I think I asked you some posts ago to explain me again that matter about the COM because I haven't understood it.

I can't also understand what happends to the Source Voltage! Also the resistor we should simulate is of 330 ohms, so How can I calculate C through that formula that Jony130 said?

hi,
I posted a clip from LTS Help in my post #56, it a Node which you can use as a reference in order to to measure using the two probes method.

You know the Vpeak and Vrms of the V1 source.

Calculate [Vpeak - the two diode voltage drops] thats the V for Jony's Eq:

Knowing the Vpk, you can calculate the Ipeak thru the 330R

You also know the required ripple voltage percentage.

So thats C in Farads = Vpk * 0.01/ %Vrip

I will not give you the answer, so calc what you think it is and let me know.

Use LTS to check your calculations..
E

 

There is still some terms that confuses me.

V rms is the same as V effective? Like Vp/sqrt (2)??

If it is as I say my Vpk should be 6 - 2*0.7, correct?
V rms should be 6/aqrt(2), correct? (What for do I need this value?)

 

There is still some terms that confuses me.

V rms is the same as V effective? Like Vp/sqrt (2)??

If it is as I say my Vpk should be 6 - 2*0.7, correct?
V rms should be 6/aqrt(2), correct? (What for do I need this value?)

Consider sqrt (2) = 1.1414 so 1/1.414 = 0.707

Use 0.707 * Vpk to give Vrms.

You can use LTS to show the RMS value of a peak waveform. [ also the Average]

Place your cursor over the Vout label at the top of the plot
press down the Ctrl key on the keyboard, left click the mouse.

 

Morning... Sorry, I couldn't come back here yesterday!

I am a bit confused.

Vpk(V1) = 6V
Vrms(V1) = 6/sqrt(2) = 4.24V

So, I think the voltage I should work with should be the Vrms from V1, as the voltage source we are using, is simulating a electrical transformer (don't know the name in english).

then in your post #81 you mention to calculate the Vpk - 2*Diode Vdrop:

6-2*0.7 = 4.6V --> is this the V in Jony130 equation?

So what do I need the Vrms for?

 

OK, A question for you.

What voltage will the smoothing capacitor charge up to on the peak/crest of the full wave rectified sine wave.???

EDIT:

There is still some terms that confuses me.
V rms is the same as V effective? Like Vp/sqrt (2)??

You asked me about the Vrms, thats why I posted the formula for you.

 

Well, I'm not quite sure...

I don't know if it's (6-2*Diode Vdrop) or (6/sqrt(2) - 2*Diode Vdrop) or if any of these!

 

Well, I'm not quite sure...

I don't know if it's (6-2*Diode Vdrop) or (6/sqrt(2) - 2*Diode Vdrop) or if any of these!

OK, I have some free time this morning, so if you want to discuss the project, please do so.

It must be [Vpeak - 2*Vdiode].

Look at you LTS sims you will see that peak of the ripple voltage on the capacitor is is equal to [Vpeak -2*Vd].

So that will drive Ipeak = [Vpk-2*Vd]/Load resistor, this will then discharge the smoothing capacitor to a voltage that equals the rising edge of the next half cycle from the bridge.
This is the lowest level of the ripple voltage.

 

hi,
Look at this image from LTS, I have chosen a low value smoothing cap of 100uF in order to show more ripple.

E

EDIT:
I have added 2nd image, it shows the Absolute value plot of Vout, COM.

 

Ok, let me look to your explanations for a while and also look the the plots...

I'll be right back to you!

Thanks in advance!

 

Let me just use Vdvd for Diode Voltage Drop.

Ok, some questions now:

What would be the name for the result of [Vpk - 2*Vdvd]??? (I'm kinda lost)!
I can't figure out how to calculate Ic because I don't have enough info because I'm looking for the capacitor value!

 

Let me just use Vdvd for Diode Voltage Drop.

Ok, some questions now:

What would be the name for the result of [Vpk - 2*Vdvd]??? (I'm kinda lost)!
I can't figure out how to calculate Ic because I don't have enough info because I'm looking for the capacitor value!

Most would call [Vpk-2*Vd] = Vcap or Vpeak DC.

