high pass filter frequency calculation question

[Derringer]

In the example below, am I correct to assume that you would consider R1 and R2 to be in series and would therefore take R as being equal to the sum of R1 and R2 when solving the equation 1/(2*P*R*C) to find the frequency roll-off point?


many thanks

[PRR]

No.

The way you have drawn, R2 is not connected to ANYthing. It isn't even in the problem. So just C and R1.

In many practical cases, R2 is going to some "high" impedance, where "high" means "much greater than R1". Example: R1=1Meg, R2 say 68K, and "out" is really a vacuum tube grid. The grid impedance is over 100Meg. You should add R2 and 100Meg, then put that in parallel with R1. However (68K+100Meg)||1Meg is 0.99Meg which "IS" 1Meg for all practical purpose.

OTOH, "out" might be a simple transistor amplifier. "Out" might be about 3K base impedance. Then with R2=68K and R1=1Meg, we have (68K+3K)||1Meg or 71K||1Meg which "IS" about 68K for most practical purposes (65.4K).

Remember electrons only run in loops. You can't have dangling leads going unknown places and be able to analyze the circuit. In many practical cases we "know" what is "out" there.... but as I show above even in "guitar-electronics" that "out" could be very-high or very-low impedance.

[Derringer]

so if then R2 leads to pin3 of an lm386, then I'd have to calculate (R1)||(R2+50K) to find 'R' to plug into the equation?

[Derringer]

so then, would Example B shown here still be considered a high pass filter and still have its frequency response calculated the same way as Example A ?

thanks

[CynicalMan]

so if then R2 leads to pin3 of an lm386, then I'd have to calculate (R1)||(R2+50K) to find 'R' to plug into the equation?

Yes

so then, would Example B shown here still be considered a high pass filter and still have its frequency response calculated the same way as Example A ?

No. Your second example would work like a highpass filter with some full frequency attenuation, because the junction of the capacitor and the resistor (plus the LM386's Zin) forms the filter, and then the resistor makes a voltage divider with the LM386's Zin. So the cutoff frequency of the filter would be 1/(2piC(R+50k)), and in addition to there there would be an attenuation of 50k/(R+50k).

If you want a high-pass filter, you might as well not use another resistor at all, and just use the Zin of the LM386.

[GibsonGM]

What is the purpose of R2, anyway? Any specific reason it's there?

[Derringer]

What is the purpose of R2, anyway? Any specific reason it's there?

1st-ly, thanks for the responses folks, you've cleared up exactly what I needed.


R2 in that first schem I posted would be leading into the inverting input of an opamp (inverting gain-stage) where the positive input is tied to 1/2 Vcc.

like this



I see now that in that scenario, R1 isn't needed to create a high-pass filter.

I was confused because pretty much every example of a high-pass filter I'd seen or looked up, looked like Example A in the second schem I posted.


I have two circuits that I'm working on, one has inverting gain/filter stages using opamps

The other circuit uses a 386 and is just ROG's Grace tuned to work with bass. I started by putting a low pass filter on the output of the 386 (it was too treble-y) and then thought that maybe I should pre-filter the signal as well, between the buffer and the input of the 386 ... and that's where my other question came from. I tried the pre-filtering in conjunction with post filtering, but decided I liked having just the post-filtering best.


again, thanks guys  .... and if it seems that I'm still missing something conceptually, please enlighten me.

muchas muchas gracias!

[PRR]

> would Example B shown here still be considered a high pass filter

Remember electrons only run in loops. Draw the WHOLE loop.

> leading into the inverting input of an opamp

Like this:



The source resistance matters; we can often neglect it.

If the op-amp is really working perfectly, then the voltage across its two input pins is "Zero". Since the "+" pin is tied to (audio) ground, and the "-" pin is zero volts away, then the "-" pin is "at ground" (see "virtual ground").

So the hi-pass is 0.1uFd and 50K, about 31.85Hz.

In a real world, that "virtual ground" impedance is really your R3 divided by the open-loop gain of the amplifier. Which varies with frequency. But a TL07x at lower audio frequencies will have gain over 50,000. If your R3 is say 200K, then the "virtual ground" impedance is about 4 ohms. Then the "50K" becomes 50.004K, which is insignificant.

If your source is another opamp under feedback, its output impedance is just a few ohms and compared to 50K we may neglect it. If the source is a direct guitar pickup it is about 5K at low audio frequencies, so "50K" is more like 55K, probably not enuff diff to care about. If the source is guitar with 250K volume pot it may be 5K to 60K.... then "50K" could really be 110K and an octave lower than you figured.

Don't leave dangling "in" and "out" nodes. With experience, you may "know it does not matter". But while learning, draw the whole loop so you are sure what is in it.

[Derringer]

thank you for taking the time.


why isn't the extremely high input impedance of the opamp considered when calculating the high-pass frequency cut-off?

