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Rain Sensor Status

Discussion in 'Sensors and Actuators' started by chopnhack, Sep 9, 2014.

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  1. chopnhack

    chopnhack

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    Still thick, LOL - haven't wrapped my head around it yet. I am so used to having a reference from residential wiring where neutral is just the return path and the "hot" is the phase side. You are saying that if I connected the neutral to D1's anode the circuit would still work?

    I produced an updated layout and etched another board today. One minor glitch with a trace which was easily fixed - attributed to the degree of difficulty in producing parts in Eagle :mad: and my not catching it, again.. :oops:

    I had to substitute a 1N4148 for the 1N914. From what I read that is an acceptable substitute with the 1n914 being a more popular part from some years ago. Is this ok? I looked at the specs and the most important were the same. I broke it while soldering.... never use a hemostat to hold parts to a board! LOL
     
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes, 1N914 and 1N4148 are kind of interchangeable. I think the 4148 is normally rated for 200 mA and the 914 for only 100 mA but I think it depends on the manufacturer anyway. Not sure.

    Right, well I think it comes down to what you consider the reference point for voltages. It's true that in house wiring, Neutral is connected to Earth and therefore it's considered the reference point. The instantaneous voltage on the Phase line varies cyclically from zero, to a positive peak of about +163V (assuming 115V AC RMS), back through zero, to a negative peak of about -163V, and back through zero again.

    But this is also the nature of the voltage BETWEEN the two wires. If you took Phase as the reference point, you would see the voltage on Neutral varying cyclically in exactly the same way: from 0V, to +163V, back through 0V, to -163V, and back up through 0V. The only difference between the two ways of looking at it is the timing of the voltage. When the P-N voltage peaks at +163V, the N-P voltage peaks at -163V, so they are 180° out of phase, or inverted, from each other.

    With circuitry that's directly connected to the mains supply, if the 0V rail is connected directly to one of the mains input connections, it's a very good idea to ensure that this is the Neutral line rather than the Phase line, because that means most of the circuitry is at safe voltage. From the circuit's point of view, there's no difference, but it' safer if the 0V rail of the circuit is actually at earth potential. But you have to fully isolate both sides, and the whole circuit and anything connected to it, because it's not safe to assume that it will necessarily be connected that way to the mains. At least, that's true here, and we have had polarised plugs (with earth connections) for a lot longer than the U.S. has.

    But in the case of a low-voltage circuit, it really doesn't matter, as long as all inputs and outputs are isolated. And that's the case with your control system. It could be that neither side of the 29V AC supply is connected to earth. We don't know, and we don't need to know, as long as the inputs and outputs are isolated. Output isolation is provided by the relay, and input isolation is assumed because the contacts of the rain sensor are not connected to anything apart from that circuit. The aren't, right?
     
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  3. chopnhack

    chopnhack

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    I understand your explanation and perhaps if I had an o-scope I could experiment to make it clearer. From observation with DMM, in the US, Neutral is the return for Hot. Neutral to Ground only shows zero potential on the DMM. That is why I thought all the cycling between positive and negative voltage values occurred in the phase line solely and returned to source on the neutral. With a DMM, phase to ground shows 120v, neutral to ground shows 0. I wish I could see what you are explaining in real life as my experience with neutral colors my view of it. Thanks!

    The rain sensor switch on the roof is isolated - the two leads come from the switch through wires directly into the circuit, when closed they pass current to the relay and then goes to "neutral".
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That's right. But from the point of view of a fully isolated circuit, there is just 115V AC RMS BETWEEN Phase and Neutral. Which one is the reference point is irrelevant. The circuit COULD be made (or connected) so that its 0V rail was connected to Phase, then from its point of view, it would be the Neutral connection that would be alternating between +163V and -163V, because the circuit sees things with reference to its 0V rail.

