# Common-Emitter Amp doubts and questions

Discussion in 'General Electronics Chat' started by adam555, Nov 4, 2013.

Aug 17, 2013
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For quite some time I've been having the same problem with transistor amplifiers: I read and understand the relevant information from books, websites, etc. I'm able to do all the analysis exercises. But as soon as I attempt to apply it all I realize that I still have so many questions and doubts that I'm incapable of using any of it in my designs. And I'm not talking about complicating things by using PNP instead of NPN; or using common-base and emitter-follower... I'm talking about simple common-emitter designs with just a NPN BJT.

I think half of the problem is that most books and websites teach you how to analyze amplifiers, but they don't say a word about how to design them. I now began to study the section on transistors from this website, which at least gives some examples of practical applications, but there are still things I feel were left out, things that differ from other sources and are confusing me, and things that I still have no idea where they come from.

I listed quite a few doubts, but I think it's best if I posted them one by one to avoid further confusions.

The first one is very simple; so simple that I feel silly asking it:

Where do you put the load on a BJT common-emitter amplifier? Is it (A), (B), (C), or can all of them be the load?

By the way: can the "Rc" (the collector resistor) also symbolize the load?

Aug 17, 2013
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The second doubt is about a particular exercise that I saw in a book; which takes into account the input resistance to an amplifier, to calculate how much current goes into the base of the transistor along with the base resistor current.

My question is: how do you know in practice what's the input resistance, when you don't have a book telling you?... and, do you really need to know this?

Aug 17, 2013
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My next doubt is about the transistor's characteristic curves and the
Quiescent point. All texts I read say that you need to obtain the Q by plotting a line on the transistor's characteristic curves graph; and my questions are...

Do always need to do this when designing a class A common-emitter amplifier?

If so, is there any other way of obtaining Q without the graph (e.g. through a formula alone)?

... and, what if you don't have access to the graph? Yesterday I leaned how to make them with LTSpice, but to be honest, they don't look anything like the real ones in the datasheets.

Aug 17, 2013
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I reached the section from this website's tutorial where it answers my first question; but its example leads me again to the second question:

It says that by placing the load at "V output" instead of placing it where "R" is, you amplify voltage instead of current. I was going to ask how do you calculate the voltage gain, however, it explains it further down with this formula:
$A_{v} = \beta \frac{ R _{out} }{ R_ {in} }$

And this is where I'm back again stuck in the second questions: how do I know from that example what's "Rin"? How do you calculate the input resistance?

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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In real life amplification we connect load as you show in example C

YES.
We need to know input resistance, because we don't want put too much loading on the input source (most of a signal source will have low current capability). So we need high input resistance.
We use small signal analysis and find Rin. For example for CE amplifier without Re resistor.
Rin = (Hfe + 1) * 26mV/Ic

Yes we don't need any characteristic curves graph to obtain the Q pint in class A.

Last edited: Nov 4, 2013
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6. ### LvW Active Member

Jun 13, 2013
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Although I agree, of course, to Jonys answer, I like to add something.

1.) The CE stage provides an output voltage at the collector node (however, it cannot be treated as a voltage source because the output resistance is app. identical to Rc).
That means: The BJT is a voltage-controlled CURRENT source and Rc alone can act already as a "load" if your application contains no other "load" - or if the impedance of this load is very large in comparison with Rc. Otherwise, the effective "load resistance" is Rc in parallel to the external Rload.

2.) The knowledge of base current Ib is necessary for correctly designing the bias network, which determines the operating (Q) point of the transistor.
However, in this context it is important to distinguish betwen the DC current Ib and the small-signal current ib, which determines the signal input resistance of the circuit.
3.) Thus, the input resistance symbol should clearly show that it is a differential quantity (small signal dynamic resistance, symbol with a small letter). For example, the voltage gain formula as mentioned by you should read (applicable without signal feedback only):

Av=-beta*Reff/rin

Please note, that the selected Q point determines the gm value: gm=Ic/Vt
(Ic: DC current, Vt: temperature voltage of app. 26 mV).

