The following chip seems to be acting funny.


I have one DC motor hooked up to 1Y and 2Y and the other to 3Y and 4Y.

Could the behavior be because I connected the VCC1 (logic supply voltage) and 1,2EN and 3,4EN all together? NB. No resistor between them either (see page 6 of link). What might happen?

The weird behavior is the following:

  1. When driving a DC motor in one direction (1A = 5V, 2A =0V), I'd expect 1Y to be VCC2 (output supply voltage) and 2Y to be 0V. Instead with a VCC2 of 9V, 1Y is around 7V and 2Y is over 1V. Making the voltage drop between the DC motors leads only 7-1 = 6V. What is happening here?

  2. When activating two motors at the same time. They start to drive as normally and then slow down to a stop and twitch. This might be a software issue, but I'm not sure...

  3. Also is it normal that the inputs of the H-bridge chip are floating values. Up to 1.3V-ish if I don't hook them up to the controller and set them to zero from there. I would expect them to take the 0 from ground

  4. Finally, what is the purpose of all the diodes on the page 6 graph? I have the outputs hooked up directly to the DC motor.

Page 6 graph:

Page 6 Graph

  • \$\begingroup\$ Please provide a schematic that shows exactly how you have connected the 754410 to the motors, to power supplies, and to whatever drives the inputs of the 754410. In particular, what are the specifications of your 9V power supply. \$\endgroup\$
    – Joe Hass
    Commented Oct 23, 2013 at 19:38

2 Answers 2


The device you have chosen isn't meeting your expectations because it doesn't work very well driving up to 1A motors. Here's a the circuit: -

enter image description here

The output stage is really old hat and poor for driving an amp or more on low supplies. The upper transistor in the output stage will barely be able to get to within 0.7V of the supply and the lower transistor will probably just about get to 1.4V of ground. This is the main problem.

If you look at page 4 it tells you this: -

  • Low level output voltage can be as high as 1.8V (not 0V) for a 1A load
  • High level output voltage can be down at Vcc - 1.8V for a 1A load

As for the inputs floating up this is to be expected given the input circuits used within the device.

Regards the circuit diagram in the question - I believe the external diodes are added because it is driving a two phase motor and this might create extra problems that the internal diodes can't cope with.

Also be aware that the continuous output current for this device has an absolute max value of +/-1.1A so be aware of that when choosing motors. There are far better devices than this old dinosaur by the way.

  • \$\begingroup\$ In the spec it says "continuous output current +-1.1A". I'm not sure of the motors characteristics. What would happen if the DC motor wanted to draw more than that? Would it just fry the H-bridge ? \$\endgroup\$
    – Ollusaurus
    Commented Oct 23, 2013 at 19:23
  • \$\begingroup\$ @Ollusaurus just a helpful note: You should capitalize "I" when referring to yourself. Also, contractions such as "don't" and "I'm" require an apostrophe. \$\endgroup\$
    – JYelton
    Commented Oct 23, 2013 at 19:25
  • \$\begingroup\$ @Ollusaurus eventually it would fry - I haven't checked if the chip can self protect but gut feeling says no. \$\endgroup\$
    – Andy aka
    Commented Oct 23, 2013 at 20:00
  • \$\begingroup\$ [...] the lower transistor will probably just about get to 1.4V of ground. Why would it get 1.4 V above ground and not just Vce(sat) (aprox. 0.2 V) above ground? \$\endgroup\$
    – m.Alin
    Commented Oct 24, 2013 at 8:08
  • 1
    \$\begingroup\$ @m.Alin The lower transistor is a darlington which means a 0.9V drop assuming the first in the d-pair can saturate to 0.2V. There is no evidence to suggest the first transistor will fully saturate given the manufacturer's schematic I posted. On this basis I plucked out a figure of 1.4V and this seems to be reinforced by page 4 on the data sheet. \$\endgroup\$
    – Andy aka
    Commented Oct 24, 2013 at 8:18
  1. The outputs to the motors are connected to Vcc2 or ground through transitors, not switches, so there will be some voltage drop across those transistors when the motor is drawing current. Also, if Vcc2 is a 9 volt battery, its voltage may drop when the motor is drawing current.

  2. Is your 9 volt supply a common 9 volt battery? If so, it may not be able to supply enough current to operate two motors at once (and perhaps not enough to reliably operate one moter.)

  3. You should always connect IC inputs that will affect the function of the part as required to make the part work as you wish. It is very bad practice to assume that an unconnected input will act as if it is grounded (particularly with CMOS parts)(and TTL-like inputs will usually act as high level when not connected.)

  4. The diodes are there to trap or attenuate voltage spikes that occur when the motor is turned off (same as the "catch" diodes used on relay coils.)

  • \$\begingroup\$ thank you. i was testing it with a separate power supply in my schools lab. i limited current to 5 A and set to 9V. I dont think that is the problem. \$\endgroup\$
    – Ollusaurus
    Commented Oct 23, 2013 at 19:16

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