Chip with simple program for Toy

[Mark Zenier]

In article <2f8acf53-1cda-424c-9073-6706a1eb55ac@k37g2000hsf.googlegroups.com>,
jalbers@bsu.edu <jalbers@bsu.edu> wrote:

I am experimenting with a relaxation oscillator circuit consisting of
a NE-2 bulb wired parallel to a capacitor and this pair is wired in
series with a resistor and connected across a 150 V DC power source (a
bunch of DC wallwarts connected in series). The resistor is variable
0-1 Meg Ohm, the capacitor is a 1uF electrolytic rated at 160V. I
don't have any capacitors on hand with a higher working voltage.

The circuit seems to work. I can get the bulb to blink around 3 times
a second but I am wanting a higher flash rate and I am not getting
it. Lowering the resistance makes the bulb turn on continuously. I
don't think that the bulb is flashing faster than the eye can
distinguish. I conneded the circuit to an oscilliscope and when the
bulfb is visually flashing I see the RC discharge curve but lowering R
until the bulf truns on continuously pretty much produces a flat line
on the scope.

I was expecting to maybe be able to get around 2-100 hz with a NE-2
relaxation oscillator. Is this possible or am I expecting too much
from this type of circuit? And if so, why?

Any help would be greatly appreciated. Thanks

Digging out my GE Glow Lamp Manual, for a 5AB (the tight spec.
equivalent to an NE-2) the resistance range for oscillators runs
from .5 meg up to 15 meg. To get 100 Hz, you need to run with
a .01 uF capacitor and 1 to 5 megohm, depending on voltage
(100-150 volts). The maximum frequency, with a capacitor in the
10-30 pf range, is 3 to 30 kilohertz.

Mark Zenier mzenier@eskimo.com
Googleproofaddress(account:mzenier provider:eskimo domain:com)

 

[Paul E. Schoen]

<mrdarrett@gmail.com> wrote in message
news:832a69e0-dafd-41e3-aae9-56eb27d658d2@w34g2000prm.googlegroups.com...

On Jul 3, 11:45 pm, "Paul E. Schoen" <pst...@smart.net> wrote:
mrdarr...@gmail.com> wrote in message

news:a731e6f8-d541-4301-9959-88d1f7d9106b@q27g2000prf.googlegroups.com...



On Jul 3, 6:54 pm, "Bob Eld" <nsmontas...@yahoo.com> wrote:
mrdarr...@gmail.com> wrote in message

news:7edf1122-2064-4799-9661-4828c556a5ce@s21g2000prm.googlegroups.com...

I built an astable multivibrator with blinking lights (much simpler
than using a 556!), as a test for a circuit that will use a power
transistor (or MOSFET) to pulse a transformer primary for future
experiments.

http://mrdarrett.googlepages.com/blinkenlights002.pdf

Strangely, when I replaced the two 2N3904 transistors with TIP31As,
the circuit would not oscillate unless I briefly disconnected then
re-
connected R2 or R3. (I bumped the voltage up from 3V to 6V during
these tests.)

As a work-around, I'm considering just putting the 2N3904s back in,
and connecting the positive end of C2 to the base of a TIP31A.
Inelegant, but I think it will work.

I'm trying to figure out why TIP31As won't work, but it also
doesn't
help me any that the TIP31A data sheet does not specify a minimum
V_BE_on.

http://www.st.com/stonline/products/literature/ds/12292/tip31a.pdf

Input?

Michael

One possible reason it doesn't work with certain transistors is that
the
two
transistors come on together and lock up the operation. It fails to
flip-flop. Also, there might not be enough base current for TIP31's
with R2
and R3 at 10K. Lower these resistors. The minimum beta is 25 at 1
amp.
so
you're likely not getting enough current to drive these transistors.

To insure that an astable won't lock up, disconnect R2 and R3 from
the
positive rail and connect the junction of the two resistors to the
cathodes
of two diodes. Connect the anodes of the diodes to the anodes of the
LED's,
one on each.

With this arrangement, if the transistors both come on together, the
base
drive is reduced and the circuit will always start.

