The datasheet shows power requirement as 5.25 V max.
The typical differential output (EXC to EXC) is shown as 7.2 V p-p.
How is this possible? Am I missing something?
Thanks for your help
The excitation output is measured differentially (EXC-EXCB) and thus each output only need swing between ground (0v) and 3.6V) which is within the supply rails; note that each output is in anti-phase so the output will swing from +3.6 to -3.6V (differential) thus 7.2 Vp-p.
I suppose I have to rephrase the question.
If the maximum supply to the chip from "chip ground" to "chip supply" is 5.25 V, how can the chip produce 7.2 V ?
Does the EXC swing below ground somehow? Does the chip have some kind of boost converter inside
I'm still missing something here?
Thanks again Sean
V positive chip rail________________________________________
Regardless of waveform
peak to peak has to be between
these two rails?
Yes both waveforms (EXC, EXCB) swing between the supply rails.
The easiest way to visualize this is to consider that EXC and EXCB are defined as follows.
EXC = A*sin(wt), EXCB = A*sin(wt+180)
So if A = 3.6Vpp then when I subtract (difference) EXCB from EXC I get
Asin(wt) - (-Asin(wt) )= 2A sin(wt). Thus the differential excitation voltage across the output coil is 7.2V pp but the absolute voltage at either end of the coil never goes below ground or above the supply.
The output drive is very limited and I don't believe it is specified in the datasheet which in retrospect is unfortunate. The intention is for those pins to drive a succeeding drive stage that would drive the resolver coil. The intention is for that stage to have a relatively high impedance such that the load current doesn't exceed on the order of 1mA.