Understanding AC Filter Vout Calculations and Identifying Filter Types

In summary, the conversation revolved around understanding the formulas and components of AC filter circuits. The conversation touched on the different types of filters, such as high pass, low pass, bandpass, and band reject, and how to identify them based on their components. The conversation also discussed the concept of impedance and how it affects the transfer function of a filter. Some resources were suggested, such as the wikipedia.org page on electronic filters and the book "The Art of Electronics" by Horowitz and Hill.
  • #1
Petrucciowns
74
0
Does anyone know the formula for finding the voltage of Vout at a certain frequency in a AC filter circuit?

I have a formula it looks like Source x (XL or XC) / Resistor + i (XL or XC)

I can get it to work for some but not others. I have a feeling it needs to be rearranged depending on the situation.


Also how can you tell if a circuit is High Pass, Low Pass, Bandpass, or Band Reject. I know what they are supposed to do, but I have difficulty identifying which is which. Does it have anything to do with a capacitor being completely charged and then discharging over time or the opposite with an inductor?

I would appreciate any help.:tongue2:
 
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  • #2
Petrucciowns said:
Does anyone know the formula for finding the voltage of Vout at a certain frequency in a AC filter circuit?

I have a formula it looks like Source x (XL or XC) / Resistor + i (XL or XC)

I can get it to work for some but not others. I have a feeling it needs to be rearranged depending on the situation.


Also how can you tell if a circuit is High Pass, Low Pass, Bandpass, or Band Reject. I know what they are supposed to do, but I have difficulty identifying which is which. Does it have anything to do with a capacitor being completely charged and then discharging over time or the opposite with an inductor?

I would appreciate any help.:tongue2:

In general, a filter is a frequency-dependent voltage divider. So in the simplest form, you have a 2-element filter, with a top impedance, a bottom impedance to ground. You drive the top element with your signal source, and tap your output signal out of the junction in the middle of the two elements.

If the bottom element is a resistor, and the top element is a capacitor, what kind of filter is that?

If the top element is an inductor, what kind of filter is that?

If the bottom element is an inductor, and the top element is a capacitor, what kind of filter is that? How is it different from the answer for the CR filter above?

A bandpass filter takes more than just two elements, because it has to combine both a LPF and a HPF together. You could cascade two such 2-element filters to make a BPF, or there are other ways to do it better. Can you say what a better BPF might look like? How are the LPF and HPF parts of that BPF implemented?
 
  • #4
Hmm. I looked into the wikipedia page and read your post and am still having difficulties. I have a paper right next to me that tells me exactly what a filter is based on its components such as:
Top: Inductor Vout: Capacitor = Low pass
Top: Inductor Vout: Resistor = Low pass
Top: Resistor Vout: Capacitor= Low pass
Top: Resistor Vout: Inductor= High pass

The problem is that I have no idea how these are found. I am thinking this way, for example:
Top: Inductor Vout: Resistor = Low pass

An Inductor when plugged into a source starts low and charges based on time, soo would it be low pass ,because it starts off low? Or am I approaching this the wrong way?Edit: I also looked at the filters section on this website under the AC tab.
 
  • #5
No, think in terms of impedance versus frequency for the R, the L and the C elements.

What is the Z(f) equation for each of these elements?

Then think how the impedance divider is going to change over frequency as the elements' impedances change.

So for example, the impedance of a capacitor is infinite at DC and zero at infinite frequency, right? So how does that affect the transfer function of a C-R filter?
 
  • #6
I'm sorry for being so dumb, but you have me completely lost.:frown: I don't know what this Z(f) formula you are speaking of is. The values I have to solve for are FC (Cutoff Frequency) = 1/ (2ΩRC) AV (Voltage Gain)= VOUT/ VIN AV(db) (decibels gain)= 20 Log AV

I need to find the impedence of the capacitor or inductor and then I need to calculate the voltage across Vout at different frequencies. Finally I have to identify the type of filter.

I have no problem with anything except for the voltage across Vout at different frequencies and identifying the filter.

I'm sorry but what you tried to explain to me was way over my head.
 
  • #7
Do these formulas look familiar for the impedances of these components?

Inductor: [tex] Z(f) = 2 \pi j f L [/tex]

Capacitor: [tex] Z(f) = \frac{1}{2 \pi j f C} [/tex]

Resistor: [tex] Z(f) = R [/tex]

Where j is the square root of -1.

Now, use each of those impedance formulas in the impedance divider equation to get the output voltage in terms of the input voltage:

[tex]Vout(f) = Vin(f) \frac{Zup(f)}{Zdown(f) + Zup(f)}[/tex]

In general you will end up with a complex-valued equation for the transfer function, and in the end you plot the magnitude and phase of the complex transfer function. The Bode Plot of the AC filter generally just shows the magnitude, but can also show the phase.

See this wikipedia.org article:

http://en.wikipedia.org/wiki/Electrical_impedance

And/or Chapter 1 in "The Art of Electronics" by Horowitz and Hill, starting with the "Impedance and Reactance" section.
 
  • #8
Yes they are familiar ,but I am used to them being called XL and XC instead, and my formulas are slightly different:

XL=2πFL XC= 1÷(2πFC) and for the resistor in RLC type circuits I take "It" which is found by : Source/ZT times the resistor so...

It x resistor.
The only one that was unfamiliar to me was the last one, and I'm not really sure how to read that. I was aware that Vout and Vin frequencies were the same.I know to find XL or XC at a certain frequency you use that frequency instead of the resonant frequency.
 
  • #9
I re-arranged that equation earlier and got it to calculate the correct voltage ,but the wrong phaser.

Source x Vout / ( Component closest to source + i Vout)

So using this equation With a Low pass filter with a 10 VAC source a 10mH inductor and Vout of 50 ohms

XL at 400 HZ= 25.1 ohms


using the equation to find the voltage at Vout:

10 x 50 / (25.1 + i 50) = 8.93 with a phaser of -63.3


The actual answer should be 8.93 with a phaser of -27


So I'm doing something wrong with the phasers. I know I didn't enter XL into the equation with the phaser of 90 because it gives me a completely wrong voltage. Do you see what I'm doing wrong?
 

1. How do I calculate the output voltage of an AC filter?

To calculate the output voltage of an AC filter, you can use the formula Vout = Vin * (1 / √2), where Vin is the input voltage and √2 is the root mean square (RMS) value. This formula assumes that the AC filter is ideal and has a perfect 50% duty cycle.

2. What factors affect the output voltage of an AC filter?

The output voltage of an AC filter can be affected by various factors such as the input voltage, the type and size of the filter components, the frequency of the AC signal, and the load connected to the filter. It is important to consider these factors when designing and calculating the output voltage of an AC filter.

3. Can the output voltage of an AC filter be higher than the input voltage?

No, the output voltage of an AC filter cannot be higher than the input voltage. The purpose of an AC filter is to smooth out the fluctuations of an AC signal and provide a stable output voltage that is lower than the input voltage. However, the output voltage can be equal to the input voltage in certain cases.

4. How can I improve the output voltage of an AC filter?

To improve the output voltage of an AC filter, you can increase the capacitance and/or inductance of the filter components, reduce the frequency of the AC signal, and/or use a higher quality filter with lower impedance. Additionally, proper design and placement of the filter components can also help improve the output voltage.

5. Are there any limitations to the Vout calculation for AC filters?

Yes, the Vout calculation for AC filters is based on ideal conditions and may not accurately reflect the output voltage in real-world scenarios. Factors such as non-ideal components, temperature, and interference can affect the output voltage and may require additional calculations or testing to determine the actual Vout of an AC filter.

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