I tried simulating various forms of this circuit with both 2N3904 and
2N3055, and it always seemed to work, at least down to 2.5 volts or so.
It
seemed to woek better if I connected C1 and C2 directly to the
collectors,
which have a bit more voltage swing. I would suggest connecting a logic
level MOSFET to drive a transformer, so you will have minimal loading.
Without the LEDs, you will have plenty of voltage swing for the gate.
And
you can use an N-channel to sink a higher voltage on a transformer CT,
or
P-channel to source the voltage. You might even be able to make a full
bridge, but you need to make sure there is dead time where both the
high-side and low side are off. This is why they have dedicated circuits
for that.

Paul


Yep, that's why I thought pulsed DC would be easier.

But, also remember that you must have very little DC current in any
transformer winding, or you will get saturation. Pulsed DC can be used if
you are building a circuit where you are alternately storing and releasing
energy in the magnetic field, in which case you really have an inductor,
which may have additional windings for various reasons (especially
isolation, multiple outputs, and large differences in input to output
voltage/current).

You might have a look at the Microchip PIC16F616, which has multiple
on-board PWM outputs that can be programmed to drive various power supply
circuits ranging from simple single inductor buck or boost converters, to
center tapped transformers, and full bridge circuits (using MOSFETs, of
course). I'm using that PIC (and its similar cousin, PIC16F684) to make a
single inductor 100 kHz boost converter from 12 VDC to anything from 20 to
60 VDC at close to 1 amp, and it would only take a few changes in the power
components (MOSFET, inductor, Schottky diode, and capacitors) to scale that
up to several hundred watts. And some simple additional diodes and
capacitors can also provide a negative output that tracks the positive
output for balanced loads. I have posted the LTSpice file before. If you
want to learn PIC programming this might be a fun project.

Paul

 

[Bob Eld]

"Bob Monsen" <rcmonsen@gmail.com> wrote in message
news:NCubk.6440$LG4.3212@nlpi065.nbdc.sbc.com...

"Bob Eld" <nsmontassoc@yahoo.com> wrote in message
news:0zabk.16122$Ri.4973@flpi146.ffdc.sbc.com...

"Eeyore" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:486C4867.F6DAF534@hotmail.com...


Bob Eld wrote:

"Eeyore" wrote
Bob Eld wrote:

Insure that the max peak of the audio signal is within
the ADC range.

Diode clamp it !

Graham

Yeah you can clamp it but then you loose that and higher values as
peaks.
The whole idea is to convert the peaks to numbers not to clip them
off.

You can actively diode clamp. Just don't want to hurt the poor ADC
'just
in
case'. I doubt that for flickering lights, losing anything below say
0x240
will
hurt any


There's really no need to clamp it. The processor input will have
internal
"diodes" that will keep the voltage from exceeding the rail voltage
plus a
little.

A discussion we've had many a time before here !


Likewise the input voltage won't go below ground.

It WON'T ? In what ideal world is this ?


A series resistance should be added to limit the input current and a
shunt
resistance
added to keep the DC level from shifting up due to rectification by
these
clamps.

Uh ?

Graham

Well, if there is an internal diode from the input to the negative rail,
it
will keep the voltage on the input from going below ground, ignoring the
diode drop, of course.

Now, if you capacitively couple to the input, this diode will rectify
the
negative going portion of the audio signal and charge the capacitor,
positive towards the input. This will offset the zero base line in the
positive direction and screw up the peak detection. A judicious
resistance
to ground with an appropriate series resistance will limit the current
and
drain off this offsetting charge. It ain't rocket surgery!




One thing to remember is that cmos inputs are usually prone to latch-up
disease. So, if the output goes higher than the positive voltage input of
the IC, you can end up with a triggered SCR between Vcc and ground. Not
good.

So, having a way to prevent this is a good thing, and generally consists
of
a diode from the input to the positive rail. For digital inputs, I believe
most microprocessors use an internal diode. However, I'm not sure about
analog inputs, and would not want to risk it, particularly when a cap from
an external input can easily cause this situation, and the solution is so
simple.

See this link for more information:

http://ww1.microchip.com/downloads/en/AppNotes/00763b.pdf

Regards,
Bob Monsen

The PIC stuff has internal protection diodes on the inputs. As far as I know
all of their processors are so protected on the IO ports whether configured
analog or digital. For a typical A/D input, see the PIC16F676 Data Sheet
page 47. There should be no trouble driving a few milliamps into these
diodes clamping off a signal without any external diodes.

I've never yet seen one latch up doing this. That's probably why the diodes
are there.

