I'm designing a BLDC motor driver using the MP6532 driver. Pg. 4 of the datasheet specifies the Logic Input Current as +/- 20uA. If I'm driving the logic inputs from an STM32F103, this would mean that the driver needs a series resistance on each of its logic control pins to be 3.3V/20uA = 165Kohms This seems unusually high and I'm wondering if my calculations are correct. Or should I simply connect the pins directly and not worry about this?
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\$\begingroup\$ IN(H) means you need at worst 20uA for the signal to register as logic high. \$\endgroup\$– Jeroen3Sep 2, 2021 at 6:04
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\$\begingroup\$ STM32 has 25 ohm drivers which can handle 20 uA with ease. \$\endgroup\$– Tony Stewart EE75Sep 2, 2021 at 6:40
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2 Answers
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You can connect the pins directly- what the datasheet is saying is that the inputs will draw/source 20uA, in other words a series resistance of 165k would likely result in the inputs not getting enough voltage swing to operate. In digital electronics it’s quite normal to connect inputs and outputs directly, unless you’re impedance matching for high speed data.
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The STM32 has 25 ohm drivers which can handle 20 uA load with ease using its 3.3V logic to the 5V logic of MP6532 which is TTL compatible. (0.8~2.0 V)
answered Sep 2, 2021 at 6:41
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\$\begingroup\$ Thanks. I did see the VIH and VIL of the MP6532. I earlier had put in a buffer between the STM32 and the MP6532 logic inputs to act as a level translator. I'm now connecting the pins directly. \$\endgroup\$ Sep 2, 2021 at 7:05