Electronics problems - again

Discussion in 'Homework Help' started by PsySc0rpi0n, Jul 9, 2014.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Well, as my 1st attempt to solve the 1st exam didn't ended up well, I'm going for the 2nd attempt.

    I'm going to post here my 1st exam's circuits and translations as best as I can do and I'll ask for help to solve it. I'll start solving the problems later today when I get home!

    1 - Draw the Transfer Function Vo(Is) for the given circuit. The diodes are ideal and the current source goes from -5mA up to 5mA.

    2-The following circuit is a filtered rectifier.
    a) - Consider the circuit without C1. Draw the voltage waveform at the rectifier entry point (transformer secondary) and voltage waveform at Rload (R1).

    b) Consider now the C1 and find it's capacitance so that Vripple is no greater than 2V.

    c) For the circuit of the previous problem (b), draw the input voltage, Rload voltage and diode current waveforms.

    Consider that Vγ = 0.7V and mark all the key points in the graphic!


    3- Find the voltage at point X. Consider Vγ = 0.7.


    4- For the following circuit:

    a) Find Rb so that current across Rc is 10mA. β=120 for T1 and β=30 for T2. Both T1 and T2 are Silicon.

    b)Find the dissipated power for both T's.


    5) For the circuit below find Av, Ai, Ri and Ro. hFE = 100 ad hie = 2KΩ
    Note: Av is voltage gain, Ai is current gain, Ri is input impedance and Ro is output impedance...


    Hope I made my self clear enough so that you can help!

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    Last edited: Jul 17, 2014
  2. anhnha

    Well-Known Member

    Apr 19, 2012
    Hi, there are many problems here. I think you should post one at a time and you should post your work or what you did so far.
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    I will work each problem in a different post...
    And yes, I'll post my work as I'm doing it!

    For the problem 1, where should I consider the GND??? Should I consider it in the '-' side of the current source for simulation purposes???

    Well, the first thing I have done was to convert the current source and the parallel resistor into a Voltage source in series with a resistor.
    Then I am analyzing each situation possible regarding the diodes. OFF - OFF, OFF - ON, ON - OFF and ON - ON.

    The 2nd one is impossible, i guess. D2 can't be ON and D1 OFF.
    Then I tried to analyse the circuit for the other 3 situations but for the ON - ON situation, I'm struggling to get Vout expression!

    I'm trying to use Nodal Analysis to try to find current at R3 as the image attached, but I can't get the expression for Vout!


    I was doing i_{1}=i_{2}+i_{3} but i can't get nowhere!
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  4. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Can anyone help???
  5. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    hi Psy,
    Yes, For simulation consider the bottom line of your circuit as ground/ref.


    This image should help.

    The image shows 8b for a battery voltage, this of course should be 8v.... duh!
    Last edited: Jul 11, 2014
  6. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Hi sir Eric...

    In the problem, the picture shows the minus on the top side of the current source. That's why I asked if I should put there the GND reference.

    What about my calculations so far?
  7. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Can't open that link... It gives an Adobe PDF error!

    That is what I want. To draw the transfer function.
    In the picture I have posted above, I'm trying to get the Vout expression for each situation, but I can't get it for the ON ON situation!
  9. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    The way I read your first post is that the I1 varies from -5mA thru 5mA

    So have you calculated the 'break points' for plotting the curve.??
    You have the Start and End of the I1 sweep and there are two breakpoints at just under 5mA and 6mA.
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    I don't know if I understood your questions.

    To draw the transfer function I need to know what is Vout for each D1 and D2 states.

    the states that D1 and D2 can take if we sweep the current source from -5mA up to 5mA may be:

    D1 - OFF || D2 - OFF
    D1 - OFF || D2 - ON
    D1 - ON || D2 - OFF
    D1 - ON || D2 - ON

    From this I need to find Vout expression for each of the 3 situations as the 2nd is impossible!!!

    This is what I've done so far but I can't find Vou for situation where D1 - ON || D2 - ON.
  11. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    Does this marked up image help.

    The diodes are not ideal , ideal being Vfd=0v

  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Morning Sir Eric...

    Thanks for the effort... What I really need is to find the Vout expression and not exact values...

    I have a similar problem solved in my classes notes, but i can't understand the expressions.

    We have wrote that i1 = i2 and what i can't understand how i1 = (vin - vout)/R1 and i2 = (vout - v1)/R2.

    Can you help me understand this???
  13. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Can anyone check if the calculations in the attached file are ok???
  14. ericgibbs

    AAC Fanatic!

    Jan 29, 2010

    As its a Current source, the instantaneous current is shared thru the 3 networks in proportion to their Admittances.

    Initially the 5k will pass all the available current and Vout will rise equal to Isrc * 5k.
    During the initial rise in negative value of Isrc both the 5V and 8V networks are not conducting as the diodes are reversed biased [ I am assuming that the bottom rail is reference].

