Chip with simple program for Toy

[Larry Brasfield]

"Bill Bowden" <wrongaddress@att.net> wrote in message news:1117502772.588863.8560@o13g2000cwo.googlegroups.com...

i thought it's the current that kills not the voltage.
the voltage is just to get past the skin resistance to
your heart. I read somewhere if the current at about 0.030 amp
would even stop your heart.

No, it's the power that kills, which is the
voltage squared divided by the resistance.

So, if your skin resistance is a Megohm,
you only get 120^2/1 Meg = 14 milliwatts which is not much.

But, if your hands are wet and the skin resistance is only 50K or so,
you get E^2/R =120^2/50K = 288 milliwatts, which might be dangerous.

It's not power that does people in, except in the
sense that, without power, there is no effect.

What kills most electrocuted people is interference
with their heart's operation due to disruption of its
"electrical" control system.

Something people seem to be missing in this
discussion is that AC is much more dangerous
than DC, due to its ability to disrupt normal
heart operation at lower current levels. (This is
why defibrillators, which pass a large unipolar
current pulse thru the chest, a curative rather
than lethal.)

Of course raw power can kill, too. But not often.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Bob Monsen]

smokie wrote:

What happens to the output voltage when the 555 timer supplies higher
current? how good is the program multisim? Would it be worth the
investment?

My guess is that the only effect that drawing more current from the
output of a 555 will have is that it'll heat it up. However, the timing
will probably not be affected; the timing is based on a comparison
between a three-way internal resistive divider, and the trigger and
threshold inputs. That won't change much when the thing heats up.

As to the simulator, I'd try LTSpice first, from www.linear.com. It's free.

---
Regards,
Bob Monsen

 

[Rich Webb]

On 30 May 2005 18:32:41 -0700, "smokie" <stillbreezy@gmail.com> wrote:

What happens to the output voltage when the 555 timer supplies higher
current?

The output voltage droops (Vcc - Vout gets larger).

--
Rich Webb Norfolk, VA

 

[John Fields]

On Mon, 30 May 2005 18:39:58 -0400, John Popelish <jpopelish@rica.net>
wrote:

John Fields wrote:
On Mon, 30 May 2005 17:21:37 -0400, John Popelish <jpopelish@rica.net
wrote:


Don Klipstein wrote:
(snip)

I do not believe the risk of fibrillation drops to zero when the current
decreases to 99 or 49 mA. I have heard of someone getting killed by a 30
mA neon sign transformer.

(snip)

Which puts out about twice that on into a low impedance load, like the
human body.


---
It can't, since it's designed to saturate at 30mA, but looking at its
load line it'll put out, say, 15kV at 0A into an open, and 0V at 30mA
into a dead short.

That means that with an initial 500kohm human load across it its
output voltage will drop to 7500VRMS and it'll be forcing 15mA through
the load, and the load will be dissipating about 113 watts. OUCH!!!

You may be right, but that is not what I understand to be the case. I
thought they are rated for normal load current over the voltage range
expected when driving a gas tube. And that they have a very nonlinear
current limit, much like Sola ferro resonant constant voltage
transformers. Those hold specified voltage regulation at rated
current but the current increases only to about double rated under a
short circuit. I don't think most neon sign transformers are just a
tightly coupled step up transformer in series with a big resistor.

---
AIUI, they're designed to be ballasts. That is, to start off with a
high enough voltage to strike the arc through the tube and then to
provide the proper current (30 or 60 mA) to run the tube when the gas
ionizes and provides a more or less constant low-resistance load
through the plasma.

Using a Transco T1512, which is a 15kV 30mA NST, like this:


Iout->
120VAC>----+ +---------+ <-----+
)||( | |
)|| >-GND [R] Eout
)||( | |
120VAC>----+ +---------+ <-----+

I get the following data:

R Eout Iout
M kV pk mA RMS
-----+-------+--------
0.00 0.000 17.3
0.05 0.662 17.3
0.09 1.647 17.1
0.13 2.586 16.8
0.17 3.456 16.3
0.21 4.270 15.9
0.25 5.02 15.5
0.29 5.744 15.1
0.33 6.37 14.7
0.37 6.971 14.3
0.41 7.511 13.8

2.0 10.0 6.6
4.0 10.8 5.5
5.0 11.0 5.3
10.0 11.37 5.2
20.0 11.55 5.1

INF 15.0 0.0

Which shows there's no voltage regulation at all, but only a 3.5mA
change in current for a 7500V change in voltage across the load from
410kohms to a dead short.

