# very confused about how basic amplifiers work

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#### billion_boi

##### New Member
hello, I've managed to understand how the voltage divider allows us to set up a Iq independent of beta.
Then when it comes to introducing the A.c input signal and bypass capacitors i get lost. If we introduce A.C signal from the base loop doesnt the parallel capacitor across re short out the input signal?

##### Well-Known Member
hello, I've managed to understand how the voltage divider allows us to set up a Iq independent of beta.
Then when it comes to introducing the A.c input signal and bypass capacitors i get lost. If we introduce A.C signal from the base loop doesnt the parallel capacitor across re short out the input signal?
Well, it does short the transistor emitter to ground, that is true, but the input includes this short plus the base-emitter junction, which means that all of your input AC voltage is impressed across only that junction. This is good because now you have your transistor working its hardest.

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#### billion_boi

##### New Member
Thanks for the input Radioron. so the capacitor is infact a short for the input A.C aswell the output A.C...
But this still confuses me. Let me explain what i thought was happening without the capacitor in the first place...a simple voltage divider with a A.C input. The a.c input doesnt show in the picture

The voltage divider(r1/r2) set up a voltage at re, thus creating an Ic(q) independent of Beta.
Now this is how i believed the A.C input signal created an A.C output signal...i had two theories...
1)The A.C input was superimposed on r2, thus creating an a.c voltage at R2..voltage changes at R2 created a voltage change to Re.So we had created an A.C voltage on Re, thus creating an A.c current in the emitter/collector loop.
2)The A.C input would travel through the base-emitter junction and superimpose on Re, creating an A.C voltage on Re, thus creating an A.C current in the emitter/collector loop.

Now with the input A.C shorted at the emitter how can the A.C input being aplied only at the base-emitter junction create an output A.C.

#### Ross Craney

##### New Member
get a copy of the characteristic curve of the transistor & it all becomes apparent - you are reading too much into it for a BASIC understanding

##### Well-Known Member
But this still confuses me. Let me explain what i thought was happening without the capacitor in the first place...a simple voltage divider with a A.C input. The a.c input doesnt show in the picture

The voltage divider(r1/r2) set up a voltage at re, thus creating an Ic(q) independent of Beta.
Now this is how i believed the A.C input signal created an A.C output signal...i had two theories...
1)The A.C input was superimposed on r2, thus creating an a.c voltage at R2..voltage changes at R2 created a voltage change to Re.So we had created an A.C voltage on Re, thus creating an A.c current in the emitter/collector loop.
2)The A.C input would travel through the base-emitter junction and superimpose on Re, creating an A.C voltage on Re, thus creating an A.C current in the emitter/collector loop.

Now with the input A.C shorted at the emitter how can the A.C input being aplied only at the base-emitter junction create an output A.C.
You are looking at this using voltages, which is only part of the story. You also have to consider current. After all, the transistor is essentially a current amplifying device. As mentioned, the characteristic curves of the transistor will show this. In any case, look into how to analyze the amplifier in the case where there is no Re, with the transistor's emitter tied directly to ground. This is the case with AC in this circuit.

#### qdn

##### New Member
Hmm, I could be wrong but here's my interpretation:
Even though the Re is shorted out with the bypass cap, you should still get the small re which is ~25ohm at 1mA emitter current. So with this common-emitter amplifier, you would get a gain of -Rc/re.

#### Jony130

##### Active Member
You have to remember that AC voltage "create" change in a base current.
So when the base current is change in the rhythm of a AC-current the collector current will also be change in the rhythm of base current multiplied by a current gain of a BJT. Ic=Ib*β; ΔIC=ΔIB*Hfe.
Of course some ac-current will flow through R1 and R2 to the ground.
And ac-current of a emitter will "miss" RE resistor and go through CE.
But this ac-current equal Ie≈Ic will flow from:
V+--->Rc--->collector-emitter---CE---->GND
Of course this ac-current is "create" by a base ac-current.

#### indulis

##### New Member
...doesnt the parallel capacitor across re short out the input signal?
That's not exactly true. While the capacitor “instantaneously” is a short, it’s only for a very brief time.

Xc=1/2*pi*f*c

This (Xc) changes inversely with frequency, which in simple terms means it’s “AC resistance" (or more properly Capacitive Reactance) is changing, and it isn’t zero (a short). Now your getting into gain bandwidth...

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#### billion_boi

##### New Member
Jony130, i was able to extract alot from your answer...and to alot of other guys who responded to help, i know the transistor was a current controlled device..

I just find it wied that during D.C biasing we can use the voltage divider R1R2 to set up a quinsient Ic -WITHOUT calculating for base current or Hfe,
but when dealing with the A.C input and effects on the output current we are dealing with its current?

#### BrownOut

##### Banned
during D.C biasing we can use the voltage divider R1R2 to set up a quinsient Ic -WITHOUT calculating for base current or Hfe,
but when dealing with the A.C input and effects on the output current we are dealing with its current?
Not exactly. You can think of both the DC and AC amplification in terms of current amplification or in terms of ratios of resistances. In the DC case, however, the emitter resistor provides a form of feedback that stabilizes the operating point. And so, we normally exploit the fact that collector and emitter currents are very close to the same, and approximate the gain as the ratio of the collector and emmiter resistors. You can make the same approximation for AC as the ratio of collector and small-signal emitter resistances.

There are usually more than one way to analyze an amplifier. And the methods can be mixed and matched for AC and DC. But in the final analysis, the physics of the transistor are identical for both. At least that is true at low frequency.

#### billion_boi

##### New Member
hmm ok, so although it may apear that biasing was done without base current in mind, the neg. feedback reisister does infact set up or base current for us to establish Icq. Thanks alot guys for ur help.

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