[MOVED] Operational amplifiers and diodes

[portuguese]

Hi guys! I'm portuguese and I have an Electricity exam tomorrow. So this is really urgent. There are 2 kinds of exercises I don't know how to do. The tests usually have similar exercises to these ones:

1) Operational Amplifiers

Consider the circuit represented in this picture:

. The voltage vi(t)=1sin(wt) V and the operational amplifier saturates in (+-)12V. Calculate and plot the signal of the voltage vo(t).

2) Diodes

Consider this circuit:

. Plot the signal vo(t) (show your calculations [all of them]).

I really need your help! Thanks a lot!

 

[mister_rf]

Re: Operational amplifiers and diodes

2. Diodes
step by step circuit diagram... :grin:

- - - Updated - - -

1.
See the attached diagram, replace vi(t)=1sin(wt) V

http://en.wikipedia.org/wiki/Operational_amplifier_applications

- - - Updated - - -

See also the graphical representation:

 

[portuguese]

Re: Operational amplifiers and diodes

Thanks a lot! I understood the graphics really really well...

I have an additional question. Imagine they ask me in 2.) Diodes to find the Peak Inverse Voltage in diode D. Do you know how to calculate?

Thanks again!

 

[mister_rf]

Re: Operational amplifiers and diodes

Peak Inverse Voltage in diode D = -10V

- - - Updated - - -

Vg = -10V
Vg = i *1K + Ud + i * 1K
but
i=0
Vg = Ud
> Ud = -10V

 

[portuguese]

Re: Operational amplifiers and diodes

Is it PIV=Initial voltage*Vout(from the voltage divider)? May I do that to every similar exercises?

 

[mister_rf]

Re: Operational amplifiers and diodes

PIV= Peak Inverse Voltage?

See details in the attached picture. Use KVL.
To be more precise, the complete formula for the peak reverse voltage:
Vg = i1 *R1 + Ud + i2*R2
but
R1=1k
R2=1k
i1= i2= 0
Vg= Ud

 

[portuguese]

Re: Operational amplifiers and diodes

Yes, PIV = Peak Inverse Voltage.

Imagine this circuit:

Can I do the same thing to calculate the PIV?

 

[mister_rf]

Re: Operational amplifiers and diodes

Use KVL/KCL
Vg = i3R3 + Vd + i1 R1
Vg= i2R2 + Vz+ i1R1
I3=0, I1=i2
Vz= 0.7V, Vg= 10V
Vd= i2R2 + Vz
But
I2 = [10 -0.7]/(R1+R2)= 9.3/(R1+R2)
So
Vd = 9.3 R2 /(R1+R2) + 0.7V