The reason the short-circuit current is 17.3 mA instead of 30 is
because I loaded the entire secondary, and it's wired so you can only
get 15mA or so that way. If you take current from either end of the
secondary to the center tap you can get 15mA out of each half of the
secondary for a total of about 30mA.


--
John Fields
Professional Circuit Designer

 

[John Larkin]

On Tue, 31 May 2005 14:09:58 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Mon, 30 May 2005 18:39:58 -0400, John Popelish <jpopelish@rica.net
wrote:

John Fields wrote:
On Mon, 30 May 2005 17:21:37 -0400, John Popelish <jpopelish@rica.net
wrote:


Don Klipstein wrote:
(snip)

I do not believe the risk of fibrillation drops to zero when the current
decreases to 99 or 49 mA. I have heard of someone getting killed by a 30
mA neon sign transformer.

(snip)

Which puts out about twice that on into a low impedance load, like the
human body.


---
It can't, since it's designed to saturate at 30mA, but looking at its
load line it'll put out, say, 15kV at 0A into an open, and 0V at 30mA
into a dead short.

That means that with an initial 500kohm human load across it its
output voltage will drop to 7500VRMS and it'll be forcing 15mA through
the load, and the load will be dissipating about 113 watts. OUCH!!!

You may be right, but that is not what I understand to be the case. I
thought they are rated for normal load current over the voltage range
expected when driving a gas tube. And that they have a very nonlinear
current limit, much like Sola ferro resonant constant voltage
transformers. Those hold specified voltage regulation at rated
current but the current increases only to about double rated under a
short circuit. I don't think most neon sign transformers are just a
tightly coupled step up transformer in series with a big resistor.

---
AIUI, they're designed to be ballasts. That is, to start off with a
high enough voltage to strike the arc through the tube and then to
provide the proper current (30 or 60 mA) to run the tube when the gas
ionizes and provides a more or less constant low-resistance load
through the plasma.

Using a Transco T1512, which is a 15kV 30mA NST, like this:


Iout-
120VAC>----+ +---------+ <-----+
)||( | |
)|| >-GND [R] Eout
)||( | |
120VAC>----+ +---------+ <-----+

I get the following data:

R Eout Iout
M kV pk mA RMS
-----+-------+--------
0.00 0.000 17.3
0.05 0.662 17.3
0.09 1.647 17.1
0.13 2.586 16.8
0.17 3.456 16.3
0.21 4.270 15.9
0.25 5.02 15.5
0.29 5.744 15.1
0.33 6.37 14.7
0.37 6.971 14.3
0.41 7.511 13.8

2.0 10.0 6.6
4.0 10.8 5.5
5.0 11.0 5.3
10.0 11.37 5.2
20.0 11.55 5.1

INF 15.0 0.0

Which shows there's no voltage regulation at all, but only a 3.5mA
change in current for a 7500V change in voltage across the load from
410kohms to a dead short.

The reason the short-circuit current is 17.3 mA instead of 30 is
because I loaded the entire secondary, and it's wired so you can only
get 15mA or so that way. If you take current from either end of the
secondary to the center tap you can get 15mA out of each half of the
secondary for a total of about 30mA.

What's the DCR of the secondary?

John

 

[John Fields]

On Tue, 31 May 2005 12:14:53 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 31 May 2005 14:09:58 -0500, John Fields
jfields@austininstruments.com> wrote:

On Mon, 30 May 2005 18:39:58 -0400, John Popelish <jpopelish@rica.net
wrote:

John Fields wrote:
On Mon, 30 May 2005 17:21:37 -0400, John Popelish <jpopelish@rica.net
wrote:


Don Klipstein wrote:
(snip)

I do not believe the risk of fibrillation drops to zero when the current
decreases to 99 or 49 mA. I have heard of someone getting killed by a 30
mA neon sign transformer.

