I'm trying to design a simple BLDC motor controller using L6234 driver IC. Looking at its application reference schematic I came across some weird pull-ups (highlighted in green, I assume these function as pull-up resistors but please correct me if I'm wrong).

https://www.st.com/resource/en/application_note/cd00004062-l6234-three-phase-motor-driver-stmicroelectronics.pdf figure 12:

Figure 12

What are these pull-ups for? I understand the use of pull-up and pull-down resistors when they are used to prevent say button input from floating. But in this case control logic IC outputs PWM signals to control L6234 IC. If these are here to prevent floating enables and inputs while control logic IC is booting then why are these pull-ups instead of being pull-downs?

I'm rather inexperienced hobbyist, am I missing something more fundamental here?

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    \$\begingroup\$ It would seem that those 10K's ARE pull-downs if \$V_s\$ is missing or inactive (via J7 and the L7805). \$\endgroup\$
    – glen_geek
    Jan 31, 2021 at 16:03
  • \$\begingroup\$ I think it's because the control logic driver might need a little help on the high output voltages. I mean it meets the L6234 input high requirements but only by a few hundred milli volt and given that this sort of 3 phase motor control circuit is quite noisy, it might need to pull-ups for a little extra security on logic levels. \$\endgroup\$
    – Andy aka
    Jan 31, 2021 at 16:47
  • \$\begingroup\$ Glen_geek has the right of it. If J7 is not attached then these function as pull-downs disabling the device. Odds are rest of circuit (control logic) is powered by +5V. \$\endgroup\$ Feb 1, 2021 at 1:59

1 Answer 1


Looking in more detail at the app note it would seem that the pullups are superfluous. They are directly connected to the outputs of a GAL in the control logic diagram. The outputs of the GAL would be available within a few microseconds of power being applied.

Also the GAL outputs meet the voltage requirements of the L6234. As @MaybeJulius mentions there should be some interlock between the Vs and the 5V supply to ensure that Vs is available before voltage is applied to the control inputs. One of the options in the circuit is to make 5V available from the same supply in which case there shouldn't be a problem (but results in high dissipation in the 5V regulator).

If the control signals are being provided by a microprocessor it would also be safer to ensure that they are held low until the processor is running and has had a chance to initialize its output ports.

If the motor controller is not held in a benign state it could cause unwanted rotation or excessive dissipation in the motor coils by constant excitation.

A common way of doing that is to provide pull-ups or pull-downs to set the motor controller in an idle state. They will be overridden by the microcontroller I/O signals when it has configured itself.

In this case, however, I don't see why they are pull-ups and not pull-downs as the enable signals for the L6234 are active high. I would not consider it a good design as is.

L6234 datasheet

  • \$\begingroup\$ Application note also states that "To avoid overload of the logic INPUTS and ENABLES, voltage should be applied to Vs prior to the logic signal inputs.". So pull-ups REALLY don't make sense then, right? \$\endgroup\$ Jan 31, 2021 at 18:57
  • \$\begingroup\$ A good question, a schematic of the control logic would help. I will take a SWAG and say the control logic is open collector/drain outputs and sink but do not source voltage. The L6238 requires a Supply voltage from 7 to 52 V, 5V logic will not work. They are pull ups, it shows that on the lower right side of the green block. As far as pull up and pull down resistors there are thousands of applications for them, not just buttons. As you get more experience this will begin to make sence. I would suggest you study the data sheet, it will help you understand what is going on. \$\endgroup\$
    – Gil
    Jan 31, 2021 at 18:59
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    \$\begingroup\$ @Gil - 10k is rather a weak pull-up for being the main mechanism to provide a logic high. The transition time would be rather long. Also 5V logic will work. Anything over 2V is considered a logic high. The maximum voltage is 7V. The device was designed for 3.5/5v logic by the looks of it. \$\endgroup\$ Jan 31, 2021 at 20:07
  • \$\begingroup\$ I would suggest you read the data sheet. It states The L6238 requires a Supply voltage from 7 to 52 V, I did not do the design nor do I have access to the internals of the driver. The board has a input voltage specified of 8 to 42V per the drawing. It was a SWAG. Since you decided they are a weak pull up, explain what the designer did wrong and why? Of course the drawing could be wrong but I have no clue on that. \$\endgroup\$
    – Gil
    Feb 2, 2021 at 2:01
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    \$\begingroup\$ @Gil - I did read the datasheet. Yes, it requires 7V or more for the power stage but the logic signals are normal TTL with a 0.8V low and 2V high. They are compatible with the GAL implementation of the control logic running at 5V. \$\endgroup\$ Feb 2, 2021 at 2:04

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