Help with single supply opamp's High Pass filtering

Thread Starter

jacopo1919

Joined Apr 12, 2020
80
Hello!

I want to set up a non inverting opamp with High pass filtering. 



The opamp is a single supply and is biased with half of the VCC. 
Doesn’t matter which opamp is, since is only theoretical at the moment.



Sinewave (starts at 1V) enters into the Non inverting opamp after the *1/2 Voltage divider (R1 and R11).


After this, the non-inverting Opamp gain is A=2. So at the end, Vout should be the same as Vin.


The R4-C2 network ''should'' create an high pass filter with Cutoff point at 795.77Hz. Please correct me here if i’m mistaking.


This is why i set the Frequency of the V3 Sine at 795. 


This RC network should give me a -3dB at 795.77Hz, therefore it should halves the amplitude at the output and showing that the filter is working.


This doesn’t happen ..The Vout is around 0.7V .
 I also tried to simulate with a sort of whinenoise + FFT but i didn’t have luck since my abilities on LTspice are still very poor.




Can you push me in the right direction?

Schermata 2022-01-27 alle 23.17.57.png
 

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Papabravo

Joined Feb 24, 2006
18,102
I think it is working. Your input has an amplitude of 1.0 Volts(Peak). the output is 3 dB down, and as we all know, an output that is 3 dB down should be at 0.707 Volts(Peak) which is pretty much where it is. I don't understand the problem. It is possible that you expect the rolloff of the filter to be sharper, but that is not the way a 1st order filter works. The rolloff is 20 dB/decade. So from 795, a decade lower will be 79.5 Hz and the amplitude should be 100 mV(Peak).
 

Ian0

Joined Aug 7, 2020
4,863
-3dB is the half-power point, not the half-voltage point. Half voltage is -6dB.
When the voltage is 1/√2 the power is half.
 

crutschow

Joined Mar 14, 2008
29,508
If you are building an active filter, why not do a Sallen-Key type, which will give a nice 2nd-order -40dB/decade rolloff with a sharp corner?
 
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Papabravo

Joined Feb 24, 2006
18,102
Here is an example of a Sallen-Key filter

1643323611725.png

Should be no problem to move the corner up to 795 Hz. As @crutschow mentioned the rolloff here is 40 dB/decade. A smaller value than 1.13K will make the frequency higher. If you change both capacitor values, the you'll need a new set of resistors to cover the desired range. The analysis is contained in the TI Application Note SLOA92
Enjoy
 
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Ian0

Joined Aug 7, 2020
4,863
The R4-C2 time constant DOES create a cutoff at 796Hz, but it is in the feedback of a non-inverting amplifier, which has a minimum gain of unity.
Your filter has a gain of 2 at high frequency and a gain of 1 at low frequency. It's a shelving filter, not a high-pass filter.
If you want a proper high-pass filter then you should use the C3|R7 time constant, but that is complicated by having the bias network and the input attenuator attached.

If you use @crutschow 's circuit, (which I recommend) you can replace R1 by two resistors of twice the value, one connected to ground and one to the positive supply. You don't then need the bias supply.
 

Papabravo

Joined Feb 24, 2006
18,102
I obviously have way too much time on my hands. Here is the Sallen-Key High Pass with smaller capacitors and resistors in the 1K to 17K range for a higher frequency register.

1643333338847.png

The blue trace corresponds to a corner at 795 Hz. with C1=C2=0.1μF and R2=RK=1.4K, and R1=2*RK=2.8K
 

Thread Starter

jacopo1919

Joined Apr 12, 2020
80
Thanks for the good hints that wll keep me busy for a while.



I didn’t consider different filtering methods because i was analysing this schematic of a pedal distorsion 
and trying to figure out a way to vary the filtering action before the actual clipping. 



Even though a steeper roll-off is very welcome (and the plots look really nice), I’m not sure if a Sallen-Key configuration would suit the application since here the gain changes.


I certainly need to explore further but by changing gain to a Sallen-Key, I think I would have issues with Q.



This is what i came up with (Vin is a sine at 1V, 723 HZ):
Schermata 2022-01-28 alle 23.46.10.png

R2+R3 (variable resistor 100k) and R4 are in parallel, hence the Rg of the Non-inverting opamp is R2+R3//R4 = from 834 Ohm to 902Ohm when R3 is maximum. 




This way, if i sweep R3 from min to max, I wouldn’t alter the gain so much but I will change the cutoff frequency of R2+R3 / C1.

This should sweep between 65 HZ to 723 HZ

R4/C2's cutoff is fixed at 1170 HZ.



In fact, a sinewave at frequency of 723 hz is at 1/√2 as has been pointed out.

C2 is helping to roll but i still have to figure out how much and how to calculate it , since to me they look like two first order RC filters in //.

The concept is a simple (actually nothing is simple) filtering and if it can be achieved with a tinier amount of component, better. 
Probably a very neat and precise Sallen-Key configuration is unnecessary before the total destruction of the wave.

I would probably need to listen to the difference and judge by hear.
 

Papabravo

Joined Feb 24, 2006
18,102
Do you know about Laplace Transforms and how they can facilitate the derivation of the transfer function of the circuit? Doing so will allow you to identify the natural frequency, the Q, and the damping factor in terms of component values.
 

crutschow

Joined Mar 14, 2008
29,508
Here's the AC simulation of the circuit:
As you can see, the filter only provides about a 6dB increase in gain in the 100 to 3kHz region, and the filter changes very little with the change in the value of R3.
I can't see where the filter is doing anything that significant.
It would be only slightly audible to the ear.

1643423107157.png
 
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Papabravo

Joined Feb 24, 2006
18,102
Here's the AC simulation of the circuit:
As you can see, the filter only provides about a 6dB increase in gain in the 100 to 3kHz region, and the filter changes very little with the change in the value of R3.
I can't see where the filter is doing anything that significant.
It would be only slightly audible to the ear.

View attachment 259016
I don't see anything interesting here.
 

Thread Starter

jacopo1919

Joined Apr 12, 2020
80
I was explecting a filter's cutoff point t sweeping between very roughly 65Hz to 723 Hz by varying {R}.
The filtering curve doesn't move and to me, this seem to be dued to R4/C2.

If I want to be able to sweep the cutoff frequency before the clipping, this doesn't look like a good way.
Also a Sallen-Key configuration with variable cutoff frequency looks a bit tricky since it would require a dual gang pot with 1 gang at 1/2 the value of the other one.

a sweeping cutoff point with that paticular non-inverting opamp configuration looks not really doable.

@crutschow i meant that the wave is going to be heavily distorted since the original design is a distorsion unit

Thanks for the help
 
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