Need to confirm resistor selection

S

SklettTheNewb

Guest
I don't like to start a post with a disclaimer, but in this case I
must; I am a programmer, not a EE or hardware guy. However I can
handle some very simple tasks and do enjoy the rare chance to work
with electronics.

So, with that said I find myself in a situation where I am taking over
on a circuit design for a project we are working on. Logic problems I
can handle, but it's some of the more rudimentary electronics issues
that I stumble on. At this point I need to specify a resistor network
for use with a SSR (http://www.clare.com/home/pdfs.nsf/www/
CPC1014N.pdf/$file/CPC1014N.pdf) that draws 2mA. The SSR requires
1.2v for it's input.

The resistor networks are needed to bring the 5v source down to the
required ~1.2v range for the SSR input. An engineer that was (no
longer is) involved on this project specified a 1k ohm resistor.
Today I was faced with the question of which power rating was needed
for this use. I did some research and found the formula:
P = V2 / ohm
or
P = 25 / 1,000 = 0.025mW

Is my understanding of the calculation correct? Did I provide enough
information?

Thanks for reading,
Steve
 
SklettTheNewb wrote:
I don't like to start a post with a disclaimer, but in this case I
must; I am a programmer, not a EE or hardware guy. However I can
handle some very simple tasks and do enjoy the rare chance to work
with electronics.

So, with that said I find myself in a situation where I am taking over
on a circuit design for a project we are working on. Logic problems I
can handle, but it's some of the more rudimentary electronics issues
that I stumble on. At this point I need to specify a resistor network
for use with a SSR (http://www.clare.com/home/pdfs.nsf/www/
CPC1014N.pdf/$file/CPC1014N.pdf) that draws 2mA. The SSR requires
1.2v for it's input.

The resistor networks are needed to bring the 5v source down to the
required ~1.2v range for the SSR input. An engineer that was (no
longer is) involved on this project specified a 1k ohm resistor.
Today I was faced with the question of which power rating was needed
for this use. I did some research and found the formula:
P = V2 / ohm
or
P = 25 / 1,000 = 0.025mW
^^
Click to expand...
Not mW, Watts. P = .025W - but that's not the best way to figure
the R value. See below.

Is my understanding of the calculation correct? Did I provide enough
information?

Thanks for reading,
Steve
Click to expand...
A low wattage 1K resistor will be great for you.

The SSR uses an LED with a Vf of 1.2 and an If of 2 mA.
That If figure - 2 mA - is the maximum current needed to
guarantee that the SSR will be energized. You could run
the LED current a lot higher than that. What you want to
do is ensure that the LED gets _at least_ 2 mA and limit
the current to something a lot higher.

With a 5V supply, and ~1.2 volts dropped across the LED, you
want to drop ~3.8V across the resistor. At 3.8 mA a 1K resistor
will drop 3.8 volts. The LED inside the SSR will happily draw
the 3.8 mA - in fact, it will draw as much current as it can get.
The 1K resistor limits what it can get to 3.8 mA. The LED could
draw a helluva lot more current without damage to itself if the
resistor was a lower value, but there is no need to have it do
that, as the 1K will give it well over the 2 mA maximum it needs
for guaranteed correct operation.

The power dissipation for the resistor is computed with P = I^2*R,
or .014 watts in this case, so use .03 watts or higher for the
resistor wattage.

Ed
 
On Aug 26, 11:05 pm, SklettTheNewb <stevekl...@gmail.com> wrote:
I don't like to start a post with a disclaimer, but in this case I
must; I am a programmer, not a EE or hardware guy.  However I can
handle some very simple tasks and do enjoy the rare chance to work
with electronics.

So, with that said I find myself in a situation where I am taking over
on a circuit design for a project we are working on.  Logic problems I
can handle, but it's some of the more rudimentary electronics issues
that I stumble on.  At this point I need to specify a resistor network
for use with a SSR (http://www.clare.com/home/pdfs.nsf/www/
CPC1014N.pdf/$file/CPC1014N.pdf) that draws 2mA.  The SSR requires
1.2v for it's input.

The resistor networks are needed to bring the 5v source down to the
required ~1.2v range for the SSR input.  An engineer that was (no
longer is) involved on this project specified a 1k ohm resistor.
Today I was faced with the question of which power rating was needed
for this use.  I did some research and found the formula:
P = V2 / ohm
or
P = 25 / 1,000 = 0.025mW

Is my understanding of the calculation correct?  Did I provide enough
information?

Thanks for reading,
Steve
Click to expand...
25/1000 = 25 mW
 
On Fri, 26 Aug 2011 20:05:59 -0700 (PDT), SklettTheNewb
<steveklett@gmail.com> wrote:

I don't like to start a post with a disclaimer, but in this case I
must; I am a programmer, not a EE or hardware guy. However I can
handle some very simple tasks and do enjoy the rare chance to work
with electronics.

So, with that said I find myself in a situation where I am taking over
on a circuit design for a project we are working on. Logic problems I
can handle, but it's some of the more rudimentary electronics issues
that I stumble on. At this point I need to specify a resistor network
for use with a SSR (http://www.clare.com/home/pdfs.nsf/www/
CPC1014N.pdf/$file/CPC1014N.pdf) that draws 2mA. The SSR requires
1.2v for it's input.

