Connect with us

Diode and Capacitor [Basic Questions]

Discussion in 'General Electronics Discussion' started by Clyd, Feb 19, 2014.

Scroll to continue with content
  1. Clyd

    Clyd

    3
    0
    Feb 19, 2014
    Hi I'm a first year student and currently studying AS Electronics. I don't quite understand how capacitors charge and how diodes clamp in an astable circuit, although I know quite well how these components function independently or at least their basic characteristics. I've also done a lot of research, watched youtube videos and used educational websites and I'm now getting some of the idea, however there are still gaps being left since we touched down on this topic.

    So my questions are:

    • At which side does a diode clamp and how?

      Does voltage without current charge a capacitor and vice versa?

    Sorry, these questions might seem very stupid, maybe I just have very low technical understanding regarding circuits.
     
  2. Arouse1973

    Arouse1973 Adam

    5,164
    1,087
    Dec 18, 2013
    Do you have an example circuit. There are many astable circuit configuration.
    Adam
     
  3. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Current charges a capacitor. The voltage across it will change as it charges.

    A diode limits the forward voltage across it to something like 0.6V. In an astatable multivibrator circuit built from two bipolar NPN transitors, it is used to keep the voltage on the base from going far enough negative to damage the transistor.

    Bob
     
  4. Clyd

    Clyd

    3
    0
    Feb 19, 2014
    Ok so I'll just bring my notes here:

    [​IMG]

    On the first note, I don't understand why Vc is clamped only at 0.7V when you have 5V as feedback coming from the schmitt inverter, cause on the last part, the input became ( 5 - 0.7 = 4.3V) only ACROSS the forward biased diode.
     

    Attached Files:

    Last edited: Feb 20, 2014
  5. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Did not notice that there is a resistor between the output of the Schmitt trigger and the anode of the diode? How does that effect the voltage on the anode?

    Bob
     
  6. Clyd

    Clyd

    3
    0
    Feb 19, 2014
    But when I follow the Ohm's law at least from what I understand is that there is no voltage drop across a single resistor which means the voltage are gonna be the same as the supply? Since there's no specified resistance, let's say we've got 2 Ohms

    I = V/I
    5/2 = 2.5A

    V = IR
    2.5 x 2 = 5V

    now how does the diode affect the 5V when it is forward biased?

    and what will be the effect? would it become 4.3V?


    Sorry if things are bit shady... :(
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,500
    2,840
    Jan 21, 2010
    At a theoretical level, the diode does not affect the resistor. The resistor will continue to operate as per ohms law. In fact (ignoring the charge on the capacitor) the current flowing through the diode when it is forward biased comes from the output of the gate, through the resistor, and then through the diode.

    The diode has a constant voltage drop (theoretically), the gate has a constant output voltage (theoretically). This works out close enough in practice provided R is large
     
  8. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Yes, the diode will allow enough current to flow through the resistor to drop the other 4.3V, as a first approximation. Of course if the current required is too high, it something will burn out. If the current is very low, (say it is a 1 Meg resitstor) the current will be below the knee of the diode's IV curve, and the voltage drop across the diode will be lower than the nominal 0.7V.

    Bob
     
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

-