    As the Isrc becomes positive the Vo will continue to rise until the voltage across the 5K exceeds the 5V battery voltage. Now the Isrc current divides into the 5K and 5V networks, so the Vo rises more slowly.

    Vo continues to increase until Vo exceeds 8V and diode and that network will start conducting. This means the available current is shared by all 3 networks and so Vo rises more slowly.

    Its important to understand that its the Isrc that determines the Total current around the circuit.

    I will look at your last 2 posts and reply later.... I am a little busy on another project at the moment.:rolleyes:
  15. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Hi Sir Eric...

    Thanks for the thoroughly reply.

    I have transformed the current source paralleled with the 5k resistor into a voltage source in series with a resistor.

    I think I did it correctly.
    I'm just struggling to understand how we find the Vout expression like I have it in post #12.
  16. MrAl

    Distinguished Member

    Jun 17, 2014

    I think you may want to keep the current source intact as it makes combining elements a little easier i think. So instead of transforming the current source into a voltage source, convert the voltage sources into current sources instead after identifying the voltage breakpoints.

    Here's a quick rundown of the procedure...note i will probably have to edit this over the next few minutes for formatting of text and such...

    The first step is to identify the model we will use for the diodes.
    Since they were specified as ideal, this usually means that the diodes
    turn on when the anode voltage is greater than the cathode voltage,
    even by a very small amount.

    Next we have to figure out when the diodes turn on, and then find
    the equations for each partial response. Since there are two diodes
    the response will first rise at a certain rate with Is, then when
    D1 turns on it will rise with a lower rate with Is, then when D2
    turns on it will rise with an even lower rate with Is. This means
    we have three different slopes to find.

    Looking at the circuit, we see that D2 is biased with an 8v battery
    and D2 is biased with a 5v battery, and as Is is increased eventually
    D1 conducts and then as Is increases more D2 conducts. So first D1
    and then D2 as Is rises.

    Since D1 is the first to conduct and it does not conduct until Is
    is equal to 0.001 amps (0.001*5000=5v), the first equation is:
    and this is valid from -5ma to 1ma only, because after that D1 starts
    to conduct and that will change the equation.

    So because of the two diodes this gives us two critical points of Vout
    voltage and Is current:
    1. When D1 just starts to conduct
    2. When D2 just starts to conduct

    This means we'll have two more different slopes for Vout as Is increases,
    and we can identify these points by noting that D1 starts to conduct
    when the voltage at A reaches 5v, and that D2 starts to conduct when
    the voltage at B reaches 8v. What we have to do then is find the
    current Is when these two voltages are reached.

    To start, when Is is very low Vout increases as Vout=Is*5000. So the
    first slope then is simply 5000. This continues until we reach Vout=5v.
    Once we reach that point, D1 starts to conduct. This puts another
    source and resistance into the circuit and so the slope changes.
    We note that when this happens Is=0.001 amps so we have our first point:

    To find the equivalent circuit now, we need to combine elements.
    Since D1 is a short now, we replace that with a short.
    The Norton/Thevenin equivalent for the 2k resistor in series with the
    5v source is:
    I1=5/2000 in parallel with a 2k resistor.
    Now when we combine the source Is, the resistor 5k, the source I1,
    and the resistor 2k, we get two current sources in parallel and two
    resistors in parallel. The parallel current source is:
    and the parallel resistor is:

    The circuit now is a single current source Is+5/2000 in parallel with a
    single resistor of value 10000/7.
    Vout is now therefore:
    We know that the next breakpoint occurs when Vout=8v, so we set Vout
    equal to 8v and solve for Is:
    Is=7*8/10000-5/2000=0.0031 amps

    So now we have solved for the second point:

    Now we also have the partial circuit with Is, 5k, 2k, and D1 reduced
    to one current source and one resistor, and since now the whole circuit
    (that partial circuit with the 1k and D2) resembles the partial circuit,
    we can do the same thing again to find any other point where Vout
    is above 8v. This gives us a third point:
    where Vx is some voltage above 8v and Ix is the current we solved for.
    This will be found when all the elements are combined to form a single
    current source in parallel with a single resistor. This resistor value
    will be the parallel combination of the previous parallel resistor and
    the last resistor 1k.

    Now we can solve for the slopes.
    Since the first point is (0.001,5) and the second point is (0.0031,8)
    we can use the two point form of a line formula to find the equation
    of the second section of the transfer function:
    or better yet:
    where y is the voltage and x is the current Is.
    This can be put into form:
    m=(y2-y1)/(x2-x1), and

    So the second part of the response is:
    where m and b are calculated as above.

    To find the third part of the response, do the same thing with the
    point (Ix,Vx) and that will give you the last equation.

    If this seems too difficult, we'll take it one step at a time.
    PsySc0rpi0n likes this.
  17. MrAl

    Distinguished Member

    Jun 17, 2014
    Hi again,

    Yes i gave two "straight line" formulas:
    1. The two point form.
    2. The slope intercept form.

