3
\$\begingroup\$

I'm working on a project that uses a ULN2003A (darlington transistor array) and I need to enable/disable it. The point is that ULN2003A does not have enable/disable pin.

I think that a IC with latch would do also the job. I've found TLC59213, that could do the job but si a quite rare IC.

Does anyone knows a darlington transistor array IC with enable/disable pin or a latch more common than TLC59213?

Thanks

\$\endgroup\$
  • \$\begingroup\$ Comparing the schematics of ULN2003A & TLC59213, I found that putting !CLR=0 in TLC59213 is same as supplying 0 to inputs of ULN2003A. !CLR doesn't disable the IC, it only resets flops inside to provide zero input to the transistor pairs. \$\endgroup\$ – hassansin May 4 '14 at 12:05
  • \$\begingroup\$ I doubt there's anything in a remotely similar availability/price range otherwise it you'd see it around in products. \$\endgroup\$ – Spehro Pefhany May 4 '14 at 14:45
  • \$\begingroup\$ Mouser has several hundred of the TLC59213 in stock. \$\endgroup\$ – tcrosley May 4 '14 at 17:23
1
\$\begingroup\$

The ULN2003A does have an interesting function at pin 9 (com). If pin 9 is held low, then all outputs will be held low. Depending upon your logic useage, this could be used as a disable. If your logic permits that "disable" means that all outputs shall be held low. I have used the ULN2003A in projects where I used the pin 9 as a lamp test, and also as a disable for preventing any outputs to be high.

\$\endgroup\$
  • \$\begingroup\$ Thanks Marla! Sorry for my ingnorance but isn't pin 9 ground? Shouldn't it always be low? \$\endgroup\$ – user3589630 May 6 '14 at 9:23
  • \$\begingroup\$ @ user3589630 : pin 8 is the emitter of all transistors (ground). If you tie pin 9 as ground, none of the transistors can ever go high (open collector) \$\endgroup\$ – Marla May 6 '14 at 15:25
0
\$\begingroup\$

The MOSFET equivalent of the ULN2003, namely the TPIC2701 can possibly\$^1\$ provide this functionality if you add a MOSFET for grounding the source pin - disconnecting the source pin from ground will likely just float all the loads connected to the outputs to the positive rail i.e. the outputs are disabled. The max ratings for the gate-source voltage are +/-20V and if you are typically running stuff at 12V then I don't think this will hurt too much!

You don't get a negative voltage specified for the ULN2003 so I'm unsure it would be at all wise to try this idea on it.

It's just a tentative idea so don't go giving me grief over it!!!


\$^1\$ I'm not saying definitely because I've never used it.

\$\endgroup\$
-1
\$\begingroup\$

I think there is reason that this device doesn't have enable/disbale pin. Darlington Transitor is just a switch that has a trasistor pair instead of one. There is nothing to enable or disable a switch. The inputs (base terminal) acts as enable/disable pin for each pair. Here is a diagram of darlington transistor:

enter image description here

This transistor is useful when current at the base of the transistor is too weak to switch a normal switching transistor. Here, First transistor is turned on with small input current(~.82mA) which is then amplified and fed to the base of 2nd transistor.

So when you provide '0' at the Input, the switched is turned off and no current flows. Except some leakage current ICE within the transistors. But this is very small, practically around 100uA.

So this means to disable the IC from power consumption, just put 0V at the inputs. No current will be drained except some small leakage.

As for pin9(com), this doesn't 'disable' the device. It just shorts the output pins with the 'com' pin(see the schematic in datasheet). You'll still have power consumption through the IC.

\$\endgroup\$
  • \$\begingroup\$ This doesn't really answer the question. \$\endgroup\$ – Chris Johnson May 4 '14 at 17:20
  • \$\begingroup\$ I implied that you don't need a separate enable/disable pin for ULN2003A. When the input is zero, it is disabled, when input is 1, it is enabled. Just think of how a normal switching BJT works. \$\endgroup\$ – hassansin May 4 '14 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.