4.) Consequently, I suggest to write the equation as given by Jony as follows:
rin = (hfe + 1)*26mV/Ic

5.) The load line in the output characteristic of a BJT can visualize the optimum (symmetrical) position of the chosen Q point. But it is not necessary for each design.
The normal procedure is to select a suitable Ic value, which ensures a sufficient DC voltage Vce (for symmetrical ac voltage across the Q point).

Last edited: Nov 5, 2013
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Aug 17, 2013
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Thank you so much Jony and LvW; most of that is definitely missing from the sources I've been reading (and I rechecked today).

I understand now how to find Rin, but I'm still a bit unclear on combining stages and the signal source taking into account their Rin / Rout resistance. I guess that's studied at a more advanced level, so let's forget about that for a moment, and look at Rc and Load for moment...

I always thought the function of Rc was to limit the current to the transistor and/or the load, and that you calculated it in function of the maximum current you needed through it. For example, if the load is an LED, then you calculate Rc = Vcc/Ic(max). However, in the post you linked you calculate Rc = 0.1*Load (which I never heard of before), and also use it to set half Vcc at the collector (which I read about before, but didn't know why).

Looking at it from where I'm standing now I'm guessing I'm mixing up the EC as a voltage and as a current amplifier (as I just realized the difference yesterday). So, anything you might have to say about this would be greatly appreciated.

8. ### LvW Active Member

Jun 13, 2013
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As I have mentioned, I recommend to strictly distinguish between static and dynamic resistances. And we speak about the differential small-signal input resistance called "rin" (in small letters). This is really important, otherwise one gets confused.
No problem with rout and rin: The input resistance rin,2 of a second stage is considered as parallel to the rout,1 of the first stage (in some cases rout,1=Rout,1 which means: dynamic=static).

No, that´s a false understanding. The transistor is to be seen as a CURRENT source, Thus, it is not the task of Rc to "limit" the current Ic, but to transfer the output signal current ic (small letters!) into a corresponding output voltage v,out.

The first step in designing a transistor amplifier is to SELECT a suitable dc quiescent current Ic (based on Vcc and power consumption constraints).
Then, Rc and Re~0.1*Rc is selected (Re for Ic stabilization).
The DC voltage drop across these resistors should be approximately (0,5...0.6)*Vcc (rule of thumb).
Thus, the remaining DC voltage Vce ensures sufficient symmetrical small-signal operation around this Q point.
In case Re=0 (bad design) Rc should be chosen so that the voltage Vce smaller than Vcc/2 in order to avoid thermal run-away.

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9. ### Veracohr Well-Known Member

Jan 3, 2011
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I also found books to be less than helpful for basic design tips. Luckily, a couple of my instructors showed a nice quick practical method for designing CE amps (assumes voltage divider bias):

1. Choose the emitter voltage. Best if it's at least 1V above the negative supply (or ground if single supply). Then choose an emitter resistor for a desired collector current (since Ic ≈ Ie).

2. Select voltage divider resistors so that Vb ≈ 0.7V above your chosen Ve. Voltage divider bias resistors affect input resistance (they are in parallel with Rin), so choose values accordingly.

3. Choose Rc so that Vc is approximately centered between your supplies. Rc = Vc/Ic.

4. Set the desired gain by bypassing RE(DC) and selecting RE(AC) such that A = RC/RE(AC).

Aug 17, 2013
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Ok, I think I see it now. I didn't know there were dynamic and static resistances. So, when we talk about "rin" (in lower-case) this is the dynamic resistance to the signal entering the base of the transistor (not the internal resistance of the signal source, as I thought).

You see, the book I've read gives you this circuit to analyze:

And without explaining anything about dynamic or static resistances they give you the value for the input resistance, which they symbolize as "$r_{e}$". So, I thought that "re" had to be the internal resistance of the "Ve" source... what else, if they don't tell you transistors have a dynamic input resistance!