 

[Phil Allison]

<angus.oliver@gmail.com>

Having found dual op amp packages around a fifth the price of
comparable single packages from online distributors, I think i'll go
for the double.
(BTW, I'm using supply of +16, and ground).

I don't want the unused channel drawing current unnecessarily though,
or bouncing the output around.. - If I tie the non-inverting input to
+3.3V, and have a high resistance feedback resistor to the inverting
input (with no other input to that pin), would that be as good as
anything? I figure the output won't have to 'try' very hard to modify
the voltage of an input pin with no other voltages on it..

Am I right?

** Use the spare op-amp to do something - like buffer the output of other
one.

No extra current draw involved.


...... Phil

 

[Tom Biasi]

<void.no.spam.com@gmail.com> wrote in message
news:91a1e5b4-dd7a-4511-b82a-cd3d228ce91c@l64g2000hse.googlegroups.com...

I bought a USB drive, and it came in a small plastic shell package.
Is it safe to assume that this plastic material is anti-static?

I have not yet opened the package, and I left it sitting on the carpet
for a while. Then I did some vacuuming, and ran the vacuum cleaner
right next to the package.

Assuming that the plastic package is anti-static, would that be enough
to prevent the USB drive from being damaged by any static electricity
generated from the carpet or the vacuum cleaner?

Why don't you open it and see?
Those devices are designed to take quite a lot of abuse.
Mine has gone through the washer and dryer several times.
What you did is nothing compared to what they go through in transit and
stocking.

Tom

 

[ian field]

"Michael Black" <et472@ncf.ca> wrote in message
newsine.LNX.4.64.0807081123000.28234@darkstar.example.org...

On Mon, 7 Jul 2008, mrdarrett@gmail.com wrote:

I found four 1N2071 diodes in my junk pile. Apparently the 1N2071 is
an obsolete diode - Mouser doesn't carry it anymore.

I looked up the datasheet for the 1N2071. It seems designed to
operate at 60 Hz, but didn't find any info on the maximum speed for
how quickly the diode can operate.

http://www.alldatasheet.com/datasheet-pdf/pdf/102157/IRF/1N2071.html

Can the diode successfully operate at up to 10 kHz? I have a sinking
feeling the answer is "no".

So, how quickly CAN the diodes reliably switch? 100 Hz? 400 Hz?


What are you trying to do? You say "switch" which implies some sort
of logic or switching supply, but it's not completely obvious. What
you need a diode for helps to determine what will be suitable.

Since you're pulling things out of the junkbox, why not find a junk
computer, open up the power supply, and use the schottky power diodes
on the secondary for this project? They are readily available, and
since they are in switching supplies, they will be good at higher
frequencies.

Beware of low PIV SB rectifiers! - In a computer PSUs they are likely to be
no more than 45V, some are low as 20V. On very rare occasions 90V SB
rectifiers can be found in scrap monitors.

High voltage silicon carbide rectifiers are beginning to hit the market, but
I've yet to find any in scrap gear.

 

[ian field]

<void.no.spam.com@gmail.com> wrote in message
news:91a1e5b4-dd7a-4511-b82a-cd3d228ce91c@l64g2000hse.googlegroups.com...

I bought a USB drive, and it came in a small plastic shell package.
Is it safe to assume that this plastic material is anti-static?

I have not yet opened the package, and I left it sitting on the carpet
for a while. Then I did some vacuuming, and ran the vacuum cleaner
right next to the package.

Assuming that the plastic package is anti-static, would that be enough
to prevent the USB drive from being damaged by any static electricity
generated from the carpet or the vacuum cleaner?

My USB stick lives in a pocket with various other junk (keys etc), the
protective cap frequently falls off - its been carried about in my pocket
for over a year and still works fine.

 

[ian field]

<sg858585@gmail.com> wrote in message
news:f23c82ef-534f-4ddb-928f-6bde073d4895@d77g2000hsb.googlegroups.com...

hi thanks for ur kind informaation which u have given,but i want to
know a small detail about bridge rectifier.
DOES THIS BRIDGE RECTIFIER CAN BE CONNECTED TO THE 230VAC/50HZ
SUPPLY ,WITH OUT TRANSFORMER. DOES IT WITH STAND THAT VOLTAGE OR IT
HAS TO ISOLATED.IF IT HAS ISOLATE YELL ME THE ALTERNATE METHOD TO
ISOLATE WITH OUT TRANSFORMER. KINDLY SOLVE THIS PROBLEM


THANK U

The best way is to find the data sheet and look up the PIV rating.

