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Need help undrstanding this circuit

  1. Sep 2, 2010 #1
    Hi all, this is my very first post on the forum... seems very cool. I do a lot of work with pinball machines and jukeboxes. These are really great fun b/c they are very much cricuit driven, but also have a mechanical aspect to them and dang, pinball is fun.

    anyway, I like to understand what I am doing, not just fix it and get it working.

    In the attached documentation, this circuit board is an electronic ballast that is used to power up 4 flourescnet light bulbs in a 100cd jukebox. the bulbs all stopped working one day. I got in there and pulled out the board. R1 was shorted and one of the fuses was blown. I have ordered the 91k resistor and plan to fix it. easy, ine.

    my question is, how does this circuit work??? I just can't make heads or tails of it.

    I believe 115vac comes into the board at J1-7, J1-3. seems like D1 and D3 are acting like a bridge rectifier (I know a bridge is just 4 diodes in a nice box). I essentially know waht the other components are and how they work, but I don't understand them in this circuit. and I'm totally unsure about what the chip U1A is all about.

    please help shed some lght. thanks, Marc

    Attached Files:

  2. jcsd
  3. Sep 2, 2010 #2
    The ballast basically converts 120V AC from the wall to 600V AC or DC.

    The first stage of the circuit converts 120V AC from the wall to DC voltage via the bridge rectifier.

    The generated DC voltage is then tapped by R1 to supply power to the chip U1 which is basically an oscillator, or AC generator which drives the two mosfet transistors - that is it switches them on/off at some frequency.

    The final stage of the circuit is a boost converter where the inductors are switched on/off by the mosfets and that induces higher voltage.

    The boost converter work something like this:

  4. Sep 2, 2010 #3
    hmmm, okay the mystery cloud is starting to lift what you say makes sense to me and I looked up the IR-2155 chip. lo and behold it says it is used a typical circuit that looks very much like the one I a dealing with.


    just to back up, what purpose does the C13, C12 and C11, C1 caps serve? I'm familiar with a capicitor being used after the bridge rectifier to smooth out DC voltage, but don't understand those there. I assume C2 is being used to smooth out the new AC voltage going to the inductors, but again if that is being toggled on/off very fast to produce the high voltage as the magnetic field of the coils collapse, why would there be a capacitor there?

    also not sure from the datasheet what the Ct, Rt, Vcc, Vb and Vs pins really are.

    again thanks for the help as you can tell, I'm no engineer just an inquisitive mind wanting to understand wtf is going on with my toys. thx, Marc
  5. Sep 2, 2010 #4
    C13 and C12 are 0.01 uF Caps, and they filter any high frequency spikes on the main AC line.

    C11, C1 are your typical bridge rectifier caps used to smooth out DC voltage. In this circuit they also are referenced to earth ground, could be the chassis.

    Ct, Rt set the frequency of this oscillator. If you change them, the frequency of oscillation will change.

    Vcc is DC power supply pin for the chip

    Vb and Vs just set the proper offset voltages needed to drive the mosfets.
  6. Sep 2, 2010 #5
    okay, so on c12, c13 if here is a spike the cap will just run the spike to ground, that makes sense. C1, if there is excess voltage or current it will go through C1, thus smoothing out the voltage heading to R1? still not sure why C11 is there.

    thanks for the explanation about the chip pins. that was very helpful.

    I guess what is still confusing me are the IRF720? how are those transistors working. seems like the bottom one has a leg to ground, but the upper one has a leg to Vs (the offset voltage). and is the base power comin in off the chip through R5 and R4? and what is the NPN switch closing.

    just a little more detail fo what doing by those IRF720s and I think it makes sense... thanks, Marc
  7. Sep 2, 2010 #6
    wait, I think it just clicked... will explain.
  8. Sep 2, 2010 #7
    the top line brings DC voltage to the top IRF720, if it is energized by HO from the chip, this will allow the voltage to flow through to the flourescents if the lower IRF720 is not energized, when that is energized that connects the DC voltage to ground and the inductor coil magnetic fields collapse, sending the high voltage to the flourescents???

    so is HO always powered on? and LO is the one that is rapidly pulsing?

    very cool.
  9. Sep 3, 2010 #8
    okay, this is a lot clearer now, but I am still unclear about what the capacitors are doing and how they work in the circuit.

    I read this and it sheds some light, but still not 100% clear.


    How would C12, C13 handle a voltage spike as mentioned in an earlier reply. Is C1, C11, C8, C9 and C10 being used for circuit protection too?

    I think C2 is being used to smooth out pulsing voltage going to the inductors???

    thanks, Marc
  10. Sep 3, 2010 #9
    There are generally two cases to consider when dealing with capacitors. The first one is when a capacitor is subjected to DC voltage. It will store charge, keep it (like a battery), and release it quickly if needed.

    The second case is when a capacitor is subjected to AC voltage. Now that's a different ball game. But in short, the capacitor will behave just like a resistor. The AC resistance, or reactance as it's called, depends on the value of the capacitor, but will vary at different frequencies.

    under DC: Capacitor = Battery
    under AC: Capacitor = Resistor

    The only circuit protection here are the fuses. The C12 and C13 handle high frequency spikes, not voltage spikes.

    Basically at low frequencies, say around 60 Hz they don't do anything because their reactance, or AC resistance is very high. But if you were to induce, let's say 100,000 Hz on the mains line, then C12 and C13 would look like resistors of very low value and SHORT the frequency spike - or noise - thus stopping it from getting into the rest of the circuit, and likewise, they prevent garbage noise generated by the circuit from leaking on the mains AC line.

    C1, and C11 on the other hand smooth out DC voltage from the rectifier. In this case they work more like batteries, rather than AC resistors, although in reality they are actually filter capacitors serving same function as C12 and C13 but at different frequencies.

    Hope that helps.
  11. Sep 3, 2010 #10
    C8 is just a DC smoothing for supplying DC stable power to the chip that was tapped off rectifier bridge

    C9 and the diode D2 is part of the function of the chip, and those are recommended in the datasheet given.

    In order to understand what they do exactly, you would have to study how the chip works and how it was designed. A typical engineer designing this circuit would usually go along with what the datasheet tells you, and wouldn't question it much. But a stupid manager would tell you to get rid of this extra cap that nobody knows what it's for to save on company's revenue, but then after millions of these product have been manufactured, they blow up.:eek:

    The only left ones are C2 and C10. At the second glance they are kind of mysterious to me too. This isn't your typical boost converter. But what basically happens is: the mosfets are switched alternatively - one is on while other is off - and this creates a switched current source and sink.

    When in current source mode (top mosfet on, bottom off) , a pulse wavefront runs through C2 (remember capacitors acts like resistors at AC) and energizes the coil.

    When in current sink mode (top mosfet off, bottom on), the current in the energized coil is sucked back in to ground by the current sink while higher voltage develops.

    This is repeated on/off at some frequency, I don't know what they run like, could be 1000 times a second.

    And C10 along with R3 just maintain correct voltage levels.
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