    As I said before, the circuit would still work, but it would probably be unnecessarily dangerous, and it might be disturbed by noise and interference on the Phase connection, because the whole planet would effectively be radiating that noise (from the circuit's point of view) and there could be significant capacitance involved. Of course there can also be noise on the Neutral line as well, so that argument isn't necessarily meaningful I guess!
    And so it should. You're right, Phase is "live" and Neutral is "neutral'. But that's because Neutral is the one that's earthed, and we consider earth to be a reference point.

    But in this case we're not even talking about Phase and Neutral; we just have two wires coming from the secondary of a 60 Hz transformer, with about 28V AC RMS between them. One wire, or the other wire, or neither, could be connected to earth. The circuit doesn't care. From its point of view, those wires are interchangeable. They are only defined by the voltage BETWEEN them, and you can measure that on wire B relative to wire A, or wire A relative to wire B.

    It's true that one wire corresponds to Phase and one to Neutral in the sense that in the part of the mains cycle when Phase is positive relative to Neutral, one wire (say wire B) will be positive relative to the other wire (wire A). So if your circuit cares about the phase or timing relationship between its low-voltage AC input supply and the mains supply, you need to keep track of it. But that circuit doesn't care; it just uses the 28V AC as a power source.
    Right, good.
     
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  5. chopnhack

    chopnhack

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    I am a little further in my thoughts on AC - phase and neutral are 180 deg. out of synch with neutral carrying the opposite of phase (and vice versa). Then technically speaking if the above is true, shouldn't one get shocked if they completed the circuit between neutral and ground? It certainly is the case when you ground the phase wire!

    As for the 28VAC example, I am having difficulty envisioning the pathways each phase should take. I outlined the pathway if the indicated input is the phase in blue and what I think would happen if the phase was on the other pin (the ground plane) in green. Where there is a solid colored line, either blue or green that is where I think power stops, either is blocked by a diode or because of polarity.

    upload_2014-10-21_15-3-38.png

    Here's a blank if you want to scrawl out directions for this wayward traveller :)

    upload_2014-10-21_15-7-48.png
     
  6. chopnhack

    chopnhack

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    And in other news: Success!
    Thank you ever so much Kris, Harald, Steve and Dave!!
    A wired, wireless Hunter Rain Sensor :D :p LOL
    upload_2014-10-21_15-12-25.png
     
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  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Fantastic! Congratulations :)
     
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  8. chopnhack

    chopnhack

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    Thanks again Kris :) Would you be able to draw some arrows on post #125 when you have time to help me visualize current flow during the cycles of the mains? If you can, it would be greatly appreciated.

    The sensor is working well! I was quite surprised by the dry out time - it was quite short, I need to fiddle with the louvers on the roof mounted switch to see if I can increase the time it takes to dry.
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    upload_2014-10-21_15-7-48 annotated.png

    The green path is the current flow from the AC input, through D1, charging C1. That current keeps C1 charged, and from that point on, it's simplest to think of the rest of the circuit as being powered from C1. The blue arrows show the first part of the current flow from C1, and the return current from the relay coil and the LEDs flows back to the negative terminal of C1.

    Sorry, I must have lost track of your questions in post #125.
    I think there's something fundamental wrong in your understanding of voltage. You need to first understand that voltage is a potential DIFFERENCE. It is ALWAYS measured BETWEEN TWO POINTS.

    You can say that any point in a circuit has "a voltage on it", but that voltage is always measured RELATIVE to something else. In a typical circuit, this reference point is the 0V ("zero volt") rail, or sometimes (and inappropriately, in my opinion), "earth", "ground" or GND.

    I wish I could do animations. I could explain this so much better. But you'll have to use your imagination.

    Imagine a see-saw sitting in a ditch, so each end can be above or below the ground level. We're only interested in one end. Now, if that end is moving up and down, so it's sometimes above ground, and sometimes below ground, that is analogous to the Phase or Hot or Live connection on the wall socket. The voltage at that point is alternately positive then negative relative to ground.

    But if you sit on that see-saw and focus at a point on the ground, from your point of view it appears to be going up and down as well; sometimes it's above you, and sometimes it's below you. And when someone at ground level says the see-saw is above them, someone at the see-saw level will say the ground level is below them, and vice versa.

