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Unable to get correct results with UJT equivalent relaxation oscillator

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Jayaram
2013-05-23
2013-05-25
  • Jayaram
    Jayaram
    2013-05-23

    Hi
    i've attached the .sch file for a circuit in which a capacitor is charged. Once it is charged it is supposed to forward bias a diode and thus discharge and once it discharges the diode is supposed to be reverse biased again and the capacitor charges again. And so on. Something like a relaxation oscillator. But even though the diode is forward-biased, the capacitor doesn't discharge. Pls help

    <Qucs Schematic="" 0.0.16="">
    <Properties>
    <View=0,0,1049,800,1,0,0>
    <Grid=10,10,1>
    <DataSet=ujt.dat>
    <DataDisplay=ujt.dpl>
    <OpenDisplay=1>
    <Script=ujt.m>
    <RunScript=0>
    <showFrame=0>
    <FrameText0=Title>
    <FrameText1=Drawn By:="">
    <FrameText2=Date:>
    <FrameText3=Revision:>
    </Properties>
    <Symbol>
    <.ID -20 -16 SUB>
    <Line -20="" 20="" 40="" 0="" #000080="" 2="" 1="">
    <Line 20="" 20="" 0="" -40="" #000080="" 2="" 1="">
    <Line -20="" -20="" 40="" 0="" #000080="" 2="" 1="">
    <Line -20="" 20="" 0="" -40="" #000080="" 2="" 1="">
    </Symbol>
    <Components>
    <R R1="" 1="" 270="" 150="" 15="" -26="" 0="" 1="" "2.518k"="" 1="" "26.85"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "26.85"="" 0="" "US"="" 0="">
    <R R2="" 1="" 270="" 260="" 15="" -26="" 0="" 1="" "38.15"="" 1="" "26.85"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "26.85"="" 0="" "US"="" 0="">
    <GND *="" 1="" 270="" 410="" 0="" 0="" 0="" 0="">
    <Vdc V2="" 1="" 400="" 250="" 18="" -26="" 0="" 1="" "15"="" 1="">
    <R R5="" 1="" 120="" 150="" 15="" -26="" 0="" 1="" "2k"="" 1="" "26.85"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "26.85"="" 0="" "US"="" 0="">
    <R R3="" 1="" 270="" 60="" 15="" -26="" 0="" 1="" "1k"="" 1="" "26.85"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "26.85"="" 0="" "US"="" 0="">
    <R R4="" 1="" 270="" 360="" 15="" -26="" 0="" 1="" "33"="" 1="" "26.85"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "26.85"="" 0="" "US"="" 0="">
    <.TR TR1 1 630 220 0 65 0 0 "lin" 1 "0" 1 "40 ms" 1 "1001" 0 "Trapezoidal" 0 "2" 0 "1 ns" 0 "1e-16" 0 "150" 0 "0.001" 0 "1 pA" 0 "1 uV" 0 "26.85" 0 "1e-3" 0 "1e-6" 0 "1" 0 "CroutLU" 0 "no" 0 "yes" 0 "0" 0>
    <Diode D1="" 1="" 200="" 220="" -26="" 15="" 1="" 2="" "1e-5="" A"="" 0="" "1"="" 0="" "10="" fF"="" 0="" "0.5"="" 0="" "0.7="" V"="" 0="" "0.5"="" 0="" "0.0="" fF"="" 0="" "0.0"="" 0="" "2.0"="" 0="" "0.0="" Ohm"="" 0="" "0.0="" ps"="" 0="" "0"="" 0="" "0.0"="" 0="" "1.0"="" 0="" "1.0"="" 0="" "0"="" 0="" "1="" mA"="" 0="" "26.85"="" 0="" "3.0"="" 0="" "1.11"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "0.0"="" 0="" "26.85"="" 0="" "1.0"="" 0="" "normal"="" 0="">
    <C C1="" 1="" 120="" 290="" 17="" -26="" 1="" 3="" "10u"="" 1="" "0"="" 0="" "neutral"="" 0="">
    </Components>
    <Wires>
    <270 180 270 220 "" 0 0 0 "">
    <270 220 270 230 "" 0 0 0 "">
    <270 90 270 120 "" 0 0 0 "">
    <120 320 120 410 "" 0 0 0 "">
    <120 410 270 410 "" 0 0 0 "">
    <120 180 120 220 "out" 150 190 40 "">
    <120 30 120 120 "" 0 0 0 "">
    <120 30 200 30 "" 0 0 0 "">
    <400 10 400 220 "" 0 0 0 "">
    <200 10 400 10 "" 0 0 0 "">
    <200 30 270 30 "" 0 0 0 "">
    <200 10 200 30 "" 0 0 0 "">
    <400 280 400 410 "" 0 0 0 "">
    <270 410 400 410 "" 0 0 0 "">
    <270 290 270 330 "" 0 0 0 "">
    <270 390 270 410 "" 0 0 0 "">
    <120 220 120 260 "" 0 0 0 "">
    <120 220 170 220 "" 0 0 0 "">
    <230 220 270 220 "" 0 0 0 "">
    </Wires>
    <Diagrams>
    </Diagrams>
    <Paintings>
    </Paintings>

     
    • domispace
      domispace
      2013-05-23

      Hi,

      You should attach your file instead of copy and paste it, since it appears the editor does not accept the .sch file anymore.

       
  • Frans
    Frans
    2013-05-23

    I don't see an attachment

     
  • randyc
    randyc
    2013-05-23

    Without seeing the schematic, this is a shot in the dark:

    If the circuit consists of a voltage supply charging a capacitor through a high value resistor (i.e. current source), with a diode shunting the capacitor, you won't achieve the equivalent of a relaxation oscillator.

