3 questions(Transmittance, Diode and Transistor)

Discussion in 'Homework Help' started by adokas, Jun 29, 2014.

  1. adokas

    Thread Starter New Member

    Jun 29, 2014
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    Hello everyone, I'm doing my homework for tomorrow and there are only 3 questions that I couldn't solve. I'd be grateful if you help me to solve and understand this questions.

    1-) Calculate the spectral transmittance of the circuit below; calculate output voltage Vout while R=1K ohm, C=1 microfarad and Vin=2sin(wt+Pi/2), w=1000 rad/s.
    [​IMG]

    2-) Calculate diode current Id in the circuit below while R=1k ohm and E= 2.1 V
    (Udf=0.7V). Recalculate Id for the diode additionally connected in parallel with R.
    [​IMG]


    3-) Calculate the operating point for the transistor supplied as beside while Rb=30k ohm, Rc=100 ohm, Vcc=10V, Ube=0.7V and B(beta)=100A/A. In what mode the transistor is working?

    [​IMG]
     
  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi adokas,
    Any chance we could see your written attempt at answering the questions.?

    Eric
     
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  3. adokas

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    Jun 29, 2014
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  4. ericgibbs

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    Jan 29, 2010
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    hi,
    Downloaded your images OK, I can read your writing.:)

    I will read thru them and post later.

    E.
     
  5. adokas

    Thread Starter New Member

    Jun 29, 2014
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    Thank you very much sir! I'd be grateful if you can post it tonight, I have to show all of my solutions and explain them to my professor tomorrow :)
     
  6. ericgibbs

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    Jan 29, 2010
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    hi,
    Looking at Q2.
    Part 1.
    The voltage acting around the circuit is: E - Ufwd = .........................

    So the current Id = (E - Ufwd)/(R +R)=........................

    If you substitute the given values you should be able to calculate Id.

    Whats your answer for Id.

    Part2.
    Calc the current flowing in the 3 Series resistors E/( 3*R) =..........................

    Whats the voltage drop across the resistor in parallel with the diode. Vr = (E/3*R) * R............

    If the Vdrop across the added resistor is =< 0.7V then the diode is not conducting.... Is it.??

    E
     
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  7. ericgibbs

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    Jan 29, 2010
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    hi,
    Q3
    Part1
    Voltage across Rb = Vcc - 0.7 = ........................................?

    So current thru Rb = (Vcc - 0.7)/ Rb = Ibase = .............................?

    The Beta [gain] is 100 so Current thru Rc = ~ 100 * Ibase

    So Voltage across Rc = ~ (100 * Ib) * Rc = .............................?

    So the collector voltage Vc, is Vc - (100 * Ib) * Rc = .......................?

    Actually the collector current Ic, will be Ib * (Beta-1), but 100 is close enough.

    Part2.
    Which of the transistors pins is connected to 0v Common.???

    So its a Common ................. configuration.

    E

    Have you understood these last two posts.??
     
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  8. adokas

    Thread Starter New Member

    Jun 29, 2014
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    Yes sir, and I believe I've solved both of them, do you have any suggestion about the first question?
     
  9. ericgibbs

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    Jan 29, 2010
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    hi,
    Thats good,

    What do you calculate the Vin to be using, 2sin(wt - pi/2) ,, where w =1000 rads.??

    E
    EDIT:
    The two R resistors attenuate all frequencies from 0Hz and higher by 6dB.

    You have to calculate the frequency at which the Vout is 0.707 of its initial value.

    Xc = 1/(2 *pi* F * C) which is in parallel with the resistor value.


    The site is so slow tonight its almost unusable for some reason.:confused:
     
    Last edited: Jun 29, 2014
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  10. adokas

    Thread Starter New Member

    Jun 29, 2014
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    I'm really grateful for your answers sir. I couldn't do the last one, maybe I'm too tired, I couldn't understand it.
     
  11. anhnha

    Active Member

    Apr 19, 2012
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    Hi,
    I am a bit confused about part 2.
    Assuming that diode is model as an ideal diode with a series voltage source 0.7V.
    The current flowing through the resistor in parallel with diode is 0.7V/1K = 0.7mA
    The current flowing through two resistors in series is (2.1V - 0.7V)/2K = 0.7mA
    So, the current flowing through diode is 0.7mA - 0.7mA = 0
    Then, is the diode on or off???
    I think it should be "on", otherwise the voltage across this diode will be 2.1V.
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    If you are talking about the case of the additional resistor R in parallel with the diode then consider removing the diode from the circuit. The three series connected equal resistors will equally share the 2.1V source voltage - meaning 0.7V across each R. Hence with the ideal diode [Vf_on=0.7V] actually in parallel with the resistor it (the diode) is just at the point of switching on but not actually on. However if the diode did conduct the reduced terminal voltage would then be insufficient to maintain conduction. Consider the Thevenin equivalent seen looking into the common resistor nodes with the diode removed and this might become clearer. Also consider what would happen in the case of the ideal diode [Vf_on=0.7V] connected across a pure 0.7V DC source in the forward direction. No current could flow in that case given the external DC source emf is exactly balanced by the diode's equivalent internal idealised 0.7V DC "source".

    Of course in the case of a real diode the result would not be the same.
     
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  13. anhnha

    Active Member

    Apr 19, 2012
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    Thanks. Some time ago, I was really confused about the approximation of Vbe = 0.6V.

    Current flowing through diode changes a lot while the voltage across it only changes in a very small amount.
    Looking at the V-I characteristic of a real diode. Assuming that we approximate Vbe =0.6V while actual forward voltage across it is 0.65V.
    With Vbe = 0.6V, the corresponding current is 3mA.
    But with Vbe = 0.65V, the corresponding current is 4mA.
    (these numbers are not really measured)
    So, we have an error of 1mA that is too big!
    However, I realized that the current flowing through diode is also controlled by other components in the circuit.
    For example, let's consider a circuit including a voltage source 5V, a resistor 1K and a diode all are in series.

    With vbe = 0.6V => I = (5 -0.6)/1K = 4.4mA.
    With Vbe = 0.65V => I = (5 - 0.65)/1K = 4.35mA.

    => error here is 0.05mA that is much smaller than what I thought above.

    I hope I am getting it correct!
     
  14. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi anhnha,
    The nominal 0.7Vfwd is for Silicon diodes, it can be much lower for Schottky diodes and Germanium diodes.

    Some low current Silicon diodes can start conducting just below 0.6Vfwd.

    Also of interest is that Vfwd falls by ~ -2mV/Cdeg temp rise

    Eric
     
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  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Yes you are correct about the role of current limiting resistance.

    Ignoring any current limiting resistance for the moment ...
    You suggested a 50mV increase in forward voltage [0.6 to 0.65V] might produce a 1mA change in diode forward current. An anticipated 1mA change in diode current for a forward voltage change of 50mV is actually too conservative. One would typically expect an almost ten-fold increase in current for a modest increase of ~60mV in forward voltage at room temperature!
     
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