Bypass Capacitor sets the gain and AC

[walters]

How does the Bypass Capacitor in parrallel with the Emitter Resistor
set the gain? the gain of what AC voltage or DC voltage?

Where does the AC come from because i'll i see from Emitter to Collector is DC voltage now AC so where is the AC voltage coming from ?

AC voltage from the Base doesn't go to the emitter or collector" what i was told

So where does the AC come from then if the Bypass Capacitor sets the gain?

 

[zachtheterrible]

A capacitor passes AC. A capacitor will block lower AC frequencies and pass higher frequencies, depending on what the value of the capacitor is. So the amplifier has higher gain at higher frequencies and lower gain at lower frequencies. I believe that the formula for what capacitive reactance is:

1
------------------
2 * pi * F * C

F=frequency in HZ
C= capacitance in farads

 

[Wrighty]

yes the formula above is correct. the capacitor is a low impedance pass for AC. If you do not have this then your hfe will drop down to about 10 I think. It just provide -ve feedback.

 

[walters]

Where did the AC frequency come from because all i see is DC from
emitter to the collector ?

 

[fsahmed]

Next time u want to know something like this, please give the circuit diagram as well.

The DC from the emitter to collecter is the supply given to the amplifier, the AC is the input signal to the amplifier, it should be on the base.

 

[lord loh.]

Yes all definitely is DC.

But the DC varies as the base current varies. This out put is actuallt an AC riding on a DC or an AC with a DC offset.

The DC components can be blocked by a coupling capacitor and what you get is a pure AC.

If this AC voltage developed between Collector and emitter is not grounded away, it shall reduce the gain and therfore is bypassed via the resistor to the ground.

The resistors in amplifier circuits are not meanat for the signals. They are there only to bias the transistor to the quincent point.

 

[walters]

Its a Bypass Capacitor in parallel with the Emitter Resistor if the AC voltage or current goes in at the Base of the transitor the AC goes through the Bypass Capacitor (not a Coupling capacitor) instead of the emitter resistor to increase the Gain but how does it increase the gain ?

 

[k7elp60]

In the attached CE, NPN transistor amplifier, biased as Class A. If the capactive reactance of C1 is 1/10 the value of R3 at the lowest frequency to be amplified, it will not affect the gain, as the gain is set by RC/RE. R1,R2 and RE+R3 set the bias point, and the no signal collector current.

 

[walters]

Thanks for the picture


So how does C1 the bypass capacitor adjusts the Gain?
is it passing AC frequencys? AC voltage? AC current?
or is is passing DC?

What does C1 really do ?

It looks like to me that AC freqencys,voltage,current goes through the
C1 and goes straight to ground but how does that increase the gain?

Can someones please tell me or give me a signal Analysis on this?

 

[mstechca]

in the most simplest terms, C1 can act as the "cancelling capacitor". C1 and the resistor in parallel with it make the "miller effect" circuit. It is almost the same as connecting a wire in parallel with C1, except that the transistor will have a harder time blowing up (ohms law proves this), and the amplifier will more than likely function better.

I find that the bandwidth of a given frequency is increased, but I am not sure how bandwidth is calculated.

 

[k7elp60]

As stated by me earlier that the gain of the CE class A amplifier is = to RC/RE. If RC=10k and RE=1K the gain would be 10.
If one remembers that capacitive reactance decreases as the frequency increases and reactance is also known as AC resistance. This AC resistance is C1 which is in parallel with R3. So the gain at AC = RC/((RE+(R3//Xc1)).
Lets take a for instance: If R3 is 470 ohms and at some frequency C1 is
47 ohms, we now have 47 ohms in parallel with 470 or AC resitance of 42.7 ohms. At this frequency the actual gain = 10K/(1k +42.7ohms) or 9.59. Going one step further if at some frequency the Xc = 470 ohms, then the AC gain =10k/(1k+235ohms) or 8.1.
From this one can see that the AC gain of the stage will decrease depending upon the reactance of C1.
If RE is replaced with a short circuit and the values of R1 and R2 are recalculated for idle current, then the reactance of C1 will increase the AC gain of the stage.

 

[walters]

Thanks k7elp60 for the math Calculations and information

So are you saying if the Emitter of the transistor is shorted to ground
you get more GAIN?

But we can't short the emitter to ground directly because that will not Bias the transistor the emitter resistor keep the transistor in bias operations i think.

The bypass Capacitor is a "Frequency based" gain circuit

 

[zachtheterrible]

So are you saying if the Emitter of the transistor is shorted to ground
you get more GAIN?

Yes. Since the signal passes through the base to emitter to ground, it will be able to turn on the transistor harder since there is nothing blocking it.

Its not always a good idea though, since you risk destroying your transistor if you're letting more current flow through it than is recommended.