Common Emitter Amplifier Design

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Homework Statement


I have to design a Common Emitter Amplifier using a 2N2222 NPN transistor and an 18V DC supply.

  • The specifications are as follows:
  • Should be stable for variations in forward current gain, β.
  • The input signal to the amplifier is a 0.2 Vp sinusoidal signal, at a frequency of 50 kHz.
  • The amplifier should have an undistorted, symmetrical output with an 8 Vpp swing.
  • The output should be maintained over a bandwidth of at least 1 kHz – 10 MHz.
  • The quiescent current is to be 6 mA, at a quiescent collector-emitter potential of 8 V
  • Assume the load impedance is much larger than the output impedance of your amplifier.
  • Ensure the power dissipation of the transistor is kept below maximum operating limits.

The Attempt at a Solution


I designed the circuit to be as follows:
attachment.php?attachmentid=52008&stc=1&d=1350468856.png


Rc=(Vcc-Vce-Ve)/Ic

Vcc=18V, Vce=8V, Ve=(1/10)Vcc=1.8V, Ic=6mA

Rc=1.36kΩ

Re=Ve/Ic=1.8/6m=300Ω

R2=(1/10)β*Re=1/10*100*300=3kΩ

Vb=(Vcc*R2)/(R1+R2) -- (1)

Vb=Vbe+Ve, Vbe=0.7V, Ve=1.8V
Vb=2.5V

from (1)
R1=R2(Vcc-Vb)/Vb
R1=18.6k

so, R1=18.6kΩ, R2=3kΩ, Rc=1.36kΩ, Re=300Ω
Is this correct?


I think I have to add an un-bypassed resistor to the emittor side.
Also how do I find the values for R5, and the 3 capacitors.

Thank you for the help
 

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Answers and Replies

  • #2
CWatters
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As for the coupling capacitors.. If they are too small there will typically be a potential divider effect or a bandwidth limiting effect (eg high pass but with too high a corner frequency). So you need to identify the other component of the potential divider/high pass filter ....

C1 and the input impedance of the amplifier form a high pass filter. C1 needs to block DC but let through the low frequency AC.

For C3 it's R4. C3 should shunt all the AC to ground so the impedance of C3 << R4 over the whole bandwidth.

C5, R5 also form a high pass filter. C5 needs to block DC but let through the low frequency AC.

You can't allways just fit enormous electrolytic capacitors everywhere because they don't behave like ideal capacitors at very high frequencies. Otherwise there is usually some leeway as to the exact choice.
 
  • #3
NascentOxygen
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I haven't checked your biasing calculations, but let's assume they are okay ....

Q1: When you find a need for, for example, Rc=1.36kΩ, is it okay to leave it like that? Or does your teacher expect you to use preferred value components, in this case either 1.2kΩ or 1.5kΩ ?

Q2: After you have biased it, are you able to predict what the AC gain of the circuit will be? Because (if I read the specifications correctly), you are aiming for a gain that gives 8V output for 0.4V input. (It is worth double checking that figure to make sure you have it right before you go any further.)
 
  • #4
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I think I should only worry about the capacitors after I have correctly found all the resistor values? is that correct?

I haven't checked your biasing calculations, but let's assume they are okay ....

Q1: When you find a need for, for example, Rc=1.36kΩ, is it okay to leave it like that? Or does your teacher expect you to use preferred value components, in this case either 1.2kΩ or 1.5kΩ ?

Q2: After you have biased it, are you able to predict what the AC gain of the circuit will be? Because (if I read the specifications correctly), you are aiming for a gain that gives 8V output for 0.4V input. (It is worth double checking that figure to make sure you have it right before you go any further.)
A1: I can just leave it like 1.36kΩ
A2: The input is 0.2V sinusoidal. The output should be symmetrically 8V.

I'm not 100% sure i got the correct values for R1,R2, Re and Rc. If you could double check it would b greatly appreciated. After I have found these 4 resistor values however what is the next step to take?

