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radiance88
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So I found out that there wasn't really anything wrong with any of my stuff - it was just me being an idiot, forgetting that a transformer by itself
doesn't convert anything from AC to DC power.
I managed to solder the power supply wires onto an old pc power supply cable, and I managed to feed the 12 volt power into a breadboard and tinker
around with a couple of LEDs. It was fun seeing my first little circuit light up.
I have a problem though regarding the switch below:
I understand enough to know that each of the 5 voltage wires from the transformer connect onto their own pin on the switch, and pair up with an LED on
its complementary side. However, what do the two pins in the middle connect to?
This is the same circuit as the diagram I've posted above.
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Zombie
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Usually 2 pins off alone like that will be either an on / off type thing or a non variate- d pole. b Sometimes input to supply power to all the
rest...
I didn't see any schematic or diagram for that switch. Can you describe what it is called or what it is sold as?
They tried to have me "put to sleep" so I came back to return the favor.
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radiance88
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It actually is up there, but the thread has gotten long so I'll let me post the circuit diagram again:
Right - the kit enclosure has a slot for a power switch, and it includes a power switch.
I guess I screwed up by soldering the transformer directly onto the power cord when I should have soldered the power switch inbetween that?
Or did I do that right, and somehow I'm supposed to solder the power switch to this switch? Do these two pins in the middle need to be connected in
order to convey the electricity from on pin to its complementary pin?
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IrC
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You are going to cause yourself a disaster if you do not quit soldering before carefully studying the diagram and following it point by point. If you
"solder the power switch to this switch" are you then going to connect the secondary to the primary side of the transformer? The AC line has a hot and
neutral. Connect the neutral to the primary '0' on your transformer. Connect the hot through a switch first unless you are going to add a fuse to the
input. From power switch on to the hot side of the primary either 117 or 220, depending on where you live (what service you have). Personally I would
add a fuse so if you screw up at least you save the transformer. The DP6T switch goes on the secondary side and has no electrical connection to the
power line (primary side). The 2 center terminals are your two commons. One goes to six as does the other (other 6). Connect your meter to measure
resistance or easier diode beep to one common (center pins).
Rotate switch fully CCW. Now touch outer pins until you hear a beep (if the meter has a diode test, easier than watching ohms display you can keep
your eyes on only the switch). Mark terminal on side or wherever handy with fine tip marker. Rotate CW one click at a time and you will hear the beep
follow pin by pin around CW until you hit position 6. Now you have one side done. Do the other side after resetting full CCW, or use logic that it
just continues on around on the other center common pin. As your switch is 6T, you will have a dead position since you only have 5 taps on the
transformer and 5 LED's. The first position will be zero volts, the two corresponding outer pins leave open. The two common's connect together (do
this test to know the switch basing first, before connecting the commons together). The 5 corresponding pins on switch 2 (one single rotary body two
switches) each get a LED plus terminal, all negatives connect together and go through a single 1K to DC ground (bridge minus). As you rotate your
switch you will have 0V, 3V, 4.5V, 6V, 9V, 12V (AC) from the two joined commons to the bridge AC input (top in diagram). What they are doing is giving
you 5 LED's stepping up in brightness as your voltage goes up. Secondary 0V goes to bridge bottom (other AC input).
"Science is the belief in the ignorance of the experts" Richard Feynman
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Zombie
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I just went back to your first post on this thread, and I have to admire how far you have come in one month.
From a "hello" to how do I assemble my voltage stepping power supply. Nice!
IrC of course has it right especially in the notion, you have to slow down, just a little. Learning to use your DMM is perhaps the MOST important
aspect of what you are doing right now. It tells you every single thing you need to know. Including which wire goes where.
It's all new, and looks like a bowl of spaghetti. No arguments there. Which end goes where? You might not know but your meter does.
Make yourself a sort of "flow chart". A schematic you can draw on. As you understand a part circle it.
Not just what you can do because you were told but what you can do because you know how, and why you are doing it.
A monkey can assemble a television. It takes a trainer to show the monkey how. Here it comes now... Wax on / wax off. Be the monkey... No I meant Be
the trainer... That sounds better.
Once you follow the above post, and learn to ID the poles with your meter, that alone will open new doors to understanding.
Just wanted to say Kudos for your progress. Got long winded.
