555 timer inverter

Discussion in 'The Projects Forum' started by AdamM, Apr 18, 2010.

  1. AdamM

    Thread Starter Active Member

    Mar 7, 2009
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    Hello folks,
    My brother has been building a 12v DC to 110v AC inverter using a 555 timer circuit. I have been helping him troubleshoot it, and we've got a few questions.

    The schematic he is working from ( http://www.123circuits.com/555inverter.htm ) gives no suggestion of what transformer to use. Right now he's got the 12v AC run into the secondary of a 110 to 12v transformer. The main problem is we're only getting about 47 volts AC on the other side of the transformer.


    Is it normal that the voltage out would be this low? The 60Hz 12v AC going into the transformer is not very pretty (see the scope picture attached). Could the quality of the waveform be affecting the effectiveness of the transformer? Or is the input/output voltage ratio normally that different depending which side of the transformer the power's coming from?

    Any thoughts welcome!

    Thanks very much,
    Adam


    P.S. The oscilloscope photo shows what's coming out of the transformer. There is a very narrow peak right at the beginning of each side of the wave (almost invisible in the photo) that goes up to about 210v. Our meter reads 47v on the output, which I suppose is the RMS voltage (when the meter's plugged into the wall, it reads about 115).
     
  2. retched

    AAC Fanatic!

    Dec 5, 2009
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    Not very pretty is an understatement...Looks like my ex.. ;)

    that looks more like a dirty DC signal.

    Did you make this circuit EXACTLY? Which 555 are you using? And are you sure your PNP is working?
    Did you reverse connect the filament transformer (like you should have but wasnt listed in the 123circuit link you gave)

    Yours doesn't show R4 as a POT on the parts list..
    Check here (original circuit) and use the trouble shooter at the bottom of the page:
    http://www.sentex.ca/~mec1995/circ/555dcac.html
     
  3. Wendy

    Moderator

    Mar 24, 2008
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    That circuit has some flaws. The big one is the output swing will be about 1.2V less than power supply voltage, maybe less. What is your power supply voltage by the way? If you are feeding it with 5VDC, and the circuit is feeding the transformer 3VP-P, you will get 30VPP out.

    All standard 555 circuits have a 1.2 to 1.4V drop from Vcc, the ground side is good. Any transistor circuit will have to adjust for it.

    From the sounds of it, you may just have too low a battery voltage. Even with a 12V power supply, you will get around 10.6VP-P out. If you are using a 120VAC/12VAC (in other words, 10:1) you will have 106VP-P max.

    Have you read these?

    The 555 Projects

    LEDs, 555s, Flashers, and Light Chasers

    Bill's Index
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Looks to me like the transformer is going into saturation.

    Big difference between feeding a transformer a square wave or a sine wave.

    But, true sine wave inverters aren't very efficient.

    However, those that use PWM to generate a pseudo-sine wave are really quite good.

    To get an idea how those work, start looking at Class D audio amplifiers.
     
  5. AdamM

    Thread Starter Active Member

    Mar 7, 2009
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    Thanks for your replies! Retched, I'm not sure whether to pity you or your ex, but either way, thanks for the laugh :).

    The 555 is a Fairchild LM555CN. As far as I know the circuit is exactly according to the schematic except C4 which is 3300uF instead of 2700uF. We did find the original design from Tony van Roon (I kept finding broken links to his U of Guelph site). We've got a fixed resistor for r4, but the frequency coming out is pretty close to 60hz, so that shouldn't be an issue.

    The transformer is reversed. It is a 110 or 120VAC/12VAC transformer nominally, but when plugged into mains, which my meter measures at 117VAC, it puts out about 13.9VAC. So the ratio is about 1:8.4.

    He's running it off a 12v battery. The power going into the transformer is close to 12v.

    In theory, if we're putting around ~11VAC into the secondary, shouldn't we be seeing close to 90VAC coming out? SgtWookie's explanation that the transformer's going into saturation sounds plausible. In this case the fault would presumably lie with our lousy wave.

