# Common Emitter Amplifier (2 unknown resistors)

Discussion in 'Homework Help' started by Wraakneming, Sep 12, 2009.

1. ### Wraakneming Thread Starter New Member

Sep 12, 2009
2
0
Hi

Im in desperate need of help, Ive tried my text book, read the e-book of the site and searched the web. I have absolutely no idea how to do this. If I can just get a similar problem or someone could work it out and explain it to me I would be forever grateful.

All that I know is that the βdc is 290 and Ic(sat)= Vcc/(Re + Rc) = 16.4 mA.

Could someone help with finding R1 and R2 need to be found and plotting Q point?

Someone worked it out as R1 = 8.2KΩ and R2 = 56kΩ (preferred values). I just want to know how they did it.

Cheers
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2. ### Wendy Moderator

Mar 24, 2008
21,477
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That is a common emitter with emitter swamping resistor. Bascially, figure the voltage at the collector if the transistor were all the way on (saturated). Then figure the voltage at the collector if the transistor were all the way off (cutoff). Split the difference, then set the circuit up to create that voltage on the collector.

The emitter voltage is the base voltage - 0.6V.

This sets up a constant current, which sets up the collector voltage.

Using the above info, you should be able to come up with a base voltage that sets up that condition. Don't forget the Base to ground resistance through the transistor is in parallel with R1, and adjust your calculations accordingly.

3. ### mik3 Senior Member

Feb 4, 2008
4,846
69
First you have to decide what is Vout at Q point.

Then find the required collector current as to drop some voltage across Rc and achieve this Q point Vout.

Vout=20-Ic*Rc

Then assume that Ic=Ie

Thus, the voltage across Re will be Ve=Ic*Re

And thus Vb will be Vb=Ve+0.7V

You can find any values for R1 and R2 as to achieve Vb but to have good results ('fixed' voltage divider as Vb not to vary much with Ib) find R1 as follows:

R1=0.1*β*Re

Then find R2 according to the value of R1.

4. ### Gustav180 Member

Aug 25, 2009
17
0
Hello Wraakneming

There are some simple rules you can youse and you don´t need the diagram because bjt transtors data is very individual.

At first rule you will calculate the emitter voltage to 10% of the total supply voltage. This will bee around 2 V, lets say 2.2 V for easier calulations.

2.2 V over 220 ohms give a current of 10 mA. Assume the same current in Rc and in the transistor Ic. It is not exact thrue, but practical thrue.

ß is 290 and this will give a base current Ic/ß = 10/290=34 µA. This value is very individual.

Next rule is: The voltage divider R1 - R2 shall have a current at least 10 times the base current to bee stabile, 340 µA. (This is variant of ruel from Mike3.)

The base voltage is 0.6 V over the emitter voltage, 2.2 + 0.6 = 2.8 V.

R1 = 2.8/0.34 = 8.2 kohms (standard value)

R2 = (20 - 2.8)/0.34 =50.6 kohms. (50.6 is not standard, you have to use 47 or 56 kohm)

Now you have to do a compromiss, we can calculate a little more.

The DC-level at Vc should be at half level of (Vcc - Ve), this will give the higest signal output without distorsion. It is an important value.

Vc = 20 - (20 - 2.2)/2 = 11.1 V. The voltage over Rc will bee 8.9 V. Therefor it is better to deacrese the current through Rc and the transistor. This will have to use a 56 kohm resistor.

Now we can recalulate from our calculated values.

VB = Vcc * R1/(R1+R2) = 20 * 8.2/(8.2+56) = 2.55 V

VE = 2.55 - 0.6 = 1.95 V

Ic = 1.95/0.22 = 8.8 mA

Vc = 20 - (8.8 * 1 ) = 11.1 V.

As you see, the Vc value is correct.

The value of R1 and R2 is not critical, but the relation between them is important. Higher values give lower powe consuption, but the base current through R2 will affect more.

Best regards
Gustav

5. ### hobbyist AAC Fanatic!

Aug 10, 2008
885
87
Wraakneming

To help avoid any confusion:

He meant to write it out as

R1= 2.8 / 0.00034

R2= (20 - 2.8) / 0.00034

6. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
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If you wan't to calculate the base current with both R1 and R2, then you take the supply voltage and divide it by the top resistor. In your case it would be R2. So for example, let's say I have a supply voltage of 9V and my top resistor is 10k. Then, the base current should be 900μA.

