High pass filter or Low pass filter?

Thread Starter

rawchicken

Joined Nov 27, 2016
2
Hi

I have a quick question about something that I'm very new to and I cannot understand and I need to.

Below is the circuit I am referring to. The critical frequency is 15.9kHz. Is this a low pass or a high pass filter and why? from my understanding I think it is a low pass filter because at higher frequencies the inductor is doing more to stop the AC, is this correct? If it is culd anyone give a slightly more detailed explanation as to why please!

When doing lab work with this circuit the higher up I turned the frequency the larger the value of the output voltage Vout was and the smaller the lead of the Vin phaser (I went from 0.1fc to 2fc), but how is that correct if it is a low pass filter, or am I completely wrong?

Thanks in advance for any help!

 

shteii01

Joined Feb 19, 2010
4,644
Setup the equation Vout=something*Vin. Then plug in a few frequencies. You will see Vout increasing at certain frequencies and decreasing at other frequencies. That will tell you which frequencies are passing through and which frequencies are being filtered out.
 

DGElder

Joined Apr 3, 2016
351
An inductor inhibits AC current. Your output is not current through the inductor, it is voltage across the inductor. You should be thinking about what higher frequencies do to the reactance of the inductor and how that affects the current and voltages in the circuit. The idea "at higher frequencies the inductor is doing more to stop the AC" is too vague to be useful.
 

shteii01

Joined Feb 19, 2010
4,644
Thinking it through, Stop the AC is what it's doing. The TS just needs to think in terms of the current.
Ok. So. Convert inductor into complex impedance. The "size" of the complex impedance is dependent on the frequency. Frequency increases, impedance increases. Meaning it becomes larger and larger "resistor". What do we know about increasing resistors? They block more and more current. So the current though inductor will be decreasing, while the frequency is increasing, because complex impedance is increasing with frequency. How do we interpret this "killing of current"? Simple. The inductor is becoming an Open Circuit! The frequency keeps increasing. Eventually there will be no current because there is no pass for it because there is an open circuit in place of the inductor. What do we know about open circuit? We know that all the input voltage will appear across the open circuit. Which means that high frequency signals are passing though the filter. What kind of filter does that?
 

WBahn

Joined Mar 31, 2012
29,976
Hi

I have a quick question about something that I'm very new to and I cannot understand and I need to.

Below is the circuit I am referring to. The critical frequency is 15.9kHz. Is this a low pass or a high pass filter and why? from my understanding I think it is a low pass filter because at higher frequencies the inductor is doing more to stop the AC, is this correct?
What do you mean by "stop the AC"? Are you talking about the AC voltage (and, if so, where) or the AC current?

If it is "stopping the AC current", then that means as you go higher in frequency there is less current, right? If there is less current, what does this mean for the voltage across the resistor? What does this mean for Vout as a fraction of Vin?
 

MrAl

Joined Jun 17, 2014
11,389
Hello,

Looking at the current through the inductor is not the right thing to do unless it is explicitly stated which it is not for this problem.

For one, the OP is looking at the voltages.
 

MrChips

Joined Oct 2, 2009
30,708
Think voltage divider:



Hence Vout is lower as R2 decreases.

With a reactive element such as an inductor or capacitor, the same equation applies except we use impedance instead of resistance.



Note that we cannot do a simple calculation using the reactance formula X2 = ωL. This is because there is a phase shift between Vin and Vout.
In the voltage divider formula we have to replace R2 with the value Z2 = jωL.
The simple way to solve this is to use vector analysis.

Is it low-pass or high-pass?
Does X2 decrease or increase with increasing frequency?
 

JoeJester

Joined Apr 26, 2005
4,390
For one, the OP is looking at the voltages.
Yes the TS is. Where is the TS? Have they returned? Did they do the task recommended to calculate different frequencies, as recommended in post 2? I did, even though I knew the type of filter upon inspection, as the calculations were pretty easy in excel as was the graphing.

Z2, the inductor becomes an open and no current flows. If Z2 becomes a high enough impedance ... the source voltage will be present at the Z1/Z2 junction.

Yes, I know I broke down and showed the graphs, but I never identified the obvious filter name. I deleted the posts because they added no value.
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

Yeah the OP never came back to check for replies.
Guess it's another case of POTD (post once then disappear) :)
 
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