Need hardware advice: ULN2003A


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  1. #1
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    Luciano,

    Thanks. The current output was always clear to me. However, it appears I am using the incorrect ULN model. If pin 9 needs to be connected to V+, I need the CMOS version. The relay I am using is a 12 volt version. What is evident to me, even thought I thought I was protected by the ULN, I have NEVER used that flyback didoe all of this time! I have used this IC for 5 years and NEVER used it! I am surprised that I have never had any problems in the last 5 years.

  2. #2
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    Hi,

    There is no CMOS or TTL version for the ULN2003A.
    If the relay is 12V, connect +12V to the pin 9.
    If the relay is 48V, connect +48V to the pin 9.
    (The outputs of the ULN2003A work up to 50V).

    What you see on the datasheet (Fig. 18 and 19) is
    what you have when the input of ULN2003A is connected
    to a CMOS IC or a TTL IC. What you see on the left of
    fig. 19 represents an output inside a TTL IC. What you
    see on the left of fig. 18 represents an output inside
    a CMOS IC.

    On the first page of the datasheet you car read
    "Inputs Compatible with various types of logic".

    Best regards,

    Luciano
    Last edited by Luciano; - 28th September 2006 at 11:36.

  3. #3
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    Quote Originally Posted by Christopher4187
    However, it appears I am using the incorrect ULN model. If pin 9 needs to be connected to V+, I need the CMOS version. The relay I am using is a 12 volt version.
    You dont need to change at all.

    Pin 9 is the pin which all the clamping diodes are connected. This should be connected to the supply voltage of the relays/lamps that you are driving. Irrespective of which chip you are using this can be upto 50V.


    The TTL/CMOS versions feature different inputs for driving from those logic families. The PIC can drive either type. The supply voltage for the relays/lamps has no bearing on which input type you need.

    Just make sure that you dont accidently connect your load supply to your logic supply or the magic smoke will escape
    Keith

    www.diyha.co.uk
    www.kat5.tv

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    Among these information, I am still confused about the parallel connection and what benefits it has.

    I am sure Luciano has a point there but if I will get only 60mA more in two-parallel connection, then it makes no sense. Why would I need a parallel connection then? 500mA current path can already handle 560mA. It is only about %10 over driving.

    I am saying this because the datasheet says that higher current can be achieved by parallel connection. Only 60mA with two, not double? even if I add all arrays parallel, it makes no sense either. No benefit at all.
    "If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte

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    If I understand the data sheet correctly, you can have 7 outputs at a duty cycle of 100% but only 120mA per output (if you have the N package). Or, you can have 3 outputs at 100% duty cycle at 220mA. You can follow the curve on the data sheet to determine what you are allowed to use. As a general statement, it would appear to me that the max output at 100% duty cycle is approximately 840mA. Can someone confirm what I just wrote?

    I will connect pin 9 to the +12V supply and test it. Again, I am surprised that I have not had any issues without connecting the pin 9. I guess I have just been lucky!

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    Hi,

    @Sayzer:

    The use of a parallel connection was your suggestion.
    Why use a parallel connection when the current of the relay coil is 77mA?

    My advice is:

    There is no need for a parallel connection.
    Do not use a parallel connections with this IC.


    * * * * * * * * *

    @Christopher4187:

    N-package
    7 x 120 mA = 840 mA if the TA is <= 70°C.
    TA = Ambient temperature, the temperature inside the enclosure where the PCB resides.
    Note that you can see TA = 85°C on the figure 15, but for the ULN2003A the
    maximum TA is 70°C (See page 4).

    * * *

    D-package
    (Based on the Digikey # in your first post, this is the SMD version you are using).
    7 x 60 mA = 420 mA if the TA is <= 70°C.

    Best regards,

    Luciano

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    Quote Originally Posted by Luciano
    Hi,

    @Sayzer:

    The use of a parallel connection was your suggestion.
    Why use a parallel connection when the current of the relay coil is 77mA?

    My advice is:

    There is no need for a parallel connection.
    Do not use a parallel connections with this IC.


    * * * * * * * * *

    For sure, from now on, I will not suggest parallel connection.
    In fact, I wonder what benefit it has; why is it there? Lets leave it to the engineer who designed it.
    "If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte

  8. #8
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    The reason for the parallel connection is clear. Here is a real world example using the "N" package at 60 degrees C:

    I have a relay that takes 700mA to close and your duty cycle is 100%. If I am using one output, the maximum current supply is about 450mA and that won't be enough. If I ganged two together, the supply will be about 580mA; still not enough. If I gang three together, the supply will be about 660mA; still not enough. If I gang four together, the supply will be about 720mA; this would just be enough. For a buffer, I would gang five together and the supply would be about 750mA.

    As you can see, as you increase the number of outputs used, the overall current increases but the individual current per pin decreases.

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