Basic understanding of low pass filter

[messy]

hello, I am very new to electronics and im trying to teach myself the basics - im currently looking at a basic stepped tone generator circuit as a means to get going as I need a practical task to learn (im crap at just absorbing theory on its own).


I have made the following circuit https://www.instructables.com/id/How-to-make-the-Atari-Punk-Console/ and it doesnt work but i dont really understand how its supposed to so im going through the components bit by bit as a way of learning.


Im currently looking at capacitor basics and high pass filters make sense to me and I THINK i get low pass filters but ima bit more hazy so just wanted to clarify something.

When a capacitor is put in parallel as per the attached image, is it filtering out high frequency signals because its charging and discharging to "fill" the lows of the ac signal, effectively turning it into dc?

My other question is what purpose does the resistor serve? It seems this component is a must for a low pass filters but why do you need to increase resistance? Is is just to protect the capacitor from too much current? If so wouldn't the need for it depend on the resistance of the rest of the circuit and the power voltage etc?

 

[duke37]

The circuit you show is a potential divider. If a divider is made from two resistors, the value of the resistors will determine the output voltage.
If the 'bottom' resistor is replaced by a capacitor , then the impedance drops as the frequency rises so the voltage drops. There is significant drop when the capacitor impedance equals the resistor value.
At zero frequency, DC, the capacitor does nothing.

The source resistance will affect the apparent value of R1 and the load will affect the output voltage.

 

[(*steve*)]

Think of a capacitor as being a type of resistor for AC signals.

It differs from real resistors by having a property like resistance which changes with frequency. The higher the frequency the lower th is value has.

This thing like resistance is called reactance.

Many high pass and low pass filters operate like voltage dividers where one leg is a resistor and the other a capacitor. Because a voltage divider gives an output that is a fraction of the input, and because the capacitor's reactance varies with frequency, the output ratio varies as a function of frequency. If the capacitor is on one leg, the fraction of the output rises with increasing frequency (high pass) or is if it's in the other leg the output lowers with increasing frequency (low pass).

To understand this explanation you also need a basic understanding of resistors and then of voltage dividers (a circuit made with 2 resistors).

 

[messy]

I thought capacitor were resistors for DC ?



my level of understanding here (basic)...

a capacitors resistance changes in regards to the frequency of the signal passing through it, high resistance to to low frequencies and low resistance to high frequencies.

In a high pass filter a capacitor is put in series. If a DC signal is put in to it, it will build up charge, building up resistance until it reaches its max charge/resistance - blocking the DC signal.

An AC signal however will pass through because the alternating current will mean the capacitor is constantly charging and discharging.


A low pass filter puts the capacitor in parallel and filters high freq (AC). Looking at the diagram I was finding it hard to see (in a similar manner to how i can see it in high pass filter) how this blocks high freq and allows low.



I have literally just been reading articles about capacitor and how they work which all talk about filters but none of them mention anything about voltage dividers - should I have learned about those first? Would that clarify how this is filtering AC?

 

[kellys_eye]

Read posts #2 and #3 again. They are quite explicit and a good place to start your understanding.

You can also find many books on the basic principles of components, their function and their application - I'd recommend Horowitz's "Art of Electronics" as a VERY comprehensive guide to all you need to know.

 

[(*steve*)]

In a high pass filter a capacitor is put in series. If a DC signal is put in to it, it will build up charge, building up resistance until it reaches its max charge/resistance - blocking the DC signal.

What you're describing here is the "step response".

Adding a DC signal involves a higher frequency component (the change in voltage).

There are mathematical relationships between capacitance, current, and voltage (as there are with resistance, current and voltage). In the case of capacitance, time is also an important variable and the math involves some simple calculus.

 

[LvW]

messy - during all your readings and investigations you must realize the difference between two basic "domains": "Time domain" and "frequency domain". Otherwise, you cannot understand what`s going on....

(1) In the time domain you describe voltage and current properties of a circuit as a function of TIME (for example: What happens after switching on a signal or DC source).

(2) In the frequency domain you are investigating the behaviour of a circuit if a sinusoidal input signal it applied with a varying frequency . That means: You ask yourself, what happens if the frequency is tuned from very low to higher frequencies?

Comment: Both domains are mathematically related to each other...but that is another question.
.

 

[messy]

I was about to write further questions as I am still confused however I then googled more articles about the whole thing and found more info they just added a tone more questions and confusion so im going to have to take some considerable time going through it all. Doesn't really help that even with supposed "basic introductions" they all use a manner of writing and language that makes it difficult to follow I wish I was back at school and had a teacher I could talk to (i checked and no evening classes around).

Thanks for the replies tho, I dare say I will return at some point

 

[WHONOES]

Read posts #2 and #3 again. They are quite explicit and a good place to start your understanding.

You can also find many books on the basic principles of components, their function and their application - I'd recommend Horowitz's "Art of Electronics" as a VERY comprehensive guide to all you need to know.

I agree with the book recommendation (I have recommended it myself in other posts). Every aspiring engineer should have one.

 

[BobK]

messy,

Referring to LvW's post, you are trying to understand the operation of the filter in the time domain. It is possible to make sense of it that way, but more difficult than using the frequency domain.

A capacitor charging through a resistor has a time constant RC (the product of the resistance and capacitance). The capacitor will get to be nearly completely charged in about 5 x RC. So, consider a low pass filter with a time constant of 0.2msec, so that 5 RC is 1 msec. This corresponds to a frequency of 1 KHz.

