# Basics of Common Emitter amplifier.

1. Nov 29, 2011

### shayaan_musta

Hello experts!

I have some question related to this configuration of BJT.

1)Why we use capacitor at the base of the circuit?(Answer: To decouple DC so that our signal that is reaching to base of the transistor is not to be DC shifted that's why it is called "Coupling capacitor". If it is not then we get our signal DC shifted from the ground level). And why we use capacitor at the emitter of the circuit i.e. CE?(Answer: Because we want signal to be ground easily. If capacitor is not placed there then we will have less gain.)

2)2)Why we use resistors at collector and at emitter?(Answer: So that we could get stability if it is not here then we get distortion at the ouput signal).

2. Nov 29, 2011

### Studiot

First two are OK (unlike your numbering)

But think a bit harder about the collector and emitter resistors and ohm's law.

3. Nov 29, 2011

### niehaoma

Emitter resistor RE affects the input impedance and the gain by the same factor, in opposite magnitude, by [1+gmRE]. The smaller RE the larger the gain, but the lower the input impedance. Everything in the amplifier is a design tradeoff dependent on the functional circuit requirements.

4. Nov 29, 2011

### technician

Usually the CE amplifier is arranged so that the steady voltage at the collector is roughly half the supply voltage Vc≈ Vs/2. The output of the amplifier is Vc and this means that Vc can 'swing' symetrically up to Vs and down to 0 (roughly).
So if Vs is 10V you would want Vc to be about 5V.
If you decided on a collector current Ic of 1mA then this means that Rc would be 5kΩ.
The way to set Ic to 1mA is by means of the emitter resistance Re in conjunction with a bias voltage set at the base. If Vb is set to 2V then the voltage across Re will be (2-0.7) for a silicon transistor = 1.3V. To have an emitter current of 1mA therefore needs a resistor Re of 1.3kΩ.
The voltage at the base of the transistor can be set by means of a potential divider across the power supply (10V)
When a small AC signal voltage is applied to the base we want it to be applied between the base and the emitter and therefore an AC 'short circuit' is needed across Re.
This is the purpose of the 'bypass' capacitor across Re. It is selected to have a low resistance (reactance) compared to Re at the frequencies to be amplified.
eg if Re is 1.3kΩ and the lowest frequency to be amplified is 100Hz then we want 1/ωC to be less than (about 1/10) Re
We want Xc to be about 130Ω at100Hz means C of about 12μF

Hope this helps !!!!

5. Nov 29, 2011

### sophiecentaur

You need the transistor to be conducting current all the time if your AC signal is to be undistorted on the way through. The input signal needs to vary the current in the output about a non-zero value. A good way to ensure this is to ensure that the DC conditions (bias) and the AC conditions (the signal) are set up separately - hence the use of Capacitors to couple the signal into the amplifier and the use of resistors to set the current.

6. Nov 30, 2011

### shayaan_musta

Thanks for all who replied.

Coupling capacitor is to set the configuration for undistorted output. Is it? and if it is not there then what will we get at output?

Similarly I have known by here that resistors at emitter, collector & base is to limit current. Is it. If they are not at their places then excessive amount of current can smoke my device.

7. Nov 30, 2011

### sophiecentaur

You presumably realise that there needs to be a standing current through the transistor and the AC signal consists of a variation around this mean value. If you don't have an appropriate value for the standing current then you can saturate the transistor or it may 'turn off' before achieving a full amplitude of AC output variation. The Collector Volts are determined by the current through the Collector load resistor.

The resistor in the emitter does two things for you: it is basically a feedback mechanism because the emitter volts will 'follow' the base volts (current will flow out of the Emitter through the Emitter load until Vbe reaches about 0.7V). It ensures that the standing current is what you want and it also defines the voltage gain of the amplifier - which is given by -Rc/Re.
Without an emitter resistor (Emitter just grounded), Vbe can be very high and damage the junction. Or, if there is a high resistor in series with the base to limit the base current, the transistor can still saturate and the linearity is poor. Feedback of some kind is essential for making a linear transistor amplifier and an emitter resistor is very suitable, although not the only way it can be achieved.

