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Effect of and on ambient temperature?

  • Thread starter The little lost angel
  • Start date
T

The little lost angel

Jan 1, 1970
0
On reading spec docs, most parts appear to be specified for X ratings
at ambient temperature of 25C. Then they have a thermal derate to zero
at some temperature like 275C.

The problem now is, ambient temperature right next to the part will
get hotter too right? So does the spec mean, as long as I start with
25C ambient, it's ok to run the part up to 275C.

Or do I have to figure out the temperature rise of the ambient air and
at what temperature it will stabilize given the heat dissipated by the
part. Then see how much temperature rise am I allowed?

In addition, if I'm constantly pushing 25C air through using forced
air cooling. Does this means I can assume 25C constant temp or does it
still mean nothing since the air next to the part will always be
hotter than 25C?

Thanks!

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N

N. Thornton

Jan 1, 1970
0
[email protected] (The little lost angel) wrote in message news: said:
On reading spec docs, most parts appear to be specified for X ratings
at ambient temperature of 25C. Then they have a thermal derate to zero
at some temperature like 275C.

The problem now is, ambient temperature right next to the part will
get hotter too right? So does the spec mean, as long as I start with
25C ambient, it's ok to run the part up to 275C.

Or do I have to figure out the temperature rise of the ambient air and
at what temperature it will stabilize given the heat dissipated by the
part. Then see how much temperature rise am I allowed?

In addition, if I'm constantly pushing 25C air through using forced
air cooling. Does this means I can assume 25C constant temp or does it
still mean nothing since the air next to the part will always be
hotter than 25C?

Thanks!


Hi

If you mount the part in the way specified, in an unenclosed space,
then the mfr's deratings are as stated, and you use the room temp, not
the temp 0.01mm from the part. If its in an enclosed box you use the
temp of the air in the box.

Once you start fan cooling a part that is only specced for non-forced
cooling, the spec sheet info wont be valid for that. Consider the heat
path: the die has the max temp of 250 or whatever the derate to zero
figure is. Heat has to travel from die to part case, from part case to
heatsink, and from heatsink to air. With unforced cooling the heatsink
to air barrier is likely to be the least thermally conductive barrier.
Fan cooling reduces that resistance considerably.

Ultimately it is the temp of the die that counts.

Regards, NT
 
M

Marc Randolph

Jan 1, 1970
0
[email protected] (The little lost angel) wrote in message news: said:
On reading spec docs, most parts appear to be specified for X ratings
at ambient temperature of 25C. Then they have a thermal derate to zero
at some temperature like 275C.

The problem now is, ambient temperature right next to the part will
get hotter too right? So does the spec mean, as long as I start with
25C ambient, it's ok to run the part up to 275C.

Or do I have to figure out the temperature rise of the ambient air and
at what temperature it will stabilize given the heat dissipated by the
part. Then see how much temperature rise am I allowed?

In addition, if I'm constantly pushing 25C air through using forced
air cooling. Does this means I can assume 25C constant temp or does it
still mean nothing since the air next to the part will always be
hotter than 25C?

Without reading the spec for the particular part, I don't think anyone
will be able to help you with your exact situation. Every vendor
seems to state their temperature dependancy (and derating)
differently. Some spec their parts with respect to ambient temps
while others use case or junction.

I have to say that I've never seen 275C even mentioned for an
allowable temperature though (but I'm not in the miltary world, so
that may be why). Make sure you aren't reading the soldering
temperature. Max junction temperature of the parts I'm used to
working on range from 105C to 150C.

But to answer your question, yes, the vendor is generally referring to
the temperature of the airflow in the general area of the part, not
the temperature of the air just a fraction off the body of the IC.

However, that still doesn't necessarily mean you're off the hook. It
all depends on how much heat the part generates, and what its thermal
resistance is:

Let's assume the part has a ThetaJA (thermal resistance from junction
to ambient) of 50 deg C per Watt when airflow is moving at 100 lfm
over the part, and that it's max safe operating junction temperature
is 125C.

Assuming you provide that much airflow, the part can give off 2 Watts
[ (125C - 25C)/50] without exceeding its specs. Any more heat than
that, and you risk the long-term reliability of the part. The only
way to get more heat out of it is to lower the ambient temperature or
lower the thermal resistance (bigger package or add a heat sink or
increase airflow).

And don't forget that if you have more than one part in a row
(parallel to the airflow), each part will heat the air for the next
part. If you generate lots of heat, your 25C air could easily become
40C air (or more) by the time it reaches the last chip in the row - so
you have to redo the above calculation with higher ambient.