7. ### hobbyist AAC Fanatic!

Aug 10, 2008
885
87
I'm not sure that that is entirely right,
I think You need to take into consideration, the voltage across the emitter resistor and the base emitter voltage, Vbe. the base current is a parrallel branch with the resistor R1,
so to slove for IB you would need to take the voltage drop across R2, then divide that voltage by R2, to give total current, then take the voltage drop across R1 and divide R1 into it to get the current flow thru R1, then the remaining base current would be the total current thru R2 minus the current thru R1 to give the remainder of current thru the base junction.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,667
1,311
Well if we assume
Ic=10mA and Ib=40uA
So we get:
Ve≈Ic*Re=2.2V and VB=Vbe+Ve=2.85V
The current that is flow through R1+R2 must be 5...20 times larger then Ib.
If we assume Id=10*Ib=400uA
Now we can use Ohm's law
R1=VB/Id=2.85V/400uA=7.1KΩ
R2=(Vcc-VB)/(Id+Ib)=(20V-2.85V)/(400uA+40uA)=39KΩ
and now we check the result
Ib=[ (R1*Vcc)-Ube*(R1+R2) ] / [ Re*(β+1)*(R1+R2)+(R1*R2) ]= 44.009712uA and Ic=11mA
so the result are not so bad.

9. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
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I did a simulation, and if both R1 and R2 equal 10kΩ, then the base current is 900μA. In the simulation, it said the current going through R1 was 831.57μA and the current going through R2 was 68.43μA. Add them up and you get 900μA.

10. ### hobbyist AAC Fanatic!

Aug 10, 2008
885
87
Sorry I must have misunderstood your original post.

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But for R1=R2=10KΩ this amplifier, the BJT will be saturated.
So this circuit is no longer the amplifier.

12. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
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That is what I have a question about. For the 2N3904, what base current is necessary in order to make it in its linear region? If 900μA is too high, then I can put another resistor in series with the base from the point at the voltage divider. What base current should I have? This will be a common-emitter amp.

13. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
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That's ok, I didn't explain myself fully.

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The base current in amplifier depends on a Ic current and β.
The Ic current depends on Rc, Re, Vcc.
And remember, you cannot saturated the BJT without Rc.
So linear region mainly depends on the values of resistors (Rb, Rc, Re) and of course on Vcc, β.

15. ### hobbyist AAC Fanatic!

Aug 10, 2008
885
87
Hi.
I too am learning more about transistor amp. design.
That's why I respond to posts with questions concerning transistors, because I can share what I've been learning to maybe help others.

I'll try to give an example of how to bias a transitor into it's linear region.

The reason I choose no load at this time is just to get a feel for how a transistor operates, with biasing it.

1. choose power supply. (VCC)
2. Choose a collector resistor. (RC)
3. choose a voltage gain. (Av.)
4. calculate emitter resistor (RE) by equ. (RC / Av.)
5. make volt. collector, to be 1/2 VCC. (VC)
6. calculate collector current (IC) by equ. (VC / RC)
7. calculate volt. emitter (VE) by equ. (IC x RE)
8. calculate volt. base (VB) by equ. (0.7 + VE)
8. choose a value for RB1 (base resistor to ground) to be 10 - 20 times RE.
9. calculate divider current (ID) by equ. (VB / RB1)
10. calculate RB2 (resistor base to supply) by equ. {(VCC - VB) / ID}
11.Prototype, (breadboard, or simulater) and check for key voltages.

1a. If Vc is too low against calculations, than check VB, if VB is too low from what you calculated than keeping voltmeter at base disconnect RE, and if VB rises considerably, than you have remedied that the base resistor RB2 is too large for this bias schem. So keeping your voltmeter at the base switch in higher value resistors at the emitter, until the VB, remains stable, then check your VC, and see how close it is to your calculated value.

In that scheme. you encountered a base current loading on the voltage divider, which means is that the transistor was biased according to it's own individual parameters, not good.

So by changing out resistors at the emitter you are decreasing the amount of base current so more current exists in the divider.

Now another remedy would be to keep the RE at the same calculated value to aquire the gain you desire, so then it is a matter of experimenting with the voltage divider current ID, you want to have more current flowing in your divider so you again calculate required base resistors by lowering RB1 and reworking yhe value for RB2 until you get a satisfactory reading on VC.