8. Nov 30, 2011

### shayaan_musta

But the purpose of coupling capacitor is to separate DC from the AC signal. If it is not there then I will get output shifted from reference and may be it gets saturate. Hence, therefore we put a capacitor between collector and load. Am I right.

Now as you said that RE perform a job of a feedback. But how? I can't realize it. And two type of feedback are; degenerative feedback(+ve feedback) and regenerative feedback(-ve feedback). Does RE performs anyone of these feedback's job? If so then how?

But it is the appreciable explanation. Thanks sophiecentaur.

9. Nov 30, 2011

### Studiot

shayaan, we need to clear up a few things.

First and most important:

The input and output capacitors are there to block the DC, not to prevent 'signal shifting'.
If the capacitors were not there you would simply get a loud bang and smoke when you connected things together. The signal would not be shifted it would disappear altogether.

The capacitors allow an alternating signal to be transferred between two circuits or parts of a circuit that are operating at different direct voltages, without damage. Proper choice of capacitor values permits transfer without distortion - you can't just bung in any old capacitor you happen to have.

The input and output capacitors are called coupling capacitors, (not decoupling capacitors).

Now you have asked about the common emitter configuration and I suggested you consider the collector and emitter resistors in the light of Ohm's law.

Did you do this?

For a CE configuration the output is across the collector resistor.
The transistor tries to drive beta times the base current through the collector resistor, regardless of the value of that resistor. It can do this so long as the circuit remains within parameters - we say in the active region.
Ohm's law says that the greater the current through the collector resistor the greater the voltage across it and that voltage equals the current times the resistance.
The greater the voltage the greater the voltage gain.
Clearly if there is no collector resistor (ie it is zero) zero times the current equals zero, ie the output and gain are zero - not a useful situation.
So we make the collector resistor as high as practicable for the circuit concerned.

If you have followed this we can move on to deal with the emitter resistor and feedback.

10. Nov 30, 2011

### technician

Lets do a design exercise about what we want !!
1) specification : a voltage amplifier to produce a power output of 10mW from a 10V supply.(this is a low power amplifier.... it would drive headphones....just)
2) voltage variation is from +5V to -5V therefore current variation is +/- 2mA
3) this 2mA is to be drawn from a supply of some sort.... lets make certain it can supply 2mA......make it capable of supplying 10 x this ie 20mA
4) If we decide to build a CE amplifier using one transistor then the output is from the collector and needs to vary by +/-5V
5) The collector resistor is chosen so that Vc =5V (0.5Vs) So 5V appears across Rc.
We want Ic to be 20mA (this is where the power current is drawn from)
This gives a value of Rc = 5/20mA = 250Ω
6) How to get Ic =20mA?... well Ic is more or less equal; to Ie (the emitter current) and the voltage across any emitter resistance is more or less = voltage at the base -0.7 (for a silicon transistor). If we make the base voltage 1.7V (using a potential divider) then V across Re = 1V
we want the current to be 20mA therefore Re = 1V/20mA =50Ω
This combination of Rc and Re will give a single transistor amplifier capable of delivering 10mW. There is an endless combination of resistors for the base bias combinaton.
If we want this amplifier to amplify audio signals with a lowest frequency of say 100Hz then a bypass capacitor across Re is required. The reactance (Xc = 1/ωC) of C must be less (X10) than Re so that it behaves as a short circuit at this frequency. So Xc must be less than 50Ω at a frequency of 100Hz gives a value for C as greater than 30μF
The coupling capacitors are another matter but as long as they are there they will do the job of isolating the DC conditions in each stage of the amplifier. Basically coupling capacitors need to have as low a resistance (reactance) as possible for the frequencies to be amplified..... there is a lot of lee way in this.

11. Nov 30, 2011

### sophiecentaur

@studiot
Some of that is a bit oversimplified. There are many DC coupled amplifiers; a capacitor is only necessary when the input DC conditions are not suitable. AS we are not referring to any particular amplifier then you cannot assert "a loud bang" would occur. With large enough coupling resistors there will be no excessive base current; look at the basic RTL logic gate configuration, for instance. (Of course, a logic gate is not a good linear amplifier but that is another issue).