(snip)

Which puts out about twice that on into a low impedance load, like the
human body.


---
It can't, since it's designed to saturate at 30mA, but looking at its
load line it'll put out, say, 15kV at 0A into an open, and 0V at 30mA
into a dead short.

That means that with an initial 500kohm human load across it its
output voltage will drop to 7500VRMS and it'll be forcing 15mA through
the load, and the load will be dissipating about 113 watts. OUCH!!!

You may be right, but that is not what I understand to be the case. I
thought they are rated for normal load current over the voltage range
expected when driving a gas tube. And that they have a very nonlinear
current limit, much like Sola ferro resonant constant voltage
transformers. Those hold specified voltage regulation at rated
current but the current increases only to about double rated under a
short circuit. I don't think most neon sign transformers are just a
tightly coupled step up transformer in series with a big resistor.

---
AIUI, they're designed to be ballasts. That is, to start off with a
high enough voltage to strike the arc through the tube and then to
provide the proper current (30 or 60 mA) to run the tube when the gas
ionizes and provides a more or less constant low-resistance load
through the plasma.

Using a Transco T1512, which is a 15kV 30mA NST, like this:


Iout-
120VAC>----+ +---------+ <-----+
)||( | |
)|| >-GND [R] Eout
)||( | |
120VAC>----+ +---------+ <-----+

I get the following data:

R Eout Iout
M kV pk mA RMS
-----+-------+--------
0.00 0.000 17.3
0.05 0.662 17.3
0.09 1.647 17.1
0.13 2.586 16.8
0.17 3.456 16.3
0.21 4.270 15.9
0.25 5.02 15.5
0.29 5.744 15.1
0.33 6.37 14.7
0.37 6.971 14.3
0.41 7.511 13.8

2.0 10.0 6.6
4.0 10.8 5.5
5.0 11.0 5.3
10.0 11.37 5.2
20.0 11.55 5.1

INF 15.0 0.0

Which shows there's no voltage regulation at all, but only a 3.5mA
change in current for a 7500V change in voltage across the load from
410kohms to a dead short.

The reason the short-circuit current is 17.3 mA instead of 30 is
because I loaded the entire secondary, and it's wired so you can only
get 15mA or so that way. If you take current from either end of the
secondary to the center tap you can get 15mA out of each half of the
secondary for a total of about 30mA.



What's the DCR of the secondary?

---

+-----S1
|
7K1
|
+----CT
|
7K1
| __
+-----S2




--
John Fields
Professional Circuit Designer

 

[Padu]

I don't know about lowest power consuming, but I'm running mine out of
regular RC batteries wired in series. The batteries give me something
around
12.5V. It is connected to a voltage regulator that regulates at 12V that
is
connected to an ATX DC-DC board, that will condition the raw 12VDC to
all
voltages that the epia (or any other PC for that matter) requires.

How long do they last?

Depends on how you wire your package.

For example, if you use a 60W DC-DC @ 12VDC board, then you will need 60/12
amps = 5A (ohms law). If you want 10 hours, then you need 50Ah. With NiMH
GP3700 thats equivalent to 14 packs of 11 cells in series (160 sub-c
cells!!!). Of course 60W is the peak consumption, it will greatly depend on
what you're doing and what's the consumption of your other components... but
that's how you calculate.

With this weight, it is by no means portable. To achieve higher energy
density, you'll have to use more sophisticated sources of energy, as for
example Lithium Polymer (LiPo). You can achieve the same power with less
weight, but spending considerable more money.

 

[John Larkin]

On Tue, 31 May 2005 15:16:43 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Tue, 31 May 2005 12:14:53 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 31 May 2005 14:09:58 -0500, John Fields
jfields@austininstruments.com> wrote:

On Mon, 30 May 2005 18:39:58 -0400, John Popelish <jpopelish@rica.net
wrote:

John Fields wrote:
On Mon, 30 May 2005 17:21:37 -0400, John Popelish <jpopelish@rica.net
wrote:


Don Klipstein wrote:
(snip)

I do not believe the risk of fibrillation drops to zero when the current
decreases to 99 or 49 mA. I have heard of someone getting killed by a 30
mA neon sign transformer.

