Data sheet interpretation of "single output"

[Peter S. May]

I'm reading the data sheet for the ULN2803A Darlington array (TI:
<http://focus.ti.com/lit/ds/symlink/uln2803a.pdf>) and its collector
current is rated at 500mA, qualified by the phrase "single output".
Does this mean that the 500mA rating applies only when a single output
is on, or that each individual Darlington's Ic is 500mA regardless of
whether the others are on? If the former, would I be able to assume
that the real max Ic for each of the eight pairs is (something like)
500mA/8, given the possibility that they could all be on at once?

This data sheet seems to lack the graphs that would make this clear.

Thanks
PSM

 

[Phil Allison]

"Peter S. May"

I'm reading the data sheet for the ULN2803A Darlington array (TI:
<http://focus.ti.com/lit/ds/symlink/uln2803a.pdf>) and its collector
current is rated at 500mA, qualified by the phrase "single output".
Does this mean that the 500mA rating applies only when a single output
is on, or that each individual Darlington's Ic is 500mA regardless of
whether the others are on?

** It means neither.

500 mA is the maximum from any one output - but as the bulb says, you can
parallel sections for more current.

The maximum continuous TOTAL current from several sections combined is
limited by the TOTAL dissipation in watts - which is limited by the
maximum chip temp of 150C, the thermal impedance of the particular package
and the ambient temp.

About 1.5 watts seems to be the safe upper limit - equates to 4 sections
running at about 300 mA each, continuous.



........ Phil

 

[John Larkin]

I'm reading the data sheet for the ULN2803A Darlington array (TI:
<http://focus.ti.com/lit/ds/symlink/uln2803a.pdf>) and its collector
current is rated at 500mA, qualified by the phrase "single output".
Does this mean that the 500mA rating applies only when a single output
is on, or that each individual Darlington's Ic is 500mA regardless of
whether the others are on? If the former, would I be able to assume
that the real max Ic for each of the eight pairs is (something like)
500mA/8, given the possibility that they could all be on at once?

This data sheet seems to lack the graphs that would make this clear.

Thanks
PSM

The datasheet says 500 mA per output and 2.5 amps total. But it
wouldn't hurt to do some junction temp calculations, too.

John

 

[Peter S. May]

The maximum continuous TOTAL current from several sections combined is
limited by the TOTAL dissipation in watts - which is limited by the
maximum chip temp of 150C, the thermal impedance of the particular package
and the ambient temp.

About 1.5 watts seems to be the safe upper limit - equates to 4 sections
running at about 300 mA each, continuous.

Of course, this makes perfect sense...if I were to fix the max ambient
operating temperature at 120F (49C) then the max dissipation would be
about 1.6W for the DIP version. Thanks for getting me this far.

I'm really afraid I'm about to say something stupid, because I'm still
in the process of developing intuition about how electricity in general,
and transistors in particular, work.

So, the power being dissipated includes both base-emitter and
collector-emitter currents, I would gather. If, as you suggest, I
assume four pairs on at any given time, I'd expect the power dissipation
to be P = n*( IiVi + IcVce ), where n is 4 in this case.

The intended circuit looks something like this:


+5V <------+--+--+-- ...
| | |
V V V LEDs Vf = 3.2V
- - -
| | |
\ \ \
/ / / 100R
\ \ \
| | |
Darl./--------+--+--+-- ...
74HC logic output |/
-->o---------------------|
|\ COM
\>-----| GND

* If the input is low, then the LEDs should be off since their cathodes
aren't grounded.
* If the input is high, then Ic would be something like (5 - 3.2 -
Vce)/100 times the number of LEDs. I'm having a tough time figuring out
what Vce ought to be because I don't have a clear idea what it depends on.
* Once that Ic is figured out, is the resulting dissipated power thereon
actually IcVce?
* The listed Ii is 1.35mA max for Vi = 3.85V. If the input is a 74HC
supplied at 5V, couldn't Vi easily be higher than 3.85V?

So, I think I have a formula, but I'm not sure which figures I use to
fill in the blanks. Suggestions?

Thanks
PSM

 

[John Popelish]

Of course, this makes perfect sense...if I were to fix the max ambient
operating temperature at 120F (49C) then the max dissipation would be
about 1.6W for the DIP version. Thanks for getting me this far.

I'm really afraid I'm about to say something stupid, because I'm still
in the process of developing intuition about how electricity in general,
and transistors in particular, work.

So, the power being dissipated includes both base-emitter and
collector-emitter currents, I would gather. If, as you suggest, I
assume four pairs on at any given time, I'd expect the power dissipation
to be P = n*( IiVi + IcVce ), where n is 4 in this case.

