Doubts on Diodes: Circuit Solving

[Bling Fizikst]

Homework Statement

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Relevant Equations

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What do i need to do for transfer characteristics for the first problem (B)?
I plotted the $V_1$ and $V_2$ plot on the same graph:

Is it correct?

For the second problem , i guess the capacitor $C_1$ is just there to ensure AC flow .
But really unsure how to approach this .

 

[BvU]

Is it correct?

Can you explain your reasoning ?

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[Bling Fizikst]

Can you explain your reasoning ?

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As long as the potential at the p type end is less than $2$ , it is reverse biased . This means the output volatge is the same as input voltage . Hence , it over laps with $V_1$ till it hits $2$ . When the potential is $2$ at the p type end , then diode is forward biased again and the output voltage is also $2$ as long as it is $2$ or more . Since the input is alternating , this means the potential will again drop below $2$ and the ouptut voltage will be the same as input voltage

 

[Bling Fizikst]

Any hints for the second problem?

 

[BvU]

As long as the potential at the p type end is less than $2$ , it is reverse biased .

How come I get the idea that it's not 'less than +2 V' , but 'more than -2 V' ?

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[Bling Fizikst]

I am sorry , the graph i sent earlier is compltetly wrong . Here is my newer attempt :

 

[BvU]

Is it necessary to bring a new actor on the stage ? What is $V_m$ ?

Be sure to scale the $V$ and $t$ axes (in order not to risk losing points)...
For the same reason, don't forget parts B) and C) ...

For the second problem , i guess the capacitor $C_1$ is just there to ensure AC flow .

There is no info on the value of $C_1$. What if it is huge? What if it is really small ?
Compare with the role of the 2V battery in exercise 4)

But really unsure how to approach this .

You can start by distinguishing $v_m\sin(\omega t) > 0$ and $<0$

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[Bling Fizikst]

I am struggling in the positive half cycle case :
I get $V_1=V_2$ as $D_1$ is reversed biased and $D_2$ is forward biased so it is shorted .
Now , i end with a circuit with $C_1,C_2$ in series and AC source $V$ .
How do i find $V_2$?

For the negative half cycle case:
$D_1$ is forward biased so all the current flow through it . Hence , $V_2=0$ and $V_1=V$

 

[BvU]

I am struggling in the positive half cycle case :
I get $V_1=V_2$ as $D_1$ is reversed biased and $D_2$ is forward biased so it is shorted .
Now , i end with a circuit with $C_1,C_2$ in series and AC source $V$ .
How do i find $V_2$?

For the negative half cycle case:
$D_1$ is forward biased so all the current flow through it . Hence , $V_2=0$ and $V_1=V$

Seems to me you also struggle with the negative half cycle case
(Just so you know: I'm not trying to insult and I don't have all the answers; -- just trying to help as best I can!)

Here's a few remarks:

I don't understand your 'Hence $V_2=0$ and $V_1=V$' . You are given that $R_L C_2 > 100 T$, so, once $C_2$ is charged, $V_2$ only drops less than one percent during a cycle !

Without $C_1$ and $D_1$ the circuit is a half-wave rectifier. Maybe it's a good idea to google that and study its working (if you aren't already familiar with the subject).

The left part of the circuit (voltage source, $C_1$ and $D_1$):
I agree with you that 'For the negative half cycle case: $D_1$ is forward biased'
So if $C_1$ isn't charged already, it will be charged up to a maximum of ... ?

And with the remainder of the circuit in place: at the same time (i.e. during a negative half cycle), $D_2$ is forward biased -- under the condition that $v_1 > v_2$

What do you think ?

[edit] ready to continue with the positive half-cycle once you are back and have filled in the ...

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[BvU]

I am struggling in the positive half cycle case :
I get $V_1=V_2$ as $D_1$ is reversed biased and $D_2$ is forward biased so it is shorted .
Now , i end with a circuit with $C_1,C_2$ in series and AC source $V$ .
How do i find $V_2$?

For the negative half cycle case:
$D_1$ is forward biased so all the current flow through it . Hence , $V_2=0$ and $V_1=V$

Lost interest?

 

[BvU]

[edit] ready to continue with the positive half-cycle

Never happened, but (for posterity):

the circuit is a voltage doubler. See here or here

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