Hi...

I don't know how to say it in English but my next job is to analyse polarized limitator circuits...

But somehow I can't get the result I should get!



PS: How are this kind of circuits named to change the title????

 

Hi...

I don't know how to say it in English but my next job is to analyse polarized limitator circuits...

But somehow I can't get the result I should get!



PS: How are this kind of circuits named to change the title????

What result are you expecting.?

Post your images and asc files explaining the results you are expecting.

 

I think I should be getting the plot attached! But I can't get it in LTSpice... I should be missing something!

How you name this circuits in eglish?

 

I get this image and plot.

Please post your asc file when you ask a question.

 

That is not what I should get... The Vout should have a negative zone and it hasn't... Check the screen I posted!

 

That is not what I should get... The Vout should have a negative zone and it hasn't... Check the screen I posted!

Show me where you see a Negative zone on that waveform on the right side, that you have ticked as OK.???

ALL the waveform at Vout is at or above the 0V line.!!

PLEASE POST YOUR ASC FILE so that we can work from the same reference.

EDIT:
And why do you have a red arrow pointing to the junction of the battery and the diode when the right side plot is Vout of the circuit.???

 

I have tried to reply that same circuit but you're right. It is not supposed to be any negative area. I misunderstood the picture. Though, shouldn't the 2nd voltage source make the curve go up like if it has a DC value???

 

I have tried to reply that same circuit but you're right. It is not supposed to be any negative area. I misunderstood the picture. Though, shouldn't the 2nd voltage source make the curve go up like if it has a DC value???

hi,
The V output Sine wave voltage has a DC Offset value due the battery voltage, so it has gone up by that DC value.

Try the sim again using different battery voltages, you will see the Vo level change.

 

I got it...

I didn't noticed that the second Voltage source should be DC.

Now, my job is to analyse a double parallel limiter with zeners ad calculate the R.

I'll upload the asc in a minute!

 

hi,
Its a good idea to have 2 or 3 copies of the basic circuit on the same simulation.

Use circuit #1 as a Reference plot, so that you can quickly see how changes in the 2nd circuit produce different plots.

This will also speed up your learning curve.

E

Use R1,R2 and R3 values to suit your circuits.

 

Ok... Thanks for that tip...

But now I need to know how to calculate the R so that I can start doing some simulations.

And why do your 3rd circuit has a +5 DC battery and the curve doesn't starts at +5V?

 

Ok... Thanks for that tip...

But now I need to know how to calculate the R so that I can start doing some simulations.

And why do your 3rd circuit has a +5 DC battery and the curve doesn't starts at +5V?
I chose an arbitrary value of Vdc for demonstration purposes only.

hi,
For R1, apply the same calculations for your previous zener problem.

The R1 value chosen is to ensure that the reverse biased zener [ regulating] remains in conduction over the period that regulation is required.

I would suggest you post the English translation of the tutors question.

 

Analyse the double parallel limiter using a 5.6v (P=0.5W) and a 3.9 (P=0.25W) zener.

Determine the range of admissible values ​​for the resistor R. Using the average value, simulate the operation of the circuit.

This is the problem we have to solve!

 

Post your calculated Rs values and the LTS plots, I will check them.

Has the tutor stated the value of the battery voltage.??

 

Post your calculated Rs values and the LTS plots, I will check them.

Has the tutor stated the value of the battery voltage.??

Nope...

He just says that the Vin is a sinusoidal wave with 6v rms, so I used 8.49V.

Doesn't the battery value changes the behaviour of the circuit if it has values higher/lower than the V source?

 

Nope...

He just says that the Vin is a sinusoidal wave with 6v rms, so I used 8.49V.

Doesn't the battery value changes the behaviour of the circuit if it has values higher/lower than the V source?

The battery will change the operation of the circuit depending upon its voltage and its polarity within the circuit.
As a test to check the effect of the battery voltage, you could set the Vsrc voltage to 0V and try different battery voltages, you should get plots for different Iz values

Did the tutor also specify at which point in the circuit is considered as Vout.
example: At the junction of the battery and the top zener or the junction of the two zeners.??

In my opinion the question is very badly written and ambiguous.

 

This is what we should work with!


PS:
What is the info you think might be missing?

 

This is what we should work with!
PS:
What is the info you think might be missing?

This is a different circuit from the one you posted in your post #9, which had a battery in series.??

If its just two zeners in opposite series connection, calculate what you think is the answer, post the answer and the LTS asc.

 

I'm sorry, I forgot to tell ya that I was now analysing a double parallel zener limiter!

PS:
I still don't know what title to put at the thread's title! (I think I can't edit it now).

I was thinking in calculating the Iz via Potency, then for each semi-cycle, calculate the R that is thru by the same current as Iz, I guess!!! Am I thinking right? (probably not)

 

PS:
I still don't know what title to put at the thread's title! (I think I can't edit it now).

Try read this
https://coefs.uncc.edu/dlsharer/files/2012/04/B8.pdf