[Edited on 5-2-2015 by Zombie]
They tried to have me "put to sleep" so I came back to return the favor.
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WGTR
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Quote: Originally posted by radiance88  | | So I found out that there wasn't really anything wrong with any of my stuff - it was just me being an idiot, forgetting that a transformer by itself
doesn't convert anything from AC to DC power. |
Been there, done that.
| Quote: Originally posted by radiance88 | | I managed to solder the power supply wires onto an old pc power supply cable, and I managed to feed the 12 volt power into a breadboard and tinker
around with a couple of LEDs. It was fun seeing my first little circuit light up. |
Good job! I knew you'd figure it out. Keep one of the 1k ohm resistors in series with the LED, to keep from burning it out.
| Quote: Originally posted by radiance88 | I have a problem though regarding the switch below:
I understand enough to know that each of the 5 voltage wires from the transformer connect onto their own pin on the switch, and pair up with an LED on
its complementary side. However, what do the two pins in the middle connect to?
This is the same circuit as the diagram I've posted above. |
If you look at the little schematic I posted earlier, it has 14 pins. There are six pins for the switch positions on both switches (12 pins), plus 2
more for the rotor connections in the middle, giving a total of 14 pins. The two rotor pins, when connected together, get connected to the bridge
rectifier.
Don't take my word for it, check it like IrC said. Put the knob on the shaft and rotate it around, seeing how the switch connects in different
positions. It's actually quite an interesting switch. After connecting the two middle pins together, the switch should connect straight through from
the transformer side to the LED side as the switch is rotated. This can be verified with the "diode" setting on your multimeter. The meter should
"beep" with this setting when the switch connects. This function can be verified by touching the two meter probes together. If there's no beep, then
you're stuck using the ohmmeter. It should read 0 ohms when the switch closes (connects), and very high or "OVLD" when the switch is open.
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IrC
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Good work WGTR I never looked at page 2 to see your diagram, would have saved me a bunch of words. Your circuit is how I would wire it, fuse first
then switch with switch opening both lines. So many houses especially older ones have hot/neutral reversed from code. Although in the EU with 220 I
cannot imagine how that matters since both lines are hot making a DPDT power switch even more logical (does the EU wiring have a 3 pin or 4 pin plug,
and separate neutral and earth pins?).
Maybe a verbal description was not redundant, if one cannot read a diagram sometimes a verbal description helps. I have several Flukes because of all
my many meters the beep is loudest on the Flukes. Nothing beats keeping your eyes on the work using sound as an indicator. Especially when trying to
work out a complex switch like his DP6T.
After around 25 years of aging I have noticed one must go on a hunt for the right piezo disc as the Fluke beep gets weaker. Happened on 3 of my Flukes
so far which I bought in the early 80's. Gently burnishing the terminal plating on the piezo disc every 6 or 8 years helps only so many times.
"Science is the belief in the ignorance of the experts" Richard Feynman
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Magpie
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| Quote: Originally posted by WGTR |
| Quote: Originally posted by radiance88 | | I managed to solder the power supply wires onto an old pc power supply cable, and I managed to feed the 12 volt power into a breadboard and tinker
around with a couple of LEDs. It was fun seeing my first little circuit light up. |
Good job! I knew you'd figure it out. Keep one of the 1k ohm resistors in series with the LED, to keep from burning it out.
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I think this is a very good example of how knowing a little about the mathematics of electronics can be very helpful and satisfying. For example, I
have some red LEDS that are rated at 2.6v and 28ma. Now, if I attempt to power this LED with my 12v supply I may well get one intense red flash just
before it burns out! The way I deal with this is to use Ohm's Law, E=IR, as follows:
Compute the LED operating resistance, R, as:
R = E/I = 2.6v/0.028a = 93 ohms (Ω)
I must limit the current to a maximum of 0.028a, so for 12v,
R=E/I = 12v/0.028a = 429 Ω;
So the total resistance of the LED circuit must be no less than 429Ω. Therefore I must place a resistor in series with the LED having a value of no
less than:
429Ω - 93Ω = 336Ω.
This will light the LED at its rated voltage of 2.6v. My circuit will consist of a 12v terminal, the LED, the 336Ω resistor, and ground (O volts),
all in series.
Experienced people like WGTR and IrC, don't do these calculations, I imagine. They know to just add a 1k resistor in series with the LED, giving them
a generous safety factor.
