Shop OBEX P1 Docs P2 Docs Learn Events
Opto Results & LED Quiz — Parallax Forums

Opto Results & LED Quiz

ArchiverArchiver Posts: 46,084
edited 2004-06-18 21:52 in General Discussion
Any way, the optos came in today and installed it on the breadboard and my
schematic worked fine.

The voltage coming out of the opto is only 4.35 volts so I lowered the
value of the LED resistor to 250 ohms in order to get the 10 ma thru it. It
calculates out at 17ma but was only 10ma when I checked it. I'm sure I'm
missing some calculation somewhere.

Sorry if this stupid, but I'm trying to learn. Everything I've done in the
past with LED's was using default values copied out of the project books.

How is the best way to determine how much current to put thru an LED for the
best results. Example: The Opto's LED states Forward Voltage of 1.25 volts
typical, I_F = 50ma
Is the I_F mean forward current? Does that mean that is the Max Current I
can use on that LED "or" the amount I "should" use?

So if I see a LED rated like this: vi @ if (ma) = 1.9 @ 20 means that 20
ma is where is should run and there is a 1.9 volt drop across it?

Richard





Original Message
From: "Richard Skinner" <rwskinner@a...>
To: <basicstamps@yahoogroups.com>
Sent: Thursday, June 17, 2004 5:08 PM
Subject: Re: [noparse][[/noparse]basicstamps] Re: Quick Opto-Coupler Question


> Whoa, the timing on the messages on the list is kind of messed up.
>
> Reason for using the Opto, because I wanted to be safe in using the 24
> volts. I also looked at the Stamp PLC and it had a very simular circuit
so
> I thought it was the way to go.
>
> I guess there is a dozen ways to skin a cat.
>
>
>
>
Original Message
> From: "Dave Mucha" <davemucha@j...>
> To: <basicstamps@yahoogroups.com>
> Sent: Thursday, June 17, 2004 10:17 AM
> Subject: [noparse][[/noparse]basicstamps] Re: Quick Opto-Coupler Question
>
>
> > --- In basicstamps@yahoogroups.com, "Richard Skinner"
> > <rwskinner@a...> wrote:
> > > I revised it some: I didn't like the way the LED Indicator was so I
> > > Grounded the LED then went from the LED thru a 470 ohm resistor to
> > Pin 4 of
> > > the opto. That way when the opto goes high it lights the LED.
> > >
> > >
> >
> > .
> > .
> > I'm not sure why you are using an isolator ?
> >
> > you have the two circuits connected together.
> > Why not eleminate it and just use a voltage divider
> > with a 10k ohm to the Stamp input ?
> >
> > If you need more power, or want the LED on the Stamp side, you can
> > substitute a npn transistor.
> >
> > If you want isolation, disconnect the common ground line and the
> > resistor in it. If you can share the common ground, then the
> > circuits are not isolated and you don't need the isolator.
> >
> > Dave
> >
> >
> >
> >
> >
> > To UNSUBSCRIBE, just send mail to:
> > basicstamps-unsubscribe@yahoogroups.com
> > from the same email address that you subscribed. Text in the Subject
and
> Body of the message will be ignored.
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
> > [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
> >
> >
>
>
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the Subject and
Body of the message will be ignored.
>
> Yahoo! Groups Links
>
>
>
>
>
>
> [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
>
>

Comments

  • ArchiverArchiver Posts: 46,084
    edited 2004-06-18 04:04
    For a Great LED Tutorial.....

    www.theledlight.com/technical.html

    Original Message
    From: "Richard" <rwskinner@a...>
    To: <basicstamps@yahoogroups.com>
    Sent: Thursday, June 17, 2004 9:46 PM
    Subject: Re: [noparse][[/noparse]basicstamps] Opto Results & LED Quiz