(snip)

Which puts out about twice that on into a low impedance load, like the
human body.


---
It can't, since it's designed to saturate at 30mA, but looking at its
load line it'll put out, say, 15kV at 0A into an open, and 0V at 30mA
into a dead short.

That means that with an initial 500kohm human load across it its
output voltage will drop to 7500VRMS and it'll be forcing 15mA through
the load, and the load will be dissipating about 113 watts. OUCH!!!

You may be right, but that is not what I understand to be the case. I
thought they are rated for normal load current over the voltage range
expected when driving a gas tube. And that they have a very nonlinear
current limit, much like Sola ferro resonant constant voltage
transformers. Those hold specified voltage regulation at rated
current but the current increases only to about double rated under a
short circuit. I don't think most neon sign transformers are just a
tightly coupled step up transformer in series with a big resistor.

---
AIUI, they're designed to be ballasts. That is, to start off with a
high enough voltage to strike the arc through the tube and then to
provide the proper current (30 or 60 mA) to run the tube when the gas
ionizes and provides a more or less constant low-resistance load
through the plasma.

Using a Transco T1512, which is a 15kV 30mA NST, like this:


Iout-
120VAC>----+ +---------+ <-----+
)||( | |
)|| >-GND [R] Eout
)||( | |
120VAC>----+ +---------+ <-----+

I get the following data:

R Eout Iout
M kV pk mA RMS
-----+-------+--------
0.00 0.000 17.3
0.05 0.662 17.3
0.09 1.647 17.1
0.13 2.586 16.8
0.17 3.456 16.3
0.21 4.270 15.9
0.25 5.02 15.5
0.29 5.744 15.1
0.33 6.37 14.7
0.37 6.971 14.3
0.41 7.511 13.8

2.0 10.0 6.6
4.0 10.8 5.5
5.0 11.0 5.3
10.0 11.37 5.2
20.0 11.55 5.1

INF 15.0 0.0

Which shows there's no voltage regulation at all, but only a 3.5mA
change in current for a 7500V change in voltage across the load from
410kohms to a dead short.

The reason the short-circuit current is 17.3 mA instead of 30 is
because I loaded the entire secondary, and it's wired so you can only
get 15mA or so that way. If you take current from either end of the
secondary to the center tap you can get 15mA out of each half of the
secondary for a total of about 30mA.



What's the DCR of the secondary?

---

+-----S1
|
7K1
|
+----CT
|
7K1
| __
+-----S2

So it must be leakage inductance that limits the current.

John

 

[Dorothy Bradbury]

Power saving is an important factor in calculations...
o VIA C3 is low power
o P-M gets a lot of its laptop capability from power saving

Note P-M-Celeron lacks some of the power saving capability.

Some interesting sites to visit...
o http://www.ipcmax.com
---- they do micro-PCs in 12V power, super-small, fully usable units
o uk.adverts.computer
---- someone sells LI-Ion batteries & chargers at very low prices
---- *generic* LI-Ion external batteries for any application & chargers

A note is LI-Ion is near-explosive without the correct charger.

Otherwise, again, component wise you can do whatever you want.
It may end up quite pricey - or quite limited in scope, up to you.

Limitation will be in the "notebook packaging" - probably insurmountable.
--
Dorothy Bradbury

 

[petrus bitbyter]

"Patrick" <pdebella@aol.com> schreef in bericht
news:1117555926.885143.109690@g44g2000cwa.googlegroups.com...

Hello!

I was wondering if anyone has or knows where I can find a simple, cheap
current sensing circuit that I could use to sense AC currents. I am
currently working on a power control project that will monitor 120 VAC
outlets with a total maximum current of 15 amps. We are currently
working with a current sensor that outputs an AC voltage (actually a
varying DC voltage since it does not drop below zero) from an AC input
current from 0-50A. When the input current is zero the output is about
2.5V. At the full rated input current of 50 amps the sensor output has
a peak voltage swing of about 2.5V which ranges from zero to 5V at the
full 50 amps (swings around the 2.5V q-point).

