RC high pass filter calculation

[sM1101011]

can someone please show me how to calculate the output voltage of this circuit or give the required formula/s to perform the necessary calculation. All help will be appreciated. [simulated output ≈ 99µV ]

 

[MikeMl]

can someone please show me how to calculate the output voltage of this circuit or give the required formula/s to perform the necessary calculation. All help will be appreciated. [simulated output ≈ 99µV ]

Here you go. For different values of the pot setting....

 

[sM1101011]

Please i need to know the formula to use

 

[audioguru]

The simple calculation for a CR highpass filter does not apply to your circuit because the capacitor is not fed from zero impedance (the output of an opamp).

 

[sM1101011]

actually the 3v is the output of an opamp

 

[audioguru]

actually the 3v is the output of an opamp

But the input to the capacitor of the lowpass filter is not zero impedance. It varies with the setting of the pot. It must be fed from the output of a buffer opamp to use the simple calculation for a highpass filter. Maybe the pot should be at the input of the opamp you show.

 

[sM1101011]

The original circuit looks something like this. Can you explain how i should go about performing the calculation for the filter part of the circuit.

 

[MrAl]

Hello there,


Here is a formula for the output of your original circuit, assuming the 10uf cap is passing the frequencies of interest with little loss (a very reasonable assumption) and the input is a sine wave:

Vo=Vin*R7b*w*C3*R8/sqrt(w^2*C3^2*(R7b*R8+R7a*R8+R7a*R7b)^2+(R7b+R7a)^2)

where
Vin is the input voltage
w=2*pi*F
R7a=upper pot resistance
R7b=lower pot resistance, also equal to 10k-R7a when R7 is 10k
C3 and R8 as per your schematic.

Note that the above formula is dependent on both the input voltage Vin and the frequency F, and also the pot setting.
You can substitute R7b with 10000-R7a so you only have to set R7a (the upper part of the pot).

Example 1:

What is the output amplitude when the upper part of the pot is set to 4k and the input voltage is 3vpeak and the frequency is 100kHz?
First, R7a=4k and R7b=10k-4k=6k, F=100000, and Vin=3 (peak) so the output is:
Vo=1.189v peak

Example 2:
What is the output with the same pot setting and same input voltage, but with F=10kHz instead?
The only thing different here is the frequency F=10000, so the output is:
Vo=0.5128 volts peak.


If you do not want to assume that the 10uf cap passes all frequencies of interest, then you'll have to use the more complicated formula which also includes C2:

Vo=(Vin*w^2*R7b*C2*C3*R8)/(sqrt((-w^2*R7b*C2*C3*R8-w^2*R7a*C2*C3*R8-w^2*R7a*R7b*C2*C3+1)^2+(w*C3*R8+w*R7b*C3+w*R7b*C2+w*R7a*C2)^2))

Note that the small cap 0.001uf dominates so the simpler formula is good down to around 10Hz or a little lower. Lower than that the amplitude gets pretty small anyway though.

 

[audioguru]

What kind of mic do you have that works when its output is shorted to the positive supply voltage and when it has your 2k load (R2)?
What kind of opamp is the second one that has its input voltage the same as its negative supply voltage (0V)?
Why does the circuit have R6?
Why does the pot, C3 and R5 reduce the output at frequencies below 21.3kHz?

 

[The Electrician]

If you do not want to assume that the 10uf cap passes all frequencies of interest, then you'll have to use the more complicated formula which also includes C2:

Vo=(Vin*w^2*R7b*C2*C3*R8)/(sqrt((-w^2*R7b*C2*C3*R8-w^2*R7a*C2*C3*R8-w^2*R7a*R7b*C2*C3+1)^2+(w*C3*R8+w*R7b*C3+w*R7b*C2+w*R7a*C2)^2))

Note that the small cap 0.001uf dominates so the simpler formula is good down to around 10Hz or a little lower. Lower than that the amplitude gets pretty small anyway though.

With some simplification and using latex, the more complicated formula isn't quite so forbidding, although it remains a bit messy:

[latex]\frac{V_o}{V_{in}}=\frac{\text{ \omega^2 R7b C2 C3 R8}}{\sqrt{\omega^2 (C2 R7+C3
(R7b+R8))^2+\left(\omega^2 C2 C3 (R7a
(R7b+R8)+R7b R8)-1\right)^2}}[/latex]

 

[MrAl]

Hi there Electrician,


Thanks for taking the time to simplify and do a more orderly arrangement of that formula. I often dont take the time to simplify, only time to write the equation itself in the quickest way, then let the user decide if they want to try to simplify it more. I never know what someone will consider 'complicated' either, formulas with 100 characters or 1000? I worked with some equations in the past that took 100 lines of text to write out, and that's in something like Notepad not on real paper, so that may have easily been 60000 characters!
It's also nice to see someone else appreciate these kinds of solutions and even general stuff like this

Oh yeah, one more little thing...
You may want to put Vin back in the numerator as that seemed to get lost somehow.

 

[The Electrician]

Oh yeah, one more little thing...
You may want to put Vin back in the numerator as that seemed to get lost somehow.

Actually, I didn't just simplify your expression. I solved for the transfer function independently to make sure it was correct; it's easy to make an algebra mistake with large expressions like this one. When I got done, I just left it as the bare transfer function without the explicit Vo/Vin equality shown. :-(

I wonder if such a "formula" has helped the OP?

 

[sM1101011]

Hello there,

You can substitute R7b with 10000-R7a so you only have to set R7a (the upper part of the pot).

How do i go about setting a value for the upper part of the pot. Is there a range of values to choose from is it specific.

 

[The Electrician]

The idea is to use the "formula" to calculate the response of the circuit for any given position of the wiper on the pot.

Let's say the wiper is positioned 70% of the way up from the bottom. Then R7a is .7*10000Ω = 7000Ω, and R7b is 10000Ω - 7000Ω = 3000Ω. Substitute these values in the formula, along with a value for ω (2*pi*f) and evaluate; it's that simple.

 

[MrAl]

How do i go about setting a value for the upper part of the pot. Is there a range of values to choose from is it specific.

Hi again,


Yes it is really quite simple. Just choose a value that is less than 10k. The value corresponds to the place where you turn the real life pot shaft to.
For example, if your pot has 300 degree rotation (they are almost always less than 360)
and you want to turn the pot DOWN by 10 percent, that would mean you turned the pot by 30 degrees, which with a linear pot that would mean R7a would be 10 percent of 10k which is 1k, so R7b would be (10k-1k)=9k. That's about it.
If you turn it down by 60 degrees, that would be 2k on top and 8k on the bottom, so
R7a=2k and R7b=8k.
If you were to turn the pot shaft exactly half way, that would mean R7a=5k and R7b=5k.

If you like i can change the formula to remove R7 completely and replace it with the degree of rotation of the pot, so you can enter the degree of rotation rather than a value for R7.

Alternately, if you would prefer to enter the percent of rotation the formula can be changed for that instead.

 

[sM1101011]

That will not be necessary MrAl. Thanks for all your help

Many thanks to The Electrician as well