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Science Forum Index » Electronics - Basics Forum » Calculating DC Output Current From Unregulated AC...
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Posted: Wed Jul 09, 2008 3:26 pm |
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Guest
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I have an application in which I need to determine if a transformer is
suitable.
Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?
I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?
Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.
How much rectified DC voltage can that transformer maintain at 1.67A?
How much current can it maintain while staying at or above 12.3V, or
if there is another voltage to consider because we don't know the
other properties of this transformer, what voltage would that be and
at that voltage what is the DC current capability?
Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.
Thanks,
JC |
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Posted: Wed Jul 09, 2008 3:39 pm |
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Guest
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On Jul 9, 9:26 pm, emailaddr... at (no spam) insightbb.com wrote:
Quote: I have an application in which I need to determine if a transformer is
suitable.
Quote: Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply.
Another thing I am wondering is if a basic, typical small brick AC-DC
switching supply is stable powering this kind of load. Stripping down
the circuit to the basic topology, what if it had for example:
AC-DC SMPS -> LM317 -> 1.25 Ohm resistor (LM317 in current regulating
config) -> 1A into Battery Pack + LM317 feedback
Since the SMPS is trying to regulate to it's spec'd voltage, let's say
that is 13V even if an uncommon value, will it run stable doing so and
if so, is that at the same constant output wattage from the SMPS the
whole time, with the linear regulator simply dropping more voltage,
creating more heat at the beginning of a battery recharge cycle since
it's suppling constant current to batteries that are at a lower
initial, low state-of-charge voltage? |
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Posted: Wed Jul 09, 2008 4:32 pm |
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On Jul 9, 9:47 pm, "Phil Allison" <philalli... at (no spam) tpg.com.au> wrote:
Quote: emailaddr... at (no spam) insightbb.com
I have an application in which I need to determine if a transformer is
suitable.
Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?
I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?
** This is your error.
This what? I was looking for details.
Quote:
The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage.
Ok, but I'm not asking what it's not, I'm asking what it is. what I
am asking is what is the like-equivalent DC it should be capable of
based on the AC rating. Is 20VA really the answer, since we're
talking about winding ratios? It would seem not, since we do see
things like peak output around 16V instead of ( picking some number
out of air) 40V at (no spam) 0.5A for 20VA.
Quote: It is also a " free air " rating - meaning
there must be plenty of ventilation and no heat sources next to the
transformer.
Yes, I am looking only for a free air rating. It all starts there.
Quote:
Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.
How much rectified DC voltage can that transformer maintain at 1.67A?
** It would be overloaded delivering that amount of DC current and
overheat - unless perhaps you blew cold air over it with a fan.
I am not interested in what overloads it, I am interested in what
current it can supply at 12.3V or what voltage it would be at 1.67A
DC, what the math is to arrive at that free air rating for DC current
and DC volts give either or the other would be a variable.
Only after I know that *overloaded* maximum can I begin to derate to
account for heat.
Quote:
Normally, one applies a de-rating figure to the AC current rating of a
transforemer when it is feeding a bridge rectifier and capacitor filter.
The figure depends somewhat on the design of the particular transformer (
toroidal, E-core ) but is in the range of 0.5 to 0.7.
So, if you really do need 1.67 amps DC at 16 volts - then pick a 40 VA
transformer.
As mentioned I need to know the transformer's capability, not 1.67A at
16V. Those were example figures, I need the actual equations for any
transformer, even the 40VA one must have that.
Beyond that, I will use the actual equations to achieve at least
12.3VDC and a current depending on cost, transformer volume, etc.
I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage. |
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| Phil Allison... |
Posted: Wed Jul 09, 2008 8:47 pm |
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Guest
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<emailaddress at (no spam) insightbb.com>
Quote: I have an application in which I need to determine if a transformer is
suitable.
Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?
I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?
** This is your error.
The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage. It is also a " free air " rating - meaning
there must be plenty of ventilation and no heat sources next to the
transformer.
Quote: Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.
How much rectified DC voltage can that transformer maintain at 1.67A?
** It would be overloaded delivering that amount of DC current and
overheat - unless perhaps you blew cold air over it with a fan.
Normally, one applies a de-rating figure to the AC current rating of a
transforemer when it is feeding a bridge rectifier and capacitor filter.
The figure depends somewhat on the design of the particular transformer (
toroidal, E-core ) but is in the range of 0.5 to 0.7.
