variable inverter

Discussion in 'General Electronics Chat' started by spankey666, Dec 16, 2011.

  1. spankey666

    Thread Starter Member

    Nov 30, 2011
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    3
    Hi all,
    i have built an inverter based around a ne555 timer. i now need to make the output voltage variable. i have tried the circuit below, but cannot seem to get it to work. am i barking up the wrong tree ? is using an lm317 with a transformer not the way to do it ?
    any help will be greatly appreciated.
    [​IMG]
     
  2. jimkeith

    Active Member

    Oct 26, 2011
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    99
    The concept looks workable.
    The first thing you need is a clamp diode across the transformer primary to keep from destroying the transistor.
    The output will now be in low voltage pulses that may not be obvious unless you get a scope to trigger on them.
    The 3N3055 may have poor gain at low current--select its input resistor for the maximum sourcing current of the 555.
     
  3. spankey666

    Thread Starter Member

    Nov 30, 2011
    91
    3
    ive been running the circuit minus the lm317 (transformer connected to +ve) and it works a treat. i will put in a diode to stop the back emf. ive been using a bench top power supply and by varying the voltage from 3-7.8v get the required voltage range from the secondary . but when i put in the regulator it doesnt play ball. wondering if it doesnt like running an inductive load .
    I want to make the unit self contained to run off a 9v wall wart.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    If you simply use a diode to suppress the back-EMF from the inductor, you'll have a problem; that will cause the inductor current to recirculate through the diode and the field will collapse very slowly; so slowly that the current through the transformer will build until saturation occurs.

    When saturation occurs, the transformer will look like almost like a dead short, and things will get hot very quickly.

    Instead of a plain diode, you need a Zener back-to-back with a diode. The Zener causes the current through the primary to stop rapidly, so the magnetic field collapses rapidly, transferring power to the secondary winding - which is what you're trying to do to begin with.

    If you use a MOSFET that has an avalanche rated body diode like an IRF510 or IRFZ34, you don't have to worry about using a Zener, as the MOSFET will take care of it.

    There is no point in regulating the input to the transformer. Put a cap across the winding to the driving transistors' emitter.

    One thing you can do is use the output voltage to control the CTRL input to the 555 timer.
    Take a look at Ronald Dekkers' "Flyback Converter for Dummies" page:
    http://dos4ever.com/flyback/flyback.html
    It's a great resource.
    Look at Figure 16, about 3/4 of the way down the page.
    Note R3 thru R6 and T1; how they are connected to control the CTRL input. As voltage on the base of T1 gets to the point where T1 starts to conduct, CTRL is pulled low, which increases the frequency of the 555 timer, and also decreases it's duty cycle. The decrease in duty cycle causes the output from the transformer to decrease.

    In your circuit, you will want to replace R4 with a much lower value, otherwise your output voltage will be quite high.
     
  5. spankey666

    Thread Starter Member

    Nov 30, 2011
    91
    3
    thanks, looking at the circuit, would not just putting a pot between pin 5 and ground acheive the same result ? or am i missing something ?
     
  6. crutschow

    Expert

    Mar 14, 2008
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    You are. Your scheme has no negative-feedback regulation from the output and thus the output voltage will change significantly with a load change.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Take another look at Ronald Dekker's page. Once voltage across the resistive divider of R4, R5 and R6 has reached high enough, the voltage at the wiper of R5, therefore the base of T1, will have exceeded the cutoff voltage of T1. T1 will then start conducting current, which will bring the CTRL pin low, reducing the PWM percent of the 555's output. If the voltage across the divider falls, then T1 stops conducting, which causes the PWM percent of the 555's output to increase, causing the voltage at the output to increase.

    The divider network controlling T1, which controls the voltage on the CTRL input, causes the circuit to be self-regulating.

    If you just used a pot on pin 5, there would not be active regulation; there would be no feedback from the output voltage to the CTRL input.

    Carl gave it to you "in a nutshell", but you might need more of an explanation than he gave.

    Do you understand it now?
     
    spankey666 likes this.
  8. spankey666

    Thread Starter Member

    Nov 30, 2011
    91
    3
    another quick question. i want to make the circuit so that i push a button and the inverter comes on until the output reaches the level controlled by the fly back controlled by vr1
    [​IMG]
    will i have an issue with connecting the 2 transistor bases together so that vr1 controls both circuits ?
    the circuit has been drawn simply and the values of components are incorrect but i think you should get the idea :)
     
  9. crutschow

    Expert

    Mar 14, 2008
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    I don't understand this circuit.

    What is the purpose of the inverter?

    Where is the flyback transformer and rectifier?
     
  10. spankey666

    Thread Starter Member

    Nov 30, 2011
    91
    3
    i left them out of the circuit because this was a continuation of the above thread and sgtwookies link. its assuming its there and the high voltage is on the rhs side of the drawing. vri / Q3 is controlling pin 5 of the ne555 of the inverter circuit which is the fly back control, ic1 is a latching circuit which starts by pressing sw1 and is reset when Q2 is switched on when the voltage from the flyback circuit is reached. my lack of understanding transistors is why i ask the question. "can the bases of 2 transistors be connected to supply different parts of the circuit using 1 voltage source without any side effects ?" please bear with me on this, the last time i picked up a soldering iron was 35 odd years ago when 555's and 741's had only just become easily accessable :( so trying to remember this stuff from back then is difficult. the circuit is to charge capacitors for an extreemly large flash tube ,
     
  11. crutschow

    Expert

    Mar 14, 2008
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    You can connect the two transistors to the same source if you make allowance for the additional current through the control pot resistance from the additional transistor base current.
     
  12. spankey666

    Thread Starter Member

    Nov 30, 2011
    91
    3
    which means doing what ? in laymans terms . :) if ive got say 1.8v coming from the pot centre, adding the second transistor is likely to do what? just so that i can visualize what to expect. it took ages to get the output voltage between the ranges i required (resistor tolerances etc) but i think adding the latching part would make the unit more user friendly.
     
  13. crutschow

    Expert

    Mar 14, 2008
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    The pot and its associated resistors generate an equivalent resistance at the pot wiper. This will change the pot voltage, depending upon the amount of current drawn from the pot wiper by its load. So you want the load, consisting of the two base resistors, to be much higher resistance than this equivalent resistance (say at least 10 times) so the pot voltage is not significantly affected.

    You can Google equivalent resistance if you don't understand that.

    Is that layman's enough for you?;)
     
    Last edited: Dec 28, 2011
  14. spankey666

    Thread Starter Member

    Nov 30, 2011
    91
    3
    yes thankyou ;) i'll do some reading up
     
    Last edited: Dec 28, 2011
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