Understanding function of circuit

Discussion in 'Homework Help' started by kiwi101, May 5, 2015.

  1. kiwi101

    Thread Starter Member

    Feb 18, 2014
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    Hey guys, the following circuit was attached to a boost convertor and is used to provide a 2 MHz frequency square waves to the Boost Convertor. My issue is when looking at the circuit, I realized I don't fundamentally understand how it achieves that.
    1) So for example I understand that the second Op-amp is an integrator (since it has a capacitor in feedback) and integrates whatever signal comes to it, but the signal at the point after second opamp is triangle waveform and signal at point "out" is a square waveform. Why? How does this third op-amp further integrate the triangle waveform?
    2)Why is their a current source between the inverting and non-inverting terminal of the first two OP-Amps?
    3) What is the point of the feedback resistor of 47K between the first two op-amps?

    Please help me to understand this circuit so I can begin to read functionality of circuits.
    Thanks
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    1) The third op-amp is configured as a comparator. When the +ve input exceeds 70% of Vdd the output goes high.

    2) That is a voltage source of 2.5V, not a current source. It is simply a 2.5V reference input.

    3) The first op-amp is an amplifier with a gain of about 2. The two op-amps together create a triangular waveform feeding the comparator.
     
    kiwi101 likes this.
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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  4. MrChips

    Moderator

    Oct 2, 2009
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    Oops. The correct explanation is provided in the link given by Jony130.

    3) First op-amp is a Schmitt trigger.
     
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  5. kiwi101

    Thread Starter Member

    Feb 18, 2014
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    Thank you guys the links really helped.
    So how do I control the duty cycle of the circuit, so right now it's 30% on and 70% off so what if I wanted it to be 50% on? How do I adjust the capacitor and resistor values? Which resistor do I have to change ?
    Also what controls the frequency ? So right now it's 2Mhz how do I change it to a lower or higher value?
     
  6. MrChips

    Moderator

    Oct 2, 2009
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    Alter the R and C in the integrator to change the frequency.

    Alter the values of the 3k and 7k resistors to change the duty cycle.
     
  7. kiwi101

    Thread Starter Member

    Feb 18, 2014
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    Okay so I changed the 3k and 7k resistors to 5k each and I got a 50% duty cycle
    But the frequency I still cant figure out. Is there a formula that connects the R and C of the integrator to its frequency?
     
  8. kiwi101

    Thread Starter Member

    Feb 18, 2014
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    Because I want to change the frequency to 200 kHz
     
  9. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello there,

    Your drawing is a little unclear about what voltages are being used for the op amps or comparators.
    If you find out what voltages YOU are actually using, report back here and i can create a formula for you and possibly suggest improvements.

    For example, maybe the two op amps use plus and minus 5 volts, or just 5 volts and ground.
     
  10. MrChips

    Moderator

    Oct 2, 2009
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    Try changing the C to ten times what it is now.
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Pretty good idea, as the output frequency is inversely proportional to the capacitor value. Increasing the cap value by 10 times means the frequency goes down by 10 times what it is now.

    A simple formula assuming symmetrical power supplies plus and minus 5 volts, and the trip point set to 0v, we have:
    f=R100/(C*R130*R47)

    where R100 is the 100k resistor, R130 is the 130k resistor, R47 is the 47k resistor, and C is the capacitor value. You can chance those in the formula to see what results you would get, but the ratio of R47 to R100 should probably stay the same. It may be wise to lower those values by ten times though, which wont affect the average output frequency but will reduce circuit cycle randomness in time due to noise a little which means cycles would be more equal from one cycle to the next.
     
  12. kiwi101

    Thread Starter Member

    Feb 18, 2014
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    Hey guys thank you so much I changed the Cap value to 20 pF and it worked.
    Also my voltage was 5V and ground
     
  13. crutschow

    Expert

    Mar 14, 2008
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    A simple rule to remember is that, for a constant voltage input, an op amp integrator's output will change by a voltage equal to the negative of the input voltage in one RC time-constant.
     
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  14. kiwi101

    Thread Starter Member

    Feb 18, 2014
    38
    0
    Hey guys

    So although I am getting the duty cycle and frequency I wanted for the rectangular waveform, but I am not getting a triangular waveform. Can anyone please tell me what they think is not letting me get the triangular waveform?
     
  15. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hi,

    A quick guess is the op amp bandwidth is too limited, which is a common mistake with first time triangle wave circuit builders. Check the bandwidth vs your expected triangle frequency.
    The harmonics of a triangle go as the inverse square of the harmonic number, and you want to support as many harmonics as possible, although you have to draw the line somewhere. Lets say you shoot for a max of the fifth harmonic. If your base frequency is 10kHz that means your op amp has to be able to handle at least 50kHz without distortion or attenuation at the needed fundamental amplitude times 1/25. If your base is 100kHz, then you have to get up to at least 500kHz with no problems at an amplitude of 1/25 times the fundamental.. Up to the 7th harmonic you'd need 700kHz at a gain of 1/49 times the fundamental amplitude. If your circuit can not do this then it will produce a distorted triangle wave, and you may need it to be higher than the 7th harmonic.

    So check the gain bandwidth of your op amp and state that and your desired max triangle frequency and we can see what you might expect.
     
    Last edited: May 15, 2015
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