    We need something like this because we want a function, not a specific single result. Unfortunately we have to find three such functions because there are three different slopes depending on what diodes are conducting. By knowing the voltage trip points we can then solve for the respective currents, and for each slope we have two points and so the two point form of the line is very convenient. The two point form is then converted to the slope intercept form because that makes a simpler expression in the end. So what we will end up with is something like this:

    Vout=m1*Is+b1, {I0<=Is<=I1}
    Vout=m2*Is+b2, {I1<=Is<=I2}
    Vout=m3*Is+b3, {I2<=Is<=I3}

    These three represent the transfer function where i assumed that Vout was taken from the top of the 2k and 1k resistors, and ground was at the bottom of both batteries. We solve for the currents I1, I2, and I3, which are constants, and I0 is simply equal to -5ma and I3 should be equal to 5ma.

    To get the slope intercept form from the two point form, simply lump all the constants to get the intercept (b) and any constant that multiplies x (which is Is) becomes the slope (m).

    If you use Nodal analysis you have to modify the circuit every time a diode starts to conduct and solve for the current. This should still lead to three equations.
    Last edited: Jul 14, 2014
  18. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Hello MrAl.

    Thanks for the thoroughly reply.

    I think I understood what you have explained but not completely.

    At some point you are using the "straight line formula" (I don't know if this is the correct name) Y = mX + b. I might get lost somehow around here!

    Did you check the PDF file I've uploaded a couple of posts ago?

    Anyway, I'm going to try to follow all the steps you stated to see where how far I can go.

    That is what I did in my post couple a days ago.

    You are considering here the very first instant when D1 "turns ON" and therefore Vout is 5V, so you solve V=R*I like V1 = R1*Is <=> 5V = Is*5kΩ <=>Is = 1mA. So this should be the "Stituation B" stated in the above quotes??

    Wasn't the previous calculation for the situation when D1 "turns ON"???
    I think I got lost here!

    I'm quite confused at this point. I'm not following the transformations your doing to the circuit.

    First, you said that Is is 1mA until Vout is less than 5V. This is before D1 "turns ON", right? So we are at the situation D1 OFF and D2 OFF.

    Then you say:

    This means we are at "Situation B"??? I'm not following. Or are you converting all voltage sources in series with resistors into current sources in parallel with resistors?

    If I'm understanding you're transforming all voltage sources in series with resistors into current sources in parallel with resistors.

    Note: When you say 10000/7 is actually 10000/7000, right? the parallel of resistors????

    So this one is when only D1 is "ON", right?

    If I understand, this one is when D1 ON and D2 ON, right?

    I got completely lost.

    I didn't figured out if you where transforming all voltage sources into current sources or what transformations were done in the process!

    A simple draw for each situation/transformation could help me understanding the whole process.

    I'm sorry for not understanding your explanation!
  19. MrAl

    Distinguished Member

    Jun 17, 2014
    Hi again,

    Ok, lets take this one step at a time.

    To start with, we have no diodes conducting, so all diodes are open circuits. This means we have a circuit with ONE current source Is and ONE resistor of 5k.

    We now allow the current Is to ramp up, and it starts at -5ma and climbs higher. The output voltage is:

    When Is=-0.005, Vout is -25v. When Is reaches 0.001a, Vout is 5v. But to find this current we work backwards, knowing D1 turns on at Vout=5v, so the current can be calculated as:

    That is actually the second point we need: (0.001,5).
    The first point was (-0.005,-25).

    Now that D1 is turned on, we need to modify the circuit because that means D1 is now a short, so the 2k resistor and +5v battery are in the circuit too now. And to keep the method we are using the same again, all we need to do is convert the +5v source and 2k resistor into its Norton/Thevenin equivalent. To convert, the new current source is 5/2000 amps, and the parallel resistance is 2k. So a current of 5/2000 amps with a parallel resistance of 2k converts the +5v source and 2k resistor to a current source and parallel resistor.

    This puts the new 2k resistor in parallel with the 5k resistor, and it puts the current source in parallel with the current source Is. Combining parallel current sources is easy and same with resistors, so the single current source is:
    I=Is+5/2000 Amps
    and the single resistor is:
    Rp=2000*5000/(2000+5000)=10000/7 Ohms.

    So now we are down to one single current source I in parallel with one single resistor Rp, so the circuit looks like it did when we started with just Is and the 5k resistor, except we have a different current source and different value parallel resistor now.

    Does this seem clear now? If not i'll draw the picture.

    The next step would be to do the very same thing again: find out when D2 conducts, then combine the last voltage source and resistor so that we are left in the end with only one current source in parallel with only one resistor.

    Norton and Thevenin equivalents often allow you to attack a circuit in small chunks like this.
    Last edited: Jul 14, 2014
    PsySc0rpi0n likes this.
  20. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Yes, I got it now...

    I'll try to do the rest yet this morning...

    Hope you can say if my calcs and thoughts are correct!