What are the dotted lines? They appear to denote circuit traces (???). If so, Res 8 and Res 14 are tied together bypassing (shorting out) OP amp 3?

Also, the amplitude curve shows a max gain of -87db which is an attenuation of 87db.

Are the voltages listed, 2.7V, a dc or an ac voltage? You should be interested in the ac voltages for gain calculations.

I am not familiar with "Edwin". Is it a simulator and free or commercial. A source reference please.

Jess

 

Your calculations for the filters are wrong. Their frequencies are backwards.

You want 625Hz to pass but the lowpass filter cuts it 14db.
The highpass filter cuts 625Hz 42dB.
The total cut at 625Hz is 56dB so the output is nearly 1/1000th. The filters have opposite phase shifts that cancel the signal and reduce its level even more. -87db is less than one part in 20 thousand.

Those are the simplest filters known to man (and woman?). There is not a flat section. Use 2nd-order Sallen and Key Butterworth filters for a flat section and sharper cutoffs.

Attached Images

lowpass and highpass filters.PNG (6.8 KB, 10 views)

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Uncle $crooge

 

try this circuit.

Attached Images

BANDPASS.jpg (281.1 KB, 16 views)

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see my website: www.geocities.com/russlk

 

Those 3rd-order Butterworth filters do a very good job. They don't have any loss in the bandpass and their corners are sharp.

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Uncle $crooge

 

Hey guys thanks for the input, it is greatly appreciated.
Im sorry if my lack of electrical education is exacerbating matters,
Hopefully, I’ll get there in the end.

Edwin is an electrical schematic and PCB design package from visionics, unfortunately, it does cost..
Info at.. http://www.visionics.a.se/

I was lead to believe it’s the industry standard, and the best to use.
Yeah Jess, those dotted lines are supposed to be ‘nets’ that indicate the actual wiring of the circuit, so yeah, your right, op amp 2 has been shorted! Unfortunately it can be temperamental with re-routing, but I’ll get right on it.

Those 2.7V readings should indeed be AC readings, but not overly sure, the op-amps are powered with DC, and the VGEN is outputting AC.

Audioguru, thankyou very much for your indepth help..
Damn, I thought my calculations where correct once I’d achieved the, rough, freq response that was required.

My original calculations where
Fc = 1 / 2 x Pi x R x C

LPF = 1 / 2 x Pi x 10k x 1 nF = 15kHz (cutting higher)
HPF =1 / 2 x Pi x 500k x 1 nF = 318Hz (cutting lower)

Are these values correct?
Once I’d implemented these, and the freq response was simulating correctly,
I then was trying to tackle the gain problem, and tried to utilise gain on the first two op-amps aswell, played around with the values of the resistors, its annoying that Edwin maintained the freq response was correct when Audioguru points out they where clearly wrong..



How does changing the ratio between the resistors for gain, effect the resistance you use for the Fc derision? I admit I am unsure how you got 2k resistance value for the HPF circuit?

Any advice on how could I combat the opposing phase shifts in each filter circuit?

Russlk – thanks for the example circuit, unfortunately, something of that complexity is well beyond my realms of understanding at this point.
I really wanted to utilise a bandpass filter, that was a simple as possible, enabling me to further my knowledge. Once I got it working, I was then going to tweak and interchange components to get further to what I actually want. So for now, flatness in the passband, and slope order can wait!

 

Lots of things are still shorted.

That is how opamps work.

That is how filters work.

You had the 1nF capacitor in series with two 1k resistors. The resistors have a total resistance of 2k.

[/quote]Any advice on how could I combat the opposing phase shifts in each filter circuit?[/quote]
Filters have phase shift. You can't change the phase shift of a filter.

Your new lowpass filter does not have the correct frequency response. Its cutoff frequency looks to be much lower than 15kHz.
The graph shows a huge signal loss, instead of the gain of 500 from the 3rd opamp.

__________________
Uncle $crooge

 

Yup, unfortunately, the nets are not the only indication of connection, the majority of those nets are running beneath the wires that are present.

I thought op-amps derived their gain by the ratio between the gain resistor and the feedback resistor, for op – amp 2, and although there are two 1k resistors in series, there is still a 500k ohm feedback resistor present.

Im happy with the frequency response as it is, I just wish to see it working. And yes, I see that there is large signal loss occurring, this is what im trying to remedy. .

Im sorry, have I done something to warrant you being quick with me? If I’ve done something to wind up or offend, please let me know ..

 

Perhaps you should download Filterlab. Its an easy way to design low/high/bandpass filters, in fact, it takes a couple of clicks.

There is a video tutorial to get you started as well.

http://www.microchip.com/stellent/id...cName=en010007

Hope that helps you.

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Spency.

PIC Micro's - Your mind is the limit

PIC's and interfacing with other devices - a PIC Basic Guide @ digital-diy.net

 

Hi Therion, I think you are doing great in your *paper* experiments. I wish the simulators had been available when I started. Keep up the digging and asking questions and you will end up understanding and will be able to avoid common mistakes when building the actual hardware.