Have fun,

Marc
 
M

Mac

Jan 1, 1970
0

Yes. You might just have to guess. Calculating how much an enclosure will
heat up based on the dissipation inside is not easy.

No, if you have 25C air blowing on the part, you can use 25C ambient. But
unless you are in a climate controlled room, you should use a higher
temperature to allow for hot days. Maybe 35 or 40 deg, depending on what
seems reasonable. For example, a piece of outdoor equipment which isn't in
the shade can experience very high ambient temperatures.
Without reading the spec for the particular part, I don't think anyone
will be able to help you with your exact situation. Every vendor
seems to state their temperature dependancy (and derating)
differently. Some spec their parts with respect to ambient temps
while others use case or junction.

I have to say that I've never seen 275C even mentioned for an
allowable temperature though (but I'm not in the miltary world, so
that may be why). Make sure you aren't reading the soldering
temperature. Max junction temperature of the parts I'm used to
working on range from 105C to 150C.

Yeah, I'm with you on that. Some power MOSFET's are spec'd at 175C
junction temperature. That's the highest I've ever seen.
But to answer your question, yes, the vendor is generally referring to
the temperature of the airflow in the general area of the part, not
the temperature of the air just a fraction off the body of the IC.

However, that still doesn't necessarily mean you're off the hook. It
all depends on how much heat the part generates, and what its thermal
resistance is:

Let's assume the part has a ThetaJA (thermal resistance from junction
to ambient) of 50 deg C per Watt when airflow is moving at 100 lfm
over the part, and that it's max safe operating junction temperature
is 125C.

Assuming you provide that much airflow, the part can give off 2 Watts
[ (125C - 25C)/50] without exceeding its specs. Any more heat than
that, and you risk the long-term reliability of the part. The only
way to get more heat out of it is to lower the ambient temperature or
lower the thermal resistance (bigger package or add a heat sink or
increase airflow).

And don't forget that if you have more than one part in a row
(parallel to the airflow), each part will heat the air for the next
part. If you generate lots of heat, your 25C air could easily become
40C air (or more) by the time it reaches the last chip in the row - so
you have to redo the above calculation with higher ambient.

Have fun,

Marc

Power MOSFET's give power ratings with a case temperature of 25C. This has
to be one of the most unrealistic ratings in the world! If the part is
dissipating power, it will be nearly impossible to keep the case at 25C
unless you use liquid cooling or something.

Anyway, the typical thermal data in a datasheet are ThetaJA, as mentioned
above, and ThetaJCase (thermal resistance from junction to case), and
sometimes Theta from the case to the heatsink. Note that Theta is a greek
letter. We are spelling out the letter, but on the datasheet they have a
font which supports the real theta, which looks a bit like an 'O' with a
horizontal line through the middle of it. Actually, they might use
Rtheta. But the real clue is that the units are degrees C per Watt. You
probably already know this, but it's hard to tell, so please don't be
offended. ;-)

You can put several Rtheta's in series. For example, R theta junction to
case (Rtjc) and R theta case to heatsink (Rtch) and R theta heatsink to
ambient (Rtha). This last one would be from the heatsink manufacturer and
it will depend on the airflow. The resistances add up just like electrical
resistances. So let's say all the above resistances add up to 40 degrees C
per Watt. If you estimate that ambient will be around 60 degrees C, and
you are dissipating 2 Watts, then you can work backwards and see that the
junction will be at 60 degrees + 2 Watts * 40 deg. C/Watt = 140C. If the
maximum operating junction temperature is 175 deg. C, then you have 35
deg. of margin, which should be plenty.

Good luck.

Mac
 
T

The little lost angel

Jan 1, 1970
0
I have to say that I've never seen 275C even mentioned for an
allowable temperature though (but I'm not in the miltary world, so
that may be why). Make sure you aren't reading the soldering
temperature. Max junction temperature of the parts I'm used to
working on range from 105C to 150C.

Well, I guess I should explain that I was looking at the parts spec
for power resistors. Such as

http://www.ohmite.com/catalog/pdf/metalohm.pdf
http://www.ohmite.com/catalog/pdf/hs_hsn_series.pdf
http://www.isotekcorp.com/dataSheets/PDF/ulh.pdf
Let's assume the part has a ThetaJA (thermal resistance from junction
to ambient) of 50 deg C per Watt when airflow is moving at 100 lfm
over the part, and that it's max safe operating junction temperature
is 125C.

