# Frequency Response of CE-Amplifier

Discussion in 'Homework Help' started by RandomPhantom, Apr 3, 2014.

1. ### RandomPhantom Thread Starter New Member

Apr 3, 2014
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0
I understand that while increasing the frequency of the input signal of a CE-amplifier from zero onward, the gain increases to a point. For some time, even though the frequency is increasing, the gain remains constant or undisturbed. As the frequency still increases the gain later drops.

This is what all our textbooks explain till. I was just hoping someone could answer why the gain of a CE-Amplifier shows this pattern.

For those who didn't get what I was trying to ask:

Why does the gain increase, become constant and then drop again while frequency of the input signal is increased?

(this is without feedback by the way...well either way it's the same, I guess.)

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The answer is very simple. The capacitors with resistors forms a filter.

Current in the capacitor is proportional to the rate of change in voltage.
I=C*du/dt=C*ΔU/Δt
So, the faster the voltage change, the larger the current flow through the capacitor. So we can say that "resistance" of a capacitor (we call it reactance Xc) decreases when frequency of AC signal increases.
For F=0Hz-->Xc=∞; F=∞ -->Xc=0Ω.
Thanks to this property we can look at filters as a voltage dividers.
For example a low-pass filter consists a resistor and capacitor connected in series. The output voltage is taken from capacitor.
This circuit should be treated as a voltage divider where the role of the variable resistance is take by capacitor. For low frequencies the capacitor has a high reactance so almost the entire input voltage is transferred to the output.
When input frequency increase the reactance of a capacitor decreases and thus causes output voltage to increase.
At frequency at which reactance of a capacitor is equal to the value of resistor the output voltage is equal Vout=0.707*Vin
And this happens at frequencies equals

$F=\frac{1 }{2*\Pi*R*C}$

The divider's voltage ratio Vout/Vin of a voltage divider that contain two resistors doesn't depend on frequency, because the resistance of resistors does not change with frequency.
In our case the divider's voltage ratio change with frequency.
Because R is unchanged but capacitor reactance Xc is change with frequency.

http://en.wikipedia.org/wiki/Low-pass_filter#Electronic_low-pass_filters

JasonL likes this.
3. ### AnalogKid AAC Fanatic!

Aug 1, 2013
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Without a schematic I'm just guessing, but...

If the CE amplifier has capacitors for input coupling, output coupling, and/or emitter bypass, Each of these acts as a high-pass filter, and that covers the gain increase part. The flat part is the transistor doing its thing. Without some kind of feedback the flat part won't be very flat, but it won't have the defined slope of the high-pass filter.

The gain dropoff at high frequencies is due to the transistor itself, and the fact that a transistor's gain decreases at high frequencies due to small capacitances among the junctions that are an inevitable consequence of the stucture. The frequency at which the transistor's beta has decreased all the way doen to 1 is called the transition frequency. For a 2N2222 this is 300 MHz.

http://en.wikipedia.org/wiki/Gain%E2%80%93bandwidth_product#Transistors

ak

4. ### w2aew Active Member

Jan 3, 2012
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BytetoEat likes this.
5. ### PRS Well-Known Member

Aug 24, 2008
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I watched your video. Well done! You have excellent equipment. I have a 35 year old Tek 475 (200 MHz) O scope and it has none of those fancy gadgets yours does. I have to calculate the 3dB point.