So, if you really do need 1.67 amps DC at 16 volts - then pick a 40 VA
transformer.
....... Phil |
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| Phil Allison... |
Posted: Wed Jul 09, 2008 9:57 pm |
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Guest
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<emailaddress at (no spam) insightbb.com>
"Phil Allison" >
Quote:
I have an application in which I need to determine if a transformer is
suitable.
Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?
I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?
** This is your error.
This what? I was looking for details.
** Read the whole post before making silly replies.
Quote: The 1.67 amp ( ie 20VA) rating of the transformer is only for AC output -
not some derived DC voltage.
Ok, but I'm not asking what it's not, I'm asking what it is.
** Read the whole post before making silly replies.
what I
am asking is what is the like-equivalent DC it should be capable of
based on the AC rating.
** Read the whole post before making silly replies.
Is 20VA really the answer, since we're
talking about winding ratios? It would seem not, since we do see
things like peak output around 16V instead of ( picking some number
out of air) 40V at (no spam) 0.5A for 20VA.
** Read the whole post before making silly replies.
Quote: It is also a " free air " rating - meaning
there must be plenty of ventilation and no heat sources next to the
transformer.
Yes, I am looking only for a free air rating. It all starts there.
** Don't complain when you get free, good advice - fuckhead.
Quote: Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.
How much rectified DC voltage can that transformer maintain at 1.67A?
** It would be overloaded delivering that amount of DC current and
overheat - unless perhaps you blew cold air over it with a fan.
I am not interested in what overloads it,
** You need to be - for safety and functional reasons.
Read the whole post before making silly replies.
I am interested in what
current it can supply at 12.3V or what voltage it would be at 1.67A
DC, what the math is to arrive at that free air rating for DC current
and DC volts give either or the other would be a variable.
** There is no possible math based only on the figures you gave. The whole
topic is about ACTUAL temp rise of the windings of the particular
transformer under particular conditions.
Best way to find THAT is to *measure* it.
Only after I know that *overloaded* maximum can I begin to derate to
account for heat.
** Read the whole post before making silly replies.
Quote: Normally, one applies a de-rating figure to the AC current rating of a
transforemer when it is feeding a bridge rectifier and capacitor filter.
The figure depends somewhat on the design of the particular transformer (
toroidal, E-core ) but is in the range of 0.5 to 0.7.
So, if you really do need 1.67 amps DC at 16 volts - then pick a 40 VA
transformer.
As mentioned I need to know the transformer's capability, not 1.67A at
16V. Those were example figures, I need the actual equations for any
transformer, even the 40VA one must have that.
** I just supplied the derating figures and the reason why it cannot be
precise.
Read the whole post before making silly replies.
I can see I should not have given any details about the project
because I only want to know how to convert a AC transformer's voltage
and current rating into a DC voltage at same current, or the resulting
current at a specific DC voltage.
** Neither is possible with simple math and only the VA rating to go on.
I have written a ( non simple) program that gets fairly close to predicting
the results ( ie the DC output voltage at any specified current ) -
providing you have much more data on the transformer like the primary and
secondary resistances, the size of the filter cap and a figure for
transformer leakage inductance.
You are WAY over-analysing the problem.
You only need to be sure the voltage is high enough, but not too high, for
you charger to work and the transformer is not overloaded under any
operating condition.
...... Phil |
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| Peter Bennett... |
Posted: Thu Jul 10, 2008 12:08 pm |
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Guest
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On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress at (no spam) insightbb.com
wrote:
Quote: I have an application in which I need to determine if a transformer is
suitable.
Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?
Go to http://www.hammondmfg.com/5cpwr.htm and download their Power
Transformer Selection Guide - it is a one page .pdf that shows various
circuit configurations and resulting voltage/current relations.
--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca |
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| Paul E. Schoen... |
Posted: Thu Jul 10, 2008 12:38 pm |
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"whit3rd" <whit3rd at (no spam) gmail.com> wrote in message
news:7869a34d-5d49-4554-8829-94d770aa0a71 at (no spam) c58g2000hsc.googlegroups.com...
On Jul 9, 10:25 pm, emailaddr... at (no spam) insightbb.com wrote:
Quote: On Jul 9, 10:57 pm, "Phil Allison" <philalli... at (no spam) tpg.com.au> wrote:
emailaddr... at (no spam) insightbb.com
"Phil Allison"
** This is your error.
The 1.67 amp ( ie 20VA) rating of the transformer is only for AC
output -
not some derived DC voltage.