The bodgers method is to connect the rectifier to the mains with a 40W lamp
in series with one of the AC terminals to limit the current if the rectifier
goes S/C.

Of course this does not guarantee that the rectifier will withstand mains
voltage in the long term - you need to have a safety margin. If you put 2
capacitors in series across the + & - terminals with the junction to one of
the AC terminals, this forms a voltage doubler.

A better plan yet is to use a resistor to limit current instead of the lamp
(33k/2W for 110V or 2x 33k/2W for 230V) this should limit the current to the
point that if the PIV is exceeded, the rectifier's leakage current will halt
the voltage rise - which you can then measure.

 

[Tom Biasi]

<msadkins04@yahoo.com> wrote in message
news:5681c686-52cc-4dc6-9cc3-31a9a5873c45@b1g2000hsg.googlegroups.com...

I recall from my hobbyist days that high-voltage capacitors can keep
a
charge for a significant amount of time, especially if they are cut-
off from the circuit subsequent to charging.

I'd like to know whether it might be feasible, from a technical and
cost-effectiveness standpoint, to use a large array of specially
designed capacitors to hold excess electricity from solar plants for,
say, periods of a week or more. This would be helpful to bridge gaps
caused by overcast weather, and for other purposes.

I know that there are potentially problems with dielectric breakdown
and leakage. On the plus side, such "batteries" could be charged and
discharged very quickly, as needed, and without the complications
involved in conventional batteries using chemical electrolytes.

I agree with Michael Black, you are asking capacitors to do the job of power
cells.
Capacitors can supply large amounts of energy for short periods of time or
small amounts of energy for longer times.
Look at the discharge curves of a charged capacitor as it discharges through
your load. You can find this info if you search.
Regards,
Tom

 

[ian field]

"Eeyore" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:4874F1E4.939C3065@hotmail.com...

ian field wrote:

sg858585@gmail.com> wrote in message
news:f23c82ef-534f-4ddb-928f-6bde073d4895@d77g2000hsb.googlegroups.com...
hi thanks for ur kind informaation which u have given,but i want to
know a small detail about bridge rectifier.
DOES THIS BRIDGE RECTIFIER CAN BE CONNECTED TO THE 230VAC/50HZ
SUPPLY ,WITH OUT TRANSFORMER. DOES IT WITH STAND THAT VOLTAGE OR IT
HAS TO ISOLATED.IF IT HAS ISOLATE YELL ME THE ALTERNATE METHOD TO
ISOLATE WITH OUT TRANSFORMER. KINDLY SOLVE THIS PROBLEM


THANK U

The best way is to find the data sheet and look up the PIV rating.

Or Vrrm as they tend to call it these days.

Graham

Damn these new fangled contraptions!

 

[Tom Biasi]

"RobZ" <robert.zeilinga@standardbank.co.za> wrote in message
news:f7f33bc7-f631-473d-8286-fe642c81418c@b1g2000hsg.googlegroups.com...

hi all, perhaps slightly off-topic but you guys have been of great
help to me in the past

When I was installing a new hifi amplifier last night, I realised i
was being lightly shocked each time i touched the amplifier case while
I was lying on the floor trying to re-route some of the speaker cables
behind the cabinet.

The shock is definatly between the Amplifier case and the floor. ( no
frayed wires - new installation)

the amplifier is supplied by a 2 wire AC cord ( no earth )

what is causing this?

is this dangerous?

Can I add my own earth wire ( joined to the case, to the AC socket
Earth Plug) ? will this help?

You are experiencing a ground loop.
The potentials of the grounds that you are touching are not the same.
The amp could be defective or just has some chassis potential.
Putting everything on the same ground ("earth" across the pond) will
eliminate this unless the amp is defective.
Tom

 

[christofire]

"RobZ" <robert.zeilinga@standardbank.co.za> wrote in message
news:f7f33bc7-f631-473d-8286-fe642c81418c@b1g2000hsg.googlegroups.com...

hi all, perhaps slightly off-topic but you guys have been of great
help to me in the past

When I was installing a new hifi amplifier last night, I realised i
was being lightly shocked each time i touched the amplifier case while
I was lying on the floor trying to re-route some of the speaker cables
behind the cabinet.