[CynicalMan]

thank you for taking the time.


why isn't the extremely high input impedance of the opamp considered when calculating the high-pass frequency cut-off?

Because, in an inverting amplifier with feedback, the inverting input goes to 0V and is at virtual ground. For purposes of AC, this can be considered ground, or an input impedance of 0.

[Derringer]

and when using a non-inverting opamp stage, the input Z is determined by the bias resistor(s) coming off of the '+' input because the opamp's input Z is so high that the parallel resistance between the input Z and the bias resistor(s) is basically equal the the Z calculated from the bias resistor(s) ?


thanks

[CynicalMan]

Most of the time, yes. Some lower-end opamps have low enough input resistances to change circuit function. For example, the LM741's input resistance is typically 2M.

[Derringer]

actually ... I spoke too soon

I should have read this more closely because it tells me what ratios to look at to determine input impedance
http://courses.cit.cornell.edu/bionb440/datasheets/SingleSupply.pdf

but I look at the examples at the bottom of this page
http://www.muzique.com/lab/boost.htm
specifically one that looks like this


gain = 1 + R2/R3 according to the TI pdf.
Am I correct to assume that R1 does not affect gain here?

also, is the input Z  (R3||R2) || R1   ?



thanks for the help

[PRR]

> gain = 1 + R2/R3

Yes.

> also, is the input Z  (R3||R2) || R1   ?

No.

How does the input signal have any clue what R2 and R3 are?

Assuming the opamp aint junk, and R1 is not insanely large, it is just R1 for any practical purpose.

> Am I correct to assume that R1 does not affect gain here?

Draw the WHOLE loop. What is the Source Resistance that you are not showing?

Often it is "small" and we may overlook it.

However if the source were a 50K guitar, and R1 were 5K, R1 would load-down the source to less than a tenth its no-load voltage. Then amp boosts it up.

[Derringer]

I mis-drew something in the last schem I posted. I was referencing the gain stage on Jack's site and I mistakenly drew R1 to GND. The intended schem has R1 tied to Vb.

here's an update and I included values


before I go farther, does having R1 connected to Vb change things?

Here's my stab:
It seems to me that it would make a difference because if R1 were connected to ground, then it would be connected to a lower potential than Vb and would have more "pull" on the electrons.


In reference to the new schem ...
The '-' input will have 4.5V on it ?
And the "+" input will match this?

Gain is still R2/R3 + 1

in the non-inverting example here http://courses.cit.cornell.edu/bionb440/datasheets/SingleSupply.pdf ... there is no R1 nor P1
the input Z is listed as R2||R3

so I assume that in this post's example, Z-in would be (R2||R3)||R1 because they are all tied to 4.5v now


I will get this!

thank you so much

[CynicalMan]

Here's my stab:
It seems to me that it would make a difference because if R1 were connected to ground, then it would be connected to a lower potential than Vb and would have more "pull" on the electrons.

As long as you have a large capacitor (10u-100u should do) from Vb to ground, it can be treated as ground for AC purposes. So, no.

In reference to the new schem ...
The '-' input will have 4.5V on it ?
And the "+" input will match this?

They will both have Vb+Vin, so around 4.5V.

in the non-inverting example here http://courses.cit.cornell.edu/bionb440/datasheets/SingleSupply.pdf ... there is no R1 nor P1
the input Z is listed as R2||R3

so I assume that in this post's example, Z-in would be (R2||R3)||R1 because they are all tied to 4.5v now

The application report says "Input impedance = R1||R2 for minimum error due to input bias current". This means that you should make R1||R2 equal to the input impedance to avoid error due to input bias current, not that that's the input impedance. For your purposes, input bias current is so small, it doesn't matter. The input impedance for your design would be your R1 in parallel with 10^12, the TL072's input resistance. So, it would be very close to the value of R1.

[GibsonGM]

To further simplify: to AC, which is always changing polarity, there isn't really a "+" or "-" as relates to the power supply.  So Vref = V+ = ground, for AC ONLY. 

I mean, the only thing that keeps + and - separate are filter caps (as stated by Cynical Man above, the large cap from Vb to ground), which are essentially a short to AC (it's more complex than that, has to do with reactance, but that will do for right now).   

[Derringer]

The application report says "Input impedance = R1||R2 for minimum error due to input bias current". This means that you should make R1||R2 equal to the input impedance to avoid error due to input bias current, not that that's the input impedance. For your purposes, input bias current is so small, it doesn't matter.

Thanks for clearing that up for me ... it was a major sticking point.

alright, I think that I'm in pretty good shape now with regards to understanding how the input Z of inverting and noninverting opamp stages is derived.

and thank you GibsonGM for the insight into how AC behaves with respect to different DC voltages.

[GibsonGM]

)  Sometimes these things are simple, sometimes they're not.
And often, we can reduce the complex things into a simple thing we can understand, which will get us "close enough" to where we want to be, anyway!!