Capacitor

[Leythos]

You are not hearing me. I am trying this kind of thing every day and I am
not seeing the results you see. If you want me to try something be more
specific. What it is precisely you want me to try and what it is I will see?
What are the design goals/parameters:

Input voltage, load, max and min ambient temperature, heat sink thermal
resistance, etc?

Here's what you've said:

Based on the spec's, the LM7805 can not provide reliable output of 5v at
500MA (or even 1A) with a standard heat sink with an supply voltage of
24VDC.

I said:

Wrong, it works, is working, and has worked on many systems for at least
about 20 years that I know of and have DIRECT experience with. I also
said that I have read and understand the technical specs, but that
sometimes you have to just believe what you see in the real world and
not just on paper.

I think you can figure out what you have to test based on the two
statements above.

 

[MM]

Here's what you've said:

Based on the spec's, the LM7805 can not provide reliable output of 5v at
500MA (or even 1A) with a standard heat sink with an supply voltage of
24VDC.

I said:

Wrong, it works, is working, and has worked on many systems for at least
about 20 years that I know of and have DIRECT experience with. I also
said that I have read and understand the technical specs, but that
sometimes you have to just believe what you see in the real world and
not just on paper.

I think you can figure out what you have to test based on the two
statements above.

1. What the hell is a "standard" heatsink?
2. What is the ambient temperature?

I don't have more time to spend on this BS. I've done my labs to prove the
Ohm's law many years ago in Grade 7 or 8. There is nothing to be gained from
this experiment. The last board I designed has 5 switching regulators and
over a dozen of linear ones. Every and each of them work according to the
known laws of physics and dissipate precisely the amount of heat I
calculated. Yes, you can run semiconductors at 150C or some even higher, but
reliable things are not designed this way. Have you ever heard of derating?
Have you ever seen a curve demonstrating failure frequency depending on
junction temperature? As I can see it, you look at a datasheet, notice the
max current and max input voltage and then you plug it in a circuit and
experiment. You might be lucky and it "works" most of the times, but have
you tried to qualify your circuit to work in avionics equipment where
ambient temperatures can easily reach 70 C and where cooling air coming from
a fan is sometimes just as hot? If people were doing designs your way we
would still be in a crystal radio age. Why do you believe the output voltage
spec and don't believe power and temperature rating specs? You don't expect
to see 7V or 17V output from your favorite 7805, but somehow you expect that
other parts of the spec along with fundamental laws of physics can be
stretched to whatever lengths. Finally, back to your 500mA 24VDC PSU. I
didn't say it can't work, but it is on the edge depending on your heatsink.
If you increase the current to 1A it will most likely fail (again depending
on the heatsink, ambient temperature and fan speed).

/MM


/MM

 

[Leythos]

1. What the hell is a "standard" heatsink?

Something you can find for sale in the common electronics supply
catalogs - not something that is the size of a Refrige.

2. What is the ambient temperature?

Room temp, don't ask what room, pick one in your house, on a normal day.

I don't have more time to spend on this BS. I've done my labs to prove the
Ohm's law many years ago in Grade 7 or 8. There is nothing to be gained from
this experiment.

Sure there is, but, as I figured, you don't like to be wrong, so you
absolve yourself of it by saying that your labs and specs are right.

The last board I designed has 5 switching regulators and
over a dozen of linear ones. Every and each of them work according to the
known laws of physics and dissipate precisely the amount of heat I
calculated.

And I never suggested that your designs don't work, never suggested that
Ohms law was wrong, or anything of the sort. If you want to boast about
your abilities, I was doing the above in the late 70's, and still have
parts cabinets full of chips - but who cares.

If you are going to make a statement that the LM7805 can not operate at
5V providing 500MA in a stable configuration with a supply voltage of +
24V then you ought to be able to prove it. I did, it works, got a couple
of them right here, working for years. I'm not asking you to take my
word for it alone, I'm saying that it works, is stable, and does not
fail, and that if you don't believe it, try it yourself. If you can't be
big enough to at least try it, then quit whining about how it can't
work!

