Low pass filter not working properly

Question

This is the pass band that I'm using to filter some EMG signals through electrodes.

When I test it with multisim it works really well as you can see, but when I test it with real equipment (osciloscope and function generator,) the active low filter works fine and the active high filter doesnt work, conversely it shows up -11.8V (around 12V) which comes from the supply voltages. Any thoughts?

The output wave frequency was also distorted. When the input wave was like 5kHz the output wave was like half of 1kHz or less.

EDIT: I had checked the schematic a couple times and already replaced the opamp 3 times.

EDIT #2: The parts that I'm using are the uA741 op-amps.

Answer 1

Quick Synopsis

Let's quickly summarize what I think I understand. It appears you have the following specifications:

• \$A_v=10\$
• \$f_{_\text{L}}=20\:\text{Hz}\$
• \$f_{_\text{H}}=500\:\text{Hz}\$
• Sallen-Key equal-value component circuits (1-pole, 2nd order)

(I don't have the available voltage rails. But we can ignore that. And we don't need to worry about the opamp, yet.)

First up is to compute a few details in order to see how to arrange the band-pass. There is a serious choice to be made that will be quite different, depending upon the results. This part I think you already know. But it's worth stating, explicitly. So here goes:

• \$f_{_\text{0}}=\sqrt{20\:\text{Hz}\cdot 500\:\text{Hz}}=100\:\text{Hz}\$
• fractional bandwidth \$=\frac{f_{_\text{H}}\:-\: f_{_\text{L}}}{f_{_\text{0}}}=\frac{500\:\text{Hz}\:-\: 20\:\text{Hz}}{100\:\text{Hz}}=4.8\$

From this last item, I conclude that your approach using a combination of a low-pass and a high-pass filter is correct. Had the fractional bandwidth been a lot less, say below 1 or better still below .8, then there may be an argument instead for combining two bandpass sections, instead. But that's not the case. So you are correct to break things up the way you did.

So far, so good.

No Input Spec -- Assuming Signal Generator Only

You didn't specify what's driving the system. No, I don't mean the signal generator you used. I mean the real thing that will be driving it when you actually get to making use of it.

Sometimes, you need to AC couple into the filter. If so and if the first section is the low-pass section, you'll need to provide a DC path to ground for the (+) input. I don't see a DC blocking capacitor in your case. And obviously, you didn't need one for the simulation. But I've no idea if you do in the actual circuit. If what's driving this circuit has a DC bias to it, you may very well require DC blocking. Just FYI.

Also, you didn't specify the allowable loading that the first section presents to the actual source. Nor the source impedance, itself. So this means I'm in the dark about these details.

But let's assume for now that this is only a proof that you can do a design and that you'll be using the \$50\:\Omega\$ output of the signal generator; with the only purpose in all this is to verify design capabilities.

Section Order and Other Assumptions

I'm going to just assume you know what you are doing and like the low-pass filter as the first section and want the high-pass filter as the second section. There are considerations, such as the the noise of typical opamps related to their 1/f and bandpass regions, and so on. But I'll just assume you've already made those evaluations and I'll just move on.

I'm also going to assume you are looking for a maximally flat (Butterworth) design. This is important because Sallen-Key equal-value component designs have a fixed relationship between voltage gain and damping (\$\zeta\$ or alternately \$Q\$.) You don't get to pick the voltage gain if you are looking for Butterworth. If you choose to pick the gain, then you don't get to choose the damping.

Broadly speaking, you have to select a voltage gain less than 3, under all possible circumstances. There's no alternative to that. Since the \$\sqrt{10}\gt 3\$, you cannot get \$A_v=10\$ with two of these sections. Just can't happen. Adding some gain at the end is okay. But you will need a third section. There's no escaping it.

Low Pass Design

The low-pass filter has a cut-off at \$f_{_\text{C}}=500\:\text{Hz}\$.

Because capacitors aren't usually found in E24 steps, you usually decide on your capacitor values, first. They go in broad steps. In this case, I might select \$47\:\text{nF}\$. The nearby resistor value would then be \$6.8\:\text{k}\Omega\$. With that combination, \$f_{_\text{C}}\approx 498\:\text{Hz}\$. Which is close enough.

The gain of the stage is already set by the required damping to meet a Butterworth filter: \$A_v=3-\sqrt{2}\approx 1.586\$. It's determined. You don't get to select it. Sorry.

(The same will be true of the high-pass filter. So we can already say that you will need a final, third section with \$A_v\approx 4\$ in order to reach your final goal of \$A_v=10\$.)

From this, the low-pass section will be:

^{simulate this circuit – Schematic created using CircuitLab}

(The actual computed value is \$R_4=22.845\:\text{k}\Omega\$. But I rounded it down. The price I paid for that is a slightly lower voltage gain of \$A_v\approx 1.564\$ and a slight change in the damping. Oh, well. Acceptable.)

High Pass Design

The high-pass filter has a cut-off at \$f_{_\text{C}}=20\:\text{Hz}\$.

In this case, I might select selecting \$220\:\text{nF}\$. The nearby resistor value (from E24 series) would then be \$36\:\text{k}\Omega\$. With that combination, \$f_{_\text{C}}\approx 20.1\:\text{Hz}\$. Which is close enough.

As before, the gain of the stage is already set by the required damping to meet a Butterworth filter: \$A_v=3-\sqrt{2}\approx 1.586\$. It's determined.

From this, the high-pass section will be:

^{simulate this circuit}

(Given that both sections had slightly less gain than might have been, ideally, the final, third section will now need \$A_v\approx 4.1\$ in order to reach your final goal of \$A_v=10\$. A slight adjustment.)

Summary

At this point, you only need to add a simple section to get close to \$A_v\approx 4.1\$. A simple approach:

^{simulate this circuit}

So the final Butterworth result is:

^{simulate this circuit}

And here's LTspice's result from the above schematic (using LT1800's for the opamp choice):

Note that the peak is right at \$20\:\text{dB}\$ and where it should be, given your specs.

Answer 2

You need to study the equations for these filters in more depth. Gain has a drastic effect on Q; you cannot make it whatever you like. For an equal R and C circuit, Q=1/(3-G). So the high pass section has a negative Q. Your question also illustrates what a piece of junk most simulators are.