    This is what I meant when I was trying to explain the 180° phase difference between the voltage at Phase measured relative to Neutral, and the voltage at Neutral measured relative to Phase. They are both AC, they both have the same amplitude; in fact they ARE the same voltage, but they are half a cycle (180°) out of phase from each other.

    The point of that explanation is to try to show how important it is to understand WHAT you are measuring a voltage RELATIVE to, or WITH RESPECT TO, i.e. the 0V REFERENCE POINT against which the voltage is measured.

    In the case of mains voltage, Neutral is connected to Earth, and Earth is the default reference point. If you touch the Neutral wire, you won't get a shock, but you will if you touch the Phase wire. The reason for this is that you are standing on a floor which is (roughly) at earth potential. There is a resistive and capacitive path from the floor to the soil under your house, and if you touch the Phase wire, current will flow through that path.

    To be exact, the complete path for this current will be:

    From the Phase side of the winding on the pole transformer that provides AC to your house; through the wiring to your house; through the main switch on your switchboard; through the fuse for that circuit; through the Phase wire in the wall wiring; through the switch on the outlet; through the Phase connection on the socket; through your skin; through your finger, hand, arm, trunk, and legs; through the soles of your shoes; through the floor and other wooden parts of your house; through the soil under and around your house; through the earth stake; through the earth stake wire; through the earth busbar in your switchboard; through the link that links the Neutral busbar to the Earth busbar; through the Neutral busbar; through the Neutral wire from your house to the pole transformer; and into the other side of the secondary winding on the pole transformer.

    The mains voltage coming from the pole transformer is actually connected between Phase and Neutral. Earth is connected to Neutral and is essential for safety, of course, but current is not supposed to flow through it. The current consumed by a load always flows between Phase and Neutral; if any current flows through Earth this is normally a fault or a problem, and will trip an ELCB/RCD or whatever you call it.

    So Phase is the dangerous terminal on the wall socket. Now, if you built a second wooden room inside your room, and insulated from it, and put a wall socket on that room, and connected all of that wooden construction to the Phase side of the mains, when you stood inside that room, you wouldn't be aware of anything unusual. Your floor, walls and ceiling, and your shoes and your body, would all be at the same potential. Just like birds that perch on live wires, you wouldn't feel anything unusual. But to you, the Neutral and Earth pins on the wall socket would be dangerous. If you touched either of them, you would get a shock. And if you touched the Phase connection, you wouldn't feel anything. So it all depends on your reference point.
    Within that circuit, everything is exactly like I just described except the voltage is a lot lower. So you don't need to worry about things getting dangerous if the AC is connected "the wrong way". In fact there may not even BE a wrong way, since most probably, neither side of the secondary of the wall adapter for the irrigation controller is connected to Earth. That whole lot of circuitry, including everything inside the controller and all external wiring to the pumps, could be fully floating relative to earth. Kind of like a see saw suspended below a hot air balloon.

    The important fact is that there is 28V AC RMS ACROSS or BETWEEN the two AC input connections to that circuit. If you look at the voltage at one terminal relative to the other terminal, it is varying between about 40V positive and about 40V negative. Looking with the terminals swapped, the voltage is the same but 180° out of phase from what it was before. That's irrelevant because we're just converting it to DC and using the DC.

    So just look at the subcircuit consisting of (a) the 28V AC RMS supply, coming in on two connection points which I've called L (left) and R (right), (b) diode D1, and (c) electrolytic C1. I've drawn that circuit in two forms, which are electrically identical except for where I've put the 0V triangle marker.

    Here's version A, which is as drawn on the schematic. In this version, L is the 0V reference for the circuit, and the groundplane on your layout.
    ps two points of view 1.png

    The voltage between R and L (measured EITHER way round) alternates smoothly from +40V to -40V. At a certain point in time, R is becoming more positive than L. Let's assume that C1 has discharged to about 38V. Once R becomes about 38.6V more positive than L, D1 will start to conduct. R will continue to get more positive than L, and current will flow through D1, increasing the voltage across C1.