    > Any way, the optos came in today and installed it on the breadboard and my
    > schematic worked fine.
    >
    > The voltage coming out of the opto is only 4.35 volts so I lowered the
    > value of the LED resistor to 250 ohms in order to get the 10 ma thru it.
    It
    > calculates out at 17ma but was only 10ma when I checked it. I'm sure I'm
    > missing some calculation somewhere.
    >
    > Sorry if this stupid, but I'm trying to learn. Everything I've done in
    the
    > past with LED's was using default values copied out of the project books.
    >
    > How is the best way to determine how much current to put thru an LED for
    the
    > best results. Example: The Opto's LED states Forward Voltage of 1.25
    volts
    > typical, I_F = 50ma
    > Is the I_F mean forward current? Does that mean that is the Max Current I
    > can use on that LED "or" the amount I "should" use?
    >
    > So if I see a LED rated like this: vi @ if (ma) = 1.9 @ 20 means that 20
    > ma is where is should run and there is a 1.9 volt drop across it?
    >
    > Richard
    >
    >
    >
    >
    >
    >
    Original Message
    > From: "Richard Skinner" <rwskinner@a...>
    > To: <basicstamps@yahoogroups.com>
    > Sent: Thursday, June 17, 2004 5:08 PM
    > Subject: Re: [noparse][[/noparse]basicstamps] Re: Quick Opto-Coupler Question
    >
    >
    > > Whoa, the timing on the messages on the list is kind of messed up.
    > >
    > > Reason for using the Opto, because I wanted to be safe in using the 24
    > > volts. I also looked at the Stamp PLC and it had a very simular circuit
    > so
    > > I thought it was the way to go.
    > >
    > > I guess there is a dozen ways to skin a cat.
    > >
    > >
    > >
    > >
    Original Message
    > > From: "Dave Mucha" <davemucha@j...>
    > > To: <basicstamps@yahoogroups.com>
    > > Sent: Thursday, June 17, 2004 10:17 AM
    > > Subject: [noparse][[/noparse]basicstamps] Re: Quick Opto-Coupler Question
    > >
    > >
    > > > --- In basicstamps@yahoogroups.com, "Richard Skinner"
    > > > <rwskinner@a...> wrote:
    > > > > I revised it some: I didn't like the way the LED Indicator was so I
    > > > > Grounded the LED then went from the LED thru a 470 ohm resistor to
    > > > Pin 4 of
    > > > > the opto. That way when the opto goes high it lights the LED.
    > > > >
    > > > >
    > > >
    > > > .
    > > > .
    > > > I'm not sure why you are using an isolator ?
    > > >
    > > > you have the two circuits connected together.
    > > > Why not eleminate it and just use a voltage divider
    > > > with a 10k ohm to the Stamp input ?
    > > >
    > > > If you need more power, or want the LED on the Stamp side, you can
    > > > substitute a npn transistor.
    > > >
    > > > If you want isolation, disconnect the common ground line and the
    > > > resistor in it. If you can share the common ground, then the
    > > > circuits are not isolated and you don't need the isolator.
    > > >
    > > > Dave
    > > >
    > > >
    > > >
    > > >
    > > >
    > > > To UNSUBSCRIBE, just send mail to:
    > > > basicstamps-unsubscribe@yahoogroups.com
    > > > from the same email address that you subscribed. Text in the Subject
    > and
    > > Body of the message will be ignored.
    > > >
    > > > Yahoo! Groups Links
    > > >
    > > >
    > > >
    > > >
    > > >
    > > >
    > > > [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
    > > >
    > > >
    > >
    > >
    > >
    > >
    > > To UNSUBSCRIBE, just send mail to:
    > > basicstamps-unsubscribe@yahoogroups.com
    > > from the same email address that you subscribed. Text in the Subject
    and
    > Body of the message will be ignored.
    > >
    > > Yahoo! Groups Links
    > >
    > >
    > >
    > >
    > >
    > >
    > > [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
    > >
    > >
    >
    >
    >
    > To UNSUBSCRIBE, just send mail to:
    > basicstamps-unsubscribe@yahoogroups.com
    > from the same email address that you subscribed. Text in the Subject and
    Body of the message will be ignored.
    >
    > Yahoo! Groups Links
    >
    >
    >
    >
    >
    >
    > [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
    >
    >
  • ArchiverArchiver Posts: 46,084
    edited 2004-06-18 14:12
    'Typical' continuous LED current is around
    10 mA to 15 mA. 20 mA can be used, but you
    don't get much more light between 15 and 20 mA.