[Edited on 5-2-2015 by Magpie]
[Edited on 5-2-2015 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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Zombie
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The 1k resistor is kind of a standard when dealing w/ +12v DC.
I retro fit boats / motorcycles / cars / ect... , and have boxes full of different color LED's, and rolls of 1k resistors.
Not to take anything from the post or the gentlemen. Knowing the formulas is mandatory in electronics.
They tried to have me "put to sleep" so I came back to return the favor.
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IrC
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"They know to just add a 1k resistor in series with the LED, giving them a generous safety factor"
Very true. For 14 volts DC I use 680 but have seen them burn out after several months so 820 to 1K is going to light it as well as letting the LED
live long. Also being 12 VAC peak voltages must be considered meaning 1K is just an overall good choice. Using 5 LEDs seemed to me to be an attempt to
indicate voltage by brightness. Would make sense also to mount LEDs in arc around switch pointer one for each 'on' position from low to high voltage.
"Science is the belief in the ignorance of the experts" Richard Feynman
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aga
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There a many 'rules of thumb' to use, or simply steal the 1k value from a published design that also uses a 12v supply.
However, there is no substitute for knowing the maths behind such values.
With an LED, the worst you can get wrong is that the led blows.
With higher power stuff, the maths become a Safety tool e.g. :
Short a 12v car battery out with a spanner.
Steel is about 6x10-7 ohms per meter (depends on the composition) and the spanner is about 33cm long, so let's guestimate it has a resistance of
2x10-7 Ohms.
I = V/R, so the maximum Amps flowing will be 12 / (2x10-7) = 60 MILLION Amps !
The battery would be unlikely to supply so much, but Something would certainly explode.
With a car battery, it's usually the spanner.
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IrC
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"The battery would be unlikely to supply so much, but Something would certainly explode. With a car battery, it's usually the spanner."
Having been the same room where this happened not once but 3 times over the years, not so much. The battery explodes faster than it can destroy the
wrench. No doubt rapid large volumes of H2 with air ignited inside blowing the case apart with a loud bang spewing acid over a large area. I have seen
it happen but luckily never been in the path of the acid. Mostly because I pay attention working on such things but never trust that someone else
does. This I learned multiple times. The loudest bang was a metal battery box on a 2 ton step van where the worker let the lid fall closed with the
battery sitting on top a wooden block meant to be put in front of the battery so it did not slide around inside the box in use. The 1/4" thick
rectangular steel lid slammed home against both top terminals. In all these examples batteries were fully charged up (or new), 800 to 1,000 CCA.
Result is loud explosions blowing case apart. In none was I in the line of fire from the spraying acid lucky me. However I did learn to have great
caution being around someone working on a vehicle.
One time (4th) incident, a guy was using a long 7/16" box/open on a battery terminal. He rotated it not paying attention until the open end contacted
the plus terminal with the box end still on one of the cable bolts of the negative terminal. Battery was not fully up which saved him, his face was
right there in front of the battery. The wrench instantly welded itself between the two terminals starting to glow red in the center fairly quickly
with steam blowing out the fill caps. As he yelled from the searing burn in his hand, letting go of the wrench just not quickly enough. Lucky him he
lost a shirt and jeans but not his face or eyes. Tried washing them at home and they started disintegrating. Depending upon many factors including the
state of the battery and capacity, combined with the ability to conduct amps in the short, more often than not in my experience they just explode
launching acid all around. My ears rang on the step van incident louder than they would have if a 12 gauge was fired in the room. It was a very loud
explosion.
"Science is the belief in the ignorance of the experts" Richard Feynman
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Fulmen
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Agreed, do the math until you can make your own judgment calls. Don't just rely on others numbers, that isn't going to get you anywhere.
As for Magpies numbers I'm sure he's right, but there is another perhaps more useful way to approach the math. With semiconductors there is not
necessarily any real ohmic properties, meaning that it does not have a fixed resistance like a light bulb or an resistor. So while you can calculate
the resistance for any given conditions it is somewhat misleading. I prefer to approach it like this:
The led can withstand a maximum current of 28mA, and at this current it exhibits a forward voltage drop of 2.6V. If you are to feed it from a 12V
source you need to drop the voltage to 9.4V. At a current of 28mA this voltage drop is produced by R=U/I=9,4V/0,028A=336Ω.
