Question

A TOBB ETO 3) Ace 12000 yoke 12.5mA akom vermesi ve Rour2000 man sayacak bir yuzdevresi tasaraying R-12000 100mA001) 3 12000 23 2

Design an interface circuit that will allow i = 12.5mA current to be given to 1200 load and Rout = 2400 as follows (R = 1200 İs = 100mA g = 0.001)?

Answer 1

The interface will be a voltage divider coupling.

R1 Vi Vo R R2. R.

- Output resistance condition sets the value of R2:

R. = V. RO = R2 = 2.4K

Find the value of R1

The output voltage is:

V. = 1.2K(12.5mA) = 15V

Then, the current on R1 is:

15 iR1 = 12.5mA + 2.4K = 18.75mA

To find R1, calculate the value of voltage VS:

+ + Interface circuit gvs. f 38 un dv2 is (a V1 2R 23 13 Bin

Using voltage nodes equation on Vs:

eq 1

v. (3.6*+ 3.6K) 3.6K V1 V2 3.6K = 100m - V, - 18.75m

Notice that V1 and V2 are voltage dividers woth the same values, V1 and V2 are expected to be V1=V2, then no current flows through the resistance that connects them and:

V = V2 = 2R Vs 3R + 2R 2 Vi = 5V

In eq1

v s) 2 3.6K, - 2 Vi 3.6K 100m - gv - 18.75m Vs (3) 2 3,6K) 12 - 2 - V = 81.25m - gV 3.65

Solve for Vs:

V = 81.25m - g Vs (5.6K) - v, (6*3.6K) V. (3.8K) - v. (6 * 3.6K) + 9Vs = 81.25m (3.6%)(-- +9+3.6K) = 81.25m VS 4 5 Vs (3.6k)(2+0.001 - 3.6K) 81.25m Vs (4.8) = 81.25m * 3.6K

3 = 60.93

To find R1, apply Ohm's Law:

R1 = V. - 15 18.75mA 60.93 – 15 18.75mA = 2.45K