# RC filter - Is this variation still RC filter

Discussion in 'Homework Help' started by vol_, May 10, 2016.

1. ### vol_ Thread Starter Member

Dec 2, 2015
93
3

In the image there is on the left a high pass filter. What I want to ask is if the left circuit is the same with the right circuit.

Thanks

2. ### SLK001 Active Member

Nov 29, 2011
745
197
The filter on the left is NOT a filter. It is the RIGHT image that is a high pass filter. The C and the R form a voltage divider, with the values of Xc and Xr. The image on the left is just a shunt resistor with a coupling capacitor.

Feb 17, 2009
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4. ### vol_ Thread Starter Member

Dec 2, 2015
93
3
Good to see you Jony. As always your answer is a reason for me to study.

I assume you meant "..load resistance for the left circuit is equal to R1", or not?

The load is the input of a TDA7052A IC. I measured with a DVM the input resistance connecting the probes on the input and signal ground. I got 0R. And i think that TDA7052 input impedance is set only externally (no info about input resistance in the topology og the IC) as in the next image:

I'm going to see in detail the thread you sent me, to understand how it fits my question.

Cheers

5. ### MrAl Well-Known Member

Jun 17, 2014
2,183
432
Hi,

To be perfectly clear, neither diagram is a circuit. A circuit is a connection of elements that form a complete loop where current flows.

Why i bring this up is because the two "circuits" are incomplete and therefore there is no way to come to a conclusion about what they can do. For example, if we short circuit the output of the first diagram, does it do anything? Maybe it does, maybe it doesnt. If it really was a filter maybe then it becomes a low pass filter.
If we use the output into another circuit that has some impedance, then maybe it becomes a high pass filter.

The interpretation also sometimes depends on the application it is used in. Maybe both circuits are just used as DC blocking circuits, and although they might still be called "high pass" the better description would then be "coupling circuit" or "DC blocking circuit" or something like that.

Capacitors in series usually form something like a high pass function but the whole circuit has to be known in order to make a good judgement about what it really is being used for.

vol_ likes this.
6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
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No, I meant R2.

Externally ? No, we can find Zin in data sheet
Zi = 20kΩ typical and Rs is just a "pull down" resistor.

7. ### vol_ Thread Starter Member

Dec 2, 2015
93
3
Thanks, i like that you define the meaning of the word circuit and you count your sayings. Its the first time that i understand the word "circuit".

I wanted to know if there is a possibility the left circuit to be a filter, and so far the answer is yes. I now try to understand when and how.

Thanks

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
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C1 (470nF) together with Rload (20kΩ for TDA7052) form a coupling circuit or a high pass filter with the corner frequency equal to Fc ≈ 0.16/(RC) ≈ 17Hz . and Rs= 5kΩ prevent the input from "floating".

9. ### vol_ Thread Starter Member

Dec 2, 2015
93
3
As i can get, if i would like the left RC connection to act same as the right RC connection and to be a high pass filter in a circuit, we want input signal source to have low resistance compared with r1 not to have attenuation of the signal in the sources' output and if the load resistance of the left circuit is equal to r2 (having c1=c2 uF) we will have the same frequency dependent voltage divider in both RC connections.

And expressing my question better, the left circuit it is not and it can not be a frequency dependent voltage divider as it is. It shall have a resistor path to ground to be one. Am i right?

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, you are right.

Exactly.

11. ### MrAl Well-Known Member

Jun 17, 2014
2,183
432

Hello again,

You're welcome. I said that mainly to draw attention to the fact that circuits that are not entirely complete are harder to define as to what they do when they are that simple. Some circuits are easier though and i see you do have the labels "IN" and "OUT" so that helps a little. I also wanted to bring attention to the fact that sometimes we call them a "coupling" circuit, and the capacitor is then called a "coupling capacitor", although in a wider sense it's still a high pass.

The difference between a coupling circuit and a high pass filter is found by looking at how they are usually used:
1. The high pass filter is usually used to get a specific kind of response needed for some filtering action.
2. The coupling circuit (or just coupling capacitor) is used mainly to remove the DC content in the signal, and not as strict on the filtering action although some attention is still payed to the cutoff frequency.

So we can say that the coupling circuit is a type of high pass filter, but it is usually not tuned as exact as a high pass filter is.

The main problem with the first circuit though is that we dont know what the load is, but most likely it is either used as a regular high pass filter or a coupling circuit.

12. ### vol_ Thread Starter Member

Dec 2, 2015
93
3
Oh yes, i was wondering about this. We still some attention to the cut off frequency but what do we substitute in f=1/2pRC equation for R if we have only a coupling capacitor in the signals path? Or the paths around the coupling capacitor will always have a dc resistor path to ground or a load?

Yes it was really a general question of me, trying to understand if there was any possibility for this circuit to be a frequency filter. Jony covered about these two RC connections and you gave me a more general enlightenment about capacitors ac use.

13. ### MrAl Well-Known Member

Jun 17, 2014
2,183
432
Hello again,

Well if you had this in a circuit there would be some output impedance from the first stage and some input impedance from the last stage, so if we say they were pure resistances for simplicity and we call the output resistance of the first stage R1 and the input resistance of the last stage R2 and our capacitor goes between those two stages, then we have:
w=1/(R1*C+R2*C)

as the cutoff frequency. That's the frequency that causes the response to drop 3db from the passband response, and the passband response is:
Vout=Vin*R2/(R1+R2)

Just to note, w=2*pi*f with f in Hertz.