1
\$\begingroup\$

Here is my breadboard using an L293D breadboard. Pin 16 (VSS) is at the top right. I'm trying to have my project operate in sleep mode drawing as little current as possible. When I added the IC, my draw increased about 24mA. I was experimenting and pulled pin 16 which was connected to 5V and the draw immediately dropped to .54mA which is what I expected.

However the weird thing is that the IC still works. Pin 8 (VS) is connected to 12V which will power my motors.

Why is this circuit working?

Edit: It looks like 12V is connected to Row 57, but it is just hanging loose.

\$\endgroup\$
4
  • \$\begingroup\$ There is no Vss pin as it's a bipolar IC. I assume that by Vss you actually mean Vcc1 (logic supply). The digital signal pins probably have ESD clamping diodes from ground to the input and from the input to Vcc1, causing the arduino to power the chip via the Vcc1 clamping diode. \$\endgroup\$
    – jms
    Feb 25, 2016 at 3:40
  • \$\begingroup\$ @jms not all manufacturers use TI's VCC1/VCC2. Some use VS and VSS. Like ST arduino.cc/documents/datasheets/H-bridge_motor_driver.PDF \$\endgroup\$
    – Passerby
    Feb 25, 2016 at 3:53
  • \$\begingroup\$ Please post a schematic. Nobody is going to chase the wire on the breadboard to try to figure out what you are doing there. \$\endgroup\$ Feb 25, 2016 at 4:00
  • \$\begingroup\$ Question doesn't really need a schematic. "Can the L293D be run without the Logic Supply Voltage, just the Power/Motor Supply Voltage" \$\endgroup\$
    – Passerby
    Feb 25, 2016 at 4:12

1 Answer 1

2
\$\begingroup\$

The reason it's drawing large amounts of current is that this is an older part that uses TTL input stages, instead of CMOS input buffers: here's what the inside looks like, at least from TI:

enter image description here

There does not appear to be an ESD protection diode to clamp the input to VCC1, so that's probably not the leakage path that is allowing this to work. I think it's still working because you have some leakage path elsewhere (please post a schematic) that is supplying VCC1 -- try measuring VCC1 in your case with a DMM and see what voltage it shows. I wouldn't leave VCC1 unconnected for the actual end product -- find a switch to turn it on/off when you need too.

As an aside, you do appear to have a bunch of floating inputs; while this isn't CMOS, I would still tie off your unused inputs to known values.

\$\endgroup\$
3
  • 2
    \$\begingroup\$ I'm thinking the leakage is through the Enable pins. \$\endgroup\$
    – Passerby
    Feb 25, 2016 at 5:23
  • \$\begingroup\$ Is there a newer H bridge that uses CMOS? \$\endgroup\$ Feb 26, 2016 at 23:57
  • \$\begingroup\$ I've used the BD6221 which I'm pretty sure is BiCMOS, bipolar for power stage, but CMOS input buffers. Checking the input bias current spec will tell you for sure -- CMOS will be in nA usually / much much lower than a bipolar part which may be a few hundred uA. \$\endgroup\$ Feb 27, 2016 at 1:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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