    The 1.25 volts at 50 ma is trying to show the
    max voltage the LED drops -- you wouldn't want
    to run it at 50 mA.

    And, you can 'pulse' LED's at high current
    (50 mA? 100 mA?) for SHORT periods of time
    to get greater brightness. However, if your
    code that pulses the LED's breaks, you burn
    them out.

    To calculate LED's you need a few values.
    1. What is the constant voltage drop acrosss it,
    when it is 'on' -- lit.
    Typically 1.2 to 1.4 volts.

    2. What is the typical current to run it at.
    Many specs tell you the MAX current you COULD run
    it at -- usually you don't want to design to this,
    since if you go over EVEN A LITTLE you burn out the
    LED.

    3. What is the MINIMUM current to run through
    it before it lights at all. Few specs tell you
    this, you may have to experiment.

    There is some range of current that the LED lights
    fully. Above this it burns out. Below this it
    looks dim. You want to be in the middle to lower
    end of this range. The voltage drop across the LED
    is constant the whole time at around 1.4 volts.

    --- In basicstamps@yahoogroups.com, "Richard" <rwskinner@a...> wrote:
    > Any way, the optos came in today and installed it on the breadboard
    and my
    > schematic worked fine.
    >
    > The voltage coming out of the opto is only 4.35 volts so I
    lowered the
    > value of the LED resistor to 250 ohms in order to get the 10 ma
    thru it. It
    > calculates out at 17ma but was only 10ma when I checked it. I'm
    sure I'm
    > missing some calculation somewhere.
    >
    > Sorry if this stupid, but I'm trying to learn. Everything I've
    done in the
    > past with LED's was using default values copied out of the project
    books.
    >
    > How is the best way to determine how much current to put thru an
    LED for the
    > best results. Example: The Opto's LED states Forward Voltage of
    1.25 volts
    > typical, I_F = 50ma
    > Is the I_F mean forward current? Does that mean that is the Max
    Current I
    > can use on that LED "or" the amount I "should" use?
    >
    > So if I see a LED rated like this: vi @ if (ma) = 1.9 @ 20 means
    that 20
    > ma is where is should run and there is a 1.9 volt drop across it?
    >
    > Richard
    >
    >
    >
    >
    >
    >
    Original Message
    > From: "Richard Skinner" <rwskinner@a...>
    > To: <basicstamps@yahoogroups.com>
    > Sent: Thursday, June 17, 2004 5:08 PM
    > Subject: Re: [noparse][[/noparse]basicstamps] Re: Quick Opto-Coupler Question
    >
    >
    > > Whoa, the timing on the messages on the list is kind of messed up.
    > >
    > > Reason for using the Opto, because I wanted to be safe in using
    the 24
    > > volts. I also looked at the Stamp PLC and it had a very simular
    circuit
    > so
    > > I thought it was the way to go.
    > >
    > > I guess there is a dozen ways to skin a cat.
    > >
    > >
    > >
    > >
    Original Message
    > > From: "Dave Mucha" <davemucha@j...>
    > > To: <basicstamps@yahoogroups.com>
    > > Sent: Thursday, June 17, 2004 10:17 AM
    > > Subject: [noparse][[/noparse]basicstamps] Re: Quick Opto-Coupler Question
    > >
    > >
    > > > --- In basicstamps@yahoogroups.com, "Richard Skinner"
    > > > <rwskinner@a...> wrote:
    > > > > I revised it some: I didn't like the way the LED Indicator
    was so I
    > > > > Grounded the LED then went from the LED thru a 470 ohm
    resistor to
    > > > Pin 4 of
    > > > > the opto. That way when the opto goes high it lights the LED.
    > > > >
    > > > >
    > > >
    > > > .
    > > > .
    > > > I'm not sure why you are using an isolator ?
    > > >
    > > > you have the two circuits connected together.
    > > > Why not eleminate it and just use a voltage divider
    > > > with a 10k ohm to the Stamp input ?
    > > >
    > > > If you need more power, or want the LED on the Stamp side, you
    can
    > > > substitute a npn transistor.
    > > >
    > > > If you want isolation, disconnect the common ground line and the
    > > > resistor in it. If you can share the common ground, then the
    > > > circuits are not isolated and you don't need the isolator.
    > > >
    > > > Dave
    > > >
    > > >
    > > >
    > > >
    > > >
    > > > To UNSUBSCRIBE, just send mail to:
    > > > basicstamps-unsubscribe@yahoogroups.com
    > > > from the same email address that you subscribed. Text in the
    Subject
    > and
    > > Body of the message will be ignored.
    > > >
    > > > Yahoo! Groups Links
    > > >
    > > >
    > > >
    > > >
    > > >
    > > >
    > > > [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
    > > >
    > > >
    > >
    > >
    > >
    > >
    > > To UNSUBSCRIBE, just send mail to:
    > > basicstamps-unsubscribe@yahoogroups.com
    > > from the same email address that you subscribed. Text in the
    Subject and
    > Body of the message will be ignored.
    > >
    > > Yahoo! Groups Links
    > >
    > >
    > >
    > >
    > >
    > >
    > > [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
    > >
    > >
  • ArchiverArchiver Posts: 46,084
    edited 2004-06-18 16:29
    >
    > And, you can 'pulse' LED's at high current
    > (50 mA? 100 mA?) for SHORT periods of time
    > to get greater brightness. However, if your
    > code that pulses the LED's breaks, you burn
    > them out.
    >
    .
    <HUGE snip>
    .

