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Questions, help!!!!!

  • 07-12-2006 12:45pm
    #1
    Tar.Aldarion
    Moderators, Science, Health & Environment Moderators, Social & Fun Moderators, Society & Culture Moderators Posts: 60,110 Mod ✭✭✭✭


    Hey, I need some help with questiond for tomorrows exam in Analog electronics.

    I can do part A) and B) but I don't know if I am going about C) the right way.

    A) Given a +10v/-10v symettric power supply design a schmitt trigger with asymmetric switching voltages at 0v and 2v? Assume the op-amp output cn swing within 0.5v of the power rails.

    B) Convert your schmitt trigger design into an astable multivibrator with an asymmetric period of 100uS.

    C) How can you adapt this multivibrator to have a symmetrical period with identical rise and fall times of 50uS?


    Also, another question as regards Class A power amplifiers.

    What is the disadvantage of using a 1K resistor to implement the current source?

    Thanks.


Comments

  • #2
    Ghost Train
    Registered Users, Registered Users 2 Posts: 30,471 ✭✭✭✭Ghost Train


    The circuit you designed in (b) its switching voltages are at 0v and 2v

    If the circuit has switching voltages at -1v and +1v the period will be symmetrical, so if you put a dc bias of -1v at the input, and have the circuit with a switching voltage of -1v and +1v and period 100us. Don't know if that's the type of solution they'd be looking for though, but can't think of any other way the period would be symmetrical

    At the output of an amplifier you normally want low output impedance to drive your signals, I'm thinking having 1k resistor for the current source gives you an amplifier with high output impedance


  • #3
    Tar.Aldarion
    Moderators, Science, Health & Environment Moderators, Social & Fun Moderators, Society & Culture Moderators Posts: 60,110 Mod ✭✭✭✭Tar.Aldarion


    eolhc wrote:
    The circuit you designed in (b) its switching voltages are at 0v and 2v

    If the circuit has switching voltages at -1v and +1v the period will be symmetrical, so if you put a dc bias of -1v at the input, and have the circuit with a switching voltage of -1v and +1v and period 100us. Don't know if that's the type of solution they'd be looking for though, but can't think of any other way the period would be symmetrical
    Ah, do you think that would work too? I was thinking of messing around with values, but he may want me to change componants or something. I think he mentioned modifying the circuit which could be anything I suppose...
    On the +ve cycle it charges up from 0v to 9.5v with switching thresholds of 2v and 0v. A total of +9.5v.
    Then on the -ve it goes from 2 to -9.5v which is a total of -11v
    If it was 1v and -1v it would be total +ve of 8.5v, and total -ve of -10.5v.
    That is where I get confused, is that the period or is it just the 1v->9.5v and
    -1v->-9.5v.
    I think it is the way that does not help me. :/
    What do you think?
    Apart from that, I would also have to half the period to 50uS.
    I could change R and C to do that, if that is what he wants, and not another componant.
    At the output of an amplifier you normally want low output impedance to drive your signals, I'm thinking having 1k resistor for the current source gives you an amplifier with high output impedance
    Ah thanks, I will go with that.
    I know the 1k resistor is there as a lab thing, as with varying Vout a constant you want a contant ve or something.
    What I have come up with since my post is that maximum output swing is limited using a resistor, the extra resistor appears in parallel with RL nd power dissipation is increased. Sorted for that one. (:


  • #4
    Ghost Train
    Registered Users, Registered Users 2 Posts: 30,471 ✭✭✭✭Ghost Train


    Thinking about it a bit more,
    Just to clarify the period for (b) and (c) should be the same, 100us, cause period = rise + fall time

    Maybe the way to alter the circuit is to have two resistors with the capacitor, each with a diode(one forward one reverse). One resistor will work for the charge cycle and the other will work when the capacitor discharges. You should come up with two time equations for the charge and discharge time. Because they both equal 50us combining the equations would give you the resistor values/ratios.

    It's tricky getting my head around the way the circuit works, anyway hope you can work it out


  • #5
    Tar.Aldarion
    Moderators, Science, Health & Environment Moderators, Social & Fun Moderators, Society & Culture Moderators Posts: 60,110 Mod ✭✭✭✭Tar.Aldarion


    This is a rather tricky thing to answer about the period, although I think diodes is the best route. The question came up and I didn't even get time to attempt it!
    The 50uS thing had me confused for a while, didn't notice it wasn't total.
    Anyway, I got an A or B without finishing the paper, so I'll let this one slide. :)

    Thanks a lot for your help.


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