    Here's a line from the original instructions: "Capacitor C4 and coil L1 filter the input to T1, assuring that it is effectively a sine wave." But we're not getting anything like a sign wave out of it. In fact the whole filtering section (C4 and L1) seems to be doing exactly nothing. With the transformer disconnected, there is a nice 12v square wave both before and after C4 and L1 (see photo 2). When we attach the transformer, the post-filter waveform changes to what's in photo 1. There's a more pronounced spike at the beginning of both the low and high phases (if that's the right word), and more of a slope but still no sign of a sine wave.

    I went through Tony's troubleshooting list, but I don't think the issue is there. We're definitely getting a signal out of the 555, the transformer is reversed, and we're getting power through the transistors.

    A final theoretical question: how is this circuit supposed to produce AC in the first place? As far as I can tell, when the 555's output goes high the NPN lets v+ onto one side of the T1's coil. When the 555 goes low, the PNP connects that side of the coil to GND. But the other side of the coil is always connected to GND, so how can it be getting anything other than DC pulses?

    Thanks for your help,

    Adam
     
    Last edited: Apr 19, 2010
  6. Wendy

    Moderator

    Mar 24, 2008
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    The transistors are a dual emitter follower. Since each transistor drops 0.7V on the BE junction, and the 555 drops another 1.2V, you can have a max of 9.6VPP out the buffer. Actually it is worse, if the transformer loads it too much the voltage goes down even further.

    Running through the calculations it appears this circuit is meant to run at 120Hz, about double the transformers frequency.

    Figure 9.6VPP X 8.4 = 80VPP. 80VPP in a square wave is around 40V RMS. Sound familiar?

    A emitter follower has some self limiting properties, the input resistance is the gain of the transistor X the emitter resistance.

    http://forum.allaboutcircuits.com/album.php?albumid=26&pictureid=309


    What are you trying to accomplish?
     
    Last edited: Apr 19, 2010
  7. eblc1388

    Senior Member

    Nov 28, 2008
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    Its like saying my wife is effectively a Miss World. :)

    For sure you will not get a sine wave. I did a simulation using LTSpice and I'm getting pretty much the same waveforms as you are getting. So you are vindicated.

    Presumably the C4 & L1 & the transformer should form a series resonant circuit to let only 60Hz passing through but it depends a lot on the characteristic(inductance) of the transformer. Maybe for Van Roon's transformer it had really done the trick but apart from that I have no idea.

    Also I didn't understand from examining the schematic on how he had made the claim that it is "effectively a sine wave" in the first place.

    Perhaps someone can provide a detailed explanation of the schematic and put us all out of misery.

    P.S. Don't bother to ask the author himself about it. I saw this on his website so he might be very busy pursuing his dream, quote:
    "What am I currently working on? I'm working on and researching true water power, meaning, you stick your garden hose in the fuel tank and run on 100% water."
     
    Last edited: Apr 20, 2010
  8. Wendy

    Moderator

    Mar 24, 2008
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    :eek: :confused: :( :rolleyes: :D

    I showed the link earlier, but I guess I'll show the image. It will boost the voltage a bit. For 12V use 5V zeners.

    [​IMG]
     
  9. eblc1388

    Senior Member

    Nov 28, 2008
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    Yours NPN/PNP connection is totally different from the OP's circuit.

    If you look carefully, the NPN transistor in the OP's circuit is at the top to the Vcc rail and the PNP is at ground.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    Like L. Chung says, the characteristics of the transformer (H, R) will dictate what values need to be used for C4 and L1. As shown in Tony's schematic, L1 would have virtually no impact, as the transformer's inductance is likely in the mH range.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    If you really want a sine wave you will need a different approach. Most cheap inverters use this approach however.

    If you want more voltage for a true AC signal you will need to boost the transformer. You might be able to use a second transformer to increase the voltage, but it will double the current. Those transistors (on either design) will need serious heat sinking.

    If you read my earlier posts I'm eliminating the 0.6V BE drop, which is robbing a considerable amount of output voltage. This was discussed.
     
  12. AdamM

    Thread Starter Active Member

    Mar 7, 2009
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    Thanks again to everyone for your informative replies.

