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Original question:

UCC21750: Driving IN+ IN- with a differential signal

UCC21750: UCC21750, SN65LVDS1

Sahil Shingote
Prodigy 40 points
Part Number: UCC21750
Other Parts Discussed in Thread: SN65LVDS1, UCC23525, UCC23513

Tool/software:

1. If I have a unipolar PWM signal with 3.3V/GND and plan to use SN65LVDS1 to convert it into a differential input to feed into my gate driver IC, i.e., UCC21750DWR, can I use it directly?
2. The other option is to use a differential driver and then a differential receiver and feed the gate driver with a single-ended input?
What would be ideal for noise immunity purposes?

  • +1
    TI__Mastermind 23196 points

    Hi Sahil,

    Why don't you want to drive IN+ directly with the 3.3V/GND signal? Can it drive a small 100pF local filter capacitor at IN+ for some more noise immunity? Most GPIO pins can drive 20mA.

    The issue with a differential input is that the IN+ and IN- pins will end up in undefined states during the transistion, and you will likely have rising edge and falling edge glitches. The hysteresis is meant to work for each pin individually. A fully differential input would have more "gain" at the transition threshold to avoid the problem. 

    Ideal for noise immunity would be a single ended n input buffer right at IN+, and IN- tied to GND right at the device. The dual inputs of this device were not designed to improve CMTI, unlike fully differential inputs that you may be referencing.

    We do have optical emulated inputs, such as UCC23513 and UCC23525, and these diode inputs can be driven differentially. 

    Best regards,

    Sean