The current sensor works great, but the problem is that we need to find
out what the actual current is and this poses somewhat of a challenge
since the output of the sensor is a varying DC (sinusoidal) voltage.
We've tried some software techniques to capture the peaks, but it takes
up too much processor time and is therefore not feasible.

Anyone have any ideas or circuits I might be able to try? Any help
will be greatly appreciated.

Patrick

Patrick,

What about the electronic skills of your team? A capacitor is enough to
separate DC from AC although it might be a huge one for 50/60Hz. If that's
not applicable a simple opamp suffices to subtract the 2.5VDC from the
signal leaving the AC-part for further processing. Of course you can find
complete measuring systems on the net. Tektronix sells some great ones but I
bet they are pretty expensive in relation to your budget. Current sensing
devices with Hall effect sensors often has a half Vs offset. So if you buy
one you will have the same problem. AMPLOC makes them for instance. Some of
them however have integrated electronics to remove the offset.

petrus bitbyter

 

[Don Kelly]

<albertleng@gmail.com> wrote in message
news:1117551965.351978.28820@f14g2000cwb.googlegroups.com...

Dear all experts,


1) If i were to connect a simple circuit to determine the effects of
capacitance in Maxim Power Transfer, how will the capacitance values
affect the Maximum Power? What's the best capacitance to be used?
-------------

0, providing that the source has negligable inductance. have you been asked
to determine the best capacitance to use when the source is inductive. It
can be found -If this is homework, you will have to work it out.

2) In Nodal Analysis, how can the frequency of the signals affect the
circuit currents? How can we minimise the currents with different
frequencies instead of using resistor?
-------

Nodal analysis is at a single frequency. All that frequency will do will is
to affect L and C impedances. Hint- look up parallel resonance.
----------

3) In a simple circuit which has a load, what's the simplest way to
know its impedance? What's the difference between resistance &
impedance?
---------

a) measure it.
b) impedance includes the wL and =-1/wC terms where w =2*pi*frequency
Z=R+j(wL-1/wC)

If you are asking these questions, you are missing the basis of AC circuit
analysis and need to go back to review it.
Get a textbook or Schaum's Outlines on circuit analysis.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

Thanks... sorry for asking simple and silly questions.

Regards,
Albert

 

[Joel Kolstad]

"BradBrigade" <sizzlefist@hotmail.com> wrote in message
news:1117228206.262895.241380@f14g2000cwb.googlegroups.com...

But now I'm wondering, what is the
purpose of the transformer? If you want to convert 100V to 10V, why
not filter the output straight from a PWM with a 10% duty cycle?
What's the difference?

Your "100V input" is typically going to vary between, say, 95-105V.
Similarly, depending on just how heavy the load on the "10V" side is, the
extra current will cause various losses so that your output would tend to vary
between, say, 9.5-10.5V even with a perfect 100V input. (How well a power
supplies copes with the former problem is called "source regulation" and how
well it copes with the laters is "load regulation.")

You do occasionally see some really cheap 12V->24V or 12V->6V DC/DC converters
out there that use a fixed ~50% duty cycle and figure the output will be
"close enough."

---Joel Kolstad

 

[Joel Kolstad]

Hi W_Tom,

"w_tom" <w_tom1@hotmail.com> wrote in message
news:42979DF8.390CDDFA@hotmail.com...

That's correct and consistent with what I had posted.

Hmm... if you say so. That's sure not how I read your post.

As noted
earlier, this can cause increased energy dissipation in the
transformer and elsewhere.

Yes, certainly true, but not really relevant to what a beginner needs to know.

An unloaded power supply must not be damaged by no load.

For a bench supply or even a PC power supply, I'm tempted to agree with you.
For embedded applications in cut-throat markets (TVs are a very good example),
it's hard to ignore the cost savings that can be achieved by not bothering to
make your power supply unconditionally stable when you 'know' there will
always be a minimum load around.

