Sink or Source from Darlington Array?

M

mark risher

Guest
Hello,

I am relatively new to this; please forgive a series of beginner questions
regarding darlington arrays.

My project is to use a MSP430 microcontroller to control a high-current LED
(~300 mA). I think a Darlington array (ULN2003) can help me out here, but
I'm confused about whether to source or sink from it.

I plan to connect as follows:

-Connect an output pin from the micro to one of the darlington's inputs
-Connect the corresponding darlington output to the cathode of the LED
-Connect the GND pin of the ULN2003 to ground

Questions:
1) Is the above connection correct?
2) If the darlington is sinking, as above, is there a voltage drop to
consider? Will it "lift" the ground from 0V to 1.4V?
3) Do I set the micro's output low or high?
4) Do I need to limit the current going into the darlington to avoid melting
the micro? The datasheet doesn't seem to list the amount of current it will
draw, only the beta I'll get given a certain current.

Thank you so much,
/m
 
mark risher wrote:

Hello,

I am relatively new to this; please forgive a series of beginner questions
regarding darlington arrays.

My project is to use a MSP430 microcontroller to control a high-current
LED (~300 mA). I think a Darlington array (ULN2003) can help me out here,
but I'm confused about whether to source or sink from it.

I plan to connect as follows:

-Connect an output pin from the micro to one of the darlington's inputs
-Connect the corresponding darlington output to the cathode of the LED
-Connect the GND pin of the ULN2003 to ground

Questions:
1) Is the above connection correct?
Click to expand...
I think probably not. IIRC the ULN2003 sinks so you need to connect its
output to the LEd and the LED to +5V via a current limiting resistor:

-.- +5V
|
--
\/ LED
--
|
---
| |
| | Resistor
---
|
|
. ULN2003 OP

HTH

Ian
 
I think probably not. IIRC the ULN2003 sinks so you need to connect its
output to the LEd and the LED to +5V via a current limiting resistor:
Click to expand...
That's what I meant by "connect the darlington output to the LED cathode,"
but it's entirely possible I phrased it incorrectly. Thank you for the
confirmation on that point.

Will the Darlington's voltage drop matter when it's sinking in the
configuration you drew below? If B-E drop is 1.4 volts, does that mean a
load between +5 and the output of the darlington will see 3.6V?

Thanks,
/m


"Ian Bell" <ian@redtommo.com> wrote in message
news:c1ghug$1gv8to$1@ID-225948.news.uni-berlin.de...
mark risher wrote:

Hello,

I am relatively new to this; please forgive a series of beginner
questions
regarding darlington arrays.

My project is to use a MSP430 microcontroller to control a high-current
LED (~300 mA). I think a Darlington array (ULN2003) can help me out
here,
but I'm confused about whether to source or sink from it.

I plan to connect as follows:

-Connect an output pin from the micro to one of the darlington's inputs
-Connect the corresponding darlington output to the cathode of the LED
-Connect the GND pin of the ULN2003 to ground

Questions:
1) Is the above connection correct?

I think probably not. IIRC the ULN2003 sinks so you need to connect its
output to the LEd and the LED to +5V via a current limiting resistor:

-.- +5V
|
--
\/ LED
--
|
---
| |
| | Resistor
---
|
|
. ULN2003 OP

HTH

Ian
Click to expand...
 
Vce will be ~0.3V

mark risher wrote:
I think probably not. IIRC the ULN2003 sinks so you need to connect its
output to the LEd and the LED to +5V via a current limiting resistor:


That's what I meant by "connect the darlington output to the LED cathode,"
but it's entirely possible I phrased it incorrectly. Thank you for the
confirmation on that point.

Will the Darlington's voltage drop matter when it's sinking in the
configuration you drew below? If B-E drop is 1.4 volts, does that mean a
load between +5 and the output of the darlington will see 3.6V?

Thanks,
/m


"Ian Bell" <ian@redtommo.com> wrote in message
news:c1ghug$1gv8to$1@ID-225948.news.uni-berlin.de...

mark risher wrote:


Hello,

I am relatively new to this; please forgive a series of beginner

questions

regarding darlington arrays.

My project is to use a MSP430 microcontroller to control a high-current
LED (~300 mA). I think a Darlington array (ULN2003) can help me out

here,

but I'm confused about whether to source or sink from it.

I plan to connect as follows:

-Connect an output pin from the micro to one of the darlington's inputs
-Connect the corresponding darlington output to the cathode of the LED
-Connect the GND pin of the ULN2003 to ground

Questions:
1) Is the above connection correct?

I think probably not. IIRC the ULN2003 sinks so you need to connect its
output to the LEd and the LED to +5V via a current limiting resistor:

-.- +5V
|
--
\/ LED
--
|
---
| |
| | Resistor
---
|
|
. ULN2003 OP

HTH

Ian
Click to expand...
 
mark risher wrote:
I think probably not. IIRC the ULN2003 sinks so you need to connect its
output to the LEd and the LED to +5V via a current limiting resistor:


That's what I meant by "connect the darlington output to the LED cathode,"
but it's entirely possible I phrased it incorrectly. Thank you for the
confirmation on that point.

Will the Darlington's voltage drop matter when it's sinking in the
configuration you drew below? If B-E drop is 1.4 volts, does that mean a
load between +5 and the output of the darlington will see 3.6V?
Click to expand...
Yes roughly.

Look at the datasheet, there should be a graph of Collector-Emitter
Saturation Voltage against Collector Current. This is the voltage
between the collector and emitter when the darlington is saturated
(fully on). Looks more like 1.2V than 1.4V at 300mA. I assume you can
just adjust the resistor value to fix this problem?

--
-----------------------------------------------------------------------
To reply to me directly:

Replace privacy.net with: totalise DOT co DOT uk and replace me with
gareth.harris
 
mark risher wrote:

I think probably not. IIRC the ULN2003 sinks so you need to connect its
output to the LEd and the LED to +5V via a current limiting resistor:

That's what I meant by "connect the darlington output to the LED cathode,"
but it's entirely possible I phrased it incorrectly. Thank you for the
confirmation on that point.

Will the Darlington's voltage drop matter when it's sinking in the
configuration you drew below? If B-E drop is 1.4 volts, does that mean a
load between +5 and the output of the darlington will see 3.6V?

Thanks,
/m
Click to expand...
I assume you mean C-E drop not b-e. In a normal transistor the saturated
c-e voltage is less than 0.4V. However in a darlington there is usualy an
extra b-e drop of 0.7V to add in giving a total of around 1V. And yes it
does mean the load will see %V less this voltage drop. You need to take
this into account in calculating the resistor value needed to set the LED
current.

Ian
 
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