    Eventually that half-cycle will reach its peak instantaneous voltage of 40V (positive at R, relative to L). C1 will have been charged to about 39.4V. Now R will start to become less positive relative to L, and D1 will stop conducting. The voltage at R, relative to L, will become less and less positive, until it crosses zero, and R is now negative relative to L. During this time, D1 will be reverse-biased and no current will flow. The load current will gradually discharge C1 until the next positive half cycle comes around with R reaching its positive peak relative to L.

    So version A charges C1 to about 38V and produces a DC supply across the output terminals marked + and - which can be used by the rest of the circuit. The rest of the circuit is fully isolated from everything, so the way in which it connects to the 28VAC supply voltage is irrelevant; all it cares about is that it gets its 38V DC supply.


    Here's version B, which is shown with the right hand pin, R, as the 0V reference for all voltages within the power supply circuit. I will be describing all voltages as being measured relative to R.
    ps two points of view 2.png

    The voltage between R and L (measured EITHER way round) alternates smoothly from +40V to -40V. At a certain point in time, L is becoming more negative than R. Let's assume that C1 has discharged to about 38V. Once L becomes about 38.6V more negative than R, D1 will start to conduct. L will continue to get more negative than R, and current will flow through D1, increasing the voltage across C1. (The polarity of this voltage always matches the polarity of C1.)

    Eventually that half-cycle will reach its peak instantaneous voltage of 40V (negative at L, relative to R). C1 will have been charged to about 39.4V. Now L will start to become less negative relative to R, and D1 will stop conducting. The voltage at L, relative to R, will become less and less negative, until it crosses zero, and L is now positive relative to R. During this time, D1 will be reverse-biased and no current will flow. The load current will gradually discharge C1 until the next negative half cycle comes around with L reaching its negative peak relative to R.

    So version B also charges C1 to about 38V and produces a DC supply across the output terminals marked + and - which can be used by the rest of the circuit. The rest of the circuit is fully isolated from everything, so the way in which it connects to the 28VAC supply voltage is irrelevant; all it cares about is that it gets its 38V DC supply.


    No matter which side of the 28VAC supply you regard as your reference side, the circuit behaves the same. It produces 38V DC at its output to supply the rest of the circuit. It uses only one half of each cycle of the incoming AC supply (it's a half-wave rectifier) and it doesn't matter whether that corresponds to the half-cycle where mains Phase is positive relative to mains Neutral, or mains Phase is negative relative to mains Neutral, because the mains phase, relative to this circuit, is irrelevant.

    Does that help?
     
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  10. hevans1944

    hevans1944 Hop - AC8NS

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    Well, this has been a very interesting thread. As I was reading it, I was wondering when someone would suggest a relay solution. And of course @KrisBlueNZ did just that.

    You just keep on trucking @chopnhack! You are definitely making progress. I like the way you are not afraid to try things out, even if it means things sometimes blow up.

    So here is another story, a blast (literally) from my past: One evening my roommate and I were sitting around in our shared barracks room playing around with a component I had "liberated" from our squadron's bench-stock supply. I had built a power supply and was using this component as an output voltage ripple filter. The component consisted of two electrolytic capacitors, each packaged in rectangular aluminum cans, both held together between two short pieces of stamped heavy-gauge aluminum secured with four longish screws and nuts. There was an insulated terminal coming out of the top of each capacitor and these two terminals were what I used to "filter" the power supply. I am not sure what happened next. I think I may have decided to connect the two terminals together, thus creating two capacitors in parallel for better filtering, with the aluminum clamps serving as the common terminal. Originally they were probably designed to be used in series, so as to allow AC operation with polarized electrolytic capacitors. Anyway, this whole rig was sitting on my bottom bunk bed, turned on, while my roommate and I sat in chairs having a conversation. We were discussing who knows what (usually something to do with cars since he was a hot-rod nut) when there was a huge bang! The capacitors (both of them!) had exploded! I guess I had failed to exercise due diligence in determining their maximum working voltage. Neither one of was hurt by the flying aluminum shrapnel, but I didn't mess around with that particular item of bench stock again. What you survive makes you smarter. Or so they say.
     
