Circuits with diodes

Discussion in 'Homework Help' started by PsySc0rpi0n, Jun 1, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hello fellow electronic'ers'...

    I'm trying to solve some problems and the first one is this circuit:
    Ex-17.png

    We are asked to write down the vOUT equation in terms of vIN (diodes are ideal).

    So, I think the first thing we should do is to write down a table where we can see all possible situations:


    ....D1.......D2......D3
    1- OFF...OFF...OFF
    2-OFF....OFF...ON
    3-OFF.....ON...OFF
    4-OFF.....ON....ON
    5-ON.....OFF...OFF
    6-ON.....OFF....ON
    7-ON......ON....OFF
    8-ON......ON.....ON

    Now I need to narrow down the situations that are impossible...
    How do I have to think or where do I have to look or what questions do I have to make to myself to easily spot the impossible situations?
     
    Last edited: Jun 1, 2015
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Simple pleas consider this cases:
    1 - If D1 is OFF can a D2 or D3 be ON ?
    2 - Do both D2 and D3 diodes be ON at the same time ??
     
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I think that if D1 is OFF, then D2 will be ON because it has 6V at it's anode which is greater than the voltage at it's cathode because D3 will be OFF as it needs over 8V to be able to be forward biased, so no, D3 can't be ON

    Is this correct for situation 1 that you stated?
     
  4. Jony130

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    looks good
     
  5. WBahn

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    Mar 31, 2012
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    It's not as simple as identifying which combinations are impossible because you have an additional variable, namely Vin. So some combinations will be impossible for some values of Vin and possible for others. Still, it's worthwhile identifying combinations that are impossible for ANY value of Vin in order to narrow the field up front.

    By "ideal" diodes I'm going to assume you mean 0V drop across it when forward biased. Some authors refer to a diode that has a constant Vd drop as "ideal", too.

    The obvious one to start with is, as already noted, D1. That's because if you assume that it is OFF then the value of Vin doesn't matter. Now, the assumption that it is OFF depends on V1, but not the possible combinations of the other two diodes IF it is off. The next one to focus on, again as already noted, are the possible combinations of D2 and D3. There are four possibilities and you can look at them separately from and you have correctly

    After that you start looking for critical values of Vin that make some combinations impossible either above or below that critical value. But you don't have to start with Vin directly. In this case, the top node (Vo) is a very good candidate to look at first.
     
  6. GarryO'Keeffe

    New Member

    Jan 28, 2015
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    Did you find the final equation, if not think of it like this :

    As you said D3 is off so ignore that branch.

    Now work out a formula for the current flowing in output resistor R5 ( Io) remember Vo = Io * Ro

    Now Io is the combination of the currents coming from Vi and the 6V battery. So Io = Ii + Ib, its just math substitution from here to give Vo = ......Vi

    You should find there is a single simple formula for Vi > 0, let us know what you come up with.
     
  7. t_n_k

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    Mar 6, 2009
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    Are you stating that D3 is never conducting, regardless of the value of Vi?
     
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok guys, so coninuing, from the table I wrote in the first post, I think I can say that all possibilities that starts with D1 and D2 OFF are impossible, so:

    ....D1.......D2......D3
    1- OFF...OFF...OFF
    2-OFF....OFF...ON
    3-OFF.....ON...OFF
    4-OFF.....ON....ON
    5-ON.....OFF...OFF
    6-ON.....OFF....ON
    7-ON......ON....OFF
    8-ON......ON.....ON

    Option 3 is the one we've just seen:

    If we assume the top node as being vOUT, option 4 I think It's also impossible because if D1 is OFF, then the "controlling" voltage will be V1 which is "imposing" a voltage of a little under 6V at vOUT node, making D2 ON but D3 will have a lower voltage at it's anode than at it's cathode making it OFF.

    Option 5, I think it's possible for 6V < vOUT < 8V. At this condition, voltage at D2 cathode is higher than the voltage at it's anode making it OFF and voltage at D3 anode is lower than the voltage at it's cathode making it also OFF.

    Option 6, it's possible, I think, for vOUT > 8V. At this condition, voltage at D2 cathode is greater than V1 making it OFF and voltage at D3 anode is greater than the voltage at it's cathode making it ON.


