Need hardware advice: ULN2003A

[Christopher4187]

So, it looks like the general consensus is that I can "gang" together some outputs and the potential issue of the flyback voltage should be mitigated. Are there any objections?

[Luciano]

Hi,

When you use only one output of the ULN2003A you can drive a relay with
a coil current up to 500mA. (450 mA with 100% duty cycle, N package).
The internal protection diode will suppress the self-induction current when
you deenergise the relay. Do not forget to connect PIN9.
All unused inputs must be tied to GND.

If you use two outputs simultaneously, then the current of each
output is limited to 280 mA. (100% duty cycle, N package).
See the picture below if you use more than two outputs simultaneously.
(The picture is from the datasheet of the ULN2003A).


(Click to enlarge the picture)


(Click to enlarge the picture)

Best regards,

Luciano

[sayzer]

Nice explanations Luciano. But I am confused.

You are saying, actually, 500mA is only valid when we use one output at a time. Interesting! because if we make a parallel connection as the data sheet says, then this statement can not be correct.

In a parallel connection, we will be using two outputs at the same time.
In that case, we multiply the current capacity so we get 2x500=1000mA, but according your information, "If you use two outputs simultaneously, then the current of each output is limited to 280 mA". So we get back to where we started.
No change!

Isn't this interesting?




Also, what about the protection diodes? I checked the datasheet but very poor information there.

[Luciano]

Hi,

The IC package can only dissipate a certain amount of heat.
If you use only one output the max current for the package is 480mA.
If you use two outputs, the current is 280 mA per output so the total
current for the package is 560 mA. This small difference is because
some of the heat is dissipated by the metal of the pin connected to
the PCB.

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

Best regards,

Luciano

[Christopher4187]

Luciano,

Thanks for the information. I was looking at one of the pictures and it says that you have to connect pin 9. I have NEVER used pin 9 and if you look at figure 17, they show it as not being connected. Other figures show it connected. Did I miss something in the data sheet?

[Luciano]

Hi,

PIN 9 must be connected to V+.
(See picture below).

Figure 17 shows 7 lamps on my datasheet. Lamps are resistive loads.
There is no self-induction when you open a circuit with resistive loads
so there is no need for the protection diodes. The figure 17 is probably
a pinball circuit, when you press the button you can verify that all the
lamps are working.

Your case is the figure 19 on the datasheet.

Best regards,

Luciano

The Datasheet I have:
http://focus.ti.com/lit/ds/symlink/uln2003a.pdf
(See page 11, fig. 19)


(Click to enlarge the picture)

(About SMD package)

(Click to enlarge the picture)

[Christopher4187]

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.