Electrolytics in design

Discussion in 'General Electronics Chat' started by justininmilwaukee, Jul 4, 2013.

  1. justininmilwaukee

    Thread Starter New Member

    Apr 13, 2012
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    Having a hard time understanding a couple circuits I ran across. Basically, I have a 555 timer's output driving a transistor. In between that there is a 100uf capacitor. I am still new, but...I thought an electrolytic is to only be used as storage basically. Not in series with something? If the 555 signal goes into the cap. The cap would fill and then never pass the (Electrons/Voltage) to the transistor. Wouldn't you want this cap to be omitted? That cap would have to be an ac style (ceramic) would it not? My thought is that one is making sure the transistor does not "stick" on by using that cap. My confusion is that the cap fills then can only discharge back out the way it came in. Can't come out the negative side to the Capacitor? What the heck am I missing I Google the hell out of this I don't get it. http://lifters.online.fr/lifters/labhvps/index.htm I know that a capacitor acts like a short at higher frequencies due to the reactance being less. Looking for the more in depth answer. The electrons move into the cap. The 555 goes low. The caps negative side would never see more electrons. Therefore they would never go into the transistor? What the heck am I not getting?
     
    Last edited: Jul 4, 2013
  2. Shagas

    Active Member

    May 13, 2013
    802
    74
    A capacitor in series blocks the DC but let's through the AC.
    A capacitor passes current through until it gets charged, then it stops.
    Putting a cap in series is very desirable in cases when you only want the ac element of your signal to come through and block the dc .
    Consider a 4 volt DC signal that has a 1volt Ac signal on it . If you connect it in series with a capacitor then on the output you will only have that 1volt ac signal.
    That's called 'capacitor coupling'
     
  3. Shagas

    Active Member

    May 13, 2013
    802
    74
    Concerning part 2 of your question :
    'That cap would have to be an ac style (ceramic) would it not?'
    I'm not certain but i'm pretty sure that this is the explanation:
    If you would be using an output that is Bipolar , for example a sin wave that goes from 3volts to -3 volts then you would use a ceramic or film cap . The reason being is that they are Bi-polar is that the go both ways , like Freddy Mercury .
    When you have for example a sin wave that goes from 0 to 6 volts and has its 'center' at 3 volts then you can use a unipolar cap like a normal electrolytic.

    Consider the following : Your output is at zero volts and you have it connected to X through an electrolytic cap(positive to output and negative to X!!).
    X will be at the same voltage as your output because the cap is completely discharged so there is no voltage drop across it .
    As the output starts rising , then the cap starts charging and a voltage starts developing across it . So X in related to the output will have a voltage diffrence relative to the voltage across the cap.
    The only Thing you can't do with a normal Unipolar electrolytic cap is have the voltage at the positive terminal more negative than the negative terminal , and if your signal is positive and never dives negative then there is no issue.

    I hope I'm not just ranting sh*t of the top of my head , can someone more experienced confirm my theory ?:D
     
    Last edited: Jul 4, 2013
  4. #12

    Expert

    Nov 30, 2010
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    I'm happy to see you at this level of thinking. It's an excellent way to polish your skills!
    Using a cut down version of the schematic (presented below):

    The diode is the first thing to consider. It allows a positive voltage on the drive line to the base of the transistor but dumps any attempt at a negative voltage on the base of the transistor (give or take about half a volt). The timer can only provide a positive voltage, therefore, electrons must be getting sucked up through the transistor (emitter to base junction) and through the 1k resistor.

    When the timer output goes positive, 12 volts (-.6Vbe)/10 ohms = 1.14 amps (max) in electron flow toward the timer. The timer can not handle this amount of current. The NE555 has a limit of 225 ma, so the 10 ohm resistor (mentally) becomes almost irrelevant.

    In the first instant, the timer side of the capacitor becomes +12V and the other side becomes .6 volts. This is a pretty fast circuit compared to the capacitor value, so I'm going to ignore the "correct" formula for time/resistance/capacitance and use V = (Time x I)/C as if this is a constant current event. The voltage (11.4V) across the capacitor lasts for 33&1/3 microseconds and the current must be .225 amps or less. That means the capacitor charges to .075 volts (max) while the timer output is, "high" and 7.5 micro-coulombs of current has flowed (flown?) through the capacitor.

    Meanwhile, the transistor allows current through the transformer winding, etc.

    Now the timer output drops to 0 volts with the capability to allow 225 ma of electrons to flow. Flow where? They just got sucked up out of the ground line and are piled up on the right side of the capacitor, so they must want to go the other way now. Ground is more negative than +12 VDC, so the ground potential at pin 3 (output of the timer) must be repelling them. Where is the exit? If the voltage amounted to a few tenths of a volt, they would escape through the diode. You only have .075 volts on the capacitor, so the electrons must be escaping through the resistor. If you want to use the "real" formula, it is at: http://forum.allaboutcircuits.com/blog.php?b=485

    For this moment, I calculate that if the voltage on the capacitor never changed, .075 volts/1k times 33&1/3 usec = 2.5 nano-coulombs, but we have 3000 times that much charge on the capacitor. It will NOT discharge to zero volts through the resistor in 33 u-seconds. It will not discharge to zero volts through the diode.