Thank you
 
  • #5
NascentOxygen
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A2: The input is 0.2V sinusoidal. The output should be symmetrically 8V.
Meaning you are aiming for an AC gain of what value, precisely?
 
  • #6
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Isn't the gain:
A=Vout/Vin
A=8/0.2
A=40
 
  • #7
NascentOxygen
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You wrote for input and output, resp.
a 0.2 Vp sinusoidal signal
[...] output with an 8 Vpp swing
and I'm advising you to make sure those figures are right before going further.

There is a difference between Vp and Vpp and do the specs show this or is it a typo you have overlooked?

I'm off for now ..... http://imageshack.us/a/img840/5059/bicyclesmiley.gif [Broken]
 
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  • #8
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There is a difference between Vp and Vpp and do the specs show this or is it a typo you have overlooked?
Thats not a typo.

The input signal to the amplifier is a 0.2 Vp sinusoidal signal, at a frequency of 50 kHz.
The amplifier should have an undistorted, symmetrical output with an 8 Vpp swing.
Now i'm more confused though.. :confused:
 
  • #9
NascentOxygen
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There are various ways that a sinewave's amplitude can be specified. One way is by its peak amplitude, Vpk. Another is by its peak-to-peak amplitude, Vpp. As you can guess, Vpp is double Vpk. So there's a factor of x2 difference.

I find the wording of the specification in your exercise to be a little lackadaisical. But the only interpretation is 0.2 Vpk input -> 4.0 Vpk output. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]

With that cleared up, it brings us back to my Q2.... have you studied a small-signal model for a transistor? — a circuit that will allow you to predict the AC gain of your CE amplifier without having to build it to measure that parameter http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken]
 
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  • #10
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lackadaisical.
perfect description of my lecturer.
But the only interpretation is 0.2 Vpk input -> 4.0 Vpk output. [PLAIN]http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif[/img[/QUOTE] [Broken]
Ok so Vin=0.4Vpp (peak-to-peak), Vout=8Vpp
so Gain=8/0.4=20
Have you studied a small-signal model for a transistor? — a circuit that will allow you to predict the AC gain of your CE amplifier without having to build it to measure that parameter http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken]
don't think so.

to continue on the design.. starting from scratch. How do I know what the circuit will look like.
As described in the OP I have R1 and R2 and R3=Rc (collector) and R4=Re (emitter) and R5=Rl (load)
I know I need the bypass Capacitor (C3) and the coupling capacitors (C1, C2) to block the DC component.
I have also seen circuits that don't have C2 and R5, and also some that have another emitter resistor that is not bypassed by C3.
 
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  • #11
NascentOxygen
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Rc=(Vcc-Vce-Ve)/Ic

Vcc=18V, Vce=8V, Ve=(1/10)Vcc=1.8V, Ic=6mA

Rc=1.36kΩ

Re=Ve/Ic=1.8/6m=300Ω

R2=(1/10)β*Re=1/10*100*300=3kΩ

Vb=(Vcc*R2)/(R1+R2) -- (1)

Vb=Vbe+Ve, Vbe=0.7V, Ve=1.8V
Vb=2.5V

from (1)
R1=R2(Vcc-Vb)/Vb
All is right up until this last line. This equation is a bit too rough and ready for my liking. (Though maybe your lecturer is okay with it.)

What that equation says is that the current through R2 is equal to the current through R1. This, in effect, ignores the current into the base of Q1, the base current here equalling 6/β mA.

A more precise approach is to say the current through R2 is equal to the current through R1 minus that into the base of Q1.

VB/R2 = (VCC - VB)/R1 – 0.006/β

If you solve this for R1 you'll get a value roughly 10% different from what you calculated by ignoring the base current. The Q point using this more precise approach will be closer to what you initially aimed for (viz., 6mA and 8V).

Capacitor values can be addressed later. The next consideration is the circuit's AC gain.
 
  • #13
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considering the base current I found R1 to be 1.735kΩ.

