square root calculation for dp flow transmitter?
give me a explanation with example
Answers were Sorted based on User's Feedback
Answer / ginoop .k.m
ma (Linear) =(% *
16)/100 +4
example - ma=
(25*16)/100 +4
linear ma of 25%
=400/100 +4
linear ma of 25%=
4+4
linear ma of 25% =
8ma
ma (sq rt) = (sqrt
of %*16)/10 +4
example -
sqrt ma of 25% =
(sqrt
of25*16)/10 + 4
sqrt ma of 25% =
(5*16)/10 +4
sqrt ma of 25% =
8+4
sqrt ma of 25%
=12ma
Is This Answer Correct ? | 45 Yes | 7 No |
Answer / sampath
Sq.Root { x / 100 } * 16 + 4 = ____ mA
where, x is the % ( percenteage value )
For eg, 25% Sq.Root { 25 / 100 } * 16 + 4 = 12 mA
50% Sq.Root { 50 / 100 } * 16 + 4 = 15.31 mA
75% Sq.Root { 75 / 100 } * 16 + 4 = 17.86 mA
Is This Answer Correct ? | 51 Yes | 23 No |
Answer / jackass
So, the way you would go about doing square root extraction is by using an analog multipler. An analog multiplier has three inputs Vx, Vy & Vz. The output is given by Vo = (Vx*Vy)/Vz; If we set Vy = 1V and Vo = Vz through a feedback loop, we get Vo.^2 = Vx => Vo = sqrt(Vx).
If the interviewer further asks about an analog multiplier, you could mention that it can be designed by a combination of logarithmic/anti-logarithmic amplifier and an adder.
Hope that helps!
Is This Answer Correct ? | 2 Yes | 14 No |
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