how do i connect this ULN2308

[Nigel Goodwin]

Nigel,
on which pins do you measure the voltage across the chip when the load is connected ? the voltage between Vcc and GND, or the voltage between the output pin and the ground ?
thanks

The output pin and ground!. Vcc and ground would just measure the supply rail.

 

[Roff]

Nigel,
on which pins do you measure the voltage across the chip when the load is connected ? the voltage between Vcc and GND, or the voltage between the output pin and the ground ?
thanks

The output pin and ground!. Vcc and ground would just measure the supply rail.

But if you want to measure the voltage that results in power dissipation on chip, measure the voltage between the output and Vcc (for the UDN2891).

 

[Nigel Goodwin]

But if you want to measure the voltage that results in power dissipation on chip, measure the voltage between the output and Vcc (for the UDN2891).

Oops! - I was still thinking ULN2803, I forgot we'd changed chips :?

For a topside switch (as Ron says), you need to measure from the output to Vcc.

 

[defcon31]

and now, last question.
do i need to place some resistors ?
if yes, where ?
or is the connection below ok?

 

[kinjalgp]

You need to put appropriate resistor between UDN2981's output and lamp. This is to limit the current through IC to ~300mA otherwise you'll fry it due to excessive power dissipation.

 

[defcon31]

and is there a way to calculate the needed resistor (as my question in the topic about "pullup resistor") ???
how can i calculate this ???
or can you give me an idea about the needed resistor and how do you come to that value ???

its a 10W Lamp and it ask around 300mA

thanks in advance !!!!

 

[Nigel Goodwin]

I'm assuming that the lamp is 24V 10W (you didn't mention it, but labelled the supply to the chip as 24V).

The formula for power is W=V*I, transposing this you get I=W/V, which is 10/24=0.42A. The easiest way to solve your problem would be to parallel two (or more) of the gates, two would give you 600mA, three 900mA, and so on - bearing in mind the total limitations of the chip.

If you want to add a series resistor (and reduce the light output from the lamp), the only two formulas you need are:
V=I*R
W=V*I

 

[defcon31]

no i dont want to reduce the light of the lamp.
so do i need a resistor, yes or not? (because kinjalgp said i HAVE TO place a resistor; is he fault?)
thanks

 

[Nigel Goodwin]

no i dont want to reduce the light of the lamp.
so do i need a resistor, yes or not? (because kinjalgp said i HAVE TO place a resistor; is he fault?)
thanks

What voltage is the lamp? - you still haven't confirmed it's 24V.

If it is, you can't feed it from a single gate without reducing it's brightness.

What actually are you trying to do overall?, this threads been going a long time, we might be able to suggest easier ways of doing things if we knew.

 

[defcon31]

the lamp is 24V yes..
so my last question was, do i need a resistor at some place in the last given figure (cause kinjalgp said I HAVE TO, and confused me).
thanks for helping, and probably this is my last question on this topic (hope so)

 

[Nigel Goodwin]

the lamp is 24V yes..
so my last question was, do i need a resistor at some place in the last given figure (cause kinjalgp said I HAVE TO, and confused me).
thanks for helping, and probably this is my last question on this topic (hope so)

You have to add a resistor, and reduce the brightness of the lamp, IF YOU ONLY USE ONE GATE TO FEED THE LAMP, if you use more than one gate you won't need a resistor.