I'm not quite sure how to obtain this thermal resistance values from
those documents.

But using the isotek as an example, at the very last page, a chart
gives % of rated power vs ambient temp. Using line 2 for 200W~300W.
Assuming I want to dissipate 200W.

Assuming that I have a standalone piece and ambient goes to 40C, I am
looking at a derate to around 66% or 130W.

Now refering back to the Ohmite estimates here,
http://www.ohmite.com/catalog/pdf/appnotes_res_select.pdf (pg 7)

if I put a 700 FPM airflow on this part, the rating factor is 0.4.
Therefore I will be able to dissipate 130W using a 53W equivalent
part, or using the same 200W part, dissipate up to 325W.

However, the isotek part seems to have some kind of operating range of
-55C to 200C. So the limited temp rise factor from the Ohmite guide is
around 2.8 for 160C limit. Which brings that figure down to 116W.

Since I am planning to use it for 200W, I would either have to use a
pair of them or improve my cooling say by adding a heatsink or
increasing the airflow.

Would this method of estimate be sufficiently close to that arrived
with by using thermal junctions (which for most parts I don't see as a
given value in most docs)

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R

Roy McCammon

Jan 1, 1970
0
The said:
On reading spec docs, most parts appear to be specified for X ratings
at ambient temperature of 25C. Then they have a thermal derate to zero
at some temperature like 275C.

The problem now is, ambient temperature right next to the part will
get hotter too right? So does the spec mean, as long as I start with
25C ambient, it's ok to run the part up to 275C.

Or do I have to figure out the temperature rise of the ambient air and
at what temperature it will stabilize given the heat dissipated by the
part. Then see how much temperature rise am I allowed?

In addition, if I'm constantly pushing 25C air through using forced
air cooling. Does this means I can assume 25C constant temp or does it
still mean nothing since the air next to the part will always be
hotter than 25C?

Thanks!

unless clearly stated otherwise, assume that the blurb
has rated it in the most optimistic fashion (bolted to
a battleship sitting in the arctic ocean). You have to
hold the ambient at 25C to get the rated dissipation.
In practice, that means the ambient is going to heat up
and you must derate the part. Or something exotic like
liquid cooling or a thermoelectric cooler.
 
M

Mac

Jan 1, 1970
0
Well, I guess I should explain that I was looking at the parts spec for
power resistors. Such as

http://www.ohmite.com/catalog/pdf/metalohm.pdf
http://www.ohmite.com/catalog/pdf/hs_hsn_series.pdf
http://www.isotekcorp.com/dataSheets/PDF/ulh.pdf


I'm not quite sure how to obtain this thermal resistance values from
those documents.

You should have said you were using resistors in your original post. They
are a bit different. The resistors you are looking at are specifically
designed to dissipate lots of power and are made from heat resistant
materials. Materials being what they are, it is possible to design
resistors for higher temperatures than it is for semiconductors. It looks
like the manufacturers, at least Ohmite, use other guidelines, but you
seem to have figured it all out.

But using the isotek as an example, at the very last page, a chart gives
% of rated power vs ambient temp. Using line 2 for 200W~300W. Assuming I
want to dissipate 200W.

Assuming that I have a standalone piece and ambient goes to 40C, I am
looking at a derate to around 66% or 130W.

Yeah, sounds right. Forced air would help, though. The combination of
forced air and a heat sink will help tremendously.
Now refering back to the Ohmite estimates here,
http://www.ohmite.com/catalog/pdf/appnotes_res_select.pdf (pg 7)

if I put a 700 FPM airflow on this part, the rating factor is 0.4.
Therefore I will be able to dissipate 130W using a 53W equivalent part,
or using the same 200W part, dissipate up to 325W.

I can't fault your logic. Note that 700 feet per minute is a lot of air
flow.
However, the isotek part seems to have some kind of operating range of
-55C to 200C. So the limited temp rise factor from the Ohmite guide is
around 2.8 for 160C limit. Which brings that figure down to 116W.

Since I am planning to use it for 200W, I would either have to use a
pair of them or improve my cooling say by adding a heatsink or
increasing the airflow.

You'll have enough trouble getting to 700 feet per minute. I would go for
the heat sink and/or multiple resistors. If you use more than 1, try to
keep some distance between them.
Would this method of estimate be sufficiently close to that arrived with
by using thermal junctions (which for most parts I don't see as a given
value in most docs)

You seem to be following the worksheet on page 7 of the Ohmite document.
That seems reasonable to me.

