I got the value of 41.72 ohms. But i think it can't be done like this!

 

I got the value of 41.72 ohms. But i think it can't be done like this!

hi,

IRL = 5.6/330 = 17mA, so Iz = 1.7mA , so Irs = 18.7mA

Vmin ripple = 6.5v, so Rs has to drop 6.5v- 5.6v = 0.9v at 18.7mA.

So Rs = 48.1R, nearest pref value is 47R...

So using the above method you should be able to calc the value of Rs for your measured Vripple min.

Show me your calculations how you got 41.72R.?

 

I used those formulas above to calculate it:

Vz = 5.6V
Rl = 330 ohms

I_rl = 5.6/330= 16.9mA
I_z=16.9mA + 1.69mA=18.6mA

I_Rs=18.6mA + 16.9mA = 35.6mA

And lastelly, using the 1st formula:

35.6mA = (5.65-5.6)/R_s <=> R_s=41.72ohms...

 

I used those formulas above to calculate it:

Vz = 5.6V
Rl = 330 ohms

I_rl = 5.6/330= 16.9mA
I_z=16.9mA + 1.69mA=18.6mA

I_Rs=18.6mA + 16.9mA = 35.6mA

And lastelly, using the 1st formula:

35.6mA = (5.65-5.6)/R_s <=> R_s=41.72ohms...

hi,
You dont add these values together again.!
I_Rs=18.6mA + 16.9mA = 35.6mA

Itotal thru Rs = Irl + Iz = 17mA + 1.7mA = 18.7mA

Try again, almost there.
E

 

Why not? By the Kirchoff laws i can say that I_rs = Iz + I_rl

 

Why not? By the Kirchoff laws i can say that I_rs = Iz + I_rl

Izener is only 10% of Irl , your question gives you that information

Irl = 17mA so Iz must be 1.7mA , giving a total of 18.7mA thru the Rs resistor.

So why would you want to add that value to the value of Irl again.??

Your post says: I_rs = Iz + I_rl , which is what I am posting.

Irs= 1.7mA + 17mA.

EDIT:
Look at this sim.

Remember Irl is the current thru the 330R

 

Hum... But why are you simulating that without a bridge... I didn't got your point now... Sorry!

 

Hum... But why are you simulating that without a bridge... I didn't got your point now... Sorry!

I used a simplified circuit that only shows the components that we are discussing, its set to 6.5V as thats the ripple minimum voltage from the previous bridge.

Do you now understand the way I calculated the Rs etc.??

 

Nope, not yet!

What formula have you used to calculate R_s???

 

Ok, i got it...

But the simulation confuses me a lot...

I have calculated an R_s for exact values this way:

I_rl = 16.9mA
I_z = 1.69mA (this is where I think I might be doing something wrong because i don't know if this info is correct, i mean if i got this correctly from the teacher)

I_rs = 16.9 + 1.69 = 18.59mA

R_s = 79.43 ohms

 

hi,
That 79.43R resistor is much too high a value, its bringing the zener out of conduction, so the 5.6V Vout is no longer regulated.

Look at your asc file run by me to show the Iz and Vout.

Note The Iz and Vout.

Who choose that value for Rs.???

EDIT:
Formula for Rs = [VcMin- Vz]/[Iz+Irl]

Where VcMin = lowest value of voltage due to the ripple on Vc

Its basically Ohms Law

 

Jesus, i can't get any of that!
I have added the directive to standard.dio and i get nothing like thaT!

What i get is this:

 

Thats exactly the same as mine.!

Plot the Vout on its own plot and the Iz on its own plot and post what you see..

EDIT:
I have changed my plot to match yours, they are the same

 

Where did you get that formula for R_s?

From those 3 formulas i told you about, i have calculated R_s.

 

I can't see the same... At least with the same aspect!

 

You have removed the 10mSec delay from the .tran 0 50m 10m, so your plot will be less accurate.!!
ie: lower resolution.

You do not need to plot the first half cycle.

 

Ok, i got it... You suited the scale of YY axxis!

But what about the formulas?

 

I have told you the formula many times in previous posts


Formula for Rs = [VcMin- Vz]/[Iz+Irl]

Where VcMin = lowest value of voltage due to the ripple on Vc

Its basically Ohms Law

 

May I ask what technical course you are attending at college or university.?

E

 

I got this!

But with exact values, i get R_s = 79.43 ohms!