 

[Barry Watzman]

Re: "The onboard PSU is normally not a switching unit, the Switching
PSU's are the large metal units that you connect AC to."

That is totally wrong. The Vcore power supply for the CPU, which is on
the motherboard, most definitely IS a switching power supply.

The correct advice here is to remove AND REPLACE the bad capacitors.
Frankly, I'd replace any other SIMILAR capacitors at the same time.
About 2-3 years ago, the industry was flooded with several hundred
million bad electrolytic capacitors. It happened because a chemical
company that was going bankrupt, in an attempt to save money, left a
critical but expensive chemical out of their production of electrolyte
used by the capacitor makers in Taiwan. I'm guessing that you have some
of those caps.

There really is no solution other than to remove and replace them, which
will require soldering. You don't need an exact replacement. If in
doubt, get parts with higher capacitance and/or higher voltage. I have
an oscilloscope that I built in 1980 that I was into to replace a pilot
lamp, and I noticed a 3000 uF 15 volt capacitor blown completely apart,
the scope worked fine without it. I temporarily replaced it with a 1000
uf 25 volt part (closest I could come at radio shack), but later
installed a 3000 uf 30 volt part.

 

[Barry Watzman]

The first poster was correct. Linear regulators, including 3-terminal
regulators like 7805's, drop the voltage by WASTING the excess as heat.
Essentially, they drop the voltage by putting an electronically
variable resistance between the higher voltage and the desired lower
voltage.

Say you are using a 7805 to drop 12 volts (which might be unregulated as
well) down to 5 volts. The extra 7 volts is across the 7805. If the
current draw at 5 volts is 100ma (0.1 amps), then you are "wasting" .7
watts in the regulator (7 volts across the regulator x .1 amps through
the regulator). That might not sound like much, but the load is only
using .5 watts (5 volts, 0.1 amps), so you are "wasting" more than you
are using. The wasted power appears as heat (the regulator gets hot,
and may require a heat sink).

3 terminal regulators can only dissipate about 2-3 watts total at most,
and only then with a good heat sink. So while a 7805 can supply 1 amp,
you can forget about it if you want to drop 24 volts to 5 volts, because
it would have to dissipate 19 watts (1 amp, 19 volts across the
regulator). On the other hand, if you were working from an 8 volt
supply, it could handle it, because the power dissipation would now only
be 3 watts (near the limit of the regulator, but manageable with a good
heat sink).

The whole reason for using a switching regulator is that they have high
efficiency, they don't waste power. That's because they don't "drop"
the output voltage from the input voltage, rather they generate the
output voltage "from scratch".

 

[Barry Watzman]

Leythos, you are just totally wrong, and have no understanding of how
these parts work.

 

[Barry Watzman]

The cap is probably neither open nor shorted, in fact it's quite
possibly still working .... as a capacitor.

The problem with the defective electrolyte that was used in a few
hundred million capacitors made a couple years ago was that a chemical
that reduced "gassing" was left out. The capacitors then could produce
excessive gas pressure and eventually the "can" would ruture somewhere.

While this can destroy the capacitor in either an open or a short mode,
sometime all that happens is that the can ruptures, electrolyte leaks
out, but the capacitor basically isn't catastrophically destroyed (at
least not immediately).

The loss of electrolyte will reduce the effective capacitance of the
capacitor, but this normally isn't critical until towards the end (when
either most of the electrolyte has leaked out or dried out). Also, as
was noted, the electrolyte is corrosive and can "eat through" the traces
on the circuit board. The capacitor should be replaced, but IN THIS
INSTANCE, it may well have neither shorted nor opened, and may, for the
moment, still be functioning "adequately".

[One other possibility is that it IS open, but these circuits often have
quite a few capacitors in parallel, and the remaining capacitors (which
have not failed YET) may be adequate to keep things working ok.]

 

[Barry Watzman]

First, you are wrong both in theory and in practice. A 7805 dropping 12
volts to 5 will dissipate more than twice as much power (heat) as one
dropping 8 volts to 5 (assuming the same current draw). And the
difference is VERY, VERY real.