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IrC
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| Quote: Originally posted by Fulmen | Agreed, do the math until you can make your own judgment calls. Don't just rely on others numbers, that isn't going to get you anywhere.
As for Magpies numbers I'm sure he's right, but there is another perhaps more useful way to approach the math. With semiconductors there is not
necessarily any real ohmic properties, meaning that it does not have a fixed resistance like a light bulb or an resistor. So while you can calculate
the resistance for any given conditions it is somewhat misleading. I prefer to approach it like this:
The led can withstand a maximum current of 28mA, and at this current it exhibits a forward voltage drop of 2.6V. If you are to feed it from a 12V
source you need to drop the voltage to 9.4V. At a current of 28mA this voltage drop is produced by R=U/I=9,4V/0,028A=336Ω. |
Except of course everyone is ignoring the fact that it is 12 volts AC not DC, nor considering the actual LED. Is the 12 volts being considered as an
RMS value and if so, running it at 28 ma based upon this voltage will cause what peak current in the LED? One should also consider the led specs as 28
ma is arbitrary itself. Is it a 5,000 or 25,000 mcd LED? Neither does 28 ma consider the LED lifetime and I can say from experience running the LED at
it's maximum ratings it will burn out sooner than you think.
Resistance can be determined with a semiconductor given its specifications such as Hfe (for a bipolar device as example), knowing where it is on the
curve for given conditions. For an amplifier one would bias it in the middle of the curve in the linear portion which would allow one to calculate the
Collector load impedance. I won't go into all the various parameters in this casual conversation. Links have been given for very good sites which
cover this in depth. In any case I can tell you from my experience you can trust the 1K mentioned under the conditions the circuit presents, which
also allows some de-rating for a long LED life even with the AC being considered.
"Experienced people like WGTR and IrC, don't do these calculations, I imagine."
Yes and no depending upon the case under consideration. I don't even want to recall the mind numbing hours running calculators over the years when
designing a circuit. Very important in any high power situations. In fact one cannot escape the time consuming work especially designing say an
amplifier where linearity and bandwidth are important. If the OP has a goal of learning electronics then he must start studying the math in depth,
learning to use Spice or something similar, and practicing soldering skills. Just never overload yourself all things take time and patience.
"Science is the belief in the ignorance of the experts" Richard Feynman
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Zombie
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This may be the single most valuable quote in this entire thread. Ask a zombie...
Quote IrC:
" I pay attention working on such things but never trust that someone else does."
I live my LIFE by this one simple rule.
They tried to have me "put to sleep" so I came back to return the favor.
Zom.
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WGTR
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Most of what I do in electronics is math-based. However, I suggested using a 1k resistor for two very simple reasons:
a. That is the value used in the schematic that radiance88 posted, so the assumption is that it is a "safe" value.
b. He already has that value, as it is part of the kit.
The current that I use in LEDs: as little as possible. If it lights up well enough with 0.5mA, then that's all I use. If the requirement's
more than a few mA, then I check the datasheet to make sure the part can handle it. Usually 10mA is "safe" for the standard ones,
and 2-3mA for the miniature leaded ones.
When applying 60Hz AC voltage to an LED, the current isn't particularly straightforward. The LED only conducts on one-half cycle, and the
applied voltage is a sine wave. The average current is one-half the average of the integral of Isinx with respect to x,
over the interval
from 0 to pi; where I = peak current. This gives an average current that is about 0.318 times the peak current. For a 12V(RMS) transformer
with an LED and a series 1k resistor, this should give about 5mA of average current.
BTW, does anyone know how to enter math equations in this forum? Is it possible?
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IrC
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Polverone installed Mathjax but I don't know anything about it other than it vanished my long post one day as I was writing it when he set it up.
Mathjax causes my windoze 2K to reboot the moment it loads. He provided the entry I added to my hosts file which blocks it loading allowing me to
still be here. So I cannot help you other than to point you in the direction of studying how to use Mathjax since your request is its purpose.
"Science is the belief in the ignorance of the experts" Richard Feynman
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radiance88
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Ok. There were definitely a few things that confused me but the more I put some focus on it bit-by-bit the things seem to become a bit clearer. Still
fuzzy, but there is progress.
