ULN2803A (darlington array) anyone know what I'm doing wrong?

[Ben]

Applying 5V to the inputs (power supply is also 5V), I'm only getting
0.6V at the outputs. I thought it was a faulty chip, or perhaps damaged
while soldering, but I've tried another one in a holder and got the same
result. What could be causing this?

 

[Eeyore]

Applying 5V to the inputs (power supply is also 5V), I'm only getting
0.6V at the outputs. I thought it was a faulty chip, or perhaps damaged
while soldering, but I've tried another one in a holder and got the same
result. What could be causing this?

How is it connected ?

What voltage did you expect to see ?

Graham

 

[Phil Allison]

"Ben"

Applying 5V to the inputs (power supply is also 5V), I'm only getting 0.6V
at the outputs.

** That is correct operation for a transistor in common emitter mode.

Input high causes output to go low.

I thought it was a faulty chip, or perhaps damaged while soldering, but
I've tried another one in a holder and got the same result. What could be
causing this?

** What the hell did you expect ??

The ULN 2803A darlington array is intended to drive relays, lamps, LEDs etc.

YOU connect one side of the load to + supply and the array will connect the
other to 0 volts when driven on.




........ Phil

 

[Peter Bennett]

Applying 5V to the inputs (power supply is also 5V), I'm only getting
0.6V at the outputs. I thought it was a faulty chip, or perhaps damaged
while soldering, but I've tried another one in a holder and got the same
result. What could be causing this?

A ULN2803 is just a bunch of Darlington transistors, emitter to
ground, and collector to the output pin. They are inverters, so a
high (+5V) input will produce a low (near zero) output. Also, the
things don't have anything to pull the outputs high, so, unless you
supply an external pull-up (such as a resistor to +5V), a low input
will also produce a low output.



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[Phil Allison]

"Peter Bennett"

A ULN2803 is just a bunch of Darlington transistors, emitter to
ground, and collector to the output pin. They are inverters, so a
high (+5V) input will produce a low (near zero) output.

** It won't be near zero volts.

The voltage drive for the base of second device in the darlington pair comes
from its own collector - so it will be at least 0.6 volts even at very low
current.

The data sheet gives saturation values for Vce up to 1.6 volts at 350 mA.

http://www.datasheetcatalog.com/datasheets_pdf/U/L/N/2/ULN2803.shtml




........ Phil

 

[Ben]

A ULN2803 is just a bunch of Darlington transistors, emitter to
ground, and collector to the output pin. They are inverters, so a
high (+5V) input will produce a low (near zero) output. Also, the
things don't have anything to pull the outputs high, so, unless you
supply an external pull-up (such as a resistor to +5V), a low input
will also produce a low output.

Thanks for the explanation, I think I'd misunderstood what a Darlington
array actually is. I was expecting a high input to produce a high
output, with enough current to drive relays. Does such an IC exist?

What value of pull-up resistor would you recommend for the Darlington array?

 

[[email protected]]

Thanks for the explanation, I think I'd misunderstood what a Darlington
array actually is. I was expecting a high input to produce a high
output, with enough current to drive relays. Does such an IC exist?

This is a perfect chip to drive your relays, the coil goes between the
output and posative supply. You may find a circuit somewhere on the
data sheet or do a search.

 

[Eeyore]

Thanks for the explanation, I think I'd misunderstood what a Darlington
array actually is. I was expecting a high input to produce a high
output, with enough current to drive relays. Does such an IC exist?

Do you know what a transistor does ?

You're fixated on 'ICs' aren't you ?

What value of pull-up resistor would you recommend for the Darlington array?

What do you need a pull-up for ? The load *is* the 'pull-up'.

Graham

 

[BobG]

Do you know what a transistor does ?
You're fixated on 'ICs' aren't you ?

=========================================
Be nice. Or just dont say anything. Some patient person will explain
politely.

 

[Ben]

Do you know what a transistor does ?

You're fixated on 'ICs' aren't you ?

I'd need 8 transistors, so a single chip that does the job seemed like a
good idea.

 

[Ben]

This is a perfect chip to drive your relays, the coil goes between the
output and posative supply.

Yeah, this worked great - thanks

 

[Jamie]

Thanks for the explanation, I think I'd misunderstood what a Darlington
array actually is. I was expecting a high input to produce a high
output, with enough current to drive relays. Does such an IC exist?

What value of pull-up resistor would you recommend for the Darlington
array?

you don't need a pull up resistor..
use the Array outputs as the Common/ground source when on.
they are simply what is called open collector outputs.
you can supply the relay coil with what ever voltage is required by
the relay up to the max stand off voltage of the array of course.
the + side of the relay coil does not need to be the same voltage
as the Vcc (+ rail) of the chip..

 

[Jamie]

Ben wrote:




Do you know what a transistor does ?

You're fixated on 'ICs' aren't you ?





What do you need a pull-up for ? The load *is* the 'pull-up'.

Graham

Graham:
the last time i looked at the name on this NG, it said "BASICS"!

where i come from that means something. maybe you should show a little

consideration for others that are at least trying, and not criticize

at every turn.


(Just an opinion from the peanut gallery)

 

[Don McKenzie]

I'd need 8 transistors, so a single chip that does the job seemed like a
good idea.

Spot on Ben,
good work.

Don...


--
Don McKenzie
E-Mail Contact Page: http://www.dontronics.com/e-mail.html

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[jasen]

Applying 5V to the inputs (power supply is also 5V), I'm only getting
0.6V at the outputs.

that's about right.

I thought it was a faulty chip, or perhaps damaged
while soldering, but I've tried another one in a holder and got the same
result. What could be causing this?

probably you're using it incorrectly.
the load should be connected between the output and a positive supply.