Op amp with TL072 (project)

Discussion in 'The Projects Forum' started by Cukrai, Nov 9, 2013.

  1. Cukrai

    Thread Starter New Member

    Nov 9, 2013
    13
    0
    Hello!!
    I am new in this forum :)

    I have an assignment to build an audio amplifier with TL072.
    Spec:
    20Hz - 20 000Hz
    Au - 100 (40 dB)
    I drawn diagram, but i am not sure did i do it in the right way.
    Please, tell me all opinions.

    Thank you!

    http://i43.tinypic.com/2vcyq1y.png
     
  2. alexfreed

    Member

    Oct 8, 2012
    72
    10
    Looks reasonable except for the value of C1. If you HAVE to add a high pass filter, for 100 Hz and 100 Kohm it should be around 15 nF, i.e. 20 times more than you have. And I don't see the purpose of C2. If you want a low pass filter for 20 KHz, the value id 1000 times too high.
    BTW unless there are stability issues, you don't normally add filters to audio amps. The range is usually "at least as good as" 20 to 20K.
     
  3. tindel

    Active Member

    Sep 16, 2012
    568
    193
    At first look, I too thought that this wouldn't work, and it certainly doesn't satisfy all of your requirements.

    You do have a amplifier gain of 100, but note that you have a high-pass filter on the input and a low-pass filter on the output, so all of that gain from 20Hz-2kHz is lost because of your input and output filters. What you ultimately have is a 0gain filter from 20Hz to 2kHz. What could you do to keep the amplifier gain, but only have a filter at 20Hz?

    I'm assuming you also have to drive a 8 ohm load - this won't drive an 8ohm load

    You're close... look at it a little longer and think about your cutoff frequencies are at each stage... we can figure out your final gain stage when you have your filter set correctly.
     
  4. Cukrai

    Thread Starter New Member

    Nov 9, 2013
    13
    0
    Thank you!
    Now i see where was my mistake..
    I changed C1 and C2 values. Now looks for me that everything is correct.
    And one more basic question :) is there any different if i use 1k and 9k resistors to make amplification or 10k and 90k? if it is, so what is the different?

    http://i41.tinypic.com/10i6ubm.png
     
  5. alexfreed

    Member

    Oct 8, 2012
    72
    10
    With ideal op-amps there is not much difference between 1k/9k and 10k/90k on the feedback loop. Real parts have non-zero offset currents, so in general using similar values resistors on + and - inputs helps to keep better balance.
     
  6. tindel

    Active Member

    Sep 16, 2012
    568
    193
    There is a difference between 1k/9k and 10k/90k... The primary difference is 10 times the power dissipation using the lower value resistors (not to mention lower power dissipation in the op-amp). The secondary difference, as alexfreed alluded to, is due to effects on the output due to non-ideal properties of the amplifier. The third difference is the larger your resistors, the larger your noise out of the amplifier. Every resistor has thermal noise present and that noise is amplified in the amplifier. Depending on your application you will weigh these three variables and try to choose resistors that give you the best performance for your application...

    My guess is that for your application you could use either unless you have tight power dissipation or noise requirements, which you didn't mention earlier.
     
  7. nigelwright7557

    Senior Member

    May 10, 2008
    487
    71
    You can work out the RC filter components yourself from:
    1
    --------- = f
    2 pi R C
     
  8. Cukrai

    Thread Starter New Member

    Nov 9, 2013
    13
    0
    thank you all, lot of new stuff for me..

    one more question about op amp in general.

    If i want to make amplification for the given frequency, does it mean, that cut off points are there Output voltage is -3dB.. Like an example: given high pass filter cut off frequency is 20Hz and i need to make 100 Au. For the input i have 20Hz 20mV peak-peak, on the output is 1.4V.. does it mean that it function right? or it has to be exactly 100 amplification for the cut off frueqency points?
     
  9. alexfreed

    Member

    Oct 8, 2012
    72
    10
    Not sure I understand the question but here is an answer to what I think it is.
    If the bandwidth of an amp is specified as 20 to 20,000 it usually means that the amplification drops by 3 dB at 20 Hz and 20 KHz. Then naturally a 20 Hz signal will be amplified by a factor of about 70 rather than 100. So if you need an amplifier with a flat frequency response in the 20 to 20K range to say 0.1 dB, as is often the case for audio equipment, your LP and HP filters should have cutoff frequencies well outside the range.
     
  10. alexfreed

    Member

    Oct 8, 2012
    72
    10
    Hard to argue but let's put things into perspective. With +/-12 V power supply the output swing will be about +/- 10 V max. So for the output amp the peak current with 1K / 9K resistors will be 1 mA with an average of about 0.6 mA if driven with a maximum amplitude sine wave. TL072 is rated 2.5 mA idle current, so using 10K/90K resistors on the output will save up to 20%. May be important for a battery powered device. Now the input half operates at 10 times lower signal voltages, so saving 60 uA is usually not an issue. On the other hand higher noise can only impact the input stage, so if both power consumption and noise level are critical, it would be best to use lower resistor values at the input and higher at the output. But for the problem at hand it's nitpicking.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    1k and 10k are standard E24 values, but 9k and 90k are not.
    Use 3k and 27k instead; as they ARE standard values while maintaining the desired ratio.
     
  12. Cukrai

    Thread Starter New Member

    Nov 9, 2013
    13
    0
    I have to use E12 standart.. So, to make 9k I will use 2.2k and 6.8k resistors in series..
     
  13. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You could do that, or you could use pairs of 1.5k in series to make 3k, and 27k is a standard E12 value.
     
  14. tindel

    Active Member

    Sep 16, 2012
    568
    193
    hmmm... last I recall P=V^2/R... if you increase the resistance by 10, your power dissipation goes down by a factor of 10. The maximum power dissipated in the op-amp is also a factor of your output resistance ((Vsource^2/4)*(1/Rload), so you will also dissipate less power in your op-amp output stage.

    Of course, I wasn't including quiescent current or a sine wave input, admittedly... and you've factored it in... just depends on how you look at it. I tend to look at things from the worst-case due to applications I've done... You're right, your total power saved will be much lower than 10x when you consider those things.

    I agree with SgtWookie's suggestion of using 3k/27k resistors. It's a good compromise for balancing large value resistors with noise and power. I was going to make a similar suggestion and forgot.
     
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