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Topic: need some assistant on the op-amp circuit attached in the thread. (Read 15086 times)previous topic - next topic

Grumpy_Mike

You can't do anything about it it is a board issue that affects us all, not just you. I also saw you sent me a PM but that is down at the moment as well so I could not read it.

Alright. Thanks.

dc42

Why is it that I cannot upload anything to forum? It says my upload folder full. How can I empty it?

Yesterday I found that I was able to upload an image after I shrunk it to 8K bytes.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

yaantey

#63
Jan 10, 2012, 03:14 pmLast Edit: Jan 10, 2012, 07:31 pm by yaantey Reason: 1
I have attached a bandpass filter I have constructed and simulated. The frequency range of the band pass filter is 704 Hz to 2040 Hz. But the problem is even when I put 60Hz I am still getting the same input wave at the out. So, what's wrong with the circuit. (Please explain in a simple and easy way to understand, cuz I am a newbie). If its wrong can you correct it for me.

Here is the link for the circuit and output.

Bandpass Filter Circuit
Bandpass Filter Output

Thanks.

Grumpy_Mike

Quote
The frequency range of the band pass filter is 704 Hz to 2040 Hz

Do you know what that means?
It is the frequencies that will have the output cut down by half.
What it does not mean is that any frequency outside that range will be reduced to zero.

Quote
even when I put 60Hz I am still getting the same input wave at the out.

Yes if you put a sin wave in you will get a sin wave out. However the amplitude of the output will be smaller that you put in.
You need to display both the input and the output to see the change in waveform amplitude.

yaantey

Alright. So it will show you sine wave at the input and output, which I am getting. By how much should the amplitude be lowered? Will the output amplitude be half of input amplitude? Also is the arrangement of both high pass and low pass correct with bias voltage divider? When calculating the high pass filter should I take the total resistance of the parallel 113 ohms or just the one connected to ground?

Grumpy_Mike

Quote
By how much should the amplitude be lowered?

No idea, and it is a lot of sums to work it out exactly and there is not much point as this circuit is not practical at all. Anyway it won't be much lower maybe just half.
It is not practical for several reasons:-
1) The component values are not real - that is you can't buy capacitors and resistors at those values.
2) The absolute values of the components are too low. Meaning the input impedance will be very low requiring a lot of current to drive it.
3) The filters only have a very small roll off so there is little difference in the frequencies you want and those you don't want.
4) As I told you a long time ago in this thread the over all concept of what you are trying to do simply will not work. Others disagreed but it doesn't make them right.

Quote
should I take the total resistance of the parallel 113 ohms

Yes

ajofscott

If your filter is used in a negative feedback loop, this stage will have to be operated as an inverting stage, as a non inverting stage has that 1+ factor in the gain equasion. As far as the filter goes the low pass section must pass the highest frequency on the passband and the highpass filter must pass the lowest frequency in the pass band. When a notch filter is used in the negative feeback loop, the result is band reject due to the fact the filter basses in the band ergo the negative feedback is greater while in the band, and gains are lowered. Capacitances should be calculated based on resistive elements that are in the realm of what the op amp is expecting in its feedback loop ie 10K to 1M. As Grumpy_Mike pointed out your resistive elements are off the chart low, almost to the point of being a voice coil impedance range.

Techone

@yaantey

Can your electronics simulator do a Bode plot ?  ( a Bode plot is a frequency responce of a device - active or passive and display the result in dB. This formula :  dB = 20 log Vout / Vin

Man, you guys, you let me look / review at my old DeVry books so I can check things up.

yaantey

Attached is the link for an amplification circuit. I want to know whether I need to put the diode D2 to cut-off the voltage. Or just the diode D1 would be enough considering that only single supply is used?

Circuit: Circuit

Techone

@yaantey

In my opinion, you don't need D1 and D2. If you are using 5 V for Vcc for the op-amp, the voltage of the output signal will not go above 5 V and not below 0 V anyway, the signal will simply be "clipping" <--- Think of a sine wave with the top / bottom clip with a sissor.

yaantey

@Techone, exactly true. Been bit sick lately. Anyway I did simulation and found that it clips at 4V. Why does it clip? Is it that the supply cannot give 5V.  How can I improve it? For example, if I use a supply voltage greater than 5V. Which diode should I use,D1 alone would be enough, right?

Attached is the output of simulation showing 4V clipping.

Thanks.

Techone

@yaantey

I did simulate your circuit using Circuit Wizard. The LM358 did clip at 3 to 4 V. I add some components ( extra voltage divider and capacitor ) <-- between op-amp one and two ( It look like the input )  and I change the op-amp with a LM339. I better improvement. Almost no clipping. You have to change the ratio of the voltage divider - bias. I add 1 uf, 62K and 47K.

yaantey

#73
Jan 17, 2012, 04:46 pmLast Edit: Jan 17, 2012, 05:37 pm by yaantey Reason: 1
unfortunately I don't have that op-amp. So, if you use 62K and 47Km, that means you are biasing at 2.1559 V right. Doesn't it clip in either top or bottom of the cosine wave? Can you please attach the circuit and simulation for me.

Can I get somewhere close to 5V if I change the supply voltage from 5V to 9V. And adjusting the gain accordingly to maintain the output in between 0 to 5V without getting the signal clipped off? Also if I use 9V i need to use a voltage regulator to convert to 5V in order to supply arduino uno, right?

Thanks.

Grumpy_Mike

It is weird but when ever I look at that circuit the first thing I see is a picture of my house. Next door is a holiday home for rent and it shows a picture of it, but it is quite an odd experience. Maybe other adverts pop up in other countries.

To reduce clipping reduce the size of the input signal or reduce the gain of the amplifier or get an op amp that can handle signals closer to the rail or increase the supply voltage.

However, why is clipping bothering you? It is not as if you are going to listen to the sound anyway. Aren't you going to put it to a peak detector circuit next? Therefore do you not want to clip negative excursions of the signal by 100% to give you more dynamic range for the sound you are looking for?

What ever you do if the input is too loud it will clip, if it is not loud enough you won't get the full swing. You could use a VOGAD (Voice Operated Gain Adjusting Device) http://en.wikipedia.org/wiki/VOGAD I have not had too much success with them but it is about 20 years since I last played with one so maybe they are better now. But then if you want to measure the peak voltage of the sound this sort of defeats the object.

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