Positive Biased Clamper Help

Discussion in 'The Projects Forum' started by kahafeez, Jan 27, 2009.

  1. kahafeez

    Thread Starter Active Member

    Dec 2, 2008
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    Hello Folks, this is supposed to be very easy..... bt i'm still having problems with this.... plz help.... i also want to retain the shape of the signal.....
     
  2. Ron H

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    Is this a homework problem? If not, what is it for?
     
  3. kahafeez

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    Dec 2, 2008
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    its nt really a homework..... its just that one block is producing 0-5 square and the next needs a different level..... the next block needs a 200mV signal that has a 1.1V as low and 1.3V as high..... bt i could do that attenuation through a simple voltage divider..... the problem is abt clamping that 200mv signal to 1.1v.....
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    Does it need to be accurate?
    What is the device which needs 1.1-1.3V ?
     
  5. kahafeez

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    Dec 2, 2008
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    its a VCO.....

    i just used this schematic.... it gave me better results in the simulation.... will try it in the lab tomorrow.... can anyone plz tell me whats the difference... like how do they work.....
     
  6. Ron H

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    You don't need a diode. You can do it with 3 resistors and a power supply.
    I have calculated the resistor values. Simple algebra will allow you to do the same. See attachment.
     
  7. kahafeez

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    Dec 2, 2008
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    i couldnt see any resistor values :-/

    and i need to clamp a 200mV signal to 1.1V.....
     
  8. kahafeez

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    Dec 2, 2008
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    i tried the new schematic too but that brings noise in the ciruit..... i dont know why its coming.... plz tell me a way to clamp a signal to any voltage level bcz my new application requires clamping the voltage to 1.6V.... thanks in advance
     
  9. Wendy

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    Mar 24, 2008
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    How accurate does the 1.6 V have to be? Would 2 diodes, 1.2V, work?

    Restating the specs, a sine wave, 0 to 5V. You want 1.3 to 6.3V out?
     
  10. kahafeez

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    Dec 2, 2008
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    thanks for replying.... no a square wave 0 to 5V and i need to convert it to a 200mV signal(which i've achieved using a simple voltage divider) bow the new square wave should have its low at 1.6V and hi at 1.8V..... i need it to be accurate.... nt like 100% even 80% would suffice...
     
  11. kahafeez

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    Dec 2, 2008
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    Hey Bill have u worked with 74HC4046's VCO???
     
  12. Wendy

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    Mar 24, 2008
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    Nope, but I think I know what you want now. Working on a drawing...
     
  13. Ron H

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    As I said, you can calculate them. Below is the solution.

    EDIT: This circuit will give you an output of 1.1V-1.3V. Your new range can use the same circuit, but with different values.
    You don't need a clamp, as long as your input is always 0 to 5V, DC coupled. All you need is an attenuator and level shifter, which is what this circuit does. It will be more stable than a clamp.
     
    Last edited: Jan 28, 2009
  14. kahafeez

    Thread Starter Active Member

    Dec 2, 2008
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    thanks Ron.... bt can u plz tell me which one is the input.... V3 or V5???? and which resistor is deciding the DC level shifting... plz explain ..... i need to know....
     
  15. Ron H

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    Apr 14, 2005
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    Dude... I edited my last post, Read it again.
    Concerning equations, can you do the algebra to calculate voltage dividers and attenuators? If not, I can post them.
     
  16. kahafeez

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    Dec 2, 2008
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    plz post the algebra...... i'll be very thankfulllllllllll
     
  17. Wendy

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    Mar 24, 2008
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    OK, as a starting point, would this do what you want? You could also substitute a 1.5V battery for the 1.6V shown.

    [​IMG]

    **********************

    Ron, you must be an early bird. When he PMed me I was the only one up and about, so I thought.
     
  18. kahafeez

    Thread Starter Active Member

    Dec 2, 2008
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    i simulated both the methods.....

    both work perfect.....

    bt i'd like to ask a few questions:
    @ Ron H:

    sir plz give me the algebra etc..... cz i need to knw the reason so that i can do it the next time myself.....

    @ Bill:

    Sir i knw the two resistors are for attenuation.... i knw how to calculate their value bt i need to know that i can get any level by only changing the DC battery level??? and do i need to change capacitor's value for different levels????


    thanks
     
  19. Wendy

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    Mar 24, 2008
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    On mine, think simple voltage divider. 5 VDC to .2 VDC, it is a pure ratio problem. The power supply sets the low voltage, no calculation needed.

    Ron's is somewhat similar. You solve it in two steps. Figure out the resistors needed to create your wanted voltage when the input is 0V (the low on the square wave).

    Then figure out the resistors needed to create your desired output with the input is +5VDC (the high on the square wave).

    You'll find it is another voltage divider problem.
     
  20. Ron H

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    OK, here are the calculations. Documenting it was a royal pain in the a$$. :eek:
    If you tell me you already know Thevenin's theorem, I'll hunt you down and tickle you to death.:D
     
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