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Well not completely but go on ..On Thu, 25 Nov 2004 10:32:02 -0600, "news" <stacy@123.com> wrote:
"tempus fugit" <toccata@no.spam.ciaccess.com> wrote in message
news:thmpd.545$rL.103@fe51.usenetserver.com...
From the link you sent:
"The value of C1 can be increased to increase the amount of current the
circuit can supply. With the values shown, the circuit can supply up to
about 15mA. Remember to increase the size of C2 also. "
You can no doubt increase the cap to a more commonly available size,
like
0.47uF, without really affecting the circuit too much. This would allow
a
greater amount of current, so you could raise C2 to 330uF or so to help
out,
if it were necessary.
Thanks. I knew I could increase it somewhat but didn't know by how much.
Im
cautious around this kind of circuit.
I'm guessing that if I Reduce the Caps I will get less Amperage.
---
Yes, you will, but for its seeming simplicity, this circuit is
devilishly complex.
In this circuit, C1 is being used used like a "lossless resistor",
with its reactance at the mains frequency determining how much current
it will allow to flow into the load and into C2. At 60 Hz, the
reactance of C1 will be:
1
Xc = ---------
2pi f C
where Xc = the reactance in ohms
f = the mains frequency in Hz.
C = the capacitance in farads
So, for 0.39ľF we have:
1 1
Xc = --------- = --------------------------- ~ 6800 ohms
2pi f C 2 * 3.14 * 60Hz * 3.9E-7F
Now, if we forget about C2 for a moment and look at the circuit like
this:
+-----+
MAINS>---[C1]--|~ +|----+
| | |
| | [RL]
| | |
MAINS>---------|~ -|----+
+-----+
FWB
all the bridge is doing is rectifying the mains, so we have,
essentially:
120VRMS>-----+-----E1
|
[6800R}
|
+-----E2
|
[RL]
|
120VRMS>-----+-----0V
Now, since the current in a series circuit is the same throughout the
circuit and, since
E = IR
you might think that if the reactance of the capacitor is equal to
6800 ohms and has the same current flowing in it that RL does, if RL
is equal to 6800 ohms it will have the same the same voltage dropped
across it, but it won't.
If you understand everything so far, and you're still interested, I'll
continue...
John Fields
Im not going to presume that I know as much as individuals who have mostOn Thu, 25 Nov 2004 16:34:08 -0600, "News" <stacy@123.com> wrote:
If you understand everything so far, and you're still interested, I'll
continue...
John Fields
Well not completely but go on ..
---
What don't you understand?
--
John Fields
Well John,
"John Fields" <jfields@austininstruments.com> wrote in message
news:iu3eq05v10grcnc0blthukb42hka3l88cs@4ax.com...
On Thu, 25 Nov 2004 16:34:08 -0600, "News" <stacy@123.com> wrote:
If you understand everything so far, and you're still interested, I'll
continue...
John Fields
Well not completely but go on ..
---
What don't you understand?
--
John Fields
---Well John,
Im not going to presume that I know as much as individuals who have most
likely gone to college for this topic or those electrical engineers in my
company that every once and a while deign to speak to me in a condescending
manor when I ask then a question. Which sometimes only encourages me to go
on and sometimes discourages me..
---But Im not sure what you mean by the numbers 3.9-E-7 .
Is that the .39 Capacitor It could only be based on your equation but what
is the -E-7
---I found a great little site that had a nice little definition of things and
caclulator see the bottome of ..
http://www.eatel.net/~amptech/elecdisc/reactnce.htm
Ahh but I've seen so many places with reverse polarity."Peter Bennett" <peterbb@nowhere.invalid> wrote in message
news:h79cq0hdj24svoe2te1rrp9ps2ntfen0pq@4ax.com...
On Thu, 25 Nov 2004 09:12:50 -0600, "news" <stacy@123.com> wrote:
This is the circuit that Im trying to do. Im just testing and learning
about
various circuits. Im putting together various circuits for the fun of
learning.
http://www.aaroncake.net/circuits/supply5.htm
That is a bad circuit to play around with - the DC output can have
hazardous voltages - there will be 11 volts between the terminals, but
they may be at full line voltage AC.
Look for a power supply circuit using a power transformer - much
safer!
--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca
GPS and NMEA info and programs:
http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq
I agree with you, this is not a circuit to play with unless you really
know what you are doing. Two other things it should have (if one still wants
to use this circuit), is a fuse (in case C1 should short out) and a
polarized plug (so that the "-" side is always near earth ground).
Brian