BTW, it's not thermal junctions. It's semiconductor junctions. Resistors
don't have semiconductor junctions. But the principal is the same. There
is a max temperature and a thermal resistance which can be affected by air
flow and heat sinking. In their own way, it seems like the documents try
to address this.

It seems like you are on the right track. If you haven't already, make
sure you read that whole Ohmite document. It seemed to have lots of good
information, although I only skimmed it. I believe that the worksheet on
page 7 should apply more or less equally to resistors from other
companies.

The only other thing I should point out is that 200 Watts is a lot of
power. You will definitely need to put a fan on the enclosure, and if the
enclosure is in a small (closet sized) space, you may need to ventillate
the space, too. Also, since the resistor or resistors will be so hot, make
sure that they are downstream from any electronics that may also be in the
enclosure.

Good luck.

Mac
 
T

The little lost angel

Jan 1, 1970
0
I can't fault your logic. Note that 700 feet per minute is a lot of air
flow.

I must have gotten something wrong then. I've got easy access to 80mm
fans rated for 40+ CFM each. I arrived at the 700 FPM by assuming I'll
put a 2x2 array of these fan at one end, to get around 160 CFM. Which
according to some online documents I've found for converting CFM/FPM
would result in a 700FPM airflow.

Roughly, 80mm radius gives 0.216 sq feet. 160 CFM / 0.216 sq ft would
give 700 fpm.
You'll have enough trouble getting to 700 feet per minute. I would go for
the heat sink and/or multiple resistors. If you use more than 1, try to
keep some distance between them.

Hehee, I was hoping to keep costs down as far as possible i.e. low
component count for this current sink / load bank. My friend said it
would cost upwards of $20~30 USD per high wattage resistor and I'm
looking at a binary ladder with values up to 32A 12V

Was hoping to use power mosfets or something but apparently I cannot
control the current sunk as their characteristics will be constantly
in flux due to the incoming voltage/temperature and I haven't figured
out how to control them using opamp feedback... few of my simulations
on EWB demo gives me what I expect when I tried :(
It seems like you are on the right track. If you haven't already, make
sure you read that whole Ohmite document. It seemed to have lots of good
information, although I only skimmed it. I believe that the worksheet on
page 7 should apply more or less equally to resistors from other
companies.

Yesh, I've been through that doc a few times, just not 100% I got it
right that's all :)
The only other thing I should point out is that 200 Watts is a lot of
power. You will definitely need to put a fan on the enclosure, and if the
enclosure is in a small (closet sized) space, you may need to ventillate

The enclosure is probably going to be the heatsink/shroud itself and
in an open room. Though 200W is just one of the piece of the whole
gadget, I'm supposed to be cooking up the design for something to
dissipate 1 to 63A at 12V, 5V and 3.3V
the space, too. Also, since the resistor or resistors will be so hot, make
sure that they are downstream from any electronics that may also be in the
enclosure.

Yup, definitely, nothing's expected to be downstream of them except
cables to ground. Though one of my friend is talking about putting the
whole resistor/heatsink part into a water tank with a water pump :ppPp

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J

John Woodgate

Jan 1, 1970
0
I read in sci.electronics.design that The little lost angel <a?n?g?e?l@l
overgirl.lrigrevol.moc.com> wrote (in <[email protected]
t.sg>) about 'Effect of and on ambient temperature?', on Fri, 19 Dec
2003:
My friend said it
would cost upwards of $20~30 USD per high wattage resistor

I very much doubt it. Even at Digikey prices. Check for yourself.
 
N

N. Thornton

Jan 1, 1970
0
[email protected] (The little lost angel) wrote in message news: said:
Hehee, I was hoping to keep costs down as far as possible i.e. low
component count for this current sink / load bank. My friend said it
would cost upwards of $20~30 USD per high wattage resistor and I'm
looking at a binary ladder with values up to 32A 12V
The enclosure is probably going to be the heatsink/shroud itself and
in an open room. Though 200W is just one of the piece of the whole
gadget, I'm supposed to be cooking up the design for something to
dissipate 1 to 63A at 12V, 5V and 3.3V

My first thought is why buy the resistors when you can use resistance
wire for the cost of about 2 peanuts.

Regards, NT
 
T

The little lost angel

Jan 1, 1970
0
My first thought is why buy the resistors when you can use resistance
wire for the cost of about 2 peanuts.