@shayaan_musta
A few issues here. You seem to have the feedback terms the wrong way round. I would have used the word regenerative feedback for positive feedback and vice versa.
Coupling Capacitors are only appropriate when you do not need to have a DC response. Many laboratory amplifiers need to have a flat response right down to DC. (Take the amplifiers in an oscilloscope for instance). It is just more convenient to use AC coupling, sometimes so you can treat the DC and AC conditions separately..
I said that Re provides negative feedback and described what happens: current will flow through Re until the Vbe stabilises at 0.7V. This feedback keeps Ve 'following' Vb-0.7V. If Ie goes too high or too low (by a small amount), then Vbe changes by a little and causes the value of Ie to change, returning Ve to Vb-0.7V again.
So the volts across Re determine the current through Re, which is the same (plus a minute amount of base current) as Ic. The volts across Rc are then equal to IcRL which is why I said the voltage gain is equal to the ratio of RL/Re (This voltage comprises the DC value due to the base bias and the AC variation of the base current) . Without the emitter resistor, the voltage gain would be much higher because the collector current would be Beta times the base current - but the output would be distorted.

12. Nov 30, 2011

### technician

I have never had any 'loud bangs' or 'smoke',
I think that I must have done something right

13. Nov 30, 2011

### sophiecentaur

HAHA - just keep all your resistors high enough and the transistor will survive anything!

Your arm waving / verbal amplifier design seems just fine to me. You have the same attitude to this as I do - start with as many simple assumptions as possible and design around them first. If you need to tidy things up then do it later.

14. Nov 30, 2011

### Studiot

Consider two CE connected transistors, both with a beta of 100 and a base current of 100μA.

Transistor 1 has a collector resistor of 1k and an emitter resistor of 100.

Transistor 2 has a collector resistor of 1k, but no emitter resistor.

What are the collector and emitter currents for each transistor? (assuming an adequate voltage rail in each case).

15. Dec 1, 2011

### sophiecentaur

How would you ensure that (and is it a relevant question at this stage)? With a constant current bias circuit? The input resistance with 100Ω would be something like 10kΩ so the base current, through whatever bias resistor was used, would be reduced accordingly. Did you not know that or were you just being 'controversial'? There's always another factor that can be introduced into a crude, practical circuit that has not actually been drawn out in full or treated rigorously. I should say that your question is more about the curves you find in a Transistor Spec Sheet than about your average circuit.

16. Dec 1, 2011

### shayaan_musta

at sudiot

As you said capacitor at input and output is to block DC. I know the purpose of capacitor is to block DC and allows AC to pass.
So now let an AC signal $\pm5V$ with 1V of DC is at collector. Now if capacitor is not present here then obviously it is DC shifted because AC signal $\pm5V$ has ground level but now without capacitor DC also adds to AC and output becomes +6V and -4V at output. Is not it?

And yes I have considered that in the light of Ohm's law and reach at the same place i.e. same to your answer.

I have just one question that I asked above after that we will move to emitter resistor and feedback.

That was really a good reply.
Thanks.

17. Dec 1, 2011

### shayaan_musta

at technician

Yes my original purpose is to design a CE amplifier.
But someone said me that before designing it you should know and understand that how every component works. And if any component is not at the place where it should be then what will be the output under these type of conditions.

18. Dec 1, 2011

### shayaan_musta

at sophiecentaur

OK. I got something i.e. RE provides negative feedback.

Now let me think how then I will discuss it more with you here.

Thanks for hint.

19. Dec 1, 2011

### sophiecentaur

You seem to have ignored any voltage gain that will be there if Rc is greater than Re.

It really is time to have a diagram that we can discuss. I am not sure where your voltages are being measured. How can you get a -4V at the output if you are operating between 0V and some Vcc that is not specified?
I have a feeling that you are making all this up on your own instead of getting the information from a book. Why not look at some practical amplifier circuits and descriptions and try to figure out how they are operating? I googled "simple transistor circuit" and found enough pictures to wallpaper my living room.