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  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I only have one interesting capacitor story. Back in the 80s when I was doing my apprenticeship at a consumer electronics repair company, a younger apprentice asked me to help with a switching power supply board from a TV set he was repairing. He powered it up and showed me how it was dead, and examined it closely, sniffing for smoke and listening for the whistle from the main transformer. He straightened up, and a second later, BANG! The output electrolytic exploded. It wasn't a big one - about 2 cm diameter and 6 cm tall, but if he had had his face buried in that power supply, he could easily have been disfigured and/or blinded!
     
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  12. chopnhack

    chopnhack

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    LOL, immeasurably! I think I got it. I was thinking at one point that neutral was PROVIDING the opposite phase... So, if I understand correctly your excellently detailed post - voltage is measured between points (i thought I understood that from earlier posts), current flows between these points and in version A of that small circuit, the top rail alternates between pos and negative values of potential TO ground. So that when the top is positive or more positive than ground, current conducts forwards through the diode, charges the cap and after the cap is charged, the charge there is sufficient to repel and thus it conducts further down the line. (Green Arrow)

    When it is more negative than ground (Yellow Arrow), no current flows because the diode prevents current from coming UP from the more positive 0V rail and through the cathode? correct??

    If this is so, wouldn't the polarity of the cap also have an effect? Would the the negative bottom of the cap be suddenly discharged by the now more positive 0V rail? Wouldn't this normally destroy an electrolytic?

    I realize that I am drawing potential arrows again instead of current!! Still trying to learn and visualize current instead of voltage. From a young age it was my common misconception that voltage from batteries was the power in toys, motors, etc. All my tinkering at a young age was typically with 9v and AA batteries so it was an easy mistake to make.

    upload_2014-10-26_8-39-52.png
     
  13. chopnhack

    chopnhack

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    Thanks guys, I appreciate your help an encouragement! And I will certainly keep on trucking :)
     
  14. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes... But forget ground. The AC voltage appears between Phase and Neutral. Ground happens to be connected to Neutral, but it is not used for anything other than safety.

    Because Neutral is grounded, it makes sense to use it as the reference point when you're describing the voltage on Phase, which varies smoothly between ~160V postive, and ~160V negative, relative to Neutral and Earth. But as I explained with the see-saw analogy, and the room-inside-a-room analogy, it's just as valid to think of Neutral and Earth as varying smoothly between ~160V positive, and ~160V negative, relative to Phase. It all depends on your reference point, and your point of view. Neither one is right or wrong; using Neutral is just conventional, because Neutral is the side that's connected to Earth, and the soil is a natural thing to use as a reference point.
    I don't use terms like "charge" and "repel". A diode's behaviour is very simple. If the anode is brought more positive than the cathode, by at least the forward voltage (typically 0.6V for a silicon diode), it conducts current. This current flows in the direction of the arrow in the diode symbol (assuming you use conventional current, which flows from positive to negative).

    When you have a diode with its cathode connected to a capacitor, and a positive voltage is applied to the anode, and this voltage is greater than the voltage presently on the capacitor, current flows through the diode and into the capacitor, increasing the voltage across the capacitor. The voltage on the capacitor (and on the cathode of the diode) follows the input voltage (on the anode of the diode) with a small voltage drop (the diode's forward voltage, typically around 0.6~0.7V for a silicon diode).

    When the voltage on the diode's anode is less than the voltage on the capacitor, the diode stops conducting and no more current flows into the capacitor. During that time, the capacitor can discharge into the load. The capacitor's capacitance is chosen so that the voltage across it doesn't drop too far during that time.

    The voltage across the capacitor in this circuit ranges between, say, 39.4V (when it has just been charged from a +40V voltage from the transformer) and, say, 38V (when it has partly discharged during the gap between positive half-cycles from the transformer). Therefore, as the voltage at R (relative to L) increases towards the positive peak of +40V, the diode will only start to conduct when that voltage reaches 38.6V. At that time, D1's anode will be 0.6V more positive than its cathode, and it will start to conduct, and feed current from the transformer into the capacitor.

