Today you'll learn to design a current divider circuit using basic current divider rule.
Design a current divider which attenuates 2 A current to 5 mA. The operating voltage is 5 V.
Design a current divider which attenuates 2 A current to 5 mA. The operating voltage is 5 V.
Let's draw the circuit of current divider:
For our present case the formula will become:
5 mA = {R1 / (R1 + R2)} * 2 A
0.005 R1 + 0.005 R2 = 2 R1
or alternatively
R2 = 399 R1... (1)
From Ohm's law:
R2 = V/I2 = 5 V / 5 mA = 1 kΩ
Putting the value of R2 in equation 1
R1 = 2.50 Ω
So a current divider whose design involves R1 = 2.50 Ω and R2 = 1 kΩ will dissipate 2 A to 5 mA.
Current Divider Design Problem 2
Design a divider which divides input current to the ratio of 3:2. So that that the equivalent resistance is 100 Ω.
The input is 3 + 2 = 5, lets assume it in amperes.
Now apply the formula with 5 A input and 2 A as output for R2 we obtain: R2 = 1.5 R1
From formula of equivalent resistance.
1/Req = 1/R1 + 1/R2.
1/100 = 1/R1 + 1/R2.
By putting R2 = 1.5 R1 we achieve R1 = 166 Ω
and R2 = 250 Ω