20. Dec 1, 2011

### Studiot

Hello shayan, I'm glad you are finding my input useful.

I wholeheartedly endorse this.

However when you first meet transistor circuit theory it can be very confusing. The purpose of each component can be hard to see since they all interact.

That is why I like to start with something simple. To see what a couple of components are doing and then see why we might need to add more. Finally to build up to a useful circuit.

IMHO what SC and T have offered are to complicated to start with and beyond your original questions.

If you would like to follow my trail I will draw some diagrams, starting from the simple and moving on to the complicated, giving reasoning.

21. Dec 1, 2011

### sophiecentaur

That is a fair point of view but the transistor is a very complicated device. It is, basically, a current amplifying / controlling device and we mostly like to think in terms of voltage amplification. It's highly non-linear and its 'curves' are pretty well irrelevant for elementary design if it was made within the last 40 years. Starting off with its behaviour as you vary the base current may be of interest to a more advanced student but, with a current gain of 100 or more, and the simple introduction of an emitter resistor, turns it into a device with a high input resistance and allows some very easy calculations to give you that resistance, suitable DC conditions and the voltage gain of the stage - just by treating it as a black box.
When I first came across transistors (1962 ish) I looked at a leaflet about the Equivalent Circuit of an early, low performance, Junction Transistor. It made this simple-to-use device about as accessible to me (a schoolboy) as a treatise on the Schroedinger Wave Equation. It was not until, five years later, (after Physics at Uni) I talked to a very clever circuit designer (seriously bright bloke with patents galore after his name) in my first job. He cut all the crap out of it and gave me a half dozen basic rules for calculating behaviour and designing with transistors - as a first stab. Suddenly they became approachable.

Designing a circuit for low frequencies, signal levels that are well above the noise and with low output power is a piece of cake and allows you to use almost any transistor 'out of the drawer'. The current gain is Irrelevant, as long as it's above about 100 and so are most of the other parameters. Once you have made basic designs work for yourself then you find your limitations and then may need to look closer.

Electronics is a very 'modular' subject and can be appreciated at many levels. I think that shayaan would really benefit from looking at practical circuits and seeing the sort of resistor values that are used. Your comment about "bangs and smoke" just doesn't apply to low voltage circuits with resistor values of 1kΩ involved.

22. Dec 1, 2011

### shayaan_musta

Hello.

Thanks for appreciable helping.

I want to know working and importance of presence of every component using in CE amplifier and then other two CB and CC are same.

at studiot

Yes. Took me through a quite basic step. So that I could easily understand each and every thing and able to teach someone else. For this I will be very great full to you. In our university, not much information is provided for any study. They just give overview and ask to do each and everything by own self.

Well here is my diagram as sophiecentaur and you(studiot) said.

Now I hope that I will easily capture things. As English is not my mother tongue. So I request to you people to use easy language.

Thanks.

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23. Dec 1, 2011

### sophiecentaur

That diagram makes things easier, thanks.
Re provides DC feedback which sets the current through the transistor. 'Looking into the base', that 220Ω will appear as approximately 220βΩ (say 22k), which will appear in parallel with R1. Almost negligible but it changes the effective value of R1 to about about 2.4k, giving a base voltage of about +1.45V (do you see the potential divider chain?). That produces an current through Re of (1.45-0.7)220 or 3mA (the 0.7V is the diode drop across the be junction). This 3mA , going through Rc gives 6V across it (the DC value).. Leaving the collector at 12V above ground.

Because of the capacitor Ce there is no AC feedback so the AC gain is not defined but, if you assume again the β is 100, then this is the best you can do. A reasonable voltage swing about the collector standing volts would be +-5V as you say. The actual voltage gain will depend on the value of RL but assume the Load is ten times the valus of Rc and you can 'ignore it'. So an input of just 100mV peak to peak will appear as a voltage swing of 10V peak to peak at the collector.

24. Dec 1, 2011

### Averagesupernova

The AC gain of an amplifier with an emitter bypass capacitor is typically Rc/R'e. I would hardly say it is undefined.

25. Dec 1, 2011

### sophiecentaur

How can it be 'defined' if there is no known value for the feedback parameter?