    You can pulse an LED with 100mA as long as the duty cycle does not
    exceed the max for the device. so a 50% duty cycle would deliver
    50mA.

    But, the LED will be signifigantly brighter at the higher currents
    and that makes it worth the effort.

    also, you can pulse them at a higher current, for shorter times and
    use less power for a brighter light.

    Dave
  • ArchiverArchiver Posts: 46,084
    edited 2004-06-18 17:29
    Okay, lets see if I understand it correctly.

    I have a green LED which states: 2.3 @ 20 Vf @ If (ma)
    The voltage into my Darlington is 5.3 volts and the output out of my
    Darlington is 4.35 volts
    I'm going to use 15 ma to be safe.....
    4.35 - 2.3 = 2.05 / 15ma = 136 ohms. Well, I actually have 150 ohms in my
    box.

    2.05 / 150 = 13 ma so a 150 ohm 1/8 watt would be fine.

    On the OptoCoupler, its ratings are If = 50ma Typical Vf = 1.25
    My input voltage is 24 vdc nominal (18-28). Design for 1/2 the ma @ 25 ma.

    24 vdc - 1.25 = 22.75 volts / 25ma = 910 ohms so in my drawer I have a 1K

    18 volts - 1.25 = 16.75 / 1000 = 17ma
    28 volts - 1.25 = 26.75 / 1000 = 27 ma
    24 volts - 1.25 = 22.75 / 1000 = 23 ma

    So, I take it a 1K resistor from 24 volts to power my Opto's LED (4N32)
    would work perfect, if 25 ma is good for long life. Whoa, that calculates
    0.600 watts so it needs a 1 watt resistor????


    Have I figured it out "Finally"?

    Richard


    Original Message
    From: "Dave Mucha" <davemucha@j...>
    To: <basicstamps@yahoogroups.com>
    Sent: Friday, June 18, 2004 10:29 AM
    Subject: [noparse][[/noparse]basicstamps] Re: Opto Results & LED Quiz