Those lecture notes look sort of familiar. The one you gave is a lot more clear though.
one question though, those describe r_be=(h_fe+1)(0.025/I_E) whereas my lecture notes simply say r_e=0.026/I_E.

anyway, I found Rl using your lecture notes:
Vout=i_c(Rc//Rl)
4=0.006((1360*Rl)/(1360+Rl))
Rl=1308Ω

So now I have found all resistor values, all that is left is the 3 Capacitors.
 
  • #14
NascentOxygen
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considering the base current I found R1 to be 1.735kΩ.
As did I. :smile:
Those lecture notes look sort of familiar. The one you gave is a lot more clear though.
one question though, those describe r_be=(h_fe+1)(0.025/I_E) whereas my lecture notes simply say r_e=0.026/I_E.
The difference there being effectively just a factor of β.

r_be accounts for the effect when we place it where it carries the base current. r_e accounts for the effect when we place it where it carries emitter current (emitter current being β times the base current). So where you place it determines whether it carries Ibase or β.Ibase
anyway, I found Rl using your lecture notes:
Vout=i_c(Rc//Rl)
4=0.006((1360*Rl)/(1360+Rl))
Rl=1308Ω
That's R5 you are calculating, is it? R5 influences DC conditions iff it is directly connected to the collector. If capacitive coupling is used, then R5 is isolated at DC and has no effect on bias conditions.
The reason I'm highlighting AC gain and the small signal equivalent circuit, is that you are aiming for an AC gain of x20. Unless you apply the small signal analysis, you have no idea whether this amplifier will be able to meet that figure; you have no idea what the AC gain will be.
So now I have found all resistor values, all that is left is the 3 Capacitors.
 
  • #15
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Yes that's R5 I calculated.

So are you saying the way I calculated it was wrong? In that case what are the things I should find in order to find R5 and what are the formulas associated?

I put some number in for the capacitors (pretty sure they are wrong though) and got an output that peaks positively at 4V but awkwardly drops to -8V.

attachment.php?attachmentid=52080&stc=1&d=1350618071.png
 

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  • #16
NascentOxygen
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Yes that's R5 I calculated.

So are you saying the way I calculated it was wrong? In that case what are the things I should find in order to find R5 and what are the formulas associated?
What you did for R5 is not valid for anything. You have a mixture of both DC and AC parameters.

I'd say you probably are not required to assign a value to R5; that's how I'd interpret this clause: "Assume the load impedance is much larger than the output impedance of your amplifier."

If you intend capacitors to have negligible AC impedance, then over the frequency range of interest you choose XC « RT where RT is the Thevenin resistance seen from the capacitor terminals. A simpler but less rigorous way would be to choose XC « R where R is the lowest of the parallel resistances in series with C.

Is that waveform the "sinewave" output? Doesn't look too healthy. That's probably because the circuit has a gain which is much greater than is needed. Can you reduce the input amplitude until the output looks sinusoidal, and determine the gain at that point?
 
  • #17
NascentOxygen
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For a sinewave input, a plot of collector voltage will be more informative.
 
  • #18
CWatters
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I agree with NascentOxygen..

If the output is 12V peak to peak and the supply voltage is 12-18V what do you think is happening ?

Look up how the gain of a common emitter amp is calculated when there is no emitter capacitor C3.

What is adding C3 doing to the gain?

There is a conflict in this area between the AC and DC analysis that I think can be solved with a small circuit change. (PS It's not as simple as removing C3!)
 
  • #19
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I'd say you probably are not required to assign a value to R5; that's how I'd interpret this clause: "Assume the load impedance is much larger than the output impedance of your amplifier."
But I need to run the simulation, so what do I do then?


where do I measure the collector voltage? is it before the capacitor C2? in that case it is the same waveshape shifted up. the same wave going between +14V and +2V
 
  • #20
NascentOxygen
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But I need to run the simulation, so what do I do then?
What you have done is okay, making it x10 the value of R3. Making it x100 might have an advantage, too.
where do I measure the collector voltage? is it before the capacitor C2? in that case it is the same waveshape shifted up. the same wave going between +14V and +2V
Collector voltage is measured between the collector and ground. What you found sounds right.

