HELPP!!! Calculating voltage drop

Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 10mA. The zener has an incremental resistance of rZ = 30

hram777 [196]

Answer:

The answer is given in the explanation.

Explanation:

The circuit is as indicated in the attached figure.

From the analytical description the zener voltage is given as

$V_z=V_z_o+I_zr_z$

Here

Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.

The equivalent model is shown in the attached figure.

From the above equation, Vzo is calculated as

$V_z_o=V_z-I_zr_z$

Here Vz is given as 7.5 V

Iz is given as 10 mA

rz is given as 30 Ω

Thus the Vzo is given as

$V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V$

The value of I_L is given as 5 mA

Now the expression of current is as

$I=I_z+I_L\\I=10mA+5mA\\I=15 mA$

Now the resistance is calculated as

$R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66$

So the value of resistance is 186.66 Ω.

Considering the supply voltage is increased by 10%

V is 10-10%*10=10+1=11 so the

$R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V$

Considering the supply voltage is decreased by 10%

V is 10-10%*10=10-1=9 so the

$R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V$

Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA

so

$R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V$

Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus

$V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V$

$R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA$

The load voltage is 7.485 V

8 0

2 years ago

A vector AP is rotated about the Z-axis by 60 degrees and is subsequently rotated about X-axis by 30 degrees. Give the rotation

Musya8 [376]

Answer:

R = $\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]$ $\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]$

Explanation:

The mappings always involve a translation and a rotation of the matrix. Therefore, the rotation matrix will be given by:

Let $\theta$ and $\alpha$ be the the angles 60⁰ and 30⁰ respectively

that is $\theta$ = 60⁰ and

$\alpha$ = 30⁰

The matrix is given by the following expression:

$\left[\begin{array}{ccc}1&0&0\\0&cos30&-sin30\\0&sin30&cos30\end{array}\right]$ $\left[\begin{array}{ccc}cos 60&-sin60&0\\sin60&cos60&60\0&0&1\end{array}\right]$

The angles can be evaluated and left in the surd form.

4 0

2 years ago

How did studying pagodas help engineers create earthquake-proof structures in modern society?A. Engineers learned that the desig

Mila [183]

D cuz i took the test

7 0

3 years ago

For precipitation hardening, how many phases (in numeric form) exist (a) following the solution heat treatment, and (b) followin

Nitella [24]

The correct answer is B

4 0

3 years ago

Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,

iren [92.7K]

Answer:

The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA, IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v

Explanation:

Solution

Given that:

V+ = 20v

Re = 2kΩ

Rc = 1kΩ

Now we will amke use of the method KVL in the loop.

= - Ve + IE . Re + VEB + VB = 0

Thus

IE = V+ -VEB -VB/Re

Which gives us the following:

IE = 20-0.7 - 10/2k

= 9.3/2k

so, IE = 4.65 mA

IB = IE/β +1 = 4.65 m /101

Thus,

IB = 0.046039 mA

IB = 46.039μA

IC =βIB

Now,

IC = 100 * 0.046039

IC is 4.6039 mA

Now,

VB = 10v

VE = VB + VEB

= 10 +0.7 = 10.7 v

So,

Vc =Ic . Rc = 4.6039 * 1k

=4.6039 v

Finally, this is the table summary from calculations carried out.

Summary Table

Parameters IE IC IB VE VB Vc

Unit mA mA μA V V V

Value 4.65 4.6039 46.039 10.7 10 4.6039

4 0

2 years ago