Ok, but I'm not asking what it's not, I'm asking what it is.
None of the information given is useful in making a determination on
this issue. Phil is telling you that the AC
rating is for 1.67A RMS, which amounts to 2.3A peak, but
that your rectifier will only conduct for a fraction of the duty
cycle, so
the 1.67A average output of the supply is all delivered in very
short time intervals (when the transformer output exceeds
the capacitor charge plus the diode forward drop). The
rectifier current could be 18A repeating peaks (even though
the average is lower), because the rectifier is almost always
turned off.
18A can melt the copper windings.
The only way copper windings could melt is if those 18 amp peaks lasted a
lot longer than one cycle. It is all related to temperature, which is
related to power and energy, and generally fusing current is given by I^t.
It also depends on how quickly the heat can get out of the hottest spot in
the windings and be dissipated by conduction, convection, and radiation.
Best bet is to put a thermistor in the deepest part of the windings and
plot temp vs time at the worst case conditions, and see if the maximum
temperature gets close to the rating of the insulation (usually about 130
C), but 90 C is safer. Another way to measure the temperature is by
measuring the winding resistance change after it has stabilized. The tempco
of copper is about 0.4% / Deg C. So a 40% increase in winding resistance is
just about your maximum.
For transformers, it is all about RMS current, and duty cycle. A
transformer rated at 1.67 amps RMS will handle 2.3 amps RMS for
intermittent duty, with 50% duty cycle, with ON times no longer than 10-20
minutes or so. It will also handle overloads of 2x for 25%, and even 10x
for 1% duty cycle, with ON times of a few cycles. We "abuse" the trannies
in our circuit breaker test sets in this way all the time. The outputs are
essentially short-circuits (several hundred microhms of breaker and
connections), and we adjust the input to get the current needed to trip the
breaker. We sometimes need 40,000 amperes to trip a 4000 amp breaker (in
onee or two cycles), and the tranny is rated at 4000 amps continuous.
Solid state devices, like the diodes in this circuit (and the SCRs we use
to control our test sets) have a sharper derating curve, and are based more
on I^t, so a 500 amp SCR or diode will usually be limited to maybe 3x its
rating before it reaches the area where the ON time is very short, as it is
for capacitive charging peaks.
In the simulation I did, with Schottkys and 2200 uF, the peak rectifier
current is 3.8 amps. Even with 22,000 uF it is not quite 4 amps. Lowering
the internal resistance of the source to 0.4 ohms makes these peaks about 5
amps. Phil is correct on this as well.
Paul |
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| Phil Allison... |
Posted: Thu Jul 10, 2008 8:49 pm |
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"Peter Bennett"
Quote: Go to http://www.hammondmfg.com/5cpwr.htm and download their Power
Transformer Selection Guide - it is a one page .pdf that shows various
circuit configurations and resulting voltage/current relations.
** There is an *unfortunate error* in several of the circuits shown, all the
ones where a cap is the filter.
The figure for " V (Avg) D.C. " is given as 64 % of " V (Peak) D.C " -
which is complete bollocks !!!
That is only the case where there is NO filter cap used.
With suitable filter cap values, the average and peak DC voltages can be a
close as you like.
...... Phil |
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Posted: Fri Jul 11, 2008 10:07 am |
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On Jul 9, 6:39 pm, emailaddr... at (no spam) insightbb.com wrote:
Quote: On Jul 9, 9:26 pm, emailaddr... at (no spam) insightbb.com wrote:
I have an application in which I need to determine if a transformer is
suitable.
Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply.
Another thing I am wondering is if a basic, typical small brick AC-DC
switching supply is stable powering this kind of load. Stripping down
the circuit to the basic topology, what if it had for example:
AC-DC SMPS -> LM317 -> 1.25 Ohm resistor (LM317 in current regulating
config) -> 1A into Battery Pack + LM317 feedback
Since the SMPS is trying to regulate to it's spec'd voltage, let's say
that is 13V even if an uncommon value, will it run stable doing so and
if so, is that at the same constant output wattage from the SMPS the
whole time, with the linear regulator simply dropping more voltage,
creating more heat at the beginning of a battery recharge cycle since
it's suppling constant current to batteries that are at a lower
initial, low state-of-charge voltage?