    >
    > >
    > > And, you can 'pulse' LED's at high current
    > > (50 mA? 100 mA?) for SHORT periods of time
    > > to get greater brightness. However, if your
    > > code that pulses the LED's breaks, you burn
    > > them out.
    > >
    > .
    > <HUGE snip>
    > .
    >
    > You can pulse an LED with 100mA as long as the duty cycle does not
    > exceed the max for the device. so a 50% duty cycle would deliver
    > 50mA.
    >
    > But, the LED will be signifigantly brighter at the higher currents
    > and that makes it worth the effort.
    >
    > also, you can pulse them at a higher current, for shorter times and
    > use less power for a brighter light.
    >
    > Dave
    >
    >
    >
    >
    >
    > To UNSUBSCRIBE, just send mail to:
    > basicstamps-unsubscribe@yahoogroups.com
    > from the same email address that you subscribed. Text in the Subject and
    Body of the message will be ignored.
    >
    > Yahoo! Groups Links
    >
    >
    >
    >
    >
    >
    > [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
    >
    >
  • ArchiverArchiver Posts: 46,084
    edited 2004-06-18 21:12
    > On the OptoCoupler, its ratings are If = 50ma Typical Vf = 1.25
    > My input voltage is 24 vdc nominal (18-28). Design for 1/2 the ma
    @ 25 ma.
    >
    > 24 vdc - 1.25 = 22.75 volts / 25ma = 910 ohms so in my drawer I
    have a 1K
    >
    > 18 volts - 1.25 = 16.75 / 1000 = 17ma
    > 28 volts - 1.25 = 26.75 / 1000 = 27 ma
    > 24 volts - 1.25 = 22.75 / 1000 = 23 ma
    >
    > So, I take it a 1K resistor from 24 volts to power my Opto's LED
    (4N32)
    > would work perfect, if 25 ma is good for long life. Whoa, that
    calculates
    > 0.600 watts so it needs a 1 watt resistor????
    >
    >
    > Have I figured it out "Finally"?
    >
    > Richard
    >

    eeeeep ! do not pass, GO, do not collect $200.00.

    the opto is rated for 20mA, not 50 or 60. check where that value
    comes from and what the graphs show as typical use.

    The opto you listed has a spec from Fairchild that lists the largest
    tranfer at something between 2 and 3 mA, with signifigant ratios down
    to about 0.7mA

    The 60mA to which you speak seems to me to be the collector/emitter
    and not the base.

    Dave
  • ArchiverArchiver Posts: 46,084
    edited 2004-06-18 21:52
    Ahh, I got you. So I'll leave the 4.7k in there which makes it more
    better. Also allows me to use allot lower wattage resister!
    Were my other LED calcs okay then?

    Original Message
    From: "Dave Mucha" <davemucha@j...>
    To: <basicstamps@yahoogroups.com>
    Sent: Friday, June 18, 2004 3:12 PM
    Subject: [noparse][[/noparse]basicstamps] Re: Opto Results & LED Quiz


    >
    > > On the OptoCoupler, its ratings are If = 50ma Typical Vf = 1.25
    > > My input voltage is 24 vdc nominal (18-28). Design for 1/2 the ma
    > @ 25 ma.
    > >
    > > 24 vdc - 1.25 = 22.75 volts / 25ma = 910 ohms so in my drawer I
    > have a 1K
    > >
    > > 18 volts - 1.25 = 16.75 / 1000 = 17ma
    > > 28 volts - 1.25 = 26.75 / 1000 = 27 ma
    > > 24 volts - 1.25 = 22.75 / 1000 = 23 ma
    > >
    > > So, I take it a 1K resistor from 24 volts to power my Opto's LED
    > (4N32)
    > > would work perfect, if 25 ma is good for long life. Whoa, that
    > calculates
    > > 0.600 watts so it needs a 1 watt resistor????
    > >
    > >
    > > Have I figured it out "Finally"?
    > >
    > > Richard
    > >
    >
    > eeeeep ! do not pass, GO, do not collect $200.00.
    >
    > the opto is rated for 20mA, not 50 or 60. check where that value
    > comes from and what the graphs show as typical use.
    >
    > The opto you listed has a spec from Fairchild that lists the largest
    > tranfer at something between 2 and 3 mA, with signifigant ratios down
    > to about 0.7mA
    >
    > The 60mA to which you speak seems to me to be the collector/emitter
    > and not the base.
    >
    > Dave
    >
    >
    >
    >
    > To UNSUBSCRIBE, just send mail to:
    > basicstamps-unsubscribe@yahoogroups.com
    > from the same email address that you subscribed. Text in the Subject and
    Body of the message will be ignored.
    >
    > Yahoo! Groups Links
    >
    >
    >
    >
    >
    >
    > [noparse][[/noparse]Scanned by AwesomeNet Anti-Virus]
    >
    >
Sign In or Register to comment.