The shock is definatly between the Amplifier case and the floor. ( no
frayed wires - new installation)

the amplifier is supplied by a 2 wire AC cord ( no earth )

what is causing this?

is this dangerous?

Can I add my own earth wire ( joined to the case, to the AC socket
Earth Plug) ? will this help?

Is it just a half-mains alternating voltage buzz? ... like you get when you
touch the metal casing of a two-wire-fed mains appliance like the Sony
DAV-S880 home theatre system, in which internal suppression/filter
capacitors connected to line and neutral appear to have their other ends
connected to the exposed metalwork!

If so, the current involved is too small to be dangerous - it's just a bit
disconcerting. You could add an earth but if the equipment is designed for
two-wire power then it probably isn't designed to have earth applied (others
will probably advise in this respect in more detail), and this could give
rise to an earth loop if anything feeding your amplifier is itself earthed.

Chris

 

[Chungho@zongazonga.zip]

hi all, perhaps slightly off-topic but you guys have been of great
help to me in the past

When I was installing a new hifi amplifier last night, I realised i
was being lightly shocked each time i touched the amplifier case while
I was lying on the floor trying to re-route some of the speaker cables
behind the cabinet.

The shock is definatly between the Amplifier case and the floor. ( no
frayed wires - new installation)
the amplifier is supplied by a 2 wire AC cord ( no earth )
what is causing this?
is this dangerous?

Can I add my own earth wire ( joined to the case, to the AC socket
Earth Plug) ? will this help?

Sounds like either your receptical is wired incorrectly or your
amp doesn't have a polarized plug and is plugged in reversed or
you have a problem inside your amp (system)....

The tingle you feel is because you aren't fully grounded yourself
as you are lying on the floor.. you are partially insulated.. But,
yes, this dangerous.. If you were to grab something that is
plugged in properly... You will complete the circuit causing a
severe shock... as it is likely that the case of something else
will/may be fully grounded.

This can also cause damage to your system if you connect one
item to a properly wired socket and another item in a socket that is
reversed. At minimum... you'll see sparks.

Inspect your wall sockets. The small blade is HOT and the large
blade is neutral/Ground (For US of course). Use a meter to measure
voltage in the socket. You should only get 120 (+ or - a few volts) ONLY if
the small blade is connected to one of your test leads. Otherwise the
socket is NOT wired correctly. Radio Shack used to sell three light
outlet testers.. (Basically three neon bulbs connected to the outlet blades
delta style...) not sure if they still do.... But try them out if your have
one near you...

One assumption that is always good to have is, if something is shocking
you, then it's a bad thing.


--
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------------------- ----- ---- -- -

 

[Phil Allison]

<emailaddress@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.

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.

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

 

[Phil Allison]

<emailaddress@insightbb.com>
"Phil Allison" >

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.

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 @ 0.5A for 20VA.

** Read the whole post before making silly replies.

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.

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.

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

 

[Mr.T]

"Joerg" <notthisjoergsch@removethispacbell.net> wrote in message
news:Whcdk.1740$zv7.200@flpi143.ffdc.sbc.com...

Some large Polynesian mussel shells supposedly do that. When you hold
them to your ear you hear the sounds of the sea.

Or the sound of your blood flow being acoustically amplified and reflected
back to your ear, at least.
Try it with a microphone in an anechoic chamber and let us know how you get
on

(and yes, I realise you were being sarcastic, just as I am)

MrT.

 

[Paul E. Schoen]

<emailaddress@insightbb.com> wrote in message
news:c66df173-a1f2-4fbe-a446-c3b0b0b353a9@e39g2000hsf.googlegroups.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?

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.

I find it easier to just simulate the circuit using LTSpice and adjust
values until they are close. I used a voltage source with 18 V peak at 60
Hz, and internal resistance of 0.8 ohms, which is probably about right for
a 20 VA transformer with about 10% regulation at full load of 1.67 A. I
added four rectifiers in FWB, a 2000 uF capacitor, and a 12 ohm load. I got
1.71 amps RMS from the voltage source, which dropped to 11.5 VRMS under
load. The output power is 13.2 watts, with 12.5 VDC as you require, and a
little over one amp. So in this case the 20 VA tranny is derated by
13.2/20, or about 65%. That's right in the range of 0.5 to 0.7 suggested by
Phil. Some of this is due to rectifier losses, which I estimate as about 1
or 2 watts. Schottkys will help a little.

