Ramp-up rise time?

Discussion in 'General Electronics Chat' started by tracecom, Jun 9, 2012.

  1. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
    How does one calculate the rise time of the ramp-up circuit at PH1 in the attachment? I think the output voltage will be approximately 11.3 V; is that correct? Thanks.
     
    Last edited: Jun 9, 2012
  2. #12

    Expert

    Nov 30, 2010
    16,261
    6,774
    Look at my blog #3

    might be as high as 11.7 volts.
     
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,386
    1,605
    Oh, about 5 time (R1*C1).

    11.3V is a good estimate.

    A useful equation for cap charge or discharge times is:

    Code ( (Unknown Language)):
    1.                  (Vss-Vi)
    2. time = RC * ln  ----------
    3.                  (Vss-Vf)
    4.  
    5. Where:
    6. RC   = Resistance times Capacitance (the 'tau')
    7. Vss  = final steady state voltage
    8. Vi   = initial voltage on cap
    9. Vf   = final voltage on cap
    10. ln   = natural log
    11.  
    12.  
    The significant voltages in your circuit are
    zero: where the cap starts off
    0.7V: when Q1 turns on and the output starts to rise ground
    12V: the steady state voltage

    Note there is a delay time between 0-0.7V, and the rise time between 0.7 and 12.

    Also note it theroetically takes an infinite time to get to 12V, and the equation will predict exactly that (so cheat a little on Vf).
     
    Last edited: Jun 10, 2012
  4. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
    Thanks for the explanation, however I am missing something. I don't see how the component values in my circuit are taken into account by the equation.

    I have breadboarded the circuit, and, using my DMM, I can watch the emitter voltage rise when the switch is closed. I can also see a change in the rise rate based on the size of C1, i.e., the larger C1 is, the slower the voltage rises.

    This little circuit is intended to provide a turn-on signal to an IC that stays inactive until a certain pin reaches 8.5 V. So, what I am actually looking for is the time it takes for the emitter of Q1 to reach 8.5 V. When I use the equation you provided, the answer seems to be 1.23 seconds, which would be okay for my practical application. However, from a learning perspective, I also want to really understand the ramp-up circuit operation, and how to do the calculations based on the component values.

    Any additional insight you can give me would be much appreciated.
     
    Last edited: Jun 10, 2012
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,386
    1,605
    Oops, my bad. I left out the RC term in that equation. I corrected my post above to reflect that.

    When you get that 1.23 seconds, it is actually 1.223*R*C. So you have to include those factors to predict your desired response.
     
    tracecom likes this.
  6. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
    OK, so in my schematic, the time is still 1.223 seconds because 10k times .0001 (100μF) is 1. Correct?

    And my final (maybe) question. Would a 2N2222A work just as well? I substituted one for the BC547B in my breadboard and didn't detect any change.
     
  7. #12

    Expert

    Nov 30, 2010
    16,261
    6,774
    Vo = dV (e^-t/RC)
    t = -RC Ln Vo/dV
    RC = -t/(Ln Vo/dV)
    R = -t/C (Ln Vo/dV)
    C = -t/R (Ln Vo/dV)
    dV = Vo/(e^-t/RC)
    e = 2.71828183

    Glad you could use one of these equasions.
     
  8. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,386
    1,605
    OK, 10K * 100 uF = 1.0, so in this case my on omission didn't matter.

    I didn't consider the transistor in my answer. A 2N2222A should be OK.

    One point: this circuit take just as long to discharge as it does to charge, so there is a lower limit as to how fast you can turn the switch off then back on again. You could change that by adding a series diode and a (smaller) resistor across R1.
     
  9. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
    I appreciate you responding to my question, but I wasn't knowledgeable enough to use any of your equations without more explanation. :)
     
  10. #12

    Expert

    Nov 30, 2010
    16,261
    6,774
    Darn. Maybe I sould put definitions of the letters in that blog. Would that fix it for you?
    Rephrasing: What would make it useful for you?
     
Loading...