LTSpice - How to add new Diode

ericgibbs

Joined Jan 29, 2010
18,766
The next question is to double the C value and see what happens to ripple voltage and find a relationship between the C value and the Vpk at diodes...
hi,

When doubling the smoothing capacitance you should also exam what increase you have in the charging current thru the diodes at and near Vpeak.

Also measure the increased V drop across the diodes at the higher pulse currents

E
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Sorry to introduce a question that is not from the same class, but, can I measure the wavelength phase gap in a low pass filter? Or how can I prove that the wavelength phase gap in a low pass filter is 90º between the Vs and the Vcap?
 

ericgibbs

Joined Jan 29, 2010
18,766
Sorry to introduce a question that is not from the same class, but, can I measure the wavelength phase gap in a low pass filter? Or how can I prove that the wavelength phase gap in a low pass filter is 90º between the Vs and the Vcap?
hi,
I am not sure what what you are asking, do you have an example circuit you post.?

Look at these two images.
 

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PsySc0rpi0n

Joined Mar 4, 2014
1,755
I need to see, for an instant of time, how many degrees 2 waves differ...

Like the Vs phase and the Vcap phase, then I need to know how many degrees they differ in time...

In your first screen, if you choose the first peak of both waves, Vout and IR1, and if you look to time axis, you can see there is maybe 5ms between both peaks, right? Then e need to know how much is that in degrees!
 

ericgibbs

Joined Jan 29, 2010
18,766
I need to see, for an instant of time, how many degrees 2 waves differ...

Like the Vs phase and the Vcap phase, then I need to know how many degrees they differ in time...

In your first screen, if you choose the first peak of both waves, Vout and IR1, and if you look to time axis, you can see there is maybe 5ms between both peaks, right? Then e need to know how much is that in degrees!

hi,
If you know that the fundamental frequency is say 50Hz, thats a period of 20mSec = 360 degrees = [a complete cycle]

IF the two plots were shifted by say, 5mSec , wouldn't that be [5/20]*360 = 90 degrees.?:rolleyes:

So what would a 2.5mSec shift be in degrees.??

E
 

ericgibbs

Joined Jan 29, 2010
18,766
Getting back to the previous circuit!

What would be the correct schematics?
hi P,
Remember what we said in an earlier post about 2 diodes being effectively shorted out by having two Gnd points, so we had to use a COM node.???

Exam the Vcap plots and you will see that one plot is only half wave rectification.

E
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Ok, thanks...

Now things changed a little because our teacher told us that we should use 6V rms and not 6V nominal. So we have to use 6*sqrt(2) V for entry voltage at Vs.

So, with higher Vpk at capacitor, the Vripple will be a bit higher too...
With the new Vs value, I have calculated a C of 337uF.

Now, doubling the capacitor value, I can observe that the vripple decreases... But don't know if I can find any numeric relationship.

The 1st screen is 337uF, the 2nd is 674uF.
The 3rd is a comparison between 6V and 8.49V with the same capacitor just for checking what happens when Vs changes!
 

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ericgibbs

Joined Jan 29, 2010
18,766
But don't know if I can find any numeric relationship.
hi,
Have you considered getting an indication of a numeric relationship by transposing
C= [I * t]/Vrip

Say, Vrip = [I *t ]/C

I and t are the same, if Vpk and the frequency are the same.
You know the values C that you are using and you can either calculate or plot Vrip.
E

EDIT:
Look at this method of comparing waveforms.
 

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Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
I don't know if i made my self clear before.
My teacher is asking to us if we can find a relationship between diode current peak and the capacitor value!

What you're telling me in the previous post is to compare Vripple with the capacitor, right??
 

ericgibbs

Joined Jan 29, 2010
18,766
My teacher is asking to us if we can find a relationship between diode current peak and the capacitor value!
I now understand what you are asking.