Why so many LM317s? They will generate heat too, and waste energy.
What are you charging? What supply voltage do you need?
Lots of surplus 12, 15 or 18VDC switching power supplies from old
laptops - see a thrift store, or a surplus electronics store
Michael |
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| ehsjr... |
Posted: Fri Jul 11, 2008 1:34 pm |
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emailaddress at (no spam) insightbb.com wrote:
Quote: On Jul 10, 1:50 am, ehsjr
e.h.s.j.r.removethespampunctuat... at (no spam) bellatlantic.net> wrote:
There is no "magic formula". You need to learn a whole
lot more about power supplies to calculate with precision
and to understand what needs to be considered and why.
Ok, but this should be resolvable to a reasonable level given the
example context of it being a typical E-core transformer and basic
silicon bridge rectified and capacitive filtered circuit. The output
at that point is the question. Even if some formula has an unknown
variable, or two or one hundred, the start would be to resolve those
which requires a working equation in which to place those variables.
I'll mention some general things.
First, with a bridge rectifier and capacitive filter,
figure the current you can supply to the load at
a little over 1/2 the transformer rating, in your
case, a bit over .9 amps.
This seems to counter the majority of small transformer examples out
there, does it not? Consider products powering just about anything
that uses a wall wart.
Consider a sub-20VA transformer rated in a wall wart for 12VDC, 1A
output. Granted, we could call that a little over 1/2 the rating but
just how much or little is the crucial issue, what variables determine
how much or little and how to express those mathematically.
Next, figure the ripple voltage, Vr. To figure Vr, seehttp://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-S...
Thank you for the link, though since I am not mass producing equipment
and so not overly concerned about small differences in component cost
or size, I'm essentially going to consider ripple current a constant
resolved later as would apply to any one design, while I am asking
about a general formula for conversion without regard to variables
that would change in different designs beyond the basic assumption of
a capacitor large enough to achieve acceptibly low ripple for example.
Next, find out how much headroom your regulator needs, usually
available from the datasheet. Then, the maximum voltage at
the output of the regulator will be
(Vsec*1.41) - 2Vf - Vr - Vheadroom = Vmax where Vsec is the voltage
at the secondary. Vf is the diode drop, which you can find
on the curve of Vf vs I on the datasheet. You should figure Vsec
based on worst case line voltage and worst case Vsec sag under load.
By the way, the 12.3 volt figure is odd - how did you arrive
at it?
This is what the example circuit would need at a bare minimum by
calculating similarly to what you have above plus other drops in the
circuit at (no spam) expected current levels, to still retain the necessary
minimum input voltage, plus or minus a margin of error as median
datasheet values where used.
However, I am not concerned about the entire circuit, not about drop
over a regulator, I am only concerned about resolving how the AC
transformer rating relates to DC output before anything further in a
circuit beyond a typical bridge rectifier and capacitor sufficient to
smooth to a hypothetical 0V ripple, and I am also accounting for the
(Often negligable) different in forward voltage over the rectifier(s)
at different current levels, but this too can be expressed
mathematically.
Essentially, I don't want information more applicable to one project
than to another. Only what remains true mathetically for all projects
which employ AC spec'd transformers of typical design and through
bridge rectification by the most common silicon diodes. In other
words, as you'd see in the vast majority of electronics already if
they're not using switching PSU.
I don't want to argue - that seems your intent.
I know you want a magic formula - you have made that
clear. There is no single formula for you. Those who
have replied have made that clear.
We've tried to show you some of the factors that are
involved. You insist on ignoring them. So, we can't
help you, no matter how hard we try. Sorry.
Ed |
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| The Phantom... |
Posted: Fri Jul 11, 2008 2:29 pm |
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Guest
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On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress at (no spam) insightbb.com wrote:
Quote: I have an application in which I need to determine if a transformer is
suitable.
Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?
I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?
Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.
How much rectified DC voltage can that transformer maintain at 1.67A?
How much current can it maintain while staying at or above 12.3V, or
if there is another voltage to consider because we don't know the
other properties of this transformer, what voltage would that be and
at that voltage what is the DC current capability?
Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.