Second, actually, the capacitor is NOT necessarily being used as a
ripple filter here, but rather as an "energy storage reservoir". There
is no ripple in the output of a linear regulator whose input is DC. And
while the output of a switching power supply would have ripple, the
frequency is VERY high (80-120 KHz is a common range) and is easily
filtered with small, tiny capacitors.

The reason for the large electrolytic caps is the same as the reason for
using 1 FARAD $200 capacitors in ultra-powerful automotive stereo
systems (the ones that you can FEEL when the guy whose car it's in parks
next to you). The reason is that the peak, instantaneous current
requirements of some modern CPU and video chips is tremendous. Some
Pentium 4's have peak Vcore current requirements in excess of 70 amps.
But these peaks last for nanoseconds or microseconds. The power
supplies, by themselves, can't supply that kind of power, and the large
electrolytic caps are present meet these demands, not to filter ripple.

 

[Barry Watzman]

The issue, within reason, isn't current draw or power dissipation but
rather junction temperature inside the device; if it gets too high,
either the device will be destroyed, or it will shut down (many of the
3-terminal regulators have thermal shutdown protection). That does,
however, come back to power dissipation and heatsink and cooling.

Leythos, I don't think you have really tried this, because its not at
all theoretical. If you try to have the regulator dissipate more than 1
or 2 watts, you WILL have SIGNIFICANT thermal issues. And further,
since these devices (except for the "LDO" parts (low drop-out)) REQUIRE
about a 2.5 volt minimum difference between their input and their output
voltage, you will hit the bottom end of that range in any situation
where the load needs 500ma (or more).

 

[Leythos]

Leythos, you are just totally wrong, and have no understanding of how
these parts work.

Um, lets see - build 24V DC output, connect to basic LM7805 circuit,
load to 500MA, leave loaded and running for weeks. Leave DVM connected
to PC tracking output in 30 second intervals over 12 hour period, record
shows stable output at each point - logged data represents real world
experience.

So, what part am I missing? It seems to me that while no one here is
arguing the specs, the physics behind it, or the heat, that what you are
failing to understand is that it's working, running fine, and has worked
for many years. I can't make it any simpler for you, it just works!

 

[seeker]

Linear regulators (eg Series 80nn) MAY waste huge amount of power
(power means watts or volt-amps or whatever you understand)..under
several situations.

If the diff. between Vin and Vout is significant...I'l use the text
book case where it specifies that Vin should be at least 3 volts over
the desired Vout.

The regulation can and will generate tons of heat...but this is reated
to the current (A) being drawn by the load connected to Vout.

If, forexample, you use a TO-220 package 7805 regulator, and feed it
say 9volts and consume ~ 1A@ 5Volts...you'll feel some real heat!!!

That's why the databooks indicate heatsink requiremnets and so
forth...Linear regulators...under real load...are not very efficient
and the lost efficiency makes heat!!

 

[MM]

If you are going to make a statement that the LM7805 can not operate at
5V providing 500MA in a stable configuration with a supply voltage of +
24V then you ought to be able to prove it.

I haven't made this statement. I believe it will work in a room temperature
with an adequate heatsink. Are you happy now? However, using a linear
regulator this way is generally a bad design practice these days unless
there are very strong reasons for it.

I think we should stop here. This is way OT for this newsgroup. It should be
moved to the sci.electronics.basics should you wish to continue.

/MM

 

[Leythos]

I believe it will work in a room temperature
with an adequate heatsink.

I agree.

 

[Barry Watzman]

I suspect that what you think is happening, really isn't.

If you are getting 5 volts from a 7805 whose input is 24 volts, there is
19 volts across the 7805. That's not speculation or theory -- it's
simple physics, e.g. it's both theory AND real-world practice.

If you are pulling 500ma through ANY device with 19 volts across it,
it's GOING to dissipate 9.5 watts. Again, that's not speculation or
theory -- it's simple physics, e.g. it's both theory AND real-world
practice.

