I can't get plot...

Anyway, why have you used 11ohms R? Are my calculations ok?

 

Hi,
Does the question from the tutor ask you to measure from the Vsrc to Vload as a plot.??

For your 11R and 220uF

tc= 11 * .00022 = 2.42mSec / 5 = 0.48mS say ~5mSec [ 5 *TC]

At 1mS from Vpeak , 2.878v/4.352 = 0.66 * 100 = 66%

Look at this plot

Tell me what you think.?

EDIT:
Read this 2nd image

 

I don't understood why 11ohms... I didn't said i have an R of 11 ohms because the R is what I want to calculate...

what you mean by tc and TC? What's the difference?

My calculations where:

5*1ms = R*C <=> 5ms = 220μF*R <=> R = 22.72Ω

I think I understood your plot.

You looked at a 1ms period in the plot. Marked the voltage at the beggining of that 1ms and the voltage at the end of that 1ms. Divided those voltages and the result was 0.66 which is roughly the 1st time constant of an RC circuit!

 

My plot was to show the voltage at, tau, ie: one time constant = R * C
[tc , TC and tau are the same ], Time Constant

Have you tried Vc = Vmax - [Vmax * exp-[t/RC]} to double check your 22R answer.?


EDIT:
Compare your 'exp' calculation with your post #61 plot,, at the 1mSec discharge time.

Are you now confident with your answer?

 

What should be t in the exponencial part of the equation?

If i use 5ms at t and 22.72Ω i get the twice the Vc at the 1st time constant. If I choose 11Ω I get 0.633 which is what it was supposed to be...

But how did you get the 11Ω value?

 

If you wanted to calculate the Vc at one time constant you would use tau = R.C [ at 1KHz the period is 1mS ], so I have assumed tau= 1mS.

BUT you say that the question states use a resistor that gives 5 * tau = 5mS

So when you use 22.72R, Vc will not decay to the same voltage at 1mSec as when using 11R.

Look at this image from my calculator
Compare the value with your plot at 1mSec, it looks ok to me.

 

Well, I'm really getting confused!

Can you tell me the calculations you have made to get the 11Ω resistor?

I tried to do what i have told you before:

5*0.001 = 220uF*R

R = 22.72Ω

I understand it's wrong because using the Vc equation, it doesn't gives my nothing like it should give if i use 11Ω. But i'm not figuring out how to get 11Ω.

 

Let us recap.
You have a 1Khz square wave, with a Period of 1mSec

tau = R * C

If you wanted to know the Vc at one tau, say a 1mSec Period.

One tau would be 0.001 = R * 0.00022, so R = 4.54R
The Vc would be at 1.6V in 1mSec


BUT the tutor said calc for 5 *tau, so 0.005 = R * 0.00022,

So R= 0.005/0.00022 = 22.7R

The Vc will at ~ 3.5V at 1mSec after the Vpeak
Note at 5*tau = 5mS the Vc is a 1.6V

 

So, that means that my calculations were correct?

Can i say that with a cap of 220μF, the resistor should be 22.72Ω?

 

So, that means that my calculations were correct?

Can i say that with a cap of 220μF, the resistor should be 22.72Ω?

Yes, its correct for a tau = 1mSec * 5.

The point I would ask your tutor is about the capacitor charging period being only one half cycle of the 1KHz square wave, so technically the discharge time should be 0.5mSec.?? before the next 0.5mSec recharge pulse.

Let me know what he says.

 

Please, let me see if I understood what you want me to ask my teacher.

If the discharging time of the capacitor is only 1 half-cycle of the 1ms period?

 

Now I still need help to solve the first exercise about Iz max and Iz min.

I'm going to recap

 

Now I still need help to solve the first exercise about Iz max and Iz min.

I'm going to recap

If its a new topic I would suggest you start a new thread.

Dá-me tanta informação quanto possível para as questões% 2C evita a confusão.

Eric

 

Now I still need help to solve the first exercise about Iz max and Iz min.

I'm going to recap:

2 zeners in oposite direction as in the asc file.

Z1 = 3.9V
Z2 = 5.6V

V_src = 8.49V (6V rms)

Pz1 = 0.25W
Pz2 = 0.5W

I_max Z1 = 0.25/3.9 = 64mA
I_maz Z2 = 0.5/5.6 = 89mA

Iz max,proj Z1 = 90% I_max Z1 = 80.36mA
Iz min,proj Z1 = 10% I_max Z1 = 8.9mA

Iz max,proj Z2 = 90% I_max Z2 = 57.6mA
Iz min,proj Z2 = 10% I_max Z2 = 6.4mA

Until here I think I'm going fine...

Now I need to know what to consider the Vin min and the Vin max that i think it should be 8.49V and -8.49V.

Last friday our teacher said us to use the highest Iz_min,proj and the lowest Iz_max,proj to calculate Rs.

Now i need to know how to use those values of Vin min and max to calculate Rs_min and Rs_max.

For the positive semi cycle should be something like:

Rs_min = ((8.49 - 0.7)-3.9)/57.6=67.53Ω
Rs_min = ((8.49-0.7)-5.6)/8.9=246.10Ω

Rs_med = 156.82Ω

Would this be ok? I have some doubts in the underlined calculations!

 

hi,

You should have,
Rs_min = ((8.49 - 0.7)-3.9)/0.0576=67.53Ω
Rs_min = ((8.49-0.7)-5.6)/.0.0089=246.10Ω
Rs_med = 156.82Ω

Have you tested your results in simulation.?

 

Just wanting to know if the signs of + and - were correctly used at those calculations...

Are they ok? I'm considering Vin max and Vin min the same? Should this be ok? Or should i change any signs there?

I haven't tried it yet because i'm trying to understand if calculations are ok following the voltage directions (positive and negative) and trying to figure out when should I use minus or plus signs!!!

 

Just wanting to know if the signs of + and - were correctly used at those calculations...

Are they ok? I'm considering Vin max and Vin min the same? Should this be ok? Or should i change any signs there?

I haven't tried it yet because i'm trying to understand if calculations are ok following the voltage directions (positive and negative) and trying to figure out when should I use minus or plus signs!!!

The Iz current for the zener [ the current due the zener operating in the breakdown region], 'conventionally flows' from Cathode to Anode, so its always a positive number.

When you say I'm considering Vin max and Vin min the same?,; as the input is a Sine wave, Vin max, I would consider as the peak of the positive going half cycle of the sine wave and the Vin min, as the negative going half cycle. So I would assume they are the same amplitude.

I can see why you are having difficulty in understanding/ answering these questions.
IMHO the questions are poorly defined and ambiguous, of course this may be intentional by the tutor, in an attempt to get you to figure out the problems.

E

 

Hi ericgibbs...

Yes they might be a little ambiguous... We have 2 teachers. One for the theorical classes and another for the laboratory. And at some point, both teachers give different answers for the same questions. I don't want to question their ability or anything like that, but sometimes we get confused...

I have one other question regarding the double zener limiter.

The output voltage wave peak suffers like an attenuation. This is the limitation. The amount of volts that are attenuated are the Vz + 0.7????

In the image attached i have the reading from the OCR. We used the cursors to see the Vpeak of the output waveform...

 

hi,
The purpose of that double zener is to 'clip' the sine peaks for positive and negative voltage input.

Clips at ~ 6.3V and ~4.6V as its supposed to do.

 

Yes, but that's not what it's in the positive half-cycle... In the positive half-cycle we can see 5.2V which is + 0.6V than what it should be!