    It seems clear that the design we're using is less than ideal, and it also seems clear that we need a transformer with close to double the turns ratio of our current one. The root of the voltage problem has been pointed out by Bill, not so much in the output from the transistors (which is very close to 12v), but in the difference between RMS and peak voltage, which I'd failed to account for properly when I was testing. The peak output from T1 (aside from the spike at the beginning) is close to 100v. But with a 12v supply, we can never have more than about 8.4v RMS going into the transformer no matter how good the circuit is, if my math is right. So in order to generate 120VAC RMS from a 12v supply, we would need a turns ratio of about 14:1, not 1:8.4. The transformer from a 7VDC power supply should be about right.

    I think we might also try adding another pair of transistors so we'll have a full H-bridge, which will feed genuine AC to the transformer. I'm guessing that will be more efficient and effective than pulsing DC through a capacitor. Once we are getting a solid (albeit square) signal out of the transformer then we'll worry about the quality of the waveform.

    I've seen a design that uses one 555 to trigger another to get the alternating pulses needed to drive the h-bridge (in fact I thought that was what Bill's schematic was until I looked more closely). We have a few little PICAXE microcontrollers, and I'm wondering if it would work to use one of those instead of the two 555s. Eventually we could try sending PWM output from the controller on the beginning and end of each wave which, along with some small capacitors to smooth out the pulses, might work to generate a stepped square wave of sorts. Any thoughts on that scheme?

    What about preventing 'shoot-through' in the h-bridge? It seems like if we were using a microcontroller with 4 outputs available, we could split apart the control signals and add a slight delay between when one pair turns off and the other one turns on (separating the upper and lower halves of each leg of the H, if that makes any sense). How much of a delay do you think we would need for the power transistor to turn off?

    Bill, in the schematic you posted, what are the diodes for?

    I think before we start rebuilding we'll try hooking the thing up and bending the coil around with the scope on the output just in case we can hit some magic resonance point that results in a better waveform.

    Thanks for your help,
    Adam
     
    Last edited: Apr 25, 2010
  13. retched

    AAC Fanatic!

    Dec 5, 2009
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    That info should be in the particular transistors datasheet.
    The good news, when using a micro-controller you can easily change the delay time by the millisecond (and faster) by just changing a number. With the 555, it will be a mathematical practice and fairly tolerant resistors to get it good.
     
  14. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    For Bill: The diodes prevent the dreaded shoot through.
     
  15. Wendy

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    Mar 24, 2008
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    If you work through the voltage drops on my layout you will see (in theory) only one transistor should be on at one time. You pick diode values that make sure this is what happens. The transistors each drop 0.6V on the base emitter, and the 555 itself drops 1.2V (both are worst case). The remaining voltage voltage is divided in half, then add a little to guarantee that only one can be on at a time. Since the 555 outputs a square wave a large dead space is not a big deal.

    A minor negative of the design is it is also an inverter. Depending on the design this can be a positive, depending on what you needed..
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    The inductor in that circuit is much too small to be a filter. It is just another resistance that reduces the output voltage. If its value is high enough to filter 60Hz or resonate with the coupling capacitor then its resistance would be so high that the output voltage will be very low when the circuit is loaded.
     
  17. Wendy

    Moderator

    Mar 24, 2008
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    I tend to favor a PWM approach to sine wave outputs. The resemblance between a high power audio amp and a high power inverter is not so far apart, other than frequency response.

    Using simple PWM is not very hard, if you want to give it a go.
     
  18. Audioguru

    New Member

    Dec 20, 2007
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    I think you are talking about an efficient class-D switching audio amp since a linear amp wastes a lot of power making heat.
     
  19. AdamM

    Thread Starter Active Member

    Mar 7, 2009
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    Good day,
    My brother's rebuilt the inverter, using a PIC controller and full H-bridge. We're still having trouble with it though! The output from the bridge is good when there's no load, but when the transformer's attached it gets very weird.

    Since the project no longer uses a 555, I've started a new thread with a more appropriate title, and posted a schematic and some scope pictures. It's at http://forum.allaboutcircuits.com/showthread.php?t=39126. If anyone has a suggestion about what we could be doing wrong, we would really like to hear about it!

    Thanks for your help,
    Adam
     
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