The intended circuit looks something like this:


+5V <------+--+--+-- ...
| | |
V V V LEDs Vf = 3.2V
- - -
| | |
\ \ \
/ / / 100R
\ \ \
| | |
Darl./--------+--+--+-- ...
74HC logic output |/
-->o---------------------|
|\ COM
\>-----| GND

* If the input is low, then the LEDs should be off since their cathodes
aren't grounded.
* If the input is high, then Ic would be something like (5 - 3.2 -
Vce)/100 times the number of LEDs. I'm having a tough time figuring out
what Vce ought to be because I don't have a clear idea what it depends on.
* Once that Ic is figured out, is the resulting dissipated power thereon
actually IcVce?
* The listed Ii is 1.35mA max for Vi = 3.85V. If the input is a 74HC
supplied at 5V, couldn't Vi easily be higher than 3.85V?

So, I think I have a formula, but I'm not sure which figures I use to
fill in the blanks. Suggestions?

The biggest problem you are going to have with this approach
is the rather large (and variable) voltage drop of the
darlington switches. If you used logic level mosfets or
high gain transistors, instead, that saturation voltage
could be so low that the resistors would much better
determine the LED current.

An example of each:
http://www.fairchildsemi.com/ds/ND/NDS355N.pdf

 

[Peter S. May]

The biggest problem you are going to have with this approach is the
rather large (and variable) voltage drop of the darlington switches. If
you used logic level mosfets or high gain transistors, instead, that
saturation voltage could be so low that the resistors would much better
determine the LED current.

To rephrase the problem, then:

I have eight distinct 74HC logic outputs. Each of them corresponds to
the on/off state of 10 high-brightness LEDs of Vf = 3.2V, preferably run
15-20mA. The HC itself doesn't have nearly the sink/source necessary,
so I'll be using transistors. I'd considered using discrete
transistors, but I'd like to avoid the requirement for all the base
resistors, and having all of the emitters connected together would be a
help, since they're all going the same place anyway.

The topology of the ULN2803A seemed to fit the bill; there's a common
cathode, internal resistors tuned for TTL-level input, and 8 inverters
to a device. But apparently I don't know how to use it as well as I
thought. Is there a different array I can use instead or am I stuck
going the discrete route?

 

[John Popelish]

The biggest problem you are going to have with this approach is the
rather large (and variable) voltage drop of the darlington switches. If
you used logic level mosfets or high gain transistors

with a base resistor

, instead, that
saturation voltage could be so low that the resistors would much better
determine the LED current.

An example of each:
http://www.fairchildsemi.com/ds/ND/NDS355N.pdf

Sorry, clicked on the wrong spot.

http://www.zetex.com/3.0/pdf/ZTX688B.pdf
This will pass 500 mA with a 100 mV drop and 2.5 mA of base
current.

 

[Phil Allison]

"Peter S. May"

I have eight distinct 74HC logic outputs. Each of them corresponds to
the on/off state of 10 high-brightness LEDs of Vf = 3.2V, preferably run
15-20mA. The HC itself doesn't have nearly the sink/source necessary,
so I'll be using transistors. I'd considered using discrete
transistors, but I'd like to avoid the requirement for all the base
resistors, and having all of the emitters connected together would be a
help, since they're all going the same place anyway.

The topology of the ULN2803A seemed to fit the bill; there's a common
cathode, internal resistors tuned for TTL-level input, and 8 inverters
to a device. But apparently I don't know how to use it as well as I
thought. Is there a different array I can use instead or am I stuck
going the discrete route?

** The data sheet has the Vce figure for a 200 mA load

- says 1 volt typical ( and 1.3 worst case. )

8 x 0.2 x 1 = 1.6 watts.

1.6 x 63 = 100.8 degress C rise above ambient for the chip inside the N
pack.


Suggest you glue a small heatsink to the package.



....... Phil

 

[ehsjr]

To rephrase the problem, then:

I have eight distinct 74HC logic outputs. Each of them corresponds to
the on/off state of 10 high-brightness LEDs of Vf = 3.2V, preferably run
15-20mA. The HC itself doesn't have nearly the sink/source necessary,
so I'll be using transistors. I'd considered using discrete
transistors, but I'd like to avoid the requirement for all the base
resistors, and having all of the emitters connected together would be a
help, since they're all going the same place anyway.

The topology of the ULN2803A seemed to fit the bill; there's a common
cathode, internal resistors tuned for TTL-level input, and 8 inverters
to a device. But apparently I don't know how to use it as well as I
thought. Is there a different array I can use instead or am I stuck
going the discrete route?

You can use the ULN2803, if the exact current through the
LEDs is unimportant, so long as nothing is damaged. However,
in your description, you did not include the Vcesat drop in
the ULN2803. That's what John Popelish referred to. If you
compute the LED current without considering that drop,
you get about 18 mA with your circuit. However, the actual
figure could be a lot closer to ~8 mA, due to that drop.

Ed