplifier stage if you would have formulas for every single component. Even a school boy could do it.

Thus, please realize that in case of circuit ANALYSIS you have one single solution only.
In contrary, for circuit DESIGN - in principle - there is an infinite number of "solutions". Here, "solution" means: One of several circuit alternatives that are able to meet your requirements. This explains why you have to choose some parts values and parameters.
And this makes that circuit design is a really challenging task:
To find the "best" solution as a trade-off between several (often conflicting) requirements (technical, economical, reliability, ..)

 

I see what you mean. It just surprised me to leave such a common component down to a matter of choice; specially when all I've studied so far was based on precise calculations. It's also true that none of the material I've touched so far dealt with design, just with analysis, and that's surely the reason why I have been approaching this in the wrong way; I was, as you said, expecting just one solution.

In any case, I realize that when they say "choose" they expect you to have an in depth knowledge of the subject, enough to take all considerations into account, and this is something that way surpasses the abilities of a beginner like me. I'm just too curious and impatient to take for granted all these little things without wanting to know all about them... and I did learn a lot along this thread.

Though I still have a list with more questions.

 

Though I still have a list with more questions.

We are waiting for you...

 

We are waiting for you...

Thanks a lot...

Well, one thing I don't have fully clear is the input/output resistance/impedance. For example: in the example we've been discussing, if the guitar outputs peaks of up to 500mV; how do I know the output resistance/impedance and the current it will send to the base of the transistor, if there is no base resistor?... Can I use "rin" to calculate the base current?

I never know whether I should use a base resistor before hand; I always end up doing this by trial and error.

 

.. how do I know the output resistance/impedance and the current it will send to the base of the transistor, if there is no base resistor?... Can I use "rin" to calculate the base current?
.

two questions:
* Why do you think that there is "no base resistor"? What do you mean?
External parts (resistors) or the internal base input resistor?
* Why do you want to know the input curent? This is interesting only if your signal source has a remarkable signal output resistance.

 

two questions:
* Why do you think that there is "no base resistor"? What do you mean?
External parts (resistors) or the internal base input resistor?
* Why do you want to know the input curent? This is interesting only if your signal source has a remarkable signal output resistance.

I'm referring to a simple hypothetical scenario similar to this one:

I read in a book that if the source is too big, then you put a base resistor (the external resistor at the base of the transistor in the circuit above); but if the source is small, then you don't use the base resistor. Of course, they don't explain when a source is big or small (I assume because it depends on each particular case), nor how to calculate the current that would go into the base without Rb.

I thought that to decide whether I would need a base resistor or not in my designs, I would first need to find a way to calculate the current that the source would output; so I make sure the transistor is in the active region.... or is it enough to just know the voltage?

 

OMG. I give you a tip, stop reading this book. Find a better one.
In our amplifier we have base resistor RB = 820R. And this resistor along with Rc determines whether BJT work in active region or not. Next we add input capacitor to ensure that our AC input signal don't affect our DC bias point.
Read this
http://forum.allaboutcircuits.com/sh...875#post494875

And our AC signal input source will see this impedance

zin = Xc + (Rb||rin) ≈ RB||((hfe + 1)*re)

 

OMG. I give you a tip, stop reading this book. Find a better one.
In our amplifier we have base resistor RB = 820R. And this resistor along with Rc determines whether BJT work in active region or not. Next we add input capacitor to ensure that our AC input signal don't affect our DC bias point.
Read this
http://forum.allaboutcircuits.com/sh...875#post494875

And our AC signal input source will see this impedance

zin = Xc + (Rb||rin) ≈ RB||((hfe + 1)*re)

Yes, I know that... I'm now talking about a different case scenario. I'm not building that amplifier, it was just a random practice exercise (I got a similar test amp working months ago by trial and error); and I already understood the doubts I had regarding it -thanks to you-.

Now what I'm after is determining when the source is big enough to need a base resistor and small enough to go without it (without biasing)... but not in the previous guitar amp; in general using the configuration of the last circuit posted.

 

I'm glad to hear that.

PS. We always need RB resistor

 

The point is, I understand the whole chapter on CE amps, how to analyze them, biasing, voltage dividers, and so on and so forth.... I just have some doubts regarding some small points that were not explained in enough depth (e.g. choosing Rc -which you already explained... that's done and dusted-).

So, it's not that I don't know how to get one working; I'm just inquiring about the little details that were left unexplained because they were beyond the scope of the book.