    During this final stage of rising voltage on R (relative to L), the voltage across C1 will follow the voltage on R, with a voltage drop of 0.6V (D1's forward voltage).

    Once the R voltage reaches its final peak of +40V and starts to decrease, D1 will again be reverse-biased and no current will flow.

    So the flow of current from the transformer, through D1, and into C1, will only occur for a short time on each positive half-cycle, when C1's voltage needs to be "topped up". This is how all half-wave and full-wave rectifiers work, when operating into a smoothing capacitor.

    The depth of the ripple across the smoothing capacitor (how much the capacitor discharges between "top-ups" from the input supply) determines how long the recharge period is. In some cases, when the load is very small and/or the capacitance is very high, the actual amount of time that the diode (or bridge rectifier) spends conducting can be extremely short.

    Not sure whether that will help you, or confuse you even more! LOL
    Yes, but current through D1 has stopped flowing a long time before the zero-crossing point.

    Look at things from D1's point of view. All it knows is the voltage on its anode, relative to its cathode. It doesn't know, and doesn't care, about any of the other voltages in the circuit. It is connected between two points in the circuit: the right side of the transformer input (point R) and the positive plate of C1. When its anode is more positive than its cathode, by 0.6V, it starts to conduct. Current flows through it, and this current charges C1 so that C1's positive terminal follows point R (with a 0.6V voltage drop). Once point R starts to fall, D1 is no longer forward-biased and no current flows through it.
    No, no, and no. C1 is charged with the right polarity because the cathode of D1 is connected to its positive plate. Current can flow through D1 in the direction of the arrow, and can charge C1 so that its positive plate becomes more positive than its negative plate.

    Once the voltage at point R (relative to point L, the 0V rail, which we use as as a reference) starts to fall, so D1's anode is no longer more positive than its cathode (which is still being held up at about 39.4V by C1's ability to store charge), D1 stops conducting, and current stops flowing. Then it is up to C1 to keep the voltage across it constant (reasonably constant) during the time period between top-ups (via D1).

    These top-ups only occur when D1's anode is at least 0.6V more positive than its cathode; at this time, D1 will conduct, and current will flow from R, through D1, into C1, topping up its voltage.
    And I'm not sure what "potential arrows" mean. Arrows indicate flow of current. Voltages are measured between two points.
    I think you were right. You just need to combine that understanding with the explanation of a diode charging a capacitor.

    Voltage, or potential difference, is needed to make current flow. Back in the dim dark days, voltage used to be called "electromotive force", i.e. the force that "makes electrons move" (causes current to flow).

    When my Dad first taught me Ohm's Law, in the 70s, voltage was called "E", for EMF, electromotive force, not "V" for voltage. Ohm's Law was "I = E / R". Hop and others will remember this, I'm sure.

    Without voltage, no current can flow. Now you need to look at things from the diode's point of view, and see that it conducts only when its anode is more positive than its cathode. This happens near the positive peak of the incoming AC voltage, when the instantaneous voltage on its anode is more than 0.6V higher than the voltage on its cathode, which is the voltage on the smoothing capacitor (C1) that has drooped (due to the load current) since the last time it was charged.
    You WILL understand this. I'm absolutely sure of it. Take your time and don't feel dumb about it. You are NOT dumb. I know you are a smart cookie. It just takes the right explanation to "make it click", as Mr. McDonald said :)

    Read and re-read my explanations. I DO understand this, and YOU WILL TOO. It's not rocket science - believe me! I'm no rocket scientist. Not even a brain surgeon!

     
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  15. chopnhack

    chopnhack

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    Oh, LOL :D:D Thanks Kris!!
    Someone needs to fix the like button.... it only works once on this thread! What gives!!!
    LOL That clip is hysterical :)

    I will get it, I already get some parts... I just need to break myself of some old ways of thinking of these things from my tinkering days.
    I am pretty sure this sums up Kris's feelings on my tutelage about now...



    LOLZ
     
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