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Diode,dc & ac voltage source

  1. Nov 8, 2014 #1
    Hello guys.I am really having a problem with a circuit that includes a dc source,an ac source,2 resistances and a diode(not ideal).In the pdf that i uploaded is my way of thinking for this.I will really value any help and guidance.

    Thnx in advance
     

    Attached Files:

  2. jcsd
  3. Nov 8, 2014 #2

    rude man

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    In determining the diode dynamic resistance r, where did you get the 0.7V? The 4 mA?
    1. Determine the diode dynamic impedance r = ∂Id/∂Vd at the 2Vdc input level
    of the diode when Vac in = 0. Use the standard I-V diode relationship for a junction diode.

    2. Then it's a linear problem if you neglect the small changes in r with input ac voltage. The dc diode current and dynaminc impedance do not change significantly with ac input voltage (why?). The diode ac current is determined by r and the 100 ohm resistors.
    Have fun!
     
  4. Nov 8, 2014 #3
    Sir at first let me thank you for your reply...In my attempt to understand what you told me i am pretty sure i burnt many brain cells..So Mr Rudeman(or any1 who has the mood) i have many questions so let me start from the beginning:

    <<In determining the diode dynamic resistance r,where did you get the 0.7V?The 4 mA?
    The as source gives maximum Vm=0.1V,so it is not possible to get 0.7 just from that.The main source is the dc source which gives steady 2V.But the AC source adds as well,does it not?
    <<1. Determine the diode dynamic impedance r = ∂Id/∂Vd at the 2Vdc input level
    of the diode when Vac in = 0. Use the standard I-V diode relationship for a junction diode>>


    ok why i should take that AC source is 0?In order to use this equation dv/di=R,must i have only the DC source?
    Is it because R is a constant only when the as source is 0?Only when there is a linear relation?Do i get from the graph that AC must not be taken into account?I am just posting some thoughts i have into mind...

    <<
    2. Then it's a linear problem if you neglect the small changes in r with input ac voltage. The dc diode current and dynaminc impedance do not change significantly with ac input voltage (why?). The diode ac current is determined by r and the 100 ohm resistors.
    Have fun!
    >>


    I take that AC source is 0,because its voltage is too low so that it doesnt really make any difference?That is why in the graph there is a linear relation between id and vd?

    And now the thing i do not have a clue about.<<The diode ac current is determined by r and the 100 ohm resistors>>.If i regarded AC source is 0,why should i find the ac current?Wouldnt it be
    id=idc+iac ~ idc(where idc is the current that is <<produced>> only by the dc source).

    Do id and vd change through time just because of the ac source?


    I know these questions are many and too simple for most of you,but it is the only way to build myself.
    Feel free guys to answer whenever you want and can.
     
  5. Nov 8, 2014 #4

    rude man

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    Not the way to determine ac diode impedance.
    Yes.
    r is very nearly always constant and is computed when ac input = 0. r does not vary appreciably with ac voltage if ac current << dc current, here the case.

    The basic junction diode equation is Id = Isexp(Vd/26 mV) at room temperature. By taking 1/r = ∂Id/∂Vd you eliminate Is so you don't need it to compute r.
    You are trying to find the diode dynamic impedance r about your operating point. Your operating ("bias") point is when the ac input = 0. The bias point is negligibly disturbed with the ac input voltage since ac << dc inputs. AC diode current is a function of r only.
    That is basically right. Your ac amplitude is << dc voltage at the diode anode so r is always constant.
    I believe the problem wants you to find iac and Idc. iac is determined by including r in your equations, along with the two 100 ohm resistors. Idc is determined by a dc analysis with the input ac voltage = 0. These are basically totally different computations.

    Electronics is full of circuits where the ac current is 10-6 or even much less of the dc current. The dc current level is typically set to give maximum ac gain, or minimum ac noise, or some other criterion. The interesting component is usually the ac component since that is typically the signal of interest. So you must not ignore the ac current even though it is << dc current.
    Only very slightly. You are to ignore that variation. The reason is that the ac amplitude is << dc input voltage.
     
  6. Nov 8, 2014 #5
    You sir,honestly speaking,made my day.Some things are more clear now and i thank you for that!!
     
  7. Nov 8, 2014 #6

    rude man

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    You're welcome!
     
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