Any power supply that is damaged by a no load condition is
typical of something bought by a bean counter - the enemy of
innovators, responsible manufacturers, and those educated in
computer electronics.

It's the consumer who pushes the bean counters to skimp on parts count and/or
quality. Although the US is a wealthy country that can readily afford to pay
a couple bucks extra for a better computer power supply, this isn't true in
all parts of the world (e.g., China). It's not the kind of engineering work I
want to do, but I can see the justification for designing these really awful
PC power supplies that cost literally no more than $10 but can readily blow-up
if you look at them crosseyed.

But a power supply
designer better damn well understand the principles of filter
design.

Mnay people get along just fine with being able to analyze single section
filters (both the main L-C 'power' filter and something cheesy like an R-C
feedback loop filter). Now, you may consider than the "principles of filter
design," but personally I would say that someone well versed in such
"principles" is more like to be able to spout off about Chebychev filters,
group delay, Butterworth pole positions -- that sort of thing -- than just
what simple LC and RC filters do. (I realize many power supplies do have
"fancy" filters in the feedback path, I'm just saying plenty of them don't and
I don't think you really need to know that much about filter design to make a
workable switcher.)

---Joel

 

[DaveC]

On Sun, 29 May 2005 16:45:33 -0700, Bill Bowden wrote
(in article <1117410333.066350.174860@g44g2000cwa.googlegroups.com&gt:

I'm not reading a news source, just stating experience.
I've been shocked many times from line voltages with
no ill effects. Once, I was shocked by 10KV from
an aviation radar system, and it threw me across the room,
but I got up and went back to work.

Your experience notwithstanding, people die from 110 vac. You're healthy and
lucky. Not everybody is.
--
Please, no "Go Google this" replies. I wouldn't
ask a question here if I hadn't done that already.

DaveC
me@privacy.net
This is an invalid return address
Please reply in the news group

 

[John Fields]

On 31 May 2005 18:27:50 -0700, "tom" <tmcmurtrie@keystone.otago.ac.nz>
wrote:

Hello All

I was wondering if anyone can help with or knows where I can find a
random ON/OFF timer circuit that will switch a 6-12v DC relay ON
between a random period of say 5 to 20secs and then turn OFF for
another random period of say 5 to 15secs.

---
So, you want it to turn ON for a random interval of not shorter than 5
and not longer than 20 seconds, and then at the end of the ON period
you want it to turn OFF for a random interval of not shorter than 5
and not longer than 15 seconds, and then you want the cycle to repeat
forever?


--
John Fields
Professional Circuit Designer

 

[w_tom]

Somewhere was an 85(?) page introduction (from linear.com)
to switching power supply design. Up front was a flow chart
that asked if one wants to design a supply? The Yes path lead
to a message that asked, "Are you Nuts?" Power supply design
is complex. As demonstrated here, one must have knowledge of
filters, feedback systems, principles of power systems, EMC,
UL approval, FCC regulations, etc. A beginner must first
decide how much will be learn before his eyes disappear into
the back of his head.

If consumer pushed bean counters to reduce costs, then Honda
and Toyota would not be dominant and growing car companies.
Bean counters typically increase product costs - at least in
the long term. It explains why GM has so many problems. Cost
controls increase costs. To reduce costs, obtain more
customers, create new markets, increase market share, create
jobs, create wealth, increase product efficiency, etc... all
require the most important thing - innovation. The US is a
miser as are most every other nation. To sell in the US or
anywhere else, the power supply manufacturer must innovate -
not cost control. Meanwhile the US does have standards that
some other nations do not have - for power supply performance,
reliability, safety, emissions, and .... Europe has even
tougher standards.

Most supplies, to be profitable, must be designed to operate
anywhere in the world - either as a universal supply or with
different options for different regions. Just another example
of innovation; the alternative being bankruptcy. Just another
reason why (outside of niche markets), the power supply
designs must meet fairly universal world standards.

Tell us about harmonics? Do we solve this problem with
filters, or what else? Will a power supply create too much
harmonics AND will it operate when line harmonics are high?
I recall the Intel spec that even demands output transient
response - another factor in a feedback control system
design. Power supply design is not about costs. It is about
innovation - where costs are only one small part of a
profitable design. The only way to cut costs and remain
competitive - innovation.