When you get the proper voltage reading especially VC, then switch out about 5 transistors in place of the original to see if there is any significant change in VC, that would cause problems in later stages. If VC is close to the proper value than the transistor is biased by external resistors and not by it's own parameters. In other words you have more or less programed the bias for that stage, the transistor does it's part by going along for the ride on it's input and controlling ONLY the output results.

Practice this until you fully understand why you had to change out a resistor value different than what you calculated, and be able to explain why this new value needed to be in place.

Than learn to bias design from the input side, set a input impedance that you cannot change, and than biass the amp by using that resistor value for RB1, and then working your way to the output stage.

Again get a thoro knowledge of why each resistor that needed to be changed had to be implemented.

Than from there practice biasing a circuit with a specific input impedance and voltage gain, choose values that are very stringent, than try to meet the bias requirements, then you will learn what works and what cannot work, again work at it until you have full understanding of why these requirements cannot be met. Yhe pro's call it (cannot be realized) than you can get a grasp at why certain values won't work.

Than a final practice where I'm at now is bias a stage, than replace one of the bias resistors with another biased transistor stage, to learn how to design stages interacting with eachother on a DC level.

The only reason I'm sharing this is because it seems like you realy want to learn this stuff from tyhe ground up, so since I am a hobbyist at electronuics I can give design insight at the most BASIC level while others can take you to the next higher levels.

16. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
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Thanks Hobbyist! I don't have time right now to look over all of that, but I will do it soon enough.

Jony,

I looked in the specs of the 2N3904 and it says for the "On Characteristics," that you need Ic=10mA and Vce=1V. This will acheive an hFE of 100. Since I know what Rc should be according to Ic, what should Ib and Ie be to put this guy in its linear region?

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,667
1,311
Forget about those "On Characteristics", Vce=1V its almost saturation not a linear region.
In general, each value of the Ic current (for example from 100uA to 50mA) may lie in linear region or in saturation region. We as designers decide whether given Ic current is a linear region or saturation region.
See, for example a amplifiers with almost the same voltage gain but different Ic current (100uA; 1m; 10mA). So linear region is "everywhere" and does not depend on Ic current that we choose for our operating point Vce=0.5*Vcc.

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18. ### Audioguru Expert

Dec 20, 2007
10,797
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Then the transistor is saturated (turned on hard) and cannot amplify anything.

19. ### Audioguru Expert

Dec 20, 2007
10,797
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No.
The minimum current gain is 100 when the Ic is 10mA and the Vce is 1V. The max current gain is 300.
The transistor in this excercise has a gain of 290.

20. ### Wraakneming Thread Starter New Member

Sep 12, 2009
2
0
Thanks for all the help.

This is what my lecturer says its suppose to look like:

a) Ie = Ic + Ib
= Ib (β + 1)
= (1 + 1/β)
= 1 + (1/290)
= 1.0034 A

b) Ic (sat) = Vcc / (Rc + Re)
= 20 / (1000 + 220)
= 16.39 mA

c) Vceq = Vcc / 2
= 20 / 2
= 10 V

d) Icq = Vceq / Rc + (IeRe)
= 10 / (1000 + (1.0034 * 220)
= 8.19 mA

e) Ib = Icq / β
= (8.19 * 10^-3) / 290
= 28.24 μA

f) Ibleed = 10 * Ib
= 10 * (28.24 * 10^-6)
= 282.4 μA

g) Ve = (Icq + Ib) * Re
= (8.218 * 10^-3) * 220
= 1.808 V

h) Vb = Ve + 0.7
= 1.808 + 0.7
= 2.508 V

i) R1 = Vr1 / Ir1 (Vr1 = Vb & Ir1 = Ibleed)
= 2.508 / (282.4 * 10^-6)
= 8881 Ω

Preferred Value => 8K2

j) Ir = Vb / R1 (Ir has to be redone because the resistor value differs)
= 2.508 / 8200
= 305 μA

k) Vr = Vcc - Vb
= 20 - 2.508
= 17.492 V

l) Ir2 = Ir + Ib
= 305 + 28.24
= 333 μA

m) R2 = Vr / Ir2
= 17.492 / (333 * 10^-6)
= 52528.53 Ω

Preferred Value => 56K

The Q points would be Ic = 8.19 mA and Vce = 5 V