Drawing Output waveform in ac/dc diode circuit for specific point (theveninish)

Discussion in 'Homework Help' started by alexchala, Mar 19, 2018.

  1. alexchala

    Thread Starter New Member

    Feb 8, 2018
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    0
    Ok. Im working through a tutorial, this is not work any marks apart from person education and satisfaction. PS solutions are given. i will attach files as proof so people don't think i hunting for an answer for an assignment. Im struggling to find a comprehensive video on you tube or article to read giving a good explanation. so im here.

    My question is regarding the last 2 questions ( last page), where it asks to draw the Vo waveform from the circuit. PS it isnt stated in the question but during the tutorial video the lecturer states that on the last question (the one with the capacitor) the RC time constant it to large relative to the waveform frequency for it to have an effect on Vo waveform.

    So i understand that with diodes Va>=Vk for the diode to be on ( using idea diode models in analysis) and if this is not a true statement then the diode is off. And with the AC and DC involved in the circuits you analyses the DC first (superposition) then analyses the AC waveform (pos, neg cycles or important points of voltage on the waveform) taking into account any effect that the DC voltage has on the Vk or Va of the diode. but im having trouble putting this understanding into systematic math to give answers of with i can simply drawn a rough graph relative to my math awnser.

    I have watched the lecture explain the solution although it is not clear, mabye to much assumed knowledge and shortcuts taken.

    Cheers
     
  2. wayneh

    Expert

    Sep 9, 2010
    14,708
    5,227
    OK, well, that means you can simplify the circuit by eliminating the RC, right? Can you solve it now?
     
  3. alexchala

    Thread Starter New Member

    Feb 8, 2018
    16
    0

    i dont think you can eliminate the cap because there is no resistance between the cap and the dc source when the diode is on. Therefore time constant when dc hits capacitor is zero and cap charges instantly.

    So the cap still plays a role? Just relative to kvl when diode is off or on?
     
  4. wayneh

    Expert

    Sep 9, 2010
    14,708
    5,227
    I misread the schematic and probably your question also. Sorry. Never mind.
     
  5. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,130
    1,111

    Hello,

    Are those triangle things supposed to be diodes? Why are they not drawn like standard diodes?

    Anyway, for your last circuit sometimes the RC matters and sometimes it does not, depending on where you are in the wave at the time. The 5v and diode makes up a "clamp" circuit and that just limits the voltage, but also the cap charges up too so you have to consider the cap being charged sometimes.

    We can go through this if you are still interested. It's easier than it may look at first but it may be a little tricky because sometimes with diode circuits you have to solve for a time value as well as a voltage and/or current value.
     
  6. alexchala

    Thread Starter New Member

    Feb 8, 2018
    16
    0

    Hey thanks for the reply. Yes the triangle things are standard diodes, no i have no idea why they are drawn missing the bar. I might just do more research myself at the moment starting to frustrate me. cant see to find logic methods to solutions other then the really basic diode circuits. Thanks
     
  7. MrAl

    AAC Fanatic!

    Jun 17, 2014
    5,130
    1,111
    Hi,

    You're welcome.

    Some diode circuits are rather difficult to solve because their switch state acts like just that, a switch, and because of that the circuit topology changes once the diode turns on, and then changes again when the diode turns off. Depending on what happens in between the circuit could have to go through several on/off cycles before we see anything we might call steady state.
    Take a full wave rectifier circuit for example, with capacitor filter and with some ESR in the cap, and of course a load RL. The solution takes many cycles to get to near a constant changing waveform that looks the same for every input cycle, as the average output voltage builds up over several cycles.

    The main idea though is that the diode acts as a switch so you have to solve for the times that it turns on, and turns off, and these could change cycle by cycle because one of the governing factors is the voltages themselves, which too are changing as the circuit response evolves.
    So the idea is to determine if the diode is on or off from the startup conditions and how the node voltages force the state of the diode, then let the circuit evolve given that on/off state, then solve for the next switch state. If it was off then it will turn on, and if it was on then it will turn off, unless of course the circuit had reached some point where the voltages have all stabilized and so no more switching.

    Once the diode changes state, the topology changes.. What this means is that we are effectively working with TWO circuits not just one. One circuit is when the diode is on, and the other circuit is when the diode is off., and the initial conditions for the next topology come from the final conditions of the previous topology, and we just keep doing that until we reach some kind of result where we just see the voltages repeat over and over again, or reach a complete DC steady state.

    So conceptually it's easy to understand, but as a practical matter it can get a little hairy because we have to solve for on/off times as well as the node voltages. If we have more than one diode we have to consider the fact that they may be turning on and off at different times too, so we may end up with four or more circuit topologies we have to deal with just for that one circuit.

    Another way to put this is we have to solve for the node voltages on BOTH sides of the diode, and see when it becomes forward biased and when it becomes reversed biased. That tells us when it turns on and when off. We may also have to solve for the zero current level through the diode so we can tell when it turns off, and then the reverse bias will hold it off until it again becomes forward biased.
    If we consider an ideal diode, that means when the cathode becomes EQUAL to the anode voltage then we say the diode turns on, and if the anode voltage is less than the cathode voltage, then we say it turns off. So if the voltage on both sides of the diode is changing, we just either watch the change and wait, or we try to come up with an expression that solves for when Va=Vc (anode and cathode voltages) and then we say we have a short circuit between those two nodes. We then have to solve for a very low current value through the diode (0 amps) so we can say that it turned off.
    It can get a little tricky though because the two voltages are changing when off, and when on the current is changing.

    Once you do a few it gets easier to understand, although the numerical calculations usually dont get any simpler. That's because we effectively have two independent variables (voltage and time) when we usually only have one (time). Voltage isnt really independent, but it looks like it is because we have to use that in order to determine the switch time.

    A simple example is a half wave rectifier. As the input sine goes higher the cap charges up. On the next cycle however, the diode does not start to conduct until the source voltage becomes greater than the cap voltage, and remember the cap voltage is also falling due to the load resistance. Once we find the switch point though then we see the cap start to charge again, and then as the source voltage falls we see the cap voltage stay higher than that so the cap starts to discharge again.
    This is a very informative circuit if you'd like to look at it in more detail we can do that too.
     
    Last edited: Mar 24, 2018
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