KCC's Quizzes AQQ278 about an integrated Resistor

if you are measuring along the length  60 squares long (each of 2u) x 20 Ohm/sq => 1200 ohms.

If you double L and W then the number of 4um squares stays at 60 and resistance is unchanged.   If you measure the resistance across the width the resistance is 60 squares in parallel or 1/3 ohms. 

1. You need to calculate the number of squares which is n=L/W=60. Resistance is R=(sheet resistance)*n=1200ohm

2. If you double both W and L, the number of squares doesn't change so you get R=1200ohm again (sheet resistance is resistance/square no matter how big the square is)

1. It is 120/2=60 squares, therefore resistance is 60*20=1200 Ohm

2. Resistance remains equal, but current capability doubles. 

Number of squares is 120/2 = 60 squares

So resistance is 60 x 20 = 1200 ohms or 1.2 kohms

If you double both the width and length of the trace, the number of squares is the same, so the resistance is the same.

120 um x 2 um is 60 squares, so that's a total resistance of 20 ohm/sq * 60 sq= 1.2 kOhm. 

If you double both width and length, you still have the same number of squares, so the resistance is unchanged. Though I guess you're a little more immune to process variation, so... hooray?

Doubling the width is like having two (discrete) resistors in parallel; doubling the length is like having two (discrete) resistors in series.

So, for question 2:  The total resistance does not change  (but the strip is 4 times larger)
Question 1:  R = 20 Ohm  *   120um / 2um    = 1200 Ohm.

1) R = 20xL/W = 20x120um/2um = 1,200 ohms

2) R = 20x2L/2W = 20xL/W = 1,200 ohms

1. Number of squares is L/W or 60.  So resistance is 60*20 or 1200 ohms.

2. Double both dimensions and 240/4 is still 560 squares, so still 1200 ohms.

1. 1200 ohm
2. It won't change - but might let you dissipate more power!

1. With L = 120 um and W = 2 um, a square is 2 x 2. The number of squares from A to B is then 120/2 = 60 squares. At 20 Ohm/sq we get 1,200 Ohm.
2. If you double L and W (240, 4) the square is 4 x 4. The number of squares from A to B is then 240/4 = 60 squares, so no change in resistance, but presumably an increase in the power handling capacity. ;-)