Another thing that confuses me about that exercise, is that it places "RL" (the load resistor) where Rc should be, and not at "Vs". Isn't the Load the resistance across the output?

So, when you see the typical example of Rc limiting to 20mA the current of an LED connected to the collector of a transistor, I assume that's just because the transistor is being used as a switch and you expect the current to the base to saturate the transistor (e.g. you expect Ib to be greater than 20mA/β, which would allow more than 20mA through Ic).

Thanks for all that, I'll see if I can try it with practical example today night... I'll be back with the results.

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11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Only when we use BJT as a switch (BJT in saturation) Rc resistor limit (set) the collector current. But in active region Rc resistor has a completely different task. Without Rc we wouldn't have a voltage gain.

Transistor without any resistor is nothing more then a base current controlled collector current source Ic = β * Ib .

So as you can see we have a "current gain", but with no voltage gain.
What we can do about that ? Let as try to add Rc resistor and see what we get.
Now we give Rc resistor very important task. His job will be to covert Ic current into voltage. Rc will act just like a current to voltage converter thanks to Ohms law Vrc = Ic*Rc.

So finally we have a voltage gain. Our amplified voltage is of-course
taken between collector and emitter (Vce).
Also notice that we can change voltage gain by changing Rc resistor value.
The larger the Rc resistor value the larger the voltage gain. Why you may ask? Well because now we need smaller change in Ic current to get the same change in Vce. For example for Rc = 1K we need change in Ic ≈ 10mA to change Vce voltage from 10V to 0V. But now if we increase Rc to 2KΩ we need ΔIc = 5mA to get the same change in Vce. So we need smaller change in input voltage to get the same output voltage. So this means that our amplifier has larger voltage gain. I hope you understand this and the important role which lies on Rc resistor.

Also see this example of a common emitter amplifier with emitter degeneration resistor.

At first we bias the transistor into linear region. The DC voltage at base is equal Vb = 2.6V, and emitter DC-voltage is Ve = 2V and collector DC bias voltage is Vc = 6V.
Next we connect the AC input signal with amplitude 2V peak to peak.
This AC voltage at input will cause that the base voltage will change from 3.6V to 1.6V in "rhythm" of a AC input signal. Those changes will result that the emitter voltage will also change from 3V to 1V. The emitter resistor RE will "convert" this change in Ve voltage into Ie current. This will result the change in emitter current by 0.9mA (3V) to 0.3mA (1V). And now this changes in Ie and Ic current in "rhythm" with a AC input signal will once again be "convert" in to voltage in Rc resistor. And this change in collector current will cause change in VRc voltage, between 9V to 3V. We have a three times larger change in VRc voltage because Rc is three times larger then Re resistor. So the voltage gain is equal to
Av = RC/RE

Even without Re resistor, transistor still have emitter resistance built-in into emitter. The ideal transistor with intrinsic emitter dynamic resistance "re = 26mV/Ic".
And this is why Av = RC/(RE + re)[/B]

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Aug 17, 2013
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OMG... I can't believe it was that easy. Why have I done 2 electronics courses (many years ago, I have to confess) and now just read 2 books and none of them explain it in such a straight forward way???

I was about to post my reply to the posts from yesterday about calculating Rc because I couldn't get it to work in a practical example. But this changes everything. Thanks a lot.

13. ### Veracohr Well-Known Member

Jan 3, 2011
566
78
Probably because the simple explanations, while useful, are technically incomplete. Books tend to be more in depth.

Aug 17, 2013
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This is what I was going to post... I'm posting it to show you where I was stuck.

--------------------------------------------

I tried to apply all this to a practical example, but it's not going well; I can't get it right following any of the three set of instructions.

This is the example I tried... just a very simple common-emitter guitar amplifier with a small speaker (no Re, filtering capacitors or voltage divider to keep as simple as possible).

