P
paul
Guest
can anybody tell me how to calculate the size of resistor required to drop a
29.4v dc supply rail down to 24.5v?
29.4v dc supply rail down to 24.5v?
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P
paul
Guest
the voltage needs to be below 25v as it is feeding an tda2050 amp the
current required would be aprox 2amps
"crooksie" <burbeck@hotmail.com> wrote in message
news:G_vOa.10196$4O4.973440@newsfep2-win.server.ntli.net...
current required would be aprox 2amps
"crooksie" <burbeck@hotmail.com> wrote in message
news:G_vOa.10196$4O4.973440@newsfep2-win.server.ntli.net...
hi Paul
how accurate does the voltage need to be? is the load on this supply
constant? how much current does the load require? these questions need to
be
answered to do the calculation.
another method would be to drop the excess voltage with 7 series
connected
silicone diodes but again the max load current will be needed to determine
the ratings of these diodes.
regards bob
"paul" <p@p.k> wrote in message
news:bednjr$6tj15@eccws12.dearborn.ford.com...
can anybody tell me how to calculate the size of resistor required to
drop
a
29.4v dc supply rail down to 24.5v?
P
petrus bitbyter
Guest
"paul" <p@p.k> schreef in bericht
news:bednjr$6tj15@eccws12.dearborn.ford.com...
Ohms law treats the relations between voltage, current and resistance. You
need to know two of them to calculate the third. So you need to know the
current that's required for your 24.5V asking load. If that current is not
constant, the voltage on your serial resistance will not be constant either.
Unless you're sure about that current, I advise to use an voltage regulator.
An LM317 may be a good choice.
pieter
news:bednjr$6tj15@eccws12.dearborn.ford.com...
can anybody tell me how to calculate the size of resistor required to drop
a
29.4v dc supply rail down to 24.5v?
Paul,
Ohms law treats the relations between voltage, current and resistance. You
need to know two of them to calculate the third. So you need to know the
current that's required for your 24.5V asking load. If that current is not
constant, the voltage on your serial resistance will not be constant either.
Unless you're sure about that current, I advise to use an voltage regulator.
An LM317 may be a good choice.
pieter
R
Robert Lacoste
Guest
R=U/I=(29.4V-24.5V)/2A=2.45ohm
However if I is not stable, then U will not be stable !
However if I is not stable, then U will not be stable !
H
happyhobit
Guest
LM317 has a max current of 1.5 amps.
I believe the LM350 can handle 3 amps
Jay
I believe the LM350 can handle 3 amps
Jay
Ohms law treats the relations between voltage, current and resistance.
You
need to know two of them to calculate the third. So you need to know the
current that's required for your 24.5V asking load. If that current is not
constant, the voltage on your serial resistance will not be constant
either.
Unless you're sure about that current, I advise to use an voltage
regulator.
An LM317 may be a good choice.
pieter
Guest
paul wrote:
where R is the resistor size in ohms and I is the current.
After figuring R, use the formula P = I^2R to figure the
wattage that will be dissipated - and then double that figure
to provide a good safety margin for the resistor wattage
rating.
That will give you what you asked for - but not what you need.
Build a regulated supply, using the 29.4 volts as the input.
Use an LM317 (or similar) as pieter indicated, with sufficient
pass transistors/heat sinks to create a regulated supply at the
voltage and current you need.
R = 4.9/Ican anybody tell me how to calculate the size of resistor required to drop a
29.4v dc supply rail down to 24.5v?
where R is the resistor size in ohms and I is the current.
After figuring R, use the formula P = I^2R to figure the
wattage that will be dissipated - and then double that figure
to provide a good safety margin for the resistor wattage
rating.
That will give you what you asked for - but not what you need.
Build a regulated supply, using the 29.4 volts as the input.
Use an LM317 (or similar) as pieter indicated, with sufficient
pass transistors/heat sinks to create a regulated supply at the
voltage and current you need.