The ASCII file follows. You can take it from there.

Paul

==========================================================================

Version 4
SHEET 1 880 680
WIRE 240 144 128 144
WIRE 288 144 240 144
WIRE 384 144 352 144
WIRE 416 144 384 144
WIRE 512 144 416 144
WIRE 528 144 512 144
WIRE 528 160 528 144
WIRE 128 192 128 144
WIRE 416 192 416 144
WIRE 240 256 240 144
WIRE 288 256 240 256
WIRE 416 256 352 256
WIRE 416 304 416 256
WIRE 528 304 528 240
WIRE 528 304 416 304
WIRE 640 304 528 304
WIRE 640 336 640 304
WIRE 128 368 128 272
WIRE 288 368 128 368
WIRE 384 368 384 144
WIRE 384 368 352 368
WIRE 128 480 128 368
WIRE 288 480 128 480
WIRE 416 480 416 304
WIRE 416 480 352 480
FLAG 640 336 0
FLAG 512 144 V+
SYMBOL diode 288 160 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D2
SYMATTR Value MUR460
SYMBOL polcap 400 192 R0
SYMATTR InstName C2
SYMATTR Value 2000ľ
SYMATTR Description Capacitor
SYMATTR Type cap
SYMATTR SpiceLine V=63 Irms=2.51 Rser=0.025 MTBF=5000 Lser=0 ppPkg=1
SYMBOL voltage 128 176 R0
WINDOW 3 -11 133 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 24 44 Left 0
SYMATTR Value SINE(0 18 60 0 0 0 100)
SYMATTR SpiceLine Rser=.8
SYMATTR InstName V1
SYMBOL res 512 144 R0
SYMATTR InstName R1
SYMATTR Value 12
SYMBOL diode 352 272 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D3
SYMATTR Value MUR460
SYMBOL diode 288 384 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D1
SYMATTR Value MUR460
SYMBOL diode 352 496 M270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName D4
SYMATTR Value MUR460
TEXT 184 528 Left 0 !.tran 1

 

[Phil Allison]

<emailaddress@insightbb.com>

"Phil Allison"
** This is your error.

Phil, isn't this a bit of irony since I clearly asked for something
specific

** And impossible.

Cos you know NOTHING about the topic - fuckhead.

** Don't complain when you get free, good advice - fuckhead.

Phil, how do I put it so you understand?

** Do not ever try that shit on - fuckhead.

I am not interested in what overloads it,

** You need to be - for safety and functional reasons.

This is not a fixed plan for a one-off design Phil, this is a generic
question about transformers.

** You still have not read the whole post - have you ?

What an utter ass you are.

** There is no possible math based only on the figures you gave.

False.

** You have no clue whatever.

Given all the parameters it can be solved.

** So what did I write above ?

The whole
topic is about ACTUAL temp rise of the windings of the particular
transformer under particular conditions.

This is getting amusing.

** You are one, colossally stubborn fool.

** I just supplied the derating figures and the reason why it cannot be
precise.

You used the term "derating", are you saying it is not a derating
figure but rather a conversion figure?

** The figure was fully explained.

It derates the AC current figure to give the available DC current while not
exceeding the VA rating.


I ask because you have gone on
and on about derating for temp, when I am very specifically looking
for only a mathematical equation for maximum transformer capability, a
theoretical one in a model that ignores temperature because the answer
is not going to be directly used to build something!


** You Q is just plain absurd.

Transformer current and VA ratings are ALL about temp rise !!!!!!

The regulation curve ( whether AC or DC plus rectifier /filter ) is a
function of VA and a number of other parameters plus the type of
construction used.


So are you claiming that if I short circuited the rectified output of
a transformer


** Never do that and I have never mentioned any such idiocy.

You are hee hawing like some barnyard ass now - pal.


If the shorted output power is not 20VA

** The power delivered into a short is always zero - fuckhead.

But massive heat build up in the transformer will destroy it - pronto.



because I am not asking about
implementation details for any transformer, only how to convert
mathematically to actual peak, not prudent design values.


** Utterly stupid question, only asked by fools.

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.

If it's complex math so be it.

** You do not need any complex math to size a tranny to do a given job.

Problem here is, you cannot even specify what the job is.