Well, that idea was thought of. But the following question arises

1) How do we know we've got exactly 0.11 ohm worth of resistance when
the best multimeter we have, has a base probe resistance of 1.5 ohm
apparently.

2) What kind of material to use for the wire so that the resistance
won't change drastically as temperature goes up?

3) Assuming we found the materials, and wound the wire... how do we
cool it since we can't attach a heatsink to a coil of wire... or can't
see how yet.

The problems won't be so bad if we could figure out how to produce a
compensating circuit that monitors the current draw and switches
on/off another load. But none of my experiments on an opamp circuit is
working out on EWB :(

--
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Standard HTML, SHTML, MySQL + PHP or ASP, Javascript.
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J

John Popelish

Jan 1, 1970
0
The said:
Well, that idea was thought of. But the following question arises

1) How do we know we've got exactly 0.11 ohm worth of resistance when
the best multimeter we have, has a base probe resistance of 1.5 ohm
apparently.

You use a combination of an amp meter and a millivolt meter, apply a
large current ot the resistor and measure both the applied voltage and
the current.
2) What kind of material to use for the wire so that the resistance
won't change drastically as temperature goes up?

Manganin. Nichrome is not as stable but fairly common and cheap. A
coating of silver solder makes it possible to soldr copper wires on.
Copper wire has a higher thermal coefficient but is really cheap and
available. Iron and brass wire are also usable and people hang
pictures on them.
3) Assuming we found the materials, and wound the wire... how do we
cool it since we can't attach a heatsink to a coil of wire... or can't
see how yet.

Weave the wire across a frame and blow the air through the web of
wire.
The problems won't be so bad if we could figure out how to produce a
compensating circuit that monitors the current draw and switches
on/off another load. But none of my experiments on an opamp circuit is
working out on EWB :(

Is this a question?
 
T

The little lost angel

Jan 1, 1970
0
You use a combination of an amp meter and a millivolt meter, apply a
large current ot the resistor and measure both the applied voltage and
the current.

But isn't the current determined by the resistor itself? The closest
to this I've thought of was using the 1mOhm current sense resistor in
series with the load resistor and measuring the millivolt drop across
the sense resistor. But again, the accuracy of the method was a big
question mark in my own head.
Weave the wire across a frame and blow the air through the web of
wire.

Hmm, what would be the thermal characteristics of such a setup? As in,
how high can the temperature be allowed to go up to and such?
Is this a question?

heehee, sorta, I did post a schematic a few days back. Only one person
replied and said it wouldn't do what I expect but when I asked why,
there was no response.

The circuit I did had a current sense resistor in line with the load
bank, e.g.

+ve ---+-Rcs-+----+--/ -R1---+
| | | |
| | +--/ -R2---+--- Gnd
| | | |
| | +--/ -R3---+
OA+ OA-

The idea was to feed OA+ into the input of the OpAmp and the OA- into
the -Vcc of the opAmp, thinking that that should make the OpAmp read
the drop over the current sense Rcs. Then I would apply a reference mV
to the +input of the OpAmp. Except of course, it didn't work.

There was also the issue of somehow tying the reference voltage to the
switches for R1 to R3. I thought a voltage adder might be the idea but
I'm stumped on exactly how to do it. EWB doesn't work the way I
expect, I can't use it as a test and try thing since I have no idea
why it's producing the voltages it is (generally full +Vcc or -Vcc)

Would greatly appreciate it if you could give me some hints about it
too! Thanks!!!

--
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Standard HTML, SHTML, MySQL + PHP or ASP, Javascript.
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But keep in mind you pay extra bandwidth for their bloated code
 
J

John Popelish

Jan 1, 1970
0
The said:
But isn't the current determined by the resistor itself? The closest
to this I've thought of was using the 1mOhm current sense resistor in
series with the load resistor and measuring the millivolt drop across
the sense resistor. But again, the accuracy of the method was a big
question mark in my own head.

The current will certainly be influenced by the resistance, unless the
current comes from a current regulated supply (which is a fine thing
to use). The point is that you do not assume some current, you
measure it by putting an amp meter in series with the resistor. At
the same time to put a separate volt meter across the resistor, and
you calculate its resistance (under those conditions, it you don't
want to make assumptions about temperature coefficient) by dividing
the voltage across the resistor by the current through it. You may
need to limit the current to a safe value for the source by also
having some other resistance in the circuit, like a large incandescent
lamp or a toaster.
Hmm, what would be the thermal characteristics of such a setup? As in,
how high can the temperature be allowed to go up to and such?

