Do Gate Drivers That Look Like Comparators Have Comparator Inputs?

Question

I do mostly VHDL & signal processing for work, so please forgive if this question is dumb.

Consider Texas Instruments' part number UCC27526DSDR. When I look up its datasheet on Digikey (1st page, lower right), the chip has two little comparator symbols inside. I want to control it using LVDS, with a 100 ohm resistor between its inverting and non-inverting terminals. The resistor will not directly be connected to ground. The common mode voltage will not be large (less than, say, 2*Vcc), so I'm not expecting the thing to reject arbitrarily large common mode voltage. Long wires (say, between 10 cm and 1 m) will deliver the differential current signal to the resistor, which is the motivation for doing all of this in the first place.

Here are my questions:

1. When gate drivers use the comparator signal, do they typically have comparator inputs? Or are they really two digital inputs and the manufacturer decided to use the op amp symbol?
2. How about this part in particular? It's not obvious to me from reading the data sheet, but I work in VHDL and I'm not 100% sure what to look for either.

Answer 1

Consider the block diagram.

Source: UCC2752x Dual 5-A High-Speed, Low-Side Gate Driver | TI, page 13

Granted, TI play notoriously fast and loose with their block diagrams, but it is a safe assumption that these are ordinary LVCMOS/TTL compatible logic receivers, independent per pin.

(In fact, given the independent input threshold (doesn't depend on VDD), they are probably open-drain NMOS, driving a hysteresis circuit; so you're actually seeing Vgs(th) for the input thresholds, more or less. This is actually shown in some other gate drivers, TC4420 for example. It's quite possible TI is using a similar structure.)

Furthermore, the input thresholds are specified, and consistent with this interpretation. If the input were differential, it would say so; it would also be impossible for the pulled-low/high inputs to provide such input thresholds if they were.

Placing an LVDS to CMOS receiver/translator in front is a perfectly fine solution.

I would however caution against using LVDS over long distances and cabling, especially in the potentially extraordinarily noisy environment of a switching converter. RS-422 is the safer bet and I would encourage its use. Transformer or optical isolation are also viable options.

(Yes yes, LVDS can be carried effectively on shielded cable, and USB and HDMI are excellent examples of it. If not TIA/EIA-644 per se, then low voltage differential signals generally, I mean. But shielding and signal quality are topics I do not expect a beginner to know how to work with; lord knows there's dozens of erroneous answers on the topic here already. The simpler, more robust solution is the better solution.)

Answer 2

1. The symbol is arbitrary. It simply has two digital inputs per channel, an active high and active low input, so you can simply choose which pin you want to drive as an input and use the other as enable pin.

2. You can't use LVDS as input for this. The inputs are not differential and they have voltage thresholds not compatible with LVDS. This chip has two single ended inputs per channel and requires less than 1V for logic low and more than 2.3V for logic high.