    A relaxation oscillator requires a very high impedance switching device that transitions to a low impedance state when a sharply-defined threshold voltage is exceeded - like a unijunction transistor when the standoff voltage is exceeded.

    A diode does not meet that criterion because the "knee" of the device is not very well-defined and the diode is not a very high impedance, even below the diode contact potential.

    As the capacitor charge approaches the contact potential, the diode begins to conduct slightly and bleeds off the capacitor charge. The net result is that the capacitor will only charge to a voltage approximating the diode contact voltage. Once equilibrium is achieved, the capacitor remains in that state - no oscillation can occur.

    Cheers,
    randyc

     
    Last edit: randyc 2013-05-24
  • Jayaram
    Jayaram
    2013-05-24

    Hi @randyc

    Thanks for your suggestion. I'm attaching the sch file. I used the diode because qucs doesn't have ujt. So I rigged up an equivalent UJT network using diode and used it for the relaxation oscillator.

    "As the capacitor charge approaches the contact potential, the diode begins to conduct slightly and bleeds off the capacitor charge. The net result is that the capacitor will only charge to a voltage approximating the diode contact voltage. Once equilibrium is achieved, the capacitor remains in that state - no oscillation can occur."

    But just to clarify if the diode conducts the capacitor should discharge in qucs which doesn't happen

    Thanks in advance

    Cheers

     
    Attachments
  • Frans
    Frans
    2013-05-24

    Dear Jayaram,
    Try building your "oscillator" with the diode in real life, and you will see that it will work exactly like simulated. Removing R1 from the circuit will further discharge your capacitor.
    If you put a DC simulation block in your schematic and press "F8" you will see all the DC levels on your wires, which will better explain the behaviour of your schematic.
    Frans

     
  • randyc
    randyc
    2013-05-24

    Hi Jayaram,

    I'm a new user of QUCS and far from being adept. Additionally, I've been traveling for almost a month now and so have no library or reference material at hand so I cannot make any suggestions regarding your UJT circuit model.

    But it seems to me that the feature in QUCS that allows one to describe the behaviour of a non-linear device with an equation would be appropriate for your purpose. I would want to try that before attempting to develop a circuit model.

    Cheers,
    randyc

     
  • randyc
    randyc
    2013-05-24

    "But just to clarify if the diode conducts the capacitor should discharge in qucs which doesn't happen"

    I'm sorry that I didn't answer your question but I thought the first post would explain how the voltage at the capacitor terminal reaches a steady-state value. Let's try a mechanical analogy for better understanding:

    Assume that a metal plate is subjected to a fixed thermal load representing the current source. The metal plate represents the capacitor, which can store heat over time, the amount determined by the heat source.

    Now assume that the metal plate is suspended in free air and radiates a certain amount of the stored heat into space.

    The free air and the radiation characteristics represent the diode (although the diode is much more non-linear, it DOES conduct as the capacitor voltage approaches the diode contact potential).

    Consider what happens to the metal plate as thermal power is added and stored heat is radiated. Thermal equilibrium occurs at the temperature when the radiated heat is equal to the additive thermal load. When radiated heat = additive heat this is the thermal analog of the diode contact potential.

    At no time can the metal plate "oscillate" because this would imply that the radiation path had a threshold, below which no heat exchange would take place. Just as the diode is not an infinite resistance below the contact potential.

    Likewise, as the diode starts to conduct, equilibrium is reached when the diode resistance conducts the same amount of current that the current source can provide. This current establishes the contact potential of the diode and is therefore the steady-state voltage of the capacitor.

    Does this make any sense ?

     
  • Jayaram
    Jayaram
    2013-05-25

    Dear randyc!!
    I loved your explanation! I got the part about thermal equilibrium (albeit after multiple reads). I have a question. I beg to differ. See, what I feel should happen is this. Capacitor charges to a level equal to the diode threshold voltage. This results in the diode being forward biased. So far so good. Once this happens, the capacitor has a low resistance path through which it can start discharging compared to the high resistance path through which it was charging till now. So I feel the capacitor should start discharging at this point of time instead of charging. Thus there will be no or less charging compared to the capacitor discharging. This will not allow equilibrium. Hope I got my point through :)

     
  • randyc
    randyc
    2013-05-25

    Hello again,

    Yes, I understand what you mean. The problem with that visualization is that the diode does not have a sharp threshold whereby it transitions from non-conduction to conduction, despite what one might infer from examing the characteristic curve of current versus contact voltage.

    The diode gradually commences conducting, behaving like a logarithmic potentiometer, not "snapping" into conduction like a UJT, a step-recovery diode or a thyristor.

    The steady-state solution of the current source + diode (assuming that the capacitor is lossless) is as I described previously: the current flowing through the diode is exactly equal to the current that the supply is able to source at equilibrium.

    The voltage across the diode (and the capacitor) is then the contact potential at that specific current, following the diode equation (with coefficients and constants appropriate to the specific diode used).

    I have no technical information with me (I'm on the road) but you should be able to substantiate the above with your own resources :)

    (As an aside, perhaps you could model your UJT circuit as an op-amp comparator. That may be a simple means of providing a precise trigger threshold, using ideal op-amps, which have infinite input impedance. The op-amp output could then be used to discharge the capacitor.)

    edited to add: maybe your equivalent circuit would function properly with an "ideal" diode, did you try that ? A perfect diode would seem to suit since it has a perfect threshold at which current flows. Just a thought ...

    Cheers,
    randyc

     
    Last edit: randyc 2013-05-25