Now, look at a low pass filter and put a signal that is much lower than 1 KHz on the input. Let's say 10 Hz. The capacitor can fully charge to the input voltage about 1 msec after the input signal changes. So, for a 100 Hz signal, it will charge in about 1/100 of the period of the signal. Which means the voltage on the capacitor will follow the voltage on the input pretty well. It will lag a little and will never get quite to the level of the input, but it will be close. since the output of the low pass filter is the voltage across the capacitor, the output signal nearly follows the input signal, i.e. it is "passed".

Now consider what happens if you put a signal on the input that changes much faster than the 5 RC period. Let's say 100 KHz, which is a period of 1/100th of a msec. The voltage on the capacitor barely changes before the input signal has reversed. So the voltage never gets anywhere near the voltage of the input signal, in other words, the signal is attenuated and not passed.

For frequencies in between, the level of the signal gradually becomes less as the frequency gets higher.

Now, lets look at a high pass filter the same way. In this case, we have the capacitor between the input and the output. So the output is the input minus the voltage on the capacitor. In the case of the 10 Hz signal, once again, the voltage on the capacitor nearly follows the signal. But, output is the input minus the voltage on the capacitor, the output is severely reduced at that frequency. And in the case of the high signal (100 KHz), once again the voltage across the capacitor barely changes, and so, the output is the input minus a very small copy of it, and it is passed nearly unchanged.

Bob

 

[messy]

messy,

Referring to LvW's post, you are trying to understand the operation of the filter in the time domain. It is possible to make sense of it that way, but more difficult than using the frequency domain.

A capacitor charging through a resistor has a time constant RC (the product of the resistance and capacitance). The capacitor will get to be nearly completely charged in about 5 x RC. So, consider a low pass filter with a time constant of 0.2msec, so that 5 RC is 1 msec. This corresponds to a frequency of 1 KHz.

Now, look at a low pass filter and put a signal that is much lower than 1 KHz on the input. Let's say 10 Hz. The capacitor can fully charge to the input voltage about 1 msec after the input signal changes. So, for a 100 Hz signal, it will charge in about 1/100 of the period of the signal. Which means the voltage on the capacitor will follow the voltage on the input pretty well. It will lag a little and will never get quite to the level of the input, but it will be close. since the output of the low pass filter is the voltage across the capacitor, the output signal nearly follows the input signal, i.e. it is "passed".

Now consider what happens if you put a signal on the input that changes much faster than the 5 RC period. Let's say 100 KHz, which is a period of 1/100th of a msec. The voltage on the capacitor barely changes before the input signal has reversed. So the voltage never gets anywhere near the voltage of the input signal, in other words, the signal is attenuated and not passed.

For frequencies in between, the level of the signal gradually becomes less as the frequency gets higher.

Now, lets look at a high pass filter the same way. In this case, we have the capacitor between the input and the output. So the output is the input minus the voltage on the capacitor. In the case of the 10 Hz signal, once again, the voltage on the capacitor nearly follows the signal. But, output is the input minus the voltage on the capacitor, the output is severely reduced at that frequency. And in the case of the high signal (100 KHz), once again the voltage across the capacitor barely changes, and so, the output is the input minus a very small copy of it, and it is passed nearly unchanged.

Bob

Thanks for the reply, it was very helpful and explains alot! Its also especially interesting as im mainly wanting to learn electronics for audio projects so knowing how to output a certain frequency is excellent.

Just one thing to clarify tho...am I right in thinking then that the output voltage is only ever what the capacitor charges to during a cycle? Ie if the capacitor is only charging half way before the current alternates then the output voltage will be half of the input voltage (or there abouts)?

 

[LvW]

messy - with reference to your last sentence: I must admit that I do not understand everything what you have written - nevertheless, I am afraid you are not on the right track. Again, you are back in the time domain (charging, half cycle,...).
This does not help at all to understand what a filter is and what a filter does.
Any electric filter is best described in the frequency domain using only IMPEDANCES.
Hence, a capacitor is considered as an impedance (Xc=1/jwC).
This way you are able to derive the most important property of any filter: The transfer function V(out)=f(Vin)

 

[BobK]

Thanks for the reply, it was very helpful and explains alot! Its also especially interesting as im mainly wanting to learn electronics for audio projects so knowing how to output a certain frequency is excellent.

Just one thing to clarify tho...am I right in thinking then that the output voltage is only ever what the capacitor charges to during a cycle? Ie if the capacitor is only charging half way before the current alternates then the output voltage will be half of the input voltage (or there abouts)?

Yes, for the low pass. Read my post again, I said that explcitly. For the high pass filter, the output is the input minus the voltage on the capacitor, which is also the voltage across the resistor.

Bob

 

[messy]

LvW, I guess I just don't fully get what people mean when they say time and frequency "domains" (im not even sure how to consider frequency without time). For example BobK mentioned the same thing right away, "think in the frequency domain" and then starts talking the input period. When i said "cycle" it may not have been the correct term or something but I was just trying to discuss his explanation -either way I think I have it at the level I need .

BobK - cheers, I felt like what I was saying was a simplification so I just wanted to make sure it was still correct.

 

[LvW]

messy - I suppose you know the trem "voltage divider".
Now - imagine such a divider consisting of a resistor R and a capacitor C..
The frequency-dependent impedance of the capacitor is Zc=1/jwC.
With a sinusoidal voltage Vin across both parts, the voltage across the capacitor is
Vc=Vin[Zc/(Zc+R)]=)Vin[1/jwC/(R+1/jwC)].
This describes a voltage divider in the frequency domain.

 

[BobK]

My explanation was in the time domain because that is what you wanted to see.

The time domain equation for a capacitor is a differential equation which applies universally. When solved for a sinusoidal input, you get the frequency domain equation, which is no longer a differential equation. But -- it does not apply universally, it applies only to a sinusoidal input. The advantage is that it is simpler than solving a differential equation.

Bob