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.

Then the equation(s) used in that would be much closer to what I was
after.

** I don't give a shit what garbage you are after - fuckhead.

What you NEED is a damn good kick up the arse.

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.

With all due respect Phil, you have no idea of the project
requirements beyond what I had told you.


** So you are now making then up as you go along - right ?


A 40VA transformer is not going to work, unless it's one rated slightly
below 12V which is an odd figure.

** Totally insane crap.


Ultimately for that particular project I will most likely use a SMPS
instead, but that project was only an example not the question asked
which is what the equation is to determine AC to DC spec conversion.
I do not need an answer based on the limited specs of the transformer
info I had given, I only need the equation as complete as it can be
for future uses.


** The detailed specs you need are NOT published by makers.

You HAVE to measure individually them after you BUY a transformer !!!

Just as quick to run a couple of bench tests if you have the damn tranny
right there in front of you.


Perhaps we got off on the wrong foot Phil, but please believe someone
when they state plainly what they don't care about and focus on what
they asked.

** Sorry - pal.

But I do not answer ridiculous Qs as specified by arrogant, bullshitting
novices

No-one on usenet EVER has to do that.


I hope I've made things clearer.


** What you have made very clear is that you are an utterly autistic, rabid
nut case.

Who is just not able to accept new information.


...... Phil

 

[Phil Allison]

"Paul E. Schoen"

I find it easier to just simulate the circuit using LTSpice and adjust
values until they are close.

** Fine - but the OP was after a math formula and will never forgive you
for offering him a LTSpice solution.

I used a voltage source with 18 V peak at 60 Hz, and internal resistance
of 0.8 ohms, which is probably about right for a 20 VA transformer with
about 10% regulation at full load of 1.67 A. I added four rectifiers in
FWB, a 2000 uF capacitor, and a 12 ohm load. I got 1.71 amps RMS from the
voltage source, which dropped to 11.5 VRMS under load. The output power is
13.2 watts, with 12.5 VDC as you require, and a little over one amp. So in
this case the 20 VA tranny is derated by 13.2/20, or about 65%. That's
right in the range of 0.5 to 0.7 suggested by Phil.

** Amazing ......

But you will have 3 volts p-p ripple with that puny 2000uF cap.

Means the DC voltage falls well below 12, a lot.

Formula: I = C dv/dt ( where dt = 6mS )

Plus a 20 VA tranny has typically 15% regulation - not 10 %.

Worse if it heats a lot.

The ASCII file follows. You can take it from there.

** Never assume folk have LT Spice available or the ability to use it.



...... Phil

 

[Paul E. Schoen]

"Phil Allison" <philallison@tpg.com.au> wrote in message
news:6dlot6F37es1U1@mid.individual.net...

"Paul E. Schoen"


I find it easier to just simulate the circuit using LTSpice and adjust
values until they are close.

** Fine - but the OP was after a math formula and will never forgive
you for offering him a LTSpice solution.

Well, Spice is really a math formula, if you dig deep enough...

I used a voltage source with 18 V peak at 60 Hz, and internal resistance
of 0.8 ohms, which is probably about right for a 20 VA transformer with
about 10% regulation at full load of 1.67 A. I added four rectifiers in
FWB, a 2000 uF capacitor, and a 12 ohm load. I got 1.71 amps RMS from
the voltage source, which dropped to 11.5 VRMS under load. The output
power is 13.2 watts, with 12.5 VDC as you require, and a little over one
amp. So in this case the 20 VA tranny is derated by 13.2/20, or about
65%. That's right in the range of 0.5 to 0.7 suggested by Phil.


** Amazing ......

But you will have 3 volts p-p ripple with that puny 2000uF cap.

Means the DC voltage falls well below 12, a lot.

Formula: I = C dv/dt ( where dt = 6mS )

You're right. Schottky diodes and a 2200 uF capacitor are just barely
adequate. And then one must figure lowest line voltage as well. That's
where switchers are often the way to go.

Plus a 20 VA tranny has typically 15% regulation - not 10 %.

Worse if it heats a lot.


The ASCII file follows. You can take it from there.


** Never assume folk have LT Spice available or the ability to use it.

Why not? It's free! And fairly easy. But I'd like a version that actually
shows components smoking or blowing up (with appropriate sound effects)
when they are overloaded.

Paul