Understanding High-Pass Filters and CE Amplifiers

[Grim Arrow]

So I wanted to make something clear. I really posted too much threads this week, but that's for the purpose of learning.
So a CE Amplifier has a blocking capacitor which together with the input impedance forms a high pass filter. But what I think is this filter is a high pass one only when the polarity of the signal source is such that the capacitor is atop the resistor and connected directly to the positive voltage. When the polarity reverses, now the resistor will be atop the capacitor and connected to the positive voltage and this is then a low pass filter. So why do we call it High pass?

Thanks in advance, I know I am getting quite vexing;)

 

[berkeman]

So I wanted to make something clear. I really posted too much threads this week, but that's for the purpose of learning.
So a CE Amplifier has a blocking capacitor which together with the input impedance forms a high pass filter. But what I think is this filter is a high pass one only when the polarity of the signal source is such that the capacitor is atop the resistor and connected directly to the positive voltage. When the polarity reverses, now the resistor will be atop the capacitor and connected to the positive voltage and this is then a low pass filter. So why do we call it High pass?

Thanks in advance, I know I am getting quite vexing;)

The HPF is formed by the input capacitor and the parallel combination of the base biasing resistors. The instantaneous polarity of the input signal does not come into play...

http://macao.communications.museum/images/exhibits/2_16_3_1_eng.png

 

[Jony130]

Why would input signal polarity change the filter type ?? Is it not the output voltage always taken across the resistor for HPF ??

 

[Grim Arrow]

Why would input signal polarity change the filter type ?? Is it not the output voltage always taken across the resistor for HPF ??

But isn't the point of HPF that if the frequency decrease, Xc will increase and more voltage will fall across the capacitor and progressively less across the resistor which results in a progressively decreasing amplitude of the output signal. Thus HPF won't pass signals below the cutoff frequency. While if the signal polarity was such, that it forms LPF, as you devrease the frequency, Xc increases and more voltage falls across the capacitor thus in this case more voltage for the output.

 

[berkeman]

But isn't the point of HPF that if the frequency decrease, Xc will increase and more voltage will fall across the capacitor and progressively less across the resistor which results in a progressively decreasing amplitude of the output signal. Thus HPF won't pass signals below the cutoff frequency. While if the signal polarity was such, that it forms LPF, as you devrease the frequency, Xc increases and more voltage falls across the capacitor thus in this case more voltage for the output.

Using the sample schematic I posted above, can you describe why the input polarity would matter?

 

[Jony130]

The output voltage for HPF is always the voltage across the resistor no matter what input voltage polarity is.
And the AC signal polarity change the current direction only (and the output voltage polarity ), but we still have a output across resistor.

 

[Grim Arrow]

Oh, so now as I think of it, "and the output voltage polarity" that might be the case. Normaly this would mean that when the signal shifts polarity, the output voltage will fall(because it forms Low Pass filter) but I think V+ decreasing is like to say the negative voltage is increasing and this is what we need at this half of the cycle.

 

[berkeman]

Oh, so now as I think of it, "and the output voltage polarity" that might be the case. Normaly this would mean that when the signal shifts polarity, the output voltage will fall(because it forms Low Pass filter) but I think V+ decreasing is like to say the negative voltage is increasing and this is what we need at this half of the cycle.

I'm not able to parse what you just wrote but just think in terms of the envelope of the sine wave signal before and after the HPF. The amplitude of the envelop will depend on the frequency of the sine wave compared to the breakpoint frequency of the HPF.

 

[Jony130]

I'm not able to parse what you just wrote

The same here

 

[Grim Arrow]

Basicaly I stated that when the polarity is reversed the HPF will act as a HPF, but for V-
(Looking at a sine wave this V- would be the minimum voltage)

 

[Jony130]

So you really don't understand what negative voltage is ?? If Vin is a AC signal +10V at positive peak and -10V at negative peak. So for exampel if a signal frequency is equal to 1Khz and Xc is 1K and R is also 1K (conner frequency) the voltage across resistor will be equal to +7.07V at positive peak and -7.07V at negative peak.

 

[Genji Shimada]

Look, Arrow, I don't think you understand how High pass or Low pass filters work. It's like what you are thinking is that HPF is intended to serve in DC circuits, but in the case of Common emitter amplifier it is under AC conditions. Think on that. There is no frequency filter without polarity shift. If the frequency is 0 (DC current) the capacitor acts like an open circuit and no current flows through it and all the voltage falls across C. If the frequency is infinite, then Xc=0 and all the voltage is across R.
It's an AC voltage divider.

 

[Genji Shimada]

One more thing. I looked twice at your question and I think I get it where you are wrong.
It doesn't matter where the capacitor is positioned with respect to the AC source, it matters where it is positioned with respect to whatever you want to connect to the output of the HPF.

 

[Jony130]

Maybe this will help you.
Let us see how the AC input current flow in CE amplifier supply from a single supply.
First we bias the BJT in the "linear region" via RB, RE resistors and 5V voltage source to be able to amplifier any AC voltage.

Vc = 6V (red plot ); Ve = 2V (green plot) ; Vinput = 1V (blue plot)
The DC voltage at base is equal Vb = 2.6V, and emitter DC voltage is 2V (Ie ≈ 600μA; Ib = 6μA).
Our AC source "produces" 2 peak to peak voltage.
And this AC input voltage will "modulate" the voltage at our transistor in the "rhythm" of a AC input signal.
The base voltage will change from 3.6V to 1.6V in "rhythm" of a AC signal.
These changes will result that the emitter voltage will also change. From 3V to 1V. This will result the change in emitter and in collector current by (3V) 0.9mA to 0.3mA (1V).
But what path the AC current flow ? Let us see:
For positive half-cycle (+1V peak) we have this situation:

As you can see 5V source provide 3.5μA of current, but the base needed to be equal to 9μA, so this additional current will come from the AC signal source. As I show in the diagram.

For Negative half-cycle situations looks like this:

As you can see this time the base need only 3μA, but our 5V bias source provide 8.5μA. This implies that our AC signal source need to "sink" this excess current 5.5μA.
Also notice that the Rin = Vin/Iin = (1V - (-1V))/(5.5μA - (-5.5μA)) = 2V/11μA ≈ 180kΩ which is true because Rin ≈ RB||(β+1)*RE ≈ 400kΩ||330kΩ ≈ 180kΩ
I encourage you to carefully analyze this.

 

[Grim Arrow]

Can't believe it. Can't believe how blind I was.. Offcourse. I mean, I know this, I was just blinded somehow. Too much thought in your head can make you dumb enough to not see the truth. It is just an ordinary HPF!
Thank you, Genji! Thanks, Jony! Thanks, Berkeman!