A large market for inferior supplies exists that create
problems such as computer system damage and intermittent
failures. All power supply outputs must even be shorted and
still not damage the supply. This too has been defacto
standard for many decades.

We have demonstrated how much is basic information on power
supplies. IOW, "your nuts" in that flow chart should be
appreciated. 85(?) pages in that introduction paper
demonstrates how complex a switching power supply really is
and why a properly constructed supply selling for only $65
retail is a marvel of free market economics.

However when purchasing a $40 retail supply, then ask what
critical functions were forgotten to sell that supply only on
price. Yes, some sell power supplies only using cost
controls.

Joel Kolstad wrote:

Hi W_Tom,
"w_tom" <w_tom1@hotmail.com> wrote in message
news:42979DF8.390CDDFA@hotmail.com...
That's correct and consistent with what I had posted.

Hmm... if you say so. That's sure not how I read your post.

As noted
earlier, this can cause increased energy dissipation in the
transformer and elsewhere.

Yes, certainly true, but not really relevant to what a beginner needs to know.

An unloaded power supply must not be damaged by no load.

For a bench supply or even a PC power supply, I'm tempted to agree with you.
For embedded applications in cut-throat markets (TVs are a very good example),
it's hard to ignore the cost savings that can be achieved by not bothering to
make your power supply unconditionally stable when you 'know' there will
always be a minimum load around.

Any power supply that is damaged by a no load condition is
typical of something bought by a bean counter - the enemy of
innovators, responsible manufacturers, and those educated in
computer electronics.

It's the consumer who pushes the bean counters to skimp on parts count and/or
quality. Although the US is a wealthy country that can readily afford to pay
a couple bucks extra for a better computer power supply, this isn't true in
all parts of the world (e.g., China). It's not the kind of engineering work I
want to do, but I can see the justification for designing these really awful
PC power supplies that cost literally no more than $10 but can readily blow-up
if you look at them crosseyed.

But a power supply
designer better damn well understand the principles of filter
design.

Mnay people get along just fine with being able to analyze single section
filters (both the main L-C 'power' filter and something cheesy like an R-C
feedback loop filter). Now, you may consider than the "principles of filter
design," but personally I would say that someone well versed in such
"principles" is more like to be able to spout off about Chebychev filters,
group delay, Butterworth pole positions -- that sort of thing -- than just
what simple LC and RC filters do. (I realize many power supplies do have
"fancy" filters in the feedback path, I'm just saying plenty of them don't and
I don't think you really need to know that much about filter design to make a
workable switcher.)

---Joel

 

[Don Klipstein]

In article <1kbp911o7sf0bgf1v4jjdstaltfm8gn80q@4ax.com>, John Fields wrote:

On Mon, 30 May 2005 18:39:58 -0400, John Popelish <jpopelish@rica.net
wrote:

John Fields wrote:
On Mon, 30 May 2005 17:21:37 -0400, John Popelish <jpopelish@rica.net
wrote:


Don Klipstein wrote:
(snip)

I do not believe the risk of fibrillation drops to zero when the current
decreases to 99 or 49 mA. I have heard of someone getting killed by a 30
mA neon sign transformer.

(snip)

Which puts out about twice that on into a low impedance load, like the
human body.

I failed to track who said this, but a 30 mA neon sign transformer
pushes only a little more than 30 mA into a short circuit in my
experience.

- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In article <1kbp911o7sf0bgf1v4jjdstaltfm8gn80q@4ax.com>, John Fields wrote:

On Mon, 30 May 2005 18:39:58 -0400, John Popelish <jpopelish@rica.net
wrote:

John Fields wrote:
On Mon, 30 May 2005 17:21:37 -0400, John Popelish <jpopelish@rica.net
wrote:


Don Klipstein wrote:
(snip)

I do not believe the risk of fibrillation drops to zero when the current
decreases to 99 or 49 mA. I have heard of someone getting killed by a 30
mA neon sign transformer.

