Hi, i tried to switch on/off a fan using this IC but used it as a current source it didnt work , i usd it as a current sink it worked, why that happend ? and why only current sink worked ?

 

hi,
The ULN is a open collector output, there is no internal pull up transistor or resistor.

It can only sink current not source it.

Does this help.?

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Eric
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yea got it thanks , and if it was for current source what would be the difference or the change ?

 

You cant use the ULN as current source.

You would choose a PMOS FET or PNP transistor.

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Eric
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thanks got it ! , yea PNP would do it because when it saturates it will close on the load and the load will draw current from the supply correct ?

 

hi,
Thats correct.
You have not said what you are going to drive or whats driving it.?

Give us more details and we may suggest ways of doing it.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
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Bill's: www.blueroomelectronics.com/

 

hi,
thanks , ah , its a uC 89S52 will drive that IC to switch on/off the fan

 

The input of the ULN2003 needs to have 150uA before doing anything.
For 100mA output the typical input current needs to be 250uA.

But the output high current from your micro-controller is only 60uA. So it won't work.
The same problem as when you tried using the TIP120 darlington in your other thread.

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Uncle $crooge

 

audioguru , cant the uC used as current sinking device and it will output 1.3mA that will feed the ULN2003 ?

by the way , i used an inverter 74HC00 after the uC then fed that to the ULN2003 and it worked.

 

One of my favorites that is UDN 2981 the source driver.

5 times bigger the price of ULNXX.

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Gayan

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hi,
The 74HC00 invertor is a CMOS device and will drive a ULN2003A.

The MCU when used as a current sink cannot drive the ULN.
In order to sink current the pin of the MCU would have to be connected to +5V via a resistor.

If the MCU can sink 1.3mA, from say +5V, then the resistor would be about R= V/I = 5/0.0013 ,, 3K9.

If you now connect the ULN to the MCU pin, the MCU should be able to drive the ULN.
If you have 74HC00 I would use that method, remember the logic inversion.

What load current is the ULN going to sink.?

Remember for the ULN to be switched ON, the MCU pin must be high.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

 

What load current is the ULN going to sink.?
0.2A

In order to sink current the pin of the MCU would have to be connected to +5V via a resistor.

why ? doesnt when i ouput zero at the pin of the MCU it doesnt act as a current sink ?
. OK, now this is for current sinking , what configuration should i design to let the MCU source current?

I dont know really why the 74HC00 solved the problem i found that the ULN doesnt work with zero Input that's why i made 74HC00 act as an inverter ,would you tell me the theory behind that ?

 

hi,
The 74HC00 is acting as 'current amplifier ' buffer between the MCU and the ULN.

OK.?

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

 

ok
thanks