That depends on the melting temperature of the wire, and how much
thermal change in the resistance you can tolerate. But you can
measure the thermal effect with the same resistance measurement method
I spoke of above.

The big advantage of this method is that there is very little air drag
compared ot an aluminum heat sink, so you can move a lot of air with a
small, axial fan. The wire is just a spider web across the air flow.
They carry a thousand watts out of an ordinary hair dryer, this way.
And you know how small those are.
heehee, sorta, I did post a schematic a few days back. Only one person
replied and said it wouldn't do what I expect but when I asked why,
there was no response.

It didn't capture my attention, and I thought you had a conversation
going on with someone, so didn't butt in.
The circuit I did had a current sense resistor in line with the load
bank, e.g.

+ve ---+-Rcs-+----+--/ -R1---+
| | | |
| | +--/ -R2---+--- Gnd
| | | |
| | +--/ -R3---+
OA+ OA-

The idea was to feed OA+ into the input of the OpAmp and the OA- into
the -Vcc of the opAmp, thinking that that should make the OpAmp read
the drop over the current sense Rcs. Then I would apply a reference mV
to the +input of the OpAmp. Except of course, it didn't work.

Linear opamp circuit make the assumption that the two inputs stay at
essentially the same voltage for the whole range of output voltage.
The input signal and the feedback have to make that happen when you
are getting the desired output.

A subtractor is one configuration that amplifies a voltage difference
and outputs that amplified difference on an arbitrary reference
voltage (like zero volts). There are also current mirror circuits
that rereference a high side current (or small voltage difference) to
another voltage rail.

Section 3 of this circuit collection shows several ways to do this
from a full, 3 opamp instrumentation amplifier (top of page 14), a two
opamp version that can have inputs outside the amplifier input voltage
range (top of page 15), and a two opamp version that has to have the
input signals within the opamp input voltage range. But since both
your voltages come from a low impedance source you can just use the 1
opamp subtractor that is all the resistors and the right amplifier out
of the I.A. at the top of page 14.
http://www.national.com/an/AN/AN-31.pdf

Here is an example of moving the current shunt signal to a different
reference with a current mirror:
http://www.zetex.com/3.0/pdf/ZDS1009.pdf
There was also the issue of somehow tying the reference voltage to the
switches for R1 to R3. I thought a voltage adder might be the idea but
I'm stumped on exactly how to do it. EWB doesn't work the way I
expect, I can't use it as a test and try thing since I have no idea
why it's producing the voltages it is (generally full +Vcc or -Vcc)

Are you trying to select the load resistor based on a measurement of
the load current, or step the load resistors to hold the current to
some setpoint (as close as the switching will allow)?
Would greatly appreciate it if you could give me some hints about it
too! Thanks!!!

If the load resistors are all equal there is a different method than
if they have a binary multiple relationship. In the first case, any
resistor is equivalent to any other so you need a circuit that just
switches them on or off in a straight sequence. In the second case,
you need ot switch them on and off with the equivalent of a binary up
down counter. But in both cases, this selection circuit has to be
connected to a decision making function that does this selection in a
stable and orderly fashion (at a speed that does not exceed the the
decision making process's ability to evaluate its previous decision
before making another one). This can all be done with opamps and
comparators and discrete logic (fast but large), or with a program
running in a microprocessor (slower but smaller and more feature
rich).
 
T

The little lost angel

Jan 1, 1970
0
Thanks for the pointers, though I'm running a flu now and it's making
my slow brain even slower, so I'll go through those documents later.
Are you trying to select the load resistor based on a measurement of
the load current, or step the load resistors to hold the current to
some setpoint (as close as the switching will allow)?

I'm stepping the load resistors to hold the current to a set point.
Essentially to be able to put a load on the voltage(current?) source
at say 1,2,4,8,16,32 amps settings to see how well the power supply is
holding up. Since the power supply is supposed to remain stable with
the load changing up to 20Khz, having the load oscillate a little
around the set point shouldn't cause it problems... if it does, the
supply's crap and I'll like to know that!
If the load resistors are all equal there is a different method than
if they have a binary multiple relationship. In the first case, any
resistor is equivalent to any other so you need a circuit that just
switches them on or off in a straight sequence. In the second case,
you need ot switch them on and off with the equivalent of a binary up
down counter. But in both cases, this selection circuit has to be
connected to a decision making function that does this selection in a
stable and orderly fashion (at a speed that does not exceed the the
decision making process's ability to evaluate its previous decision
before making another one). This can all be done with opamps and
comparators and discrete logic (fast but large), or with a program
running in a microprocessor (slower but smaller and more feature
rich).

