Question

For the 3-bit binary counter shown in the figure, the output increments at every positive transition in the clock (CLK). Assume ideal diodes and the starting state of the counter as 000. If output high is 1 V and output low is 0 V, the current I (in mA)
following through the 50$\Omega$ resistor during the 5th clock cycle is (up to one decimal place)

• A. 10

Answer 1

The correct option is A 10
During the 5th clock cycle, the combination of output bits of the counter is (101).
For this combination, the equivalent of the given circuit can be given as,

Let us assume all the three diodes are in ON state. So, for this assumption, the circuit can be equivalently drawn as shown below:

By appling KCL at node A, we get,

$\frac{1-V_A}{100}+\frac{1-V_A}{100}=\frac{V_A-0}{50}$

$\frac{1-V_A}{50}=\frac{V_A}{50}$

$V_A=\frac{1}{2}V$

$I_1=\frac{1-V_A}{100}=\frac{1}{200}A=5mA$

$I_2=\frac{1-V_A}{100}=\frac{1}{200}A=5mA$

$I=\frac{V_A}{50}=\frac{1}{100}A=10mA$

$I_1>0,I_2>0,I>0.$

So, our assumption is correct and the correct answer is,

$I=10mA$