First of all, on the enclosure, it had room for 6 adjustments, and 7 LEDs. Each was 3V, 4.5V, 6V, 7.5V, 9V, and 12V. What confused the heck out of me
until recently is the 7.5 adjustment with its corresponding LED hole, which is right in the middle of the 6 and the 9. As there are only 5 taps for me
to draw from, it's become obvious that the enclosure is a manufacturer screw-up, there is no 7.5V with this device (thank you China).
I also had no clue what a bridge rectifier was, but after it being mentioned here I looked it up, and now I have a developing understanding of that as
well.
I've learned to use my multimeter to find the complementary pins of the switch, so that is a win for me as it's no longer a problem now. I'll try to
keep your advice in mind "when in doubt, check with the multimeter".
Also the switch I have only has two pins, one hole each. So I guess that it will only connect to my negative side, right after my fuse but before my
transformer? Another source of my confusion was the fact that I didn't see this on/off switch anywhere on the diagram, so I had no idea where to place
it.
I also have a red LED, which also seems to need to be placed in a hole right above the power switch.. So this is our "power on" LED. It doesn't seem
to be on our diagram though. Where should I connect this?
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IrC
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| Quote: Originally posted by radiance88 | Also the switch I have only has two pins, one hole each. So I guess that it will only connect to my negative side, right after my fuse but before my
transformer?
I also have a red LED, which also seems to need to be placed in a hole right above the power switch.. So this is our "power on" LED. It doesn't seem
to be on our diagram though. Where should I connect this? |
Since you only have 5 taps wire it as WGTR indicated in his schematic, CCW = 0 volts (off number 2 - output side, power switch being off number 1 - AC
line input side).
There is no negative side for the AC power line. One could call the neutral such but in the interests of not giving a student more confusion to erase
from the mind later when studying DC its best to stick with 'hot' and 'neutral'. As to the LED for power connect it to the 12 volt tap on the
transformer, either way, get another 1K resistor, connect to other lead of LED. Other end of resistor goes to the '0' or bottom of the secondary
winding. This way the LED will light when the main power switch is on no matter where the rotary is set. The AC line hot side goes through the fuse
then SPST switch to the hot side of the primary, the '0' of the primary goes to the AC neutral. I do not know what service you have, am assuming USA
117 VAC. If EU or AU 220 VAC it would not matter other than being sure of your primary taps (220). In the diagram WGTR posted he is assuming your
power switch is a DPDT and was merely switching both sides of the AC line at once. I am unsure what you have but if two holes I assume SPST so only
switch the hot side, connect the neutral to the bottom of the primary direct.
"Science is the belief in the ignorance of the experts" Richard Feynman
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WGTR
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Oops. From the picture it looked like the power switch had four pins, but I suppose it was just a pair of strong shadows. In that case, add the
switch right after the fuse, as you said.
This kit is not exactly for beginners. This is not because the circuit is difficult; it is actually fairly simple. It is because the schematic only
offers "guidance", and assumes that you already know where everything goes. That's not very helpful, is it?
There is a reason that there is a 7.5V setting on the panel. Can you guess what it is?
Here is some "thinking outside of the box" for you: There isn't really a "0V" and a "12V" wire coming out of the transformer. In this context the
important thing is that there is a 12V difference (in RMS AC voltage) between the two taps, not that one is 0V and the other is 12V. It is just as
easy to flip the taps mentally, and say that the 12V tap is 0V, and the 0V tap is 12V.
What happens if we decide that our 12V tap is now the 0V tap? We end up with taps for 3V, 6V, 7.5V, 9V, and 12V. Notice that the 4.5V setting is now
missing. It just depends on your preference as to how you want to wire it up. Some people might want the 7.5V setting instead of the 4.5V one.
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IrC
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Nice thinking out of the box. I did not like using the term '0' myself because it is misleading but it is already in his mind due to the labeling on
top the transformer. Just figured he would see the '0' as his common and since it is marked that way on the actual transformer. Hopefully he will
figure it out. I would think the 4.5 volt tap is more useful than 7.5 volt tap unless maybe one is powering a circuit on the bench which is designed
for a 9 volt battery when not on the bench.
"Science is the belief in the ignorance of the experts" Richard Feynman
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WGTR
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I need to be a little more careful with my second-person pronouns. I was directing that info to radiance88, as I figured that this
is old-hat for you, IrC!
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IrC
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I knew you were I was just adding my 2 cents to the board.
