Connect with us

Load, zener, series resistors currents

Discussion in 'Electronics Homework Help' started by 381, Oct 4, 2018.

Scroll to continue with content
  1. 381

    381

    3
    0
    Oct 4, 2018
    Hi
    I am not sure if my measurements are correct. I have to calculate load current, series, zener current. I tried convert 8 voltage peak into Voltage by multiplying by .707 I got close current to my measurements. I don't know if this is right.

    Load current should be same as series resistor current? I have no idea how to get zener current. What formula i should use exactly or what steps i should take? I am really confuse any advice?

    Thank you

    ps.
    zener diode= 5.1 V
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    Why did you multiply the peak voltage by 0.707? Think carefully about this.

    Then consider the circuit without the load.

    Then determine if the load is "compatible" with that. Depending on your answer here, you have one of two paths to follow.
     
  3. 381

    381

    3
    0
    Oct 4, 2018
    so for zener current:

    Iz= Is- IL
    Is= (Vs-Vz)/Rs = (8V-5.1V)/180 or i should use (7.3V-5.1V)/180= 11.45 mAbecause diode is .7V drop
    IL= VL/RL 5.1V/6800=.75
    Iz= 12.2 mA - .75mA = 11.45
    So calculated zener current is 11.45 mA?
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    It's hard to say whether or not you should take into account the forward voltage drop of the rectifier diode.

    Have you been given any other information, or is there a precident set in other questions?

    With no other information I would assume it's a perfect diode.
     
  5. 381

    381

    3
    0
    Oct 4, 2018
    All the information I gave in pictures
     
  6. hevans1944

    hevans1944 Hop - AC8NS

    4,521
    2,104
    Jun 21, 2012
    Given the high frequency, 1 kHz, small load current, about 16 mA, and largeish capacitor, 100 μFd... calculate the voltage ripple from cycle-to-cycle assuming there is NO diode voltage drop. Is there a "significant" amount of ripple? If not, assume the capacitor charges to the 8 VDC peak of the input voltage and remains at that voltage for the purpose of solving the problem.

    The capacitor will charge to nearly the 8 V peak of each cycle, and the current at that point will be the same as if a constant 8 VDC voltage were applied. Immediately after the peak, the diode is reverse biased (why?) and the voltage across the capacitor begins to decay until, almost a full cycle later, the sinusoidal input voltage becomes greater than the remaining voltage across the capacitor plus the diode forward-conduction voltage drop.

    At this point the diode is no longer reverse biased but becomes forward biased and begins to conduct current, recharging the 100 μF capacitor and providing current to the rest of the circuit. The current through the diode, and its forward voltage drop, both continue to increase until the peak of the AC input occurs, but the rate of increase slows down because of the "rounded" peak of the half-cycle AC input. Unless a "significant" amount of current is drawn during the forward conduction phase of the diode, you can assume that the capacitor charges to the peak voltage of the AC input and ignore the forward voltage drop across the diode.

    That assumption depends on the amount of ripple you calculated earlier. If there is a heavy load, or the capacitor is too "small," the forward voltage drop across the diode becomes a significant factor and the ripple voltage will be a significant percentage of the output voltage. It becomes a "non-linear" problem then and analytical solutions become much harder to find. When this happens in the real world, I just add more capacitance until the ripple becomes insignificant. There is a "rule of thumb" that many people use to calculate the capacitance value needed for power supplies operating from the full-wave rectified 60 Hz power line. Roughly, 1000 μF per ampere of power supply output current is sufficient for most applications, depending on the amount of tolerable ripple. Here is an Internet website that provides a more in-depth explanation.
     
  7. Ratch

    Ratch

    1,075
    327
    Mar 10, 2013
    Use the worse case of 7.3 volts and be pessimistic. To continue your calculations, the dynamic resistance of the zener at that current 5.1 volts/11.45 mA = 446 Ω. In parallel with 6800 Ω is 419 Ω. In series with 180 Ω is a total of 599 Ω the capacitor "sees". The RC time constant is 100 μF x 599 Ω = 60 ms . Compared with the 1 ms period of the half-wave AC (alternating cycle) supply, the ripple is negligible and the voltage applied to the zener will always be greater than 5.1 volts. Therefore, the zener will never be "starved" and will absorb all the ripple so that the voltage across the load will be constant.

    Ratch
     
    hevans1944 likes this.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-