    Option 7, I think, is also possible for vOUT < 6V. At this condition, voltage at D2 cathode is lower than at it's anode, making it ON and volage at D3 anode is lower than voltage at it's anode is lower than the voltage at it's cathode making it OFF

    Option 8, I think, it's impossible, because for D2 to be ON, voltage at it's cathode must be lower than 6V and for D3 to be also ON, voltage at it's anode must be greater than 8V. This is impossible to have, at the same node, a voltage that is lower than 6V and greater than 8V.
     
    Last edited: Jun 2, 2015
  9. Jony130

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    Look good to me. But can you find the corresponding input voltage ??
     
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    That is the equation I need to find. It will be an equation with as many branches as possible options!

    But I think I don't know how to find it!
     
  11. Jony130

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    First assume that D1 is OFF and now try to find Vo.
    Next you can for example assume D1 ON and Vo = 6V so D2 and D3 are OFF.
    This means that the load current is 6V/5KΩ = 1.2mA
    Since D2 and D3 are OFF all this current must come from input voltage source.
    Vin = 6V + 1.2mA * 2.5kΩ = 9V
    We can do the same think for Vo = 8V.
     
  12. WBahn

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    Mar 31, 2012
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    This is where my suggestion of looking for critical voltages comes into play.

    Walk Vin from -∞ to +∞ and looking for critical points at which the changes occur.

    For instance, with Vin = -∞, determine the state of the diodes. That will give you a value of Vout for all the values of Vin from -∞ up to whatever Vout is, at which point D1 turns on. So that's your first critical value for Vin. Now ask yourself what Vout needs to become in order to change the state of the D2/D3 combination. Then you just ask what value of Vin corresponds that value of Vout.

    Notice that, for a set configuration of the diodes, you just have a linear circuit. This means that if you find (Vin,Vout) at the transition points from one diode configuration to the next, that the Vout(Vin) characteristic is going to be a straight line between those points.

    Also, each diode will transition state just once. So you will have, at most, four regions of operation. You might have fewer since it is possible that one or more diodes never change state or that two or more diodes change state at the same point.
     
    Last edited: Jun 2, 2015
  13. PsySc0rpi0n

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    Mar 4, 2014
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    Ok, if I look to the circuit for situation 3, D1 is an open-circuit and also is D3, so I would say that Vout = 5*6/10 = 3V using a voltage divider! But our teacher wants Vout in funciton of Vin!
     
  14. ericgibbs

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    Jan 29, 2010
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    Hi Psy,
    Have you considered using LTSpice to check your answers, use a Vramp , say 0V thru 10V for Vin.???
     
  15. PsySc0rpi0n

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    Mar 4, 2014
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    Hello Mr. EricGibbs...

    I didn't want to loose time with LTSpice for now... Just wanted to understand how to find out Vout in function of Vin... The answer I gave in the previous post was Vout in function of V1 and that's not what we were asked to do!
     
  16. WBahn

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    So just figure out what values of Vin allow the assumption that D1 is open to be true. Vout is 3V for all such values of Vin.
     
  17. WBahn

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    I agree with your approach, but not your reason for it. It is far better to fight with this and get an answer manually and only then check your answer with LTSpice (being aware that LTSpice may not have a diode model that is for an ideal diode having 0V forward bias, though one could be created pretty easily (or at least one that is fairly close).
     
  18. ericgibbs

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    Jan 29, 2010
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    hi WB,
    We agree, you and I have made the same point, I would not suggest he solves his problem using LTS.

    I posted.
    Eric
     
  19. WBahn

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    Oh, I wasn't trying to suggest that you were recommending that he use LTS to FIND the answer. It's just that his response indicated (and maybe I read too much into it) that that was how he was viewing the use of LTS.
     
  20. ericgibbs

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    Jan 29, 2010
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    hi WB,
    I have noticed, with all due respect to 'Psy' that quite often he is posting what he believes to be the answer to to a particular equation/problem and then asks for confirmation that the answer is correct.
    Some of his answers are way out and the mistakes he has made could be spotted if he developed a method of double checking his calculations.

    Most of us apply 'rule of thumb' techniques to check the correctness of our calculations, especially during formal written examinations.

    Eric
     
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