You can't dissipate 9.5 watts in a device the size of a 7805 without
some serious thermal considerations (and you may not be able to do it at
all, period).

My suggestion is that either the voltage or the current are not what you
say they are (or even believe that they are), or that you are just
making this up and have not really tried it.

[which, among other things, leaves me wondering if you know how to use a
multi-meter, especially to measure current]

 

[Leythos]

I suspect that what you think is happening, really isn't.

If you are getting 5 volts from a 7805 whose input is 24 volts, there is
19 volts across the 7805. That's not speculation or theory -- it's
simple physics, e.g. it's both theory AND real-world practice.

If you are pulling 500ma through ANY device with 19 volts across it,
it's GOING to dissipate 9.5 watts. Again, that's not speculation or
theory -- it's simple physics, e.g. it's both theory AND real-world
practice.

You can't dissipate 9.5 watts in a device the size of a 7805 without
some serious thermal considerations (and you may not be able to do it at
all, period).

My suggestion is that either the voltage or the current are not what you
say they are (or even believe that they are), or that you are just
making this up and have not really tried it.

[which, among other things, leaves me wondering if you know how to use a
multi-meter, especially to measure current]

Rather than argue with me, try it yourself, as I've suggested many times
in this thread. I was doing this in the early 80's for a living, have
used a scope and meter since the late 70's, and am a little tired of
people saying that something can't or doesn't work when they are only
willing to quote the spec sheet back at me. I'll say it again, 24vdc
supply to a LM7805 with heat sink, producing 5v out, and at a 500ma load
on the 5v side. Working for years. If you don't try it for yourself then
then you can't really say much about it.

 

[Tom S]

Rather than argue with me, try it yourself, as I've suggested many times
in this thread. I was doing this in the early 80's for a living, have
used a scope and meter since the late 70's, and am a little tired of
people saying that something can't or doesn't work when they are only
willing to quote the spec sheet back at me. I'll say it again, 24vdc
supply to a LM7805 with heat sink, producing 5v out, and at a 500ma load
on the 5v side. Working for years. If you don't try it for yourself then
then you can't really say much about it.

I've done this before. I found that if I didn't put a honking big series
dropping resistor between the supply and the regulator to dump most of the
heat, the regulator gets hot enough to go into thermal shutdown.

It's not Ohm's _theory_; it's Ohm's *Law*! :^)

Tom S

 

[Roger Hamlett]

I suspect that what you think is happening, really isn't.

If you are getting 5 volts from a 7805 whose input is 24 volts, there is
19 volts across the 7805. That's not speculation or theory -- it's
simple physics, e.g. it's both theory AND real-world practice.

If you are pulling 500ma through ANY device with 19 volts across it,
it's GOING to dissipate 9.5 watts. Again, that's not speculation or
theory -- it's simple physics, e.g. it's both theory AND real-world
practice.

You can't dissipate 9.5 watts in a device the size of a 7805 without
some serious thermal considerations (and you may not be able to do it at
all, period).

My suggestion is that either the voltage or the current are not what you
say they are (or even believe that they are), or that you are just
making this up and have not really tried it.

[which, among other things, leaves me wondering if you know how to use a
multi-meter, especially to measure current]

Rather than argue with me, try it yourself, as I've suggested many times
in this thread. I was doing this in the early 80's for a living, have
used a scope and meter since the late 70's, and am a little tired of
people saying that something can't or doesn't work when they are only
willing to quote the spec sheet back at me. I'll say it again, 24vdc
supply to a LM7805 with heat sink, producing 5v out, and at a 500ma load
on the 5v side. Working for years. If you don't try it for yourself then
then you can't really say much about it.

We all have, and _know_ that Ohm's _law_ works. The only way your system
would generate the same amount of heat in the regulator, is if the
incoming supply is 'drooping', and the heating is taking place somewhere
earlier in the supply train (the transformer probably). Seriously if the
'19v', is a 'no load' voltage off a small transformer/rectifier, then this
could easily be dropping to under 10v under load, and the heat on the
regulator would only be 2.5W.

Best Wishes

 

[Leythos]

Best Wishes

You too