Newbie question about capacitor discharge through BJT

Discussion in 'The Projects Forum' started by manthura_assemblies, Jan 21, 2016.

  1. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
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    Hi everyone,

    I am working on a ramp wave generator using this chain of components:

    1. an op-amp integrator (with a capacitor C), which generates the ramp
    2. an op-amp comparator, which detects if the step 1. integrator output reaches a fixed threshold and switches from 'high' to 'low'
    3. a saturated BJT. The base is connected with 2. output
    I want to pose the Vce output in parallel with the capacitor C to discharge it, but all my efforts are vain.

    Here below is an image of the schematics: synth_20160121.JPG

    Thank you in advance for your help!!!!
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,145
    3,055
    What is V3 doing? It's shorted.
     
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  3. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
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    It's the negative rail (or, at least, my personal concept of negative rail). Am I wrong? :(
     
  4. #12

    Expert

    Nov 30, 2010
    16,320
    6,818
    Your drawing has an error. Too many ground lines on the voltage source.
     
  5. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
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    Thank you for the reply.
    I've deleted a line betweeen ground and Ve.
    Talking about V3, what I want to do is draw the negative rail, which feeds my op-amps along with positive rail (expressed by V1).

    Here is the new version:

    upload_2016-1-22_9-11-3.png
     
  6. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,998
    745
    V3 is drawn wrong, it should be Negative to the op amps, and its still shorted out.
     
  7. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
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    Modified the voltage of V3 to -9V from 9V. Do I have to place a resistor in series with the V3?

    Many thanks.
     
  8. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
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    Here's the new diagram. Still I don't know how to interface the BJT with the capacitor in order to discharge it. :(

    upload_2016-1-22_9-46-11.png
     
  9. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,998
    745
    V3 is shown as as a positive supply the +/- signs are wrong,as it stands its ok the output will go from zero to positive,

    If you connect your emitter to the negative V3 rail, and take your output from ground and the collector, your output will swing negative and positive through zero.
     
  10. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
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    This is what i get with the modifications you suggested:

    upload_2016-1-22_11-42-1.png

    Maybe I am missing something... (i.e. the connection of Ve with negative rail). Still I miss how to connect BJT to capacitor...
     
  11. wayneh

    Expert

    Sep 9, 2010
    12,145
    3,055
    That looks better. What is the problem?
    Note that the base voltage of the transistor will turn it on at any voltage above -8.3V.
     
  12. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,998
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    Dont see why you need U2, if you connect the transistor to the output of U1, it should work.
     
  13. wayneh

    Expert

    Sep 9, 2010
    12,145
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    I don't see much need for R2 either.
     
  14. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
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    What if I connect Emitter to ground? Anyway, putting cpacitor in parallel with Vce seems not to work (I don't get the desired oscillation). Maybe it's a timing problem...
     
  15. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,998
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    Can you scope the output of U1 from ground, and the negative supply, what signal is it.?
     
    Last edited: Jan 23, 2016
  16. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
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    It's the same of V(Vc)

    upload_2016-1-23_15-59-51.png

    there is a small oscillation: (better zoom)

    upload_2016-1-23_16-0-35.png

    but not the oscillation that I want (from 0 to threshold, ramp style)
     
  17. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,801
    1,105
    Here you go. Note the positive feedback resistor R4. The ratio R4/R2 determines the ramp amplitude (and also affects frequency).
    RampGen.PNG
    Here's the asc file:
     
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  18. manthura_assemblies

    Thread Starter New Member

    Feb 11, 2014
    11
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    Thank you very much! The problem was the lack of 'hysteresis' in the comparator, I guess. I have to study this topic a little bit more...:p

    Thank you all again!!!
     
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