uln2004an

[brian@1234]

Can anybody provide me with a circuit for uln2004an chip. I'm trying to use this chip to take a simple 5V input signal from a pic which should output to a 5 volt 42mA relay coil. I orignally though that I could use the shared power supply of the pic to go to the E and com connection, but when testing with an LED, I find that I have to use to seperate power supplies to get the LED to light. It seems that a resistor is also needed at E connection so that the chip doesn't draw to much current. Can anybody give any advice and/or examples for this chip.

thanks,
Brian

 

[Boncuk]

Hi Brian,

to drive a load using an output of a PIC you should select an ULN2003. The ULN2004 is designed for 6 - 15V CMOS or PMOS, while the ULN2003 is designed for 5V TTL or CMOS.

They differ in built in serial resistor which is 2.7KΩ for the ULN2003 and 10.5KΩ for the ULN2004.

Here is an example of ULN2003 usage.

Note that the common pin (9) must be connected to the positive rail of the relay supply voltage.

You don't require to use a protection diode parallel with the relay coil if the common pin is connected to the relay power source.

Boncuk

 

[brian@1234]

I saw the 6-15V supply voltage on the datasheet but I was thinking that this chip would work fine because I have a seperate supply of 12 volts for this chip. My main problem with the 12 volts supply is that it draws way to much current; even when i'm not sending a signal to the chip to output. I don't have a ULN2003 chip currently and i'm trying to finish this project before Wednesday; do you think that uln2004an chip should be able to work for now? I'm very thankful for the diagram, I would not have thought to connect my relay coil so that it always has a positive voltage. It appears this chip works by supplying the common connection for the relay coil. I appreciate your help!

thanks,
Brian

 

[Boncuk]

Hi Bryan,

the ULN20XX series is not an IC but an NPN-transistor array. Connecting +12V to pin9 you just connect the free wheeling diodes (clamping diodes) inversely (cathode to +12V).

The transistors switch ground to the load. Therefor the other side of the load must be connected to V+ as shown in the circuit diagram.

If the load draws too much current you might parallel several inputs and outputs as shown with the relay connected to two inputs and outputs.

One darlington pair can sink 1W, being limited to total package power dissipation of 2.1W (ULN2004A) at 25°C. The ULN2004L can dissipate 1.4W at 25°C.

If these parameters are exceeded you should think about using power MosFet transistors instead.

Boncuk

 

[audioguru]

My main problem with the 12 volts supply is that it draws way to much current; even when i'm not sending a signal to the chip to output.

Since your relay has a 5V coil then why are you using 12V for the ULN2004? If the relay's current is 42mA with a 5V supply then its current will be 101mA with a 12V supply.

The ULN2003 or ULN2004 draws no current when it is turned off with an input of 0V. Then how can it draw way too much current?

I would not have thought to connect my relay coil so that it always has a positive voltage. It appears this chip works by supplying the common connection for the relay coil.

Didn't you look at its datasheet? We have no idea how you connected it wrong.