I'm still not quite sure which is the best method. The
buy-fixed-resistors-in-binary-steps-without-current-control seems to
be the easiest but most expensive and there's only that close I can
get to what I want.

I think I've mentioned (during my last vacation) of using mosfets to
control the resistance to a bank of same value resistors, except at
this point I realize (maybe wrongly) that it would be almost
impossible to control the mosfet Rds to the exact level I want. Once
the load bank is started up, temperature will kill the preset
calibration and again the same problem of I don't know how to control
it electronically comes in.


--
L.Angel: I'm looking for web design work.
If you need basic to med complexity webpages at affordable rates, email me :)
Standard HTML, SHTML, MySQL + PHP or ASP, Javascript.
If you really want, FrontPage & DreamWeaver too.
But keep in mind you pay extra bandwidth for their bloated code
 
N

N. Thornton

Jan 1, 1970
0
[email protected] (The little lost angel) wrote in message news: said:
On 19 Dec 2003 12:30:27 -0800, [email protected] (N. Thornton) wrote:
Well, that idea was thought of. But the following question arises

1) How do we know we've got exactly 0.11 ohm worth of resistance when
the best multimeter we have, has a base probe resistance of 1.5 ohm
apparently.

1. fix the multimeter
2. look up the resitance per metre of the wire type/gauge you have,
and cut to length - easiest option.
3. apply a known V and measure the i. V=IR.

2) What kind of material to use for the wire so that the resistance
won't change drastically as temperature goes up?

I'd suggest googling s.e.d for "resistance wire", or looking for
constantan's thermal characteristics. There are many wire types to
choose from.
3) Assuming we found the materials, and wound the wire... how do we
cool it since we can't attach a heatsink to a coil of wire... or can't
see how yet.

Look at how a fan heater is made. Wire strung across an open ceramic
frame. Look at the amount of fanning, the amount of wire and the power
dissipation to get a rough idea. 200 or 300w is easily done.

The problems won't be so bad if we could figure out how to produce a
compensating circuit that monitors the current draw and switches
on/off another load. But none of my experiments on an opamp circuit is
working out on EWB :(

I dont really know what youre trying to achieve, so cant add much
useful.


Regards, NT
 
J

John Popelish

Jan 1, 1970
0
The said:
....
I'm stepping the load resistors to hold the current to a set point.
Essentially to be able to put a load on the voltage(current?) source
at say 1,2,4,8,16,32 amps settings to see how well the power supply is
holding up. Since the power supply is supposed to remain stable with
the load changing up to 20Khz, having the load oscillate a little
around the set point shouldn't cause it problems... if it does, the
supply's crap and I'll like to know that!


I'm still not quite sure which is the best method. The
buy-fixed-resistors-in-binary-steps-without-current-control seems to
be the easiest but most expensive and there's only that close I can
get to what I want.

It takes fewer resistors for a given step size, with the binary
method, but the worst case switching spike is bigger. you can combine
a single stage of continuously variable load (linear mosfet) with the
switched blocks to be able to use pretty coarse steps and still settle
to a completely specified current, as long as the setpoint does not
end up right on the boundary between blocks.
I think I've mentioned (during my last vacation) of using mosfets to
control the resistance to a bank of same value resistors, except at
this point I realize (maybe wrongly) that it would be almost
impossible to control the mosfet Rds to the exact level I want. Once
the load bank is started up, temperature will kill the preset
calibration and again the same problem of I don't know how to control
it electronically comes in.

Lets take the simplest (to understand) case and see if that is good
enough.

Lets say you have 4 equal blocks of resistance selected by 4
comparators that have switching thresholds set in a nicely spaced
sequence of voltages by a small divider of 5 resistors with switching
thresholds of, say, 1,2,3,4 volts. There is an opamp error amplifier
that evaluates the match between the continuously measured current
(via one of the subtractor circuits we talked about last time) and a
voltage that represents the current setpoint.

The error amplifier has a capacitor feedback, so that it ramps up and
down for positive and negative errors. The output is what drives the
4 comparators that switch the 4 equal resistors on and off, in
sequence. If the desired current is half way between that drawn by
two of the combinations, the circuit will ramp back and forth between
those two number of blocks being on, producing a square wave load that
swings back and forth between the current that is a little too high
and the current that is a little too low, with the time in each state
dependent on the size of the feedback capacitor around the error
amplifier. This constitutes the coarse load selection part of the
system.