Your most recent diagram still shows Op Amp 3 shorted out, its not there. Therefore, you don't get any gain from it at all. I also noticed you changed Res 8 from 1K to 500K. If the Op Amp 3 was not shorted that would give you a gain slightly less that 1, i.e., 500k / (500K + 1K). The gain is Rfeedback / (Rinput) where Rinput is the total series resistance in the input leg. Cap 4 is also in the series input leg but is not a resistance - it causes phase shift and affects the LPF cutoff but not midband gain. Looking from the input of Op Amp 3 toward the output of Op Amp 2 you see Res 13 + Res 8 (originally 1K + 1k to give the 2K mentioned above). They make up the Rinput, the Total resistance. The output impedance of Op Amp 2 is very low so its output impedance (resistance) compared to Res 13 and Res 8 is negligible. So that's why the Rinput is calculated as 2K previously and now is501K in the new circuit. Cap 4 affects the LPF frequency calculations. So if you change Rinput, to change gain, then you must adjust Cap 4 to restore the ratio to give you the desired frequency response.

In the formula: Fc = 1 / (2 x Pi x R x C), note I added the ( ) to clarify that 1 gets divided by all the denominator terms, consider the RC product as controlling the frequency response - everything else is a constant. Hence if either R, (the total resistance as discussed above) or C, (Cap 4) change, then Fc will also change. But you probably understood all this anyway.

I really haven't looked at calculating your Hi and Lo cutoff frequences but even if the shape of the Amplitude curve is what you want, the -133dB tells you its 133dB of attenuation, not gain. When you are looking for Gain, a Negative dB, rather than a hoped for Positive dB, means it is signal attenuation. [The negative dB can also mean a 180 degree phase shift between input and output signals. Since the sign of the dB can be interpreted two ways it can get a bit confusing but if the sign (+ or -) of the expected Gain is opposite what you are expecting, then you have attenuation.] Clear as mud??? Note that for an Op Amp, signal input on the negative, (-) terminal or inverting input will give a 180 degree inversion of the signal (at MIDBAND - the frequency range where the LPF and HPF neither give any attenuation), where as a signal input on the non-inverting, (+) input will give a positive gain, slightly different in magnitude than the negative input gain.

Just as Gain is controlled by the ratio of Rfeedback and Rinput, the phase shift at higher and lower frequencies (outside *mid band*) will change as the Capacitive reactance (Xc) of the capacitor changes with frequency.

Xc = -j1/(2 x Pi x F x C) where j represents a 90 degree phase shift and the calculations represent the numerical value of Xc. So, as far as phase shift is concerned, you get what your particular circuit gives you at different frequencies. And it changes with frequency because the *capacitive effect* of the capacitor changes with frequency. The actual phase shift which occurs depends on the ratio of R to C or C to R. Again, sorry for the mud. May I suggest you read up on RC circuits, particularly the mathematical effects which lead to the phase shift (due to the present of Capacitance or in other circuits the presence of Inductance). The math will get you into the Tangent function. If you are familiar with trig functions studying some RC circuits (attenuation and phase shift) from a mathematical perspective will clear up a lot of things for you.

As for the very low *gain* (in the mid-band) in your circuit there are two possible causes.

1. Op Amp 3 (is shorted out and therefore) does not exist, electrically. Res 8 = 500K is a very high series resistance. Op Amp 2 has a mid band gain of ONE, Rf = 10K and Rin = 10K. Op Amp 6 has a mid band gain of Res 15/ (Res 8 + Res 13 + Res14) or just under ONE.

2. The other is the frequency overlap (Midband range) of the Hi pass and Low pass sections. The pass frequencies of each must overlap where neither LPF or HPF have any appreciable attenuation - this is the MIDBAND. If the pass frequencies of the LPF and HPF don't overlap then there is no Midband region. The HP filter section should come first and the LP second and the cutoff frequencies of each adjusted so that signals in the desired bandpass are not attenuated by either filter.

Edit Note: The LPF cutoff frequency, Fc(LP), should determine the highest frequency desired and the HPF cutoff frequency, Fc(HP), should determine the lowest frequency desired.

Its fun to learn but you have many different things happening simultaneously that you have to sort out several different effects which are happening together.

Hope this helps and not confuses. If you have specific questions about my comments I'll try to further explain.

Edits to correct spelling and add a couple of sentences, hopefully to clarify.
Jess

 

Therion, I just *cursored* through the on-line manual for Filter Lab 2 by Microchip which gramo recommended (thanks gramo).

Its apparently available for free download. I highly recommend you at least read through the manual (video reference) as it graphically gives a good explanation of the design concept and appears to be an easy to use tool. You could quickly design your amplifier with it and then transfer the RC values into Edwin to calculate the simulation. It does not get into the mathematics appreciably but gives graphical tools which give an equivalent understanding. As you use it to design your amp, it checks things and may give you an error pop-up indicating a potential problem which you can quickly fix.

Really seems to be a neat tool. I'd like to play for it for a while, but Taxes must get done first. Think I'll download it anyway.

Thanks again gramo.

Jess