SparkFun Forums

[MarkS]

I was looking at the ULN2803A to drive the displays on a counter I'm building. At first I thought that it was pretty clever turning a 3-pin component into a 2-pin component, but the more I thought about it, the less sense this made.

It wouldn't have worked in my case, since the 4026 chip outputs only .44mA at 5v, so each input alone wouldn't drive a single LED without a transistor, which was, BTW, the reason for looking at the 2803. I realized that all the 2803 would be is a complicated resistor. The input voltage and current is driving the gate and is coming out at a smaller current and voltage than it was when it went in, but otherwise going straight through. :think:

Why would I ever want to use a Darlington array? It *does* have a purpose, right?

[MarkS]

Hmmm. :think: Google helped clear up one issue and that is that it wouldn't act as a resistor, but an amplifier. Still, that doesn't seem like enough of a reason to use one.

[UhClem]

You use a Darlington pair when the current gain of a single transistor isn't enough. The gains multiply so that if they each have a gain of 100, the total gain is 10,000.

[Valen]

I was looking at the ULN2803A to drive the displays on a counter I'm building. At first I thought that it was pretty clever turning a 3-pin component into a 2-pin component, but the more I thought about it, the less sense this made.

It wouldn't have worked in my case, since the 4026 chip outputs only .44mA at 5v, so each input alone wouldn't drive a single LED without a transistor, which was, BTW, the reason for looking at the 2803. I realized that all the 2803 would be is a complicated resistor. The input voltage and current is driving the gate and is coming out at a smaller current and voltage than it was when it went in, but otherwise going straight through. :think:

Why would I ever want to use a Darlington array? It *does* have a purpose, right?

I think you got your parts mixed up in your explanation. Or do not understand how to connect the ULN2803A. The ULN2803A darlington array doesn't output current. It is just a dual NPN-transistor amplifier times 8. The first transistor of the pair turns on base of the other. The input switches the transistor pair on to conduct the 'output' to ground, alowing more current than just a single transistor. The load that these outputs pin are connected to must be powered by a power source. It still is a low-side switch. [edit] to add, none of the input current goes out to the output. Transistors can't do that. Well, maybe some tiny leakage in specific circumstances.

The 'common' pin is connected to the cathodes of the 8 internal clamping diodes for over-voltage protection of the outputs, and should lead back to the powersource. It is not to provide power to the device.

[KeithB]

alowing more current than just a single transistor

Not really, the driver transistor draws very little current from the load - only the base current for the output transistor. The darlington is used to greatly increase the current gain, especially since a saturated switch has very low Beta.