"Science is the belief in the ignorance of the experts" Richard Feynman
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radiance88
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Ok, a few new concepts learnt there - when dealing with AC (as they switch direction back and forth) there is no such thing as positive and negative ,
and voltage is really just the difference between two potentials.
I'm curious though as to how the current flows through this circuit. I'm quite sure my current understanding of electricity is a bit flawed so let me
ask a few questions: Are the capacitors and resistors essential in this circuit, and does current actually flow through them?
e.g. if there are multiple pathways for current to flow, doesn't it always choose the path of least resistance?
The following questions assume we're talking about the positive movement of our sine wave (12 volt tap actually does output 12 volts):
Couldn't the flow of electrons go from the tap, bypass the LEDs and 1K resistor by going through the connection in the middle of our switch, head to
the bridge rectifier, then head out directly our positive DC end on top? And on its return route, head from the negative DC end on bottom, pass
through the rectifier and head straight to our 0 tap? How did our current flow through the diodes, capacitors and resistors?
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WGTR
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I think now is a good time to introduce Ohm's Law. It is probably the most basic and important equation when starting to learn electronics. It is
simply:
V=I*R, where
V= voltage (in volts), I= current (in amps), and R= resistance (in ohms)
Using simple algebra, the equation can be rearranged. In other words, R=V/I, and I=V/R.
If you have a 9V battery, and you put a 1,000 ohm resistor across it, 0.009A of current flows through the resistor. Also, if you have a 100 ohm
resistor, and you measure 0.01A flowing through it with the meter, then that means there is 1V across the resistor. Try verifying these things with
your meter, and see if this equation makes sense to you.
A resistor resists the flow of electricity. Technically, everything has some resistance, even copper wire. Resistors, however, are designed
to have a specific amount of it. The resistance can be measured with your multimeter, or the value of a resistor can be determined by the resistor
color-code that I posted earlier. Using a resistor, you can control the amount of current that flows in a particular part of the circuit.
| Quote: | | Ok, a few new concepts learnt there - when dealing with AC (as they switch direction back and forth) there is no such thing as positive and negative ,
and voltage is really just the difference between two potentials. |
Yes, sort of...another way to say it is that voltage is relative. In most circuits, we pick one part of the circuit to be the
common, or ground. This part of the circuit is considered to be 0V. Everything other voltage in the circuit is compared against
that.
Since a transformer outputs an alternating voltage, its average voltage equals zero. In this way you are right that there is no net positive or
negative voltage output.
| Quote: | I'm curious though as to how the current flows through this circuit. I'm quite sure my current understanding of electricity is a bit flawed so let me
ask a few questions: Are the capacitors and resistors essential in this circuit, and does current actually flow through them?
e.g. if there are multiple pathways for current to flow, doesn't it always choose the path of least resistance?
The following questions assume we're talking about the positive movement of our sine wave (12 volt tap actually does output 12 volts):
Couldn't the flow of electrons go from the tap, bypass the LEDs and 1K resistor by going through the connection in the middle of our switch, head to
the bridge rectifier, then head out directly our positive DC end on top? And on its return route, head from the negative DC end on bottom, pass
through the rectifier and head straight to our 0 tap? How did our current flow through the diodes, capacitors and resistors? |
To answer your first question, yes, current flows through all of the components in one way or another, and to varying degrees. They all perform
specific functions in the circuit. The circuit would operate differently if any of them were removed.
Ohm's Law answers your next question. If there are two parallel paths for current to flow, then more current will flow in the path with the least
resistance. One path doesn't hog all of the current, though, unless that path happens to be a short circuit (0 ohms).
Capacitors are like miniature batteries. They are actually quite different from batteries, but this is a generalization. They are placed on the
output of the power supply to smooth out the output from the bridge rectifier. You know how the transformer puts out an AC wave? The bridge
rectifier folds over the AC voltage so that both halves of the AC voltage have the same polarity. While this gives a positive output voltage (in this
circuit), it is still a very noisy voltage, since it is a rectified AC voltage. The capacitors try to smooth out this noise, and supply smooth DC
voltage at the output.
I have to go to bed now, but to answer your last question, the LEDs don't get bypassed unless you tie a wire directly from the output to ground, i.e.,
you short out the output. At this point the fuse would blow (hopefully). Once the capacitors charge up, they look like an open circuit, so they do
not short out the output.
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