Question 149: Semiconductors and Logic Gates

Question

The reverse breakdown voltage of a Zener
diode is 5.6 V in the given circuit.
The current IZ through the Zener is :
![](https://imagex.cdn.examgoal.net/F2kDUVl1FpPyhcdEG/60czrBpMoGjiqC41khbxIyOkYG1Nv/TI0TvO5H83id0rpm7iuE1A/image.png)

Accepted Answer

9 = Vz + V
R
1
V
Z
= 5.6 V
V
R
1
= 9 - 5.6
V
R
1
=  3.4
${I_{{R_1}}} = {{{V_{{R_1}}}} \over R} = {{3.4} \over {200}};\,{I_{{R_1}}} = 17\,mA$
V
z
= V
R
2
=  I
R
2
(R
2
)
${{5.6} \over {700}} = {I_{{R_2}}}$;     ${I_{{R_2}}} = 7\,mA$
I
z
= (17 Ã¢ÂÂ 7) mA = 10 mA

Answer

7 mA

Answer

17 mA

Answer

15mA

Answer

10 mA

Semiconductors and Logic Gates

The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit. The current IZ through the Zener is :

Choose the correct answer: