Maker Pro
Maker Pro

load

J

JaBrIoL

Jan 1, 1970
0
All.

I have to build a load device.

specs are:

5 volts @ 38A @ .5 Ohms @ 380 watts.

I am constrained to using resistor no bigger than 50 watts each.
I figure out 10 .1 resisitors in parrallel to give me an aproximate
load.

in the words of Bill O'Reilly


What say you?
 
F

Frank Bemelman

Jan 1, 1970
0
JaBrIoL said:
All.

I have to build a load device.

specs are:

5 volts @ 38A @ .5 Ohms @ 380 watts.

I am constrained to using resistor no bigger than 50 watts each.
I figure out 10 .1 resisitors in parrallel to give me an aproximate
load.

in the words of Bill O'Reilly


What say you?

More like a load of rubbish. 5V @ 38A is not .5 ohms, and it
is not 380W and 10 resistors of 0.1 in parallel is 0.01 ohm.
 
F

Fred Bloggs

Jan 1, 1970
0
JaBrIoL said:
All.

I have to build a load device.

specs are:

5 volts @ 38A @ .5 Ohms @ 380 watts.

I am constrained to using resistor no bigger than 50 watts each.
I figure out 10 .1 resisitors in parrallel to give me an aproximate
load.

in the words of Bill O'Reilly


What say you?

I say you typify the type of "thing" that listens to O'Reilly, troll.
People like you are handicapped enough without the distraction of
political issues you can't understand- see your doctor about a pacifying
lobotomy.
 
C

CFoley1064

Jan 1, 1970
0
All.
I have to build a load device.

specs are:

5 volts @ 38A @ .5 Ohms @ 380 watts.

I am constrained to using resistor no bigger than 50 watts each.
I figure out 10 .1 resisitors in parrallel to give me an aproximate
load.

in the words of Bill O'Reilly

What say you?

O'Really? Now that you've given us the "fair & balanced" version, why don't
you let us know what you need? Outside of the "No Spin Zone",

Ohm's Law
V(volts) = I(current, amps) * R(resistance, ohms)

and

Power Equation
P(power, watts) = V(volts) * I(current, amps)

Of course, it may be different for Fox News electronics.

Questions of this type should be posted on sci.electronics.basics.
 
J

John Larkin

Jan 1, 1970
0
All.

I have to build a load device.

specs are:

5 volts @ 38A @ .5 Ohms @ 380 watts.

I am constrained to using resistor no bigger than 50 watts each.
I figure out 10 .1 resisitors in parrallel to give me an aproximate
load.

in the words of Bill O'Reilly


What say you?

I say the numbers don't make sense.

John
 
R

Ross Mac

Jan 1, 1970
0
JaBrIoL said:
All.

I have to build a load device.

specs are:

5 volts @ 38A @ .5 Ohms @ 380 watts.

I am constrained to using resistor no bigger than 50 watts each.
I figure out 10 .1 resisitors in parrallel to give me an aproximate
load.

in the words of Bill O'Reilly


What say you?

Shall I say "what a load"?
..13 ohms not .5, 190 watts not 380 and 10 .1ohm resistors in parallel is
..01ohms. I hope you have good medical insurance!
.....Ross
 
R

Ross Mac

Jan 1, 1970
0
Fred Bloggs said:
I say you typify the type of "thing" that listens to O'Reilly, troll.
People like you are handicapped enough without the distraction of
political issues you can't understand- see your doctor about a pacifying
lobotomy.
Hey Fred, he will be seeing the doctor soon....a few seconds after he plugs
this in!
 
J

JaBrIoL

Jan 1, 1970
0
Frank Bemelman said:
More like a load of rubbish. 5V @ 38A is not .5 ohms, and it
is not 380W and 10 resistors of 0.1 in parallel is 0.01 ohm.

you don't do this often, am I correct?

If you looked a this more closely, you would have noticed, that the
current was
cut by half, and the wattage doubled. It is a resitor load to run a
device at 50% or half load as some may call it.
 
J

JaBrIoL

Jan 1, 1970
0
O'Really? Now that you've given us the "fair & balanced" version, why don't
you let us know what you need? Outside of the "No Spin Zone",

Ohm's Law
V(volts) = I(current, amps) * R(resistance, ohms)

and

Power Equation
P(power, watts) = V(volts) * I(current, amps)

Of course, it may be different for Fox News electronics.

Questions of this type should be posted on sci.electronics.basics.

It wasnt a tech school question. It was a design question.
You could have said: "I don't know"
 
F

Frank Bemelman

Jan 1, 1970
0
JaBrIoL said:
you don't do this often, am I correct?

If you looked a this more closely, you would have noticed, that the
current was
cut by half, and the wattage doubled. It is a resitor load to run a
device at 50% or half load as some may call it.

More closely??? Where does the other half of the current go?
And since when goes wattage up with currents being cut in half?

P = I^2 x R. If your current is cut in half, the power is cut
in 4, certainly not doubled ;) Or should I somehow notice that
the voltage is cut in half, or the resistors are cut in half,
or the load has been cut in half ? Next time you say that it
can't be done because your budget has been cut in half.

Forget O'Reilly, you better team up with Paul Burridge for
this project ;)
 
K

Ken Taylor

Jan 1, 1970
0
JaBrIoL said:
[email protected] (CFoley1064) wrote in message

It wasnt a tech school question. It was a design question.
You could have said: "I don't know"

Everyone has said you don't know, what's your problem?

Why don't you rephrase the question in terms of what you are doing and what
result you want. You mentioned something about trying to possibly slow down
something (halve the load)?? Give us all the facts or don't be surprised
when you're told that what you've said is gibberish.

Ken
 
C

CFoley1064

Jan 1, 1970
0
It wasnt a tech school question. It was a design question.
You could have said: "I don't know"

I also could have said, "Your question, as you phrased it, makes no sense, and
that should be obvious to anyone who has a few weeks of trade school under
their belt. So, I'm going to assume you don't. Here's how to calculate what
you need."

If you would really like an answer to your question and aren't just trolling,
here are some observations...

Derating power resistors for loading to 50% of rated wattage is a standard
practice, although I'd personally think that's on the conservative side. A lot
of it depends on how conservative you want to be, what type of voltage waveform
you're applying to the load, and duty cycle considerations. You haven't seen
fit to give us any of that information, though, so I would suppose 50% power is
a good place to start.

As of when I'm reading this, no respondent to your original question can make
sense of what you're asking or figure out what you need. Possibly you might
want to take a deep breath, then drop the assumption that the respondents to
sci.electronics.design can read minds (I already get that from The War
Department, thank you), and work on the hypothesis that you could phrase your
question a little better.
 
R

Roger Gt

Jan 1, 1970
0
JaBrIoL said:
All.

I have to build a load device.
specs are: 5 volts @ 38A @ .5 Ohms @ 380 watts.

I am constrained to using resistor no bigger than 50 watts each.
I figure out 10 .1 resistors in parallel to give me an approximate
load.

in the words of Bill O'Reilly; What say you?

Not clear.

Full power load of 38 Amperes is 0.13157895 ohms,
which may be had with 19 each 2.5 ohm 10 watt resistors.
Full power is 190 watts. (Increase resistor rating to 25 watts if using
this in production, with a fan!)

Half load is 19 each 5 ohm 10 watt resistors
which is 95 watts with 190 watts of resistor rating,
so they will not be quite as hot (Will still burn your fingers)

Use a terminal strip to mount the resistors, and if this is a production
application, put a fan on them with a cover to avoid any papers being
stacked on the resistors. (FIRE!)

It the data is not correct, and you are actually trying to do something you
failed to convey. Try again!
 
J

JaBrIoL

Jan 1, 1970
0
Frank Bemelman said:
More like a load of rubbish. 5V @ 38A is not .5 ohms, and it
is not 380W and 10 resistors of 0.1 in parallel is 0.01 ohm.

The final load design would be as follow:
10 resistors 5 ohms each, in parallel

5/10 - 0.5 ohms for the total.
 
F

Frank Bemelman

Jan 1, 1970
0
JaBrIoL said:
The final load design would be as follow:
10 resistors 5 ohms each, in parallel

5/10 - 0.5 ohms for the total.

That makes more sense. Vishay has a nice series of
resistors, RPH100 series, with screw terminals.
 
R

R.Legg

Jan 1, 1970
0
All.

I have to build a load device.

specs are:

5 volts @ 38A @ .5 Ohms @ 380 watts.

I am constrained to using resistor no bigger than 50 watts each.
I figure out 10 .1 resisitors in parrallel to give me an aproximate
load.

in the words of Bill O'Reilly


What say you?

For a 5V load using 50W resistors at 38W (76%stress level), you would
need a values of 1R5. (W/V^2)

Connecting incrementally in parallel will give you 380W load, by the
time you reach ten resistors.

Make sure your interconnections are low impedance, so as not to reduce
the effectiveness of any individual element. 20 milliohms kelvin
connection loss is permitted, without a reduction in approximate power
loss, in each element.

RL
 
J

JaBrIoL

Jan 1, 1970
0
Frank Bemelman said:
More closely??? Where does the other half of the current go?
And since when goes wattage up with currents being cut in half?

P = I^2 x R. If your current is cut in half, the power is cut
in 4, certainly not doubled ;) Or should I somehow notice that
the voltage is cut in half, or the resistors are cut in half,
or the load has been cut in half ? Next time you say that it
can't be done because your budget has been cut in half.

yup.. ye are new at this... look up the term halfload for power
supplies test and development.
 
J

JaBrIoL

Jan 1, 1970
0
Frank Bemelman said:
That makes more sense. Vishay has a nice series of
resistors, RPH100 series, with screw terminals.

I will look them up... if they are 50 watts, they will not serve my application.
 
J

JaBrIoL

Jan 1, 1970
0
the power supply orignal draw is 76A, for a half load to exercise the
supply, you cut the current by 50%. some companies, would test these
at 80%, so of ciurse you do the math for that number.
 
Top