ULN2004 Darlinton array

[Tiger555]

Hello there, can anyone please give a simple example of how to connect the darlington pairs?

in the above ULN2004, pin 8 goes to ground and pin 9 is for +vcc?

so if I give a 5v on one of the input(pin1) will I get my +Vcc on output of pin1??

any simple circuit example will help me, please provide one.

From google, i got the basic working principle of darlington, but getting confused with the pin and outputs...

thanks in advance.

 

[audioguru]

Why don't you download the datasheet for the ULN2004A and read it?

Yes, pin 8 is ground.
Pin 9 is the cathodes of the output protection diodes which connect to the same positive supply voltage as the inductive load.
If the input is +5V then the datasheet says the output current will be limited to only 125mA. The circuit inverts so +5V on the input results in about +1V at the output. When the input is less than about +1.2V then the output is disconnected.

 

[Tiger555]

Thanks guru,

I read the spec sheet, I dont have a big knowledge into electronics. So got doubts..Thanks anyway..

This is what I want, I hope this should work... re-place ULN2004 for the darlington pairs

Circuit Simulator Applet

Thanks... in advance, please see the link..

Why don't you download the datasheet for the ULN2004A and read it?

Yes, pin 8 is ground.
Pin 9 is the cathodes of the output protection diodes which connect to the same positive supply voltage as the inductive load.
If the input is +5V then the datasheet says the output current will be limited to only 125mA. The circuit inverts so +5V on the input results in about +1V at the output. When the input is less than about +1.2V then the output is disconnected.

 

[audioguru]

The circuit simulator is wrong and makes no sense.
It is feeding the darlington an input of -15V that will blow it up.

You will probably feed the input of your ULN2004A darlington almost +5V to turn it on and 0V to turn it off.

 

[KeepItSimpleStupid]

That series ULN200? does logic translation too. The darlington only sinks current. The chip is a little hard to understand because of the internal protection diode clamps. To protect the semiconductors with inductive kickback, a diode is placed across the coil in a manner that it will not conduct when the coil is on. When the coil is turned off, the current flows in the opposite direction, through the diode and the coil to prevent damage to the darlingtons.

The cool part about this chip is that when the pin is unconnected, the driver is off. With TTL logic, inputs FLOAT high, thus you can get undesireable power up conditions without an interface IC such as the ULN200?.

Thus, these chips provide a NICE interface to micro processors where the I/O ports need to be configured before use. The ports are normally configured as inputs and therefore the output will be off as the power up state.

 

[Roff]

That series ULN200? does logic translation too. The darlington only sinks current. The chip is a little hard to understand because of the internal protection diode clamps. To protect the semiconductors with inductive kickback, a diode is placed across the coil in a manner that it will not conduct when the coil is on. When the coil is turned off, the current flows in the opposite direction, through the diode and the coil to prevent damage to the darlingtons.

This is a minor nit-pick.
The coil current does not change direction. When the Darlington turns off, the voltage on the collector rises very rapidly until the diode is forward-biased. The coil current is still flowing in the same direction, but through the diode, and begins to decay when the diode is conducting.

 

[KeepItSimpleStupid]

Thanks. Nit pick accepted.

 

[KeepItSimpleStupid]

Rethinking.

Assume an inductor is drawn vertically with +12 at the top and a switch at the bottom to ground.Let's label the top of the inductor (+). The diode is across the coil with the anode(-) at the top which is connected to +12.

When the coil is energised, current is flowing through the terminal that was labeled (+) through the the switch to ground.

When the switch is opened, current has to flow OUT of the inductor (+) terminal. THIS IS A REVERSAL OF DIRECTION, otherwise it cannot get through the diode.

This all stems from the equation that the voltage across the inductor is v(t)=Ldi/dt. di/dt changes direction going from 0 current at t0 to some value at tn and from some value to zero at tm where tm>tn. If you assume a 12V supply then deltaI/deltaT = (12-0)/time which is positive and deltaI/deltaT = (0-12)/time for the discharge case. The latter has a negative sign.

Thus, I don't agree anymore and stand by my original statement.

 

[Roff]

When the current begins to decay, di/dt becomes negative. Thus, the sign of the voltage across the inductor becomes negative, exhibited as the flyback spike (in the absence of the diode).
We can jaw about this all day, but, as they say, a picture is worth a thousand words. See the result of the simulation in the attachment. Note that the current in the inductor never changes direction.

 

[Tiger555]

Hi, Yep I understood. No need of ap-amp or anything. I connected the o/p of timer chip to the darlington in, out of darlington to the second pin of relay. So when ever timer generates pulse, it opens and goes to ground and relay gets active.
Thanks for your time to teach me. This is a basic circuit darlington and relay.. i saw many in google, but when searching for the first time i didnt know what is darlington and so.. thanks all who helped me.

The circuit simulator is wrong and makes no sense.
It is feeding the darlington an input of -15V that will blow it up.

You will probably feed the input of your ULN2004A darlington almost +5V to turn it on and 0V to turn it off.

 

[crutschow]

Assume an inductor is drawn vertically with +12 at the top and a switch at the bottom to ground.Let's label the top of the inductor (+). The diode is across the coil with the anode(-) at the top which is connected to +12.

The anode is the (+) terminal of the diode, not the (-). The cathode(-) would be connected to where +12V is applied to the inductor.

Roff is right. Current in an inductor can not instantaneously change direction. It will always try to keep flowing in the same direction. Thus, when the switch is opened the inductor current keeps flowing through the anode to cathode direction of the diode.

It's remarkable how many people become confused when discussing the current flow in an inductor that's switched. It's the inductor polarity that changes direction when you go from the inductor being a sink to the inductor being a source, not the current.