    As this cycle repeats, the charge on the capacitor increases until the diode can be used as an escape route. Now we can predict that the drive line will arrive at a condition where one limit is the diode forward voltage to ground (.8V @ .2A) and the other limit is the Voltage (emitter to base) of the transistor.

    We have now arrived at the graph provided (at your link) showing the positive range of the voltage on the drive line to be .6 Vbe in the positive direction and .8 Vforward (of the 1N4007) in the negative direction.

    You're welcome.
     
    Last edited: Jul 5, 2013
  5. justininmilwaukee

    Thread Starter New Member

    Apr 13, 2012
    3
    0
    Thank you very much guys. I am rereading again and again trying to digest it all. I will get it after a few more times. Exactly what I was looking for fundamental to the core.

    I just built this board...almost works...Doesn't make sense but the 4k7 resistor with the led that lights when there is power on burnt up. I am slightly higher on the voltage 51vac on tx around 70 vdc after bridge. But, thats still only 70/4700= .014 amps ? That resistor has nothing on it...figured maybe too small of a resistor...bumped it up to a 2 watt still burning up...bumped up the 4k7 to a 33k now it is ok. .014 amps and it's hot. At 33k thats .002 amps. The 4k7 potentiometer was doing the same thing...?

    Resistors to my knowledge are rated for watts meaning just do your ohm's law calcs and you should be ok. My only thinking is that resistors do have a voltage limitation at that 70 vdc?

    Thank you again so much for the great responses.
     
  6. #12

    Expert

    Nov 30, 2010
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    You must have a mistake in the wiring. The LED circuit is dead simple and couldn't possibly be acting that way if you had it wired right.

    For voltage limits on resistors, you can look them up on a vendors site. Over 100 volts is probably right for a 1/4 watt resistor, and it gets higher from there. R1 has 1 watt of power on it. A 2 watt resistor is a good thing.

    The 78L12 is in danger. It has a 35 volt input limit. If you turn up the voltage (P1) and there is no load on the 12 volt supply, the regulator will smoke. You might want to redesign that section. Put a 1k in parallel with C3?

    At 70 volts, you're putting more than 1 watt into P1. Check your wattage rating on P1.
    C2 is the wrong size, according to the datasheet for the regulator.

    Now that I'm paying attention to the 78L12 I see that it has an advertised current limit of 100ma. The truth is, it will pass 150ma to 180ma under certain conditions of low voltage difference and low heating. That is a limit on how much current the 555 chip can supply. R2 is also limiting the drive current. Judging from the oscilloscope view, the timer must be finding .2 amps available somewhere. The 100nf capacitor will empty to about 4 volts if it supplies .2 amps for 33 usecs, but C2 also contributes. The 78L12 datasheet requires that C2 be .33uf to avoid oscillation problems, but changing it to .33uf will change the current available to the timer chip. Changing C2 to the "required" size might have a bad effect on the drive current. R2 uses up all the available voltage at about 39ma. This is a very tightly designed dance.

    ps, some of these problems are about too much wattage. Watts Law: P = IE
    You need to calculate the watts on the parts with this formula to see if they are going to smoke.

    Here are some ideas:
     
    Last edited: Jul 6, 2013
  7. Wendy

    Moderator

    Mar 24, 2008
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    I suspect the capacitor / diode could be eliminated entirely, since the transistor uses DC drive and the 555 goes to a good level to turn the transistor off.

    Been a long time since since I've seen a lifter. They are interesting devices, as they generate moving air with no moving parts (corona discharge). Be careful, the high voltage is dangerous.
     
  8. #12

    Expert

    Nov 30, 2010
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    Thanks for the comment, Bill. It's good to know somebody is double checking for me. (I DO make mistakes!)

    Without the capacitor, you don't have to dump the accumulated charge on the capacitor (.2 amps for 33 usecs), just stop supplying it. So, yes, it might work without the capacitor or the diode. I can see why they exist. They don't just assure that the "off" voltage is only a few tenths, they actually reverse the polarity of the base drive voltage. That sucks any lingering base-emitter charge out quickly and sharply. The original designer might have found this important to square up the shape of the drive current to the transformer.

    I'm sure the original designer had a 'scope, and used it. There is no way I'd be sure of how the 78L12 is working without a 'scope! That's why I'm tip-toeing around C2. That capacitor makes an important contribution to the drive current and I can't see how it affects the circuit without a 'scope.
     
  9. Mike33

    AAC Fanatic!

    Feb 4, 2005
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    Another point and the original "cap in series" concern: if you're worried about using an electrolytic cap in series with an AC circuit, you can use TWO, facing opposite directions (in parallel)...each one should be about HALF the value of the cap called for, since caps in || with each other add. Maybe 220u in this case. Or just get a non-polarized electrolytic cap of the right value, they're only 10 cents :eek:)

    I've never had an issue with this...just that down the road, in a permanent circuit, that cap could fail (the dielectric gel may get dried out due to working outside its design parameters, etc). If you're planning something and ordering parts, better to use non-polarized, IMO.
     
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