Thanks,
JC
You should go to the library and make a copy of the paper:
"Analysis of Rectifier Operation", O. H. Schade, Proceedings of the IRE,
July 1943
This will give you an idea of what's involved in a detailed analysis of the
sort you're asking for. |
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| John Fields... |
Posted: Fri Jul 11, 2008 3:34 pm |
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Guest
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On Fri, 11 Jul 2008 12:29:09 -0700, The Phantom <phantom at (no spam) aol.com>
wrote:
Quote: On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress at (no spam) insightbb.com wrote:
I have an application in which I need to determine if a transformer is
suitable.
Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?
I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?
Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries. The
circuit is a bit more involved than only this (protection diodes,
charge controller, etc) but this is the subsection in question and the
rectified output of the transformer will need to stay above roughly
12.3VDC so I'm trying to figure out what constant current this
transformer (or others) can supply. I've alread accounted for other
looses like that through the regulator when coming up with the 12.3V
figure.
How much rectified DC voltage can that transformer maintain at 1.67A?
How much current can it maintain while staying at or above 12.3V, or
if there is another voltage to consider because we don't know the
other properties of this transformer, what voltage would that be and
at that voltage what is the DC current capability?
Please I'm asking to learn how to calculate this myself instead of
only the numerical answer.
Thanks,
JC
You should go to the library and make a copy of the paper:
"Analysis of Rectifier Operation", O. H. Schade, Proceedings of the IRE,
July 1943
This will give you an idea of what's involved in a detailed analysis of the
sort you're asking for.
---
There's also Motorola's Silicon Rectifier Manual, an excellent
reference and cheap at:
http://www.amazon.com/gp/offer-listing/B000HVEWGK/ref=dp_olp_3/103-7647257-4204638
JF |
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| John Fields... |
Posted: Fri Jul 11, 2008 3:37 pm |
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Guest
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On Wed, 9 Jul 2008 18:26:12 -0700 (PDT), emailaddress at (no spam) insightbb.com
wrote:
Quote: I have an application in which I need to determine if a transformer is
suitable.
Transformer is rated for 12VAC, 1.67A but I need DC. I could put a
bridge rectifier and capacitor after it easily enough but how does one
go about calculating the DC output this is capable of?
I've previously used a x 1.41 factor to convert which would give
16.92V, minus the forward drop of a couple silicon diodes in the
bridge rectifier estimated at roughly ( 2 * 0.7V, ignoring changes in
diode forward drop at different current) which would leave (16.92 -
1.4) = 15.5V, but am I correct in thinking this is peak DC output and
there is a different calculation needed to arrive at the output
voltage if the load were drawing 1.67A?
Perhaps a little more info about the project is in order. Following
the transformer there'll be a bridge rectifier, smoothing cap... then
a linear regulator suppling constant current to charge batteries.
---
Why can't you just used the raw unfiltered DC and current limit the
input to the battery?
JF |
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| Hammy... |
Posted: Fri Jul 11, 2008 4:13 pm |
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On Fri, 11 Jul 2008 15:34:47 -0500, John Fields
<jfields at (no spam) austininstruments.com> wrote:
Quote:
You should go to the library and make a copy of the paper:
"Analysis of Rectifier Operation", O. H. Schade, Proceedings of the IRE,
July 1943
This will give you an idea of what's involved in a detailed analysis of the
sort you're asking for.
---
There's also Motorola's Silicon Rectifier Manual, an excellent
reference and cheap at:
http://www.amazon.com/gp/offer-listing/B000HVEWGK/ref=dp_olp_3/103-7647257-4204638
JF
The hanbook from onsemi as well as the link I posted to algonquin
(volt-reg) provide work examples using Schade curves.
Buying a book for future refrence is a good idea though. Hell for five
bucks I might get that John. We used to have a shelf full of those
motorola data books and manuals in school. |
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| ... |
Posted: Sat Jul 12, 2008 9:43 am |
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On Jul 12, 7:40 am, emailaddr... at (no spam) insightbb.com wrote:
...snip..
Quote: However, it is fairly beside the point that I am looking for
a universal equation that ignores all of this. Many people seem to be
saying the rest matters and yes of course it does - but it is still
theoretically expressable in an equation with the factors that change
as variables.
Universal equation. Here you go.
http://en.wikipedia.org/wiki/Heat_equation
http://en.wikipedia.org/wiki/Newton%27s_law_of_cooling#Newton.27s_law_of_cooling
Also check out "Introduction to Heat Transfer" by Incropera and
DeWitt.
The game is to keep the transformer wires at only a moderate
temperature higher than ambient, or you will melt the wiring
insulation.
Knock yourself out.
Michael |
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