The resistor networks are needed to bring the 5v source down to the
required ~1.2v range for the SSR input. An engineer that was (no
longer is) involved on this project specified a 1k ohm resistor.
Today I was faced with the question of which power rating was needed
for this use. I did some research and found the formula:
P = V2 / ohm
or
P = 25 / 1,000 = 0.025mW

Is my understanding of the calculation correct? Did I provide enough
information?

Thanks for reading,
Steve
Click to expand...

Where did the 25 for V2 come from?

The important values are:
LED current to operate: 2 mA (you want to supply somewhat more
than that)
Absolute Maximum control current: 50 mA (you want to stay well
below that)

The resistor has to drop [supply voltage] - [LED "on" voltage], or 5 -
1.2 = 3.8 volts.

I'd plan for about 10 mA, which would be
R = E/I = 3.8V/.010mA = 380 ohms (390 ohms is a standard value)

The suggested 1000 ohms would give 3.8 mA, which is still OK.

The power dissipated in the 1000 ohm resistor is about .014 watts, so
the lowest-power resistor available would be much more than adequate.

--
Peter Bennett, VE7CEI
peterbb (at) telus.net
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca
 
On Aug 26, 10:03 pm, Globemaker <alanfolms...@cabanova.com> wrote:
On Aug 26, 11:05 pm, SklettTheNewb <stevekl...@gmail.com> wrote:









I don't like to start a post with a disclaimer, but in this case I
must; I am a programmer, not a EE or hardware guy.  However I can
handle some very simple tasks and do enjoy the rare chance to work
with electronics.

So, with that said I find myself in a situation where I am taking over
on a circuit design for a project we are working on.  Logic problems I
can handle, but it's some of the more rudimentary electronics issues
that I stumble on.  At this point I need to specify a resistor network
for use with a SSR (http://www.clare.com/home/pdfs.nsf/www/
CPC1014N.pdf/$file/CPC1014N.pdf) that draws 2mA.  The SSR requires
1.2v for it's input.

The resistor networks are needed to bring the 5v source down to the
required ~1.2v range for the SSR input.  An engineer that was (no
longer is) involved on this project specified a 1k ohm resistor.
Today I was faced with the question of which power rating was needed
for this use.  I did some research and found the formula:
P = V2 / ohm
or
P = 25 / 1,000 = 0.025mW

Is my understanding of the calculation correct?  Did I provide enough
information?

Thanks for reading,
Steve

25/1000 = 25 mW
Click to expand...
Ah yes, typo. Good catch.
 
On Aug 27, 6:17 pm, Peter Bennett <pete...@somewhere.invalid> wrote:
On Fri, 26 Aug 2011 20:05:59 -0700 (PDT), SklettTheNewb









stevekl...@gmail.com> wrote:
I don't like to start a post with a disclaimer, but in this case I
must; I am a programmer, not a EE or hardware guy.  However I can
handle some very simple tasks and do enjoy the rare chance to work
with electronics.

So, with that said I find myself in a situation where I am taking over
on a circuit design for a project we are working on.  Logic problems I
can handle, but it's some of the more rudimentary electronics issues
that I stumble on.  At this point I need to specify a resistor network
for use with a SSR (http://www.clare.com/home/pdfs.nsf/www/
CPC1014N.pdf/$file/CPC1014N.pdf) that draws 2mA.  The SSR requires
1.2v for it's input.

The resistor networks are needed to bring the 5v source down to the
required ~1.2v range for the SSR input.  An engineer that was (no
longer is) involved on this project specified a 1k ohm resistor.
Today I was faced with the question of which power rating was needed
for this use.  I did some research and found the formula:
P = V2 / ohm
or
P = 25 / 1,000 = 0.025mW

Is my understanding of the calculation correct?  Did I provide enough
information?

Thanks for reading,
Steve

Where did the 25 for V2 come from?

The important values are:
        LED current to operate: 2 mA (you want to supply somewhat more
than that)
        Absolute Maximum control current: 50 mA (you want to stay well
below that)

The resistor has to drop [supply voltage] - [LED "on" voltage], or 5 -
1.2 = 3.8 volts.

I'd plan for about 10 mA, which would be
        R = E/I = 3.8V/.010mA = 380  ohms (390 ohms is a standard value)

The suggested 1000 ohms would give 3.8 mA, which is still OK.

The power dissipated in the 1000 ohm resistor is about .014 watts, so
the lowest-power resistor available would be much more than adequate.

--
Peter Bennett, VE7CEI  
peterbb (at) telus.net
GPS and NMEA info:http://vancouver-webpages.com/peter
Vancouver Power Squadron:http://vancouver.powersquadron.ca
Click to expand...
The 25v came from 5^2 - I had used "v2" because my superscript didn't
work, I forgot that I could use the "^" character instead.

Thanks for providing the additional formula and explanation on how to
achieve the required current. This is all new to me, I love learning
this stuff!
 
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