Are you with me so far?
 
T

The little lost angel

Jan 1, 1970
0
But since both
your voltages come from a low impedance source you can just use the 1
opamp subtractor that is all the resistors and the right amplifier out
of the I.A. at the top of page 14.
http://www.national.com/an/AN/AN-31.pdf

That would be everything to the right of the word "balance" in the
middle, IOW, R2, R3, R4, R5, one LM107? I hope so because I'm not sure
how to fit R1 in except by connecting it in series to R2//R3.
Here is an example of moving the current shunt signal to a different
reference with a current mirror:
http://www.zetex.com/3.0/pdf/ZDS1009.pdf

Does this thing work without external power? I got confused (ok now
everybody realizes why I've got lost in my nick), because on page one,
the right hand side schematic seems to indicate that no external Vcc
needs to be applied to the IC, or rather nothing is connected to the
X1 Y1 pins. However in pg2 the spec table says max of 120V between X1
and Y1.

Also, the spec seems to indicate that it can only be used on a circuit
with no more than 10V, since Max Voltage (E1-E2, E3-E4) is 10V and no
more than 1 amp can be measured Continuous Current (E1-E4, E2-E3).

Or did you mean that I should take the IC schematic as a reference and
build one using four transistors and resistors?

Thanks!!!

--
L.Angel: I'm looking for web design work.
If you need basic to med complexity webpages at affordable rates, email me :)
Standard HTML, SHTML, MySQL + PHP or ASP, Javascript.
If you really want, FrontPage & DreamWeaver too.
But keep in mind you pay extra bandwidth for their bloated code
 
J

John Popelish

Jan 1, 1970
0
The said:
That would be everything to the right of the word "balance" in the
middle, IOW, R2, R3, R4, R5, one LM107? I hope so because I'm not sure
how to fit R1 in except by connecting it in series to R2//R3.

No R1. R2 and R3 would connect to the ends of your current shunt.
The main limitation of this circuit is that the opamp inputs can
respond only inside a certain voltage range, called the common mode
range (a range of voltage that they have in common). For the LM107
that would be a couple volts inside the supply rails. For an LM324,
that would be from 1.5 volts below the positive rail to .3 volts below
the negative rail.
so you could use this to subtract the voltages across a current shunt
that had voltage about 1.5 volts below the positive rail of the opamp
(or a little higher because of the voltage dividers) and still keep
both inputs inside the common mode range. So if you build the
subtractor with low gain (say, 1) the dividers increase the common
mode range on the inputs to almost double the positive rail.

At the moment, I don't know what opamp supply voltage you are
planning, and what the positive supply rail voltage will be at the
current shunt.
Does this thing work without external power?

The only power it needs is that the outputs be held at a negative
voltage (about 2 to 120 volts) compared to its inputs at the current
shunt end.
I got confused (ok now everybody realizes why I've got lost in my nick),
because on page one,
the right hand side schematic seems to indicate that no external Vcc
needs to be applied to the IC, or rather nothing is connected to the
X1 Y1 pins. However in pg2 the spec table says max of 120V between X1
and Y1.

Also, the spec seems to indicate that it can only be used on a circuit
with no more than 10V, since Max Voltage (E1-E2, E3-E4) is 10V and no
more than 1 amp can be measured Continuous Current (E1-E4, E2-E3).

Normally, the voltage from E1 to E2 is the low voltage produced by the
current shunt, and the voltage between E3 and E4 is the output
voltage. The voltage between the E1 E2 pair and the E3 to E4 pair is
how far from the shunt rail the error amplifier will be.
Or did you mean that I should take the IC schematic as a reference and
build one using four transistors and resistors?

The current mirror requires well matched resistors to have a
predictable accuracy. However, you can easily copy it with separate
small transistors and get it to work, as long as you don't expect it
to have exactly the specifications on this data sheet. So, either one
might work for a one off.

The purpose of this current mirror is to take a small voltage as an
input (the drop across a current shunt, R2) and turn it into a current
difference, scaled way down (by the resistor R1) to something in the
milliamp range. Then that current difference can be changed back to a
voltage difference (and enlarged) by the load resistors R3 and R4. So
you get voltage amplification by the ratio of R4/R1 and relocation of
the signal from a positive rail to a negative one.
 
D

David Dong

Jan 1, 1970
0
The final temperature is power multiply the thermic resistor which can
be found in the devices datasheet.
 
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