It wouldn't hurt to make all capacitors x10 larger, so that XC ⋘ R at all frequencies in the band of interest.

Have you measured the gain as I suggested? If it turns out that gain is higher than the value you are aiming for — and we hope that is the case :smile: — then you will need to reduce the gain. What are the different techniques you know for reducing the gain of this single transistor amplifier? Which do you think is the best to use here?
 
  • #21
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Is that waveform the "sinewave" output? Doesn't look too healthy. That's probably because the circuit has a gain which is much greater than is needed. Can you reduce the input amplitude until the output looks sinusoidal, and determine the gain at that point?
so, at Vin=0.02V Vout is 4.891V peak-to-peak. Gain is 122.275
obviously way to high.


There is a conflict in this area between the AC and DC analysis that I think can be solved with a small circuit change. (PS It's not as simple as removing C3!)
Add an unbypassed resistor to the emitter? but how does that work then


Are the calculations that I did for R1, R2, RC, RE all correct?
 
  • #22
CWatters
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Without C3 the ac gain is roughly - Rc/Re = 1360/300 = 4 which it too low.

With C3 the gain is boosted enormously because it reduces the impedance at the emitter to near zero.... Rc/0 = ∞

So what you can do is choose Re to suit the DC bias conditions then use a capacitor C3 to bypass "part" of the emitter resistor. To do that split Re into two resistors in series.

Read all of this page (eg scroll down)...

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npncegain.html

EDIT: That page shows how the gain depends on on rE the internal emmiter resistor but the same principle applies.

See this circuit..

http://www.ecircuitcenter.com/Circuits/trce/Image02.gif

Note how RE has been split into two and only the lower half is bypassed. This allows RE to be different for AC vs DC.
 
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  • #23
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ok so rE=1360/20=68
then RE=300-68=232

This gives a nice sinusoidal symmetrical output, however it is only 3.6V peak to peak.
that is only a gain of about 9

attachment.php?attachmentid=52194&stc=1&d=1350901411.png
 

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  • #24
NascentOxygen
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so, at Vin=0.02V Vout is 4.891V peak-to-peak. Gain is 122.275
obviously way to high.
That looks promising. It is not clear to me whether you are expected to calculate the component values to bring the gain back to -20, or whether you can experimentally determine it. (It is probably a good habit to be keeping the negative sign in front of the gain; this is after all an inverting amplifier, and there may be a ½ mark reserved for that detail in some future exam. :smile:)

You can adjust the gain by inserting a variable resistor in series with C3, say 500 Ohms, and reducing that resistor until the gain becomes -20. I'm not confident that you are going to find gain is constant over a full 10MHz bandwidth, but we'll see. Re-compute the value* for C3 when done.
Are the calculations that I did for R1, R2, RC, RE all correct?
You can see for yourself whether you have them right by looking at the transistor voltages and currents on your simulated circuit. If the parameters are what you designed for, then you must have got it right! :wink:
 
  • #25
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I increased RL to 13.08K and now it's a lot closer to 4Vp.

But I am a little confused. Shouldn't each value be able to be calculated in order to reach the requirement of an 8V peak-to-peak output. In the end the only real assumption was the value of Rload which I just made really large.

I haven't finished finding the values of the capacitors yet though. But if I don't come to exactly 8Vpp symmetrical output then I would have had to gone wrong somewhere. (and just adding a resistor and experimentally finding a value that works is wrong)

And what is the formula for finding the capacitor values?
Edit:
I used http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npnce.html
and http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npncecoup.html
to find the values for C1,C2,C3 to be: 2.06uF for the bypass cap. 0.22uF for C1 and 16nF for C2
how does that sound?
 
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