The values are:
Vin = 500mV (maximum peak voltage coming out of the guitar)
Vcc = 5V
RL = 8Ω (the resistance of the speaker)
PL = 0.3W (speaker power)
β = 100
The first thing I did was to find out the voltage and maximum current I would need for the load (the speaker); and I came out with:

P = V*I;
I = √(P/R) = √(0.3/8) ≈ 200mA (I'm rounding to make it simpler)

And then I got stuck calculating Rc!

If I follow Jony's instructions I come out with:
Rc = 0.1*RL
Rc = 0.1 * 8Ω
Rc = 0.8Ω

That can't be right; that's more than 6 amps through it!!!... and that only for 0.5*Vcc.
Then tried to follow LvW's instructions:
You don't say how to calculate Rc other than:
"SELECT a suitable dc quiescent current Ic (based on Vcc and power consumption constraints).Then, Rc and Re~0.1*Rc is selected (Re for Ic stabilization).The DC voltage drop across these resistors should be approximately (0,5...0.6)*Vcc (rule of thumb)."
But how do I do that exactly? I don't know what Ic should be yet -I just know I need 200mA maximum through the load-, but I'm pretty sure it's not (0.5*Vcc)/0.2; because 0.2A is the maximum current, not the quiescent current.
And finally tried Veracohr's instructions:
First I chose the emitter voltage, which I need it to be the Vbe drop plus the signal peak; which is:

Ve = 0.8 + 0.5 = 1.3V

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.

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15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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First of all speaker don't like DC-current. So you need to add a capacitor at output and at input also.
Secondly you cannot drive directly 8R speaker with CE amplifier, you need to add emitter follower stage. Why? because CE amplifier has large output impedance.

Aug 17, 2013
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I was just trying to keep it simple for calculating the resistors.

So, there was no way to get that working...

This is what I meant when I said I didn't understand how to combine stages, sources and outputs... I don't know how the input and output resistance and impedance influence the result.

For what you just said; do they need to be of an equal impedance?

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
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You want:
1 - practical example = build in real life ??
2 - practical example = calculating the resistors values and use LTspice to see if this work?

If our signal is voltage then Rout<<Rin

Choose Rc = 0.5RL or RC = RL or replace Rc with RL

Aug 17, 2013
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I was doing both; in real life and in a simulator.

I can get it to work, but using what I call the "Edison method"; that is: trying all combinations until "the light goes on".

What I want to do now is start calculating this things properly; instead of wasting hours changing the value of resistors until I get things to work.

I saw a similar example on this website's tutorial, in which they also used an 8Ω speaker (but as you say -in the place of Rc-). In the example they could only get a gain of about 1, so they changed the load (the speaker) from 8Ω to 30Ω, and since the base resistor (which I guess is the same as "Rin" [R in uppercase]) was 1K and the β 100, it gave a gain of 3.

But let's forget about all those things for a moment. What would be the quiescent Ic that got me stuck in the original example as it was; and how would you calculate it (the simplest way using the original values given above)?

Last edited: Nov 5, 2013
19. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,018
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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.

Now let as choose RC resistor value. Normally we use CE amplifier in preamp stages to amplify small signal with relatively high load resistance. So we normally select quiescent Ic current much larger than desired load current.
Icq >> ILoad. I pick Icq = 0.5A
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.

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20. ### LvW Active Member

Jun 13, 2013
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Quote Jony: "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"

Hi Jony - excellent contribution. Sometimes it is really helpful and instructing to demonstrate that and why a particular circuit should not be used in practice.

May I add another explanation to the part of your post as cited above:
This combination of current injection Ib and - at the ame time - voltage feedback via Re is counter-productive and, thus, shouldnt be used.
* Voltage feedback via Re makes sense only if a voltage is applied at the base node and if this voltage can be considered to be nearly constant (resistive base divider).
* On the other hand, using only Rb biasing with current injection, it is logical to apply voltage controlled current feedback. This is accomplished connecting Rb at the collector (instead of Vcc).
* Summary: Both principles shouldnt be mixed with each other.