Thats not going to be easy to define numerically, there are so many unknown variables.
eg:
the diode characteristics of Vfwd at different pulse current levels.
the capacitors ESR value.
in a practical circuit, the impedance of the transformer driving the bridge diodes.

I would be interested to see your numerical solution.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
I now understand what you are asking.

Thats not going to be easy to define numerically, there are so many unknown variables.
eg:
the diode characteristics of Vfwd at different pulse current levels.
the capacitors ESR value.
in a practical circuit, the impedance of the transformer driving the bridge diodes.

I would be interested to see your numerical solution.
Please let me know what is capacitor ESR value...

I think I'll ask my teacher if he really means to ask us the relationship between the Diode current peak and the capacitor value.


Now we have the following circuit to analyse... This is a bit more complex i think...

Let me know how do I calculate the value for R_S assuming R_L can be 330, 1k and 1MEG...
 

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ericgibbs

Joined Jan 29, 2010
18,766
hi,
You know that for the regulation to be viable, the 330R must have 5.6V across it, this means that the current thru the 330R must Vz/RL.

You should check the zener diode datasheet and determine the current that the zener requires to operate in the optimum breakdown region.

Knowing these two currents you should be able to calculate the value of Rs.

Post what you calc RL it to be.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
Morning my friend...

Well, last night I was at the lab trying to sort things out. I thought i had sorted them out but after all, i think my work hasen't payed out.

I have calculated R_S = 33.99 ohms for the 330 ohms R_L, but looks like I didn't calculations correctlly somehow!

I have the following intel from 1st attachment...

So I_rl = 5.6/330 = 16.9mA

But now I'm forgetting something from yesterday because I'm not remembering how we get the R_s!!!
 

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ericgibbs

Joined Jan 29, 2010
18,766
hi,
Assume that the zener requires 5mA for optimum regulation, so ~17mA + 5mA = 22mA.

The value of Vcap ranges from 7V thru 6.5V, so this means you have ~6.5Vmin - 5.6Vz = ~1V across RL.

So, 1/0.022 = ~45R

EDIT
Note:
1. you have to ensure that zener current is always greater than zero at the lowest level of the ripple, ie: the zener must always be conducting else the voltage regulation will fail.

2. at the maximum level of the ripple the zener current current must not be such that the wattage dissipation of the zener is not exceeded.
 
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ericgibbs

Joined Jan 29, 2010
18,766
In that case I would suggest a 'generic' zener current of 5mA at the lowest Vcap ripple voltage.

It would be usual to select a zener diode from datasheets that met the specification for the project.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,755
So you think that even with this intel from teacher about those 3 currents, i can't calculate the R_s????

I have the following formulas given by the teacher:

\(I_{R_{S}} = \frac{V_{in} - V_{Z}}{R_{S}}\)

\(I_{R_{L}} = \frac{V_{Z}}{R_{L}}\)

\(I_{Z} = I_{R_{L}} + 10% \) of \(I_{R_{L}}\)

and also

\(I_{R_{S}} = I_{Z} + I_{R_{L}}\)
 

ericgibbs

Joined Jan 29, 2010
18,766
So you think that even with this intel from teacher about those 3 currents, i can't calculate the R_s????

I have the following formulas given by the teacher:

\(I_{R_{S}} = \frac{V_{in} - V_{Z}}{R_{S}}\)

\(I_{R_{L}} = \frac{V_{Z}}{R_{L}}\)

\(I_{Z} = I_{R_{L}} + 10% \) of \(I_{R_{L}}\)

and also

\(I_{R_{S}} = I_{Z} + I_{R_{L}}\)
hi,
You have enough information to do the calculations.

3rd Equation on your list.

Vz = 5.6v RL = 330R ,so work out IRL

You know I RL and you are given that Iz = 10% of IRL

Knowing the combined current of RL and Iz and that Vz= 5.6V and the Vripple min = ~6.5V, calc Rs

Calc and post your result.

E
 
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