If the input voltage V_in =V_ocoswt, plot the output voltage as a function of ti

Question

If the input voltage V_in =V_ocoswt, plot the output voltage as a function of time of each circuit,  use V_mu= 0.7 V.

Assume the input voltage Vin =V_ocoswt, plot the output voltage as a function of time of each circuit, use, use an ideal diode model.

Find the current for the diode using piecewise line model V_mu= 0.6 V.

Find the current I and voltage V in the circuits below using piecewise linear model with V_mu = 0.7 V.

Solve the following problems in details. Do not solve any parts that requires computer simulation If the input voltage Vin =V0cosomegat, plot the output voltage as a function of time of each circuit, use V gamma= 0.7 V. Assume the input voltage Vin =V0cosomegat, plot the output voltage as a function of time of each circuit, use, use an ideal diode model. Find the current for the diode using piecewise line model V gamma= 0.6 V. Find the current I and voltage V in the circuits below using piecewise linear model with V gamma= 0.7 v

Explanation / Answer

1. inpart one as diode pass current in forward direction only.

so

a)

Vout = Vb - 0.7 ----------------------------------------1

and

Vout - Vin = I*R1------------------------------------------2

and

I = Io*EXP(q(Vb - Vout -0.7)/KT)-----------------------3

these three are the governing eqution. remember current will flow only when Vb > Vin.

use iterative methode or hand calcuklation for the above equation and get the result.

b.

in the second problem we have,

Vout = Vb - I*R1----------------------1

and

I = Io*EXP(q(Vout -0.7 - Vin)/KT)-----------------------2

these two are the governing equation.

here also current flows when Vb > Vin

c.

using mess analysis we can write

-Vout + Vb - I*R1 - Vin = 0 ----------------1 where Vd = 0.7 V

and

I = Io*EXP(q(Vout -0.7)/KT)-----------------------2

these two are the governing equation.

see the polarity of Vout has been canged in the 1st equation.

2.

a) here we get

Vin - I*R1 - Vb = Vout.--------------------1

and

I = Io*EXP(q(Vout -0.7)/KT)-----------------------2

b)

Vout = I*R1-------------------------------1

and

I = Io*EXP(q(Vin - Vb -0.7)/KT)-----------------------2

c) Vb + Vd1 = Vout-------------------1

and

Vin - I*R1 = Vout--------------------2

I = Io*EXP(q(Vout - Vb -0.7)/KT)-----------------------3

solve and get the result.

3.

et say the volatge in the n terminal is Vn and P terminal is Vp.

so Vd = Vp - Vn which is the diode potential

now from nodal analysis at Vn we can write,

(Vn - 3)/3 + Vn/2 - I = 0-----------------------------------1

similarly at Vp

(Vp - 3)/2 + Vp/2 + I = 0-----------------------------------2

and the other is

I = Io*EXP(q(Vp - Vn - 0.6)/KT)-----------------------3

use iterative methode to solve this.

4.

a)

the governing equations are

-7 + I*R1 = Vout------------------------------------1

I = Io*EXP(q(3 -Vout - 0.7)/KT)-----------------------2

b)the governing equations are

5 - 16*I = Vout-------------------------------1

I = Io*EXP(q(Vout + 5 -0.7)/KT)-----------------------2

c)

as u can see the two extreem terminal is 5 and -5 so V has to be inbetween them so the voltage across the diode is negative.

so for ideal diode current won't flow here.

d)similar is the case for d.

here also there is negative Vd for the diode.

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