(snip)

Which puts out about twice that on into a low impedance load, like the
human body.


---
It can't, since it's designed to saturate at 30mA, but looking at its
load line it'll put out, say, 15kV at 0A into an open, and 0V at 30mA
into a dead short.

That means that with an initial 500kohm human load across it its
output voltage will drop to 7500VRMS and it'll be forcing 15mA through
the load, and the load will be dissipating about 113 watts. OUCH!!!

What I expect:

15 mA has minimal (but not zero) electrocution risk, although by many
(but not all) accounts is able to cause humans to latch onto shocking
conductors, and become unable to let go and may die from fibrillation
resulting from prolonged shock or may die from breathing muscles being
impaired by the shock.

Main result that I expect from 113 watts of power dissipation into a
human body is burns.

With initial contact resistance of 500K ohms (highish side but fairly
typical of human contact with wires), I expect most of this resistance tio
be in small volumes of skin, that get burned through quickly with 113
watts. That means the victim of such a neon sign transformer shock
quickly gets to dissipate a small fraction of the above-mentioned 113
watts, while conducting a current close to 30 mA - which has a non-zero
chance of being fatal.

Another thing: The output impedance of neon sign transformers is mostly
inductive reactance rather than resistance. Put a 500K-ohm load across
the output of a 15KV 30mA neon sign transformer, and you get more than
7.5 KV at 15 mA, as in closer to 10 KV at 20 mA, ideally 10.3 KV at 20.6
mA.

- Don Klipstein (don@misty.com)

 

[Terry Pinnell]

"BradBrigade" <sizzlefist@hotmail.com> wrote:

Hi,

First of all, I'm trying to figure out how switching power supplies
work (the ones in PCs). I've found very basic info, but I want more
technical stuff. If anyone has some good links please let me know.
These are questions I have yet to find an answer for.

Anyway, here's my question. One thing I read was that the output
voltage of the supply is fed back to the PWM which changes it's duty
cycle accordingly to keep the output voltage constant. But I thought
that the input-to-output ratio of a transformer is fixed. If the PWM
is outputting 100V at 20KHz to a 10:1 transformer, you get out 10V at
20KHz, right? What does it matter what the duty cycle is? It's still
100V at 20KHz. What am I missing?

Second, why does a switching power supply break without a load?

Third, in all my years in electronics, I have never used a choke, now I
see them all over these power supplies. Can someone clue me in about
what they do, and why they are in these things?

I appreciate any info at all. Thanks a lot.

The following thread I started a couple of days ago may also be
helpful.

From: Terry Pinnell <terrypinDELETE@THESEdial.pipex.com>
Newsgroups: sci.electronics.design
Subject: Rating of PC power supplies?
Date: Sun, 29 May 2005 07:10:38 +0100
Message-ID: <3fmi91pcd4ig5uaighsu62dnpdk8a69l91@4ax.com>

In particular, Franc Zabkar posted this link to a detailed schematic:
http://www.pavouk.comp.cz/hw/en_atxps.html

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Terry Pinnell]

"tom" <tmcmurtrie@keystone.otago.ac.nz> wrote:

Hello All

I was wondering if anyone can help with or knows where I can find a
random ON/OFF timer circuit that will switch a 6-12v DC relay ON
between a random period of say 5 to 20secs and then turn OFF for
another random period of say 5 to 15secs.

I'm building an outdoor device and I'm trying to simulate random
movement.

Any help will be greatly appreciated.


Regards Tom McMurtrie

Pending a more elegant solution, you could adapt this circuit I
designed for my 'Surf Synthesiser' many years ago:
http://www.terrypin.dial.pipex.com/Images/Randomiser.gif

That delivers an output duration (simulating gaps between successive
breakers) chosen randomly from 10 preset values from roughly 9s to 20s
(determined by the resistor choices). It sounds like you need TWO such
4017 sections, one for ON and one for OFF.

But if I was tackling your task from cold, I'd probably start by
playing with a couple of pseudo-random generators made from shift
registers.

--
Terry Pinnell
Hobbyist, West Sussex, UK