ULN2804A

[Elias1]

Hello, I would like to know
to know, for the ULN2804A, if the logic gates invert or not.
If so, is there an equivalence without inverting gates?

[Peter BENSCH · Accepted solution]

Welcome @Elias1, to the community!

Like all drivers in its class, the ULN2804A is a so-called Darlington array. This means that it does not contain logic gates, but two npn transistors connected in series, whose total gain therefore corresponds approximately to the product of the individual gain. Such a Darlington stage can be simplified as a very high gain npn transistor (BJT) and is therefore always an inverting stage.

"Inverting" refers to the output voltage and therefore to an inappropriate view of the Darlington stages as logic gates. Instead, the stages should be considered as non-inverting current amplifiers: a small base current can result in a considerably larger collector current.

Does it answer your question?

Regards
/Peter

[Elias1]

Thank you for your reply. Can you help me understand the diagram below?

Thank you in advance

[Peter BENSCH]

In your schematics, someone has drawn the stages of the ULN2803 as digital inverters. They can be interpreted as such under certain circumstances, but this is actually incorrect, as explained above. The ULN are not digital stages, but originally output drivers for loads that a normal digital output (TTL, CMOS, etc.) cannot drive directly: relays, power LEDs etc. I would choose a representation that shows at least one transistor per stage.

[Elias1]

Currently my schematic recovers a 5V input from my 2 components. If I remove the first component and keep only the ULN and connect its input to the ground, will I be able to switch my relays?

[Peter BENSCH]

Do you realise the function of a Darlington stage after reading the Wikipedia article?
Have you looked at the internal circuit of the ULN2804 and ULN2803 stages?
Have you looked at the parameters (fig 12, max Input Current vs Collector Current)?

I assume that you have a previous stage with 5V voltage level as input signal, right?
For 500mA current to be driven, each ULN variant requires typ approx. 420µA, max approx. 650µA input current. With the internal circuit you can see that the first mentioned ULN2804 is not suitable for 5V signals, because depending on the preceding output stage (TTL, CMOS, please specify) at input resistance Rb=10.5k at worst only flow:

• (2.4V-(2*0.7V))/10.5k = 95µA (for TTL, Vout 2.4V)
• (3.3V-(2*0.7V))/10.5k = 190µA (for CMOS, Vout 3.3V)
• (5V-(2*0.7V))/10.5k = 343µA (for CMOS, Vout 5V)

With ULN2803 with Rb=2.7k it looks a little better:

• (2.4V-(2*0.7V))/2.7k = 370µA (for TTL, Vout 2.4V)
• (3.3V-(2*0.7V))/2.7k = 704µA (for CMOS, Vout 3.3V)
• (5V-(2*0.7V))/2.7k = 1333µA (for CMOS, Vout 5V)

From the diagram in fig 12 you can read off the collector current flowing, which must be sufficient for the load you want to drive (obviously a relay).

If you have a CMOS device (e.g. 74HC) with VCC=5V in front of it, then it can drive the ULN2803 directly. However, not with input to GND, but to the preceding CMOS output. An H signal would then switch on your load, an L signal would switch it off.

Hope that helps?

Regards
/Peter