Lets assume that Vpeak on the Cap is +4.5V, so with a 330R resistive load the Ipeak would be Vcap/330R = 13.6mA ~14mA

The voltage on the Cap will fall exponentially.

As the voltage on the Cap falls due to being discharged by the 330R so will the current thru the 330R.

Using the Equation C= [I*t]/VRipple [ assume ripple is 0.45V ie10%]

Thats C in Farads = [0.014* 0.01]/0.45 = 304uFarad

Run your sim with these image values to check your calculations.

 

Some questions about your answer:

1 - That Vpk DC or Vcap is the voltage that reaches capacitor and also resistor, correct?

2 - Are you assuming that the current that reaches the resistor Vpk/330 is the same that reaches the capacitor? If so, why do you do it?

3 - Why do you use 0.01 in the formula?
Tank you for replying!

 

I got this in he meantime!

Why is the current at D2 so small?

Why do we have a peak in diodes current at the first semi-cycle and after that, both currents, D1 and D2, are the same?

Why do we have a few moments of time without any current in diodes?

 

Some questions about your answer:

1 - That Vpk DC or Vcap is the voltage that reaches capacitor and also resistor, correct?

2 - Are you assuming that the current that reaches the resistor Vpk/330 is the same that reaches the capacitor? If so, why do you do it?

3 - Why do you use 0.01 in the formula?
Tank you for replying!

hi,
1. Yes.

2. The diode current that charges the capacitor on the peak of Vin is a high current pulse over shorter time period.
The current out of the capacitor is the same as current thru the 330R

3. Because the mains frequency is 50HZ which has a Period of 1/50 = 20mSec, BUT dont forget its full wave rectified so there is a pulse at 100Hz
ie: 10mSec [ its the t in the formula]

 

I got this in he meantime!

Why is the current at D2 so small?

Why do we have a peak in diodes current at the first semi-cycle and after that, both currents, D1 and D2, are the same?

Why do we have a few moments of time without any current in diodes?

hi,
Initially the capacitor Voltage is zero at switch ON. so the first half cycle charges the cap from zero.! so the current is high on that first half cycle.

After that both D1 and D2 have to charge the cap on the crest/peak of the half cycle, so the current looks like a pulse. NOTE: the value of the current on those Ipeaks compared to the current thru the 330R

The diodes will stop conducting as soon as the cap is charged OR when the voltage on the cap is greater than VPeak - 1.4V

 

Jeeezzzz, so many info to acknowledge... xD

Thanks for the info!

I need to understand better the current that goes thru the capacitor and at the resistor, because usually the current that reaches a parallel impedance, divides itself for each branch of that parallel! So its not very clear to me that that current is the same!

 

Jeeezzzz, so many info to acknowledge... xD

Thanks for the info!
I need to understand better the current that goes thru the capacitor and the resistor!

The current thru the 330R is Vcap/330R, use LTS to plot the 330R current.

I have simplified the LTS to show the effect of just one half wave cycle charging the cap to Vpeak and the resultant exponential decay of Vcap due to the current discharging the cap via a 333R.

EDIT:
I have chosen 333uF and 333R for a reason, can you figure out why.??

 

Morning...
Yesterday night at my classes I asked an old teacher of mine to explain me some basic concepts about circuits and about current and voltage flow thru components.

Now I understand why we are using (Vpk - 2*Vdvd)/R_L to calculate Ipeak or IR_L.

Regarding your question, I think you chosen the same numeric values to cap and res so that the curve of cap current discharge would be precisely the same as the voltage curve at the R_L.

 

Morning,
Thats good news.

Its important to know that the capacitor will discharge exponentially from its peak value. The short section of the discharge curve we use to measure the ripple, we consider as linear.
Look thru this link for the actual discharge formula.

http://www.electronics-tutorials.ws/rc/rc_2.html

A practical point is that the actual capacitance value of electrolytic capacitors can have a very wide tolerance, some as high as +/-40%.

So an actual circuit may not perform as plotted in simulation.

Whats the next step in your simulations.?

 

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...

After answering to that question, the next step will be a lot more complicated to me... I'll come to it later! Maybe tomorrow because today I have 2 other reports to complete and to send to my teacher (it's from another calss, not Electronics)!