Full wave rectifier with center-tap transformer and output stage of Filter and Regulator

Thread Starter

huntersmile

Joined Jun 10, 2017
34
Hello everyone,

I am in need of help with this simple power supply circuit. (Please see my attachment)
1) I don't understand how to find the R value.
2) Basically, the turns ratio is also not told, then how do we find the secondary voltage?
3) Does the Rs resistor at the primary side of transformer mean anything in the calculation?
4) What is that Risolation doing? How do I bring it into my equations?

Would some people be so kind as to share some thoughts please?
I am really stuck, I could only start by calculating Vm=Vrms*sqrt(2).
The 5.1V is assumed to be the DC value of the filter (Capacitor) also?
If yes then with maximum ripple factor=5%, we get Vripple=0.255V. (Vripple/Vdc = ripple factor)
Then I don't know how to continue.

The information on Vdc: +/- 5% is useful?

Thank you very much for your help,
I am really in need of them.
 

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#12

Joined Nov 30, 2010
18,076
This is not simple.:(
1) I don't understand how to find the R value.
Read the zener chart to find the minimum current required to maintain the zener voltage. Read the specifications to find the current maximum which might be required by the load.
2) Basically, the turns ratio is also not told, then how do we find the secondary voltage?
Decide what rectified DC voltage you need and declare the turns ratio yourself. I don't know if you need to use a transformer you can purchase, or you're designing for a factory with your own transformer shop, like I did. Just know there is a difference caused by who you're designing this for.
3) Does the Rs resistor at the primary side of transformer mean anything in the calculation?
Not much, but you have to know the alleged turns ratio, and therefore the current ratio, to find out how much Rs interferes with the primary current. Then you decide if the turns ratio needs to be altered to compensate for Rs.
4) What is that Risolation doing? How do I bring it into my equations?
Is it part of the load?
The 5.1V is assumed to be the DC value of the filter (Capacitor) also?
Is the capacitor connected to the zener, or is it connected to a different voltage?

It's all a great, fussy, balancing act where you never guess everything right the first time. Start running the numbers and you will learn as you go. That's a start. Now I'll go hide because other people on here don't like the way I do homework problems.:eek::(
 

Thread Starter

huntersmile

Joined Jun 10, 2017
34
Hellloooooo!
Thanks for your kind reply and guidance!

Let me see, so if I am not digesting your words wrongly,

I should do something like this:
120Vrms-> peak Vm=169.7V.
So let say I choose turns ratio of 10:1, the secondary peak Voltage Vm-s=16.97V.
Then, because this is full-wave rectifier,
 

Thread Starter

huntersmile

Joined Jun 10, 2017
34
This is not simple.:(

Read the zener chart to find the minimum current required to maintain the zener voltage. Read the specifications to find the current maximum which might be required by the load.

Decide what rectified DC voltage you need and declare the turns ratio yourself. I don't know if you need to use a transformer you can purchase, or you're designing for a factory with your own transformer shop, like I did. Just know there is a difference caused by who you're designing this for.

Not much, but you have to know the alleged turns ratio, and therefore the current ratio, to find out how much Rs interferes with the primary current. Then you decide if the turns ratio needs to be altered to compensate for Rs.

Is it part of the load?

Is the capacitor connected to the zener, or is it connected to a different voltage?

It's all a great, fussy, balancing act where you never guess everything right the first time. Start running the numbers and you will learn as you go. That's a start. Now I'll go hide because other people on here don't like the way I do homework problems.:eek::(


Hellloooooo!
Thanks for your kind reply and guidance!

Let me see, so if I am not digesting your words wrongly,

I should do something like this:
120Vrms-> peak Vm=169.7V.
So let say I choose turns ratio of 10:1, the secondary peak Voltage Vm-s=16.97V.
Then, because this is full-wave rectifier, Vrectified-dc=2*Vm-s/pi ==> I get Vrectified-dc=10.8V.
Then, from ripple factor information of 5%, I know ripple factor=Vripple/Vrectified-dc ==> I get Vripple= 0.54V.
Am I right so far?

Then, for full wave, Vripple=Vm-s/(2*f*RC) ==>>> I am supposed to get C from this, but I can't since the filter is not R+C in parallel.
Is this R the Risolation in my question? But is that connected in parallel with C?

Thanks a lot for your kind answer!!!
 

#12

Joined Nov 30, 2010
18,076
That's a way to start. When you find out how many watts the resistor will require, and why it requires that much power, you will change your turns ratio.;)
Like I said, it's a great, fussy, balancing act, and you never get it right the first time.
It is the act of running the math a dozen times that makes you good at this.

The p-p ripple on the cap is: radical 2 C Erip F = I
Yes. I know they didn't teach you p-p ripple. Just use my example and degrade it to what your teacher wants.

Risolation has nothing to do with the power supply design. It's as useless as a static electricity discharge resistor.
 

Thread Starter

huntersmile

Joined Jun 10, 2017
34
That's a way to start. When you find out how many watts the resistor will require, and why it requires that much power, you will change your turns ratio.;)
Like I said, it's a great, fussy, balancing act, and you never get it right the first time.
It is the act of running the math a dozen times that makes you good at this.

The p-p ripple on the cap is: radical 2 C Erip F = I
Yes. I know they didn't teach you p-p ripple. Just use my example and degrade it to what your teacher wants.

Risolation has nothing to do with the power supply design. It's as useless as a static electricity discharge resistor.

Hiii #12!
I am sorry, but I don't get the equation. what is 'radical'? is it square root?
Then what is the I here? What current is this?
 

#12

Joined Nov 30, 2010
18,076
Yes, radical2 = sqrt2= 2^1/2
You need to learn some computer languages. They make writing math easier.:p
and I = current...in amps
What current are you designing for?
1 ma or 100 ma? or is it 101 ma?
 

Thread Starter

huntersmile

Joined Jun 10, 2017
34
I can't read calculus.:(
But the diodes don't rectify average voltage. They rectify peak voltage.

Ok I think it is fine, they rectify peak voltage yes, then the average here means the average of the ripple voltage.
So I think I am fine to use that /pi equation.

Now would you please allow me to confirm whether it is radical 2 or just 2?
I found at some websites that it is just 2 : http://www.visionics.a.se/html/curriculum/Experiments/FW Rectifier/Full Wave Rectifier1.html

And the I, since we are in normal load condition, that means I = Izk (of zener) + Iload = 1mA+10.2mA=11.2mA right? 10.2 is from 5.1 divided by 500Ohm.

Then, after get C above, I now can get R from here (use I=11.2mA):
I*R= (Vrectified-dc) - 5.1V = 10.8 - 5.1
R=.....
Correct?

Thank you!
 
Last edited:

#12

Joined Nov 30, 2010
18,076
Read the written instructions in your first image. Max load is 100 ma.

radical 2 is the square root of 2 which is 2 to the 1/2 power which is 1.4142136
That's what I wrote. That's what works to find the p-p ripple of a rectifier into a capacitor.
I found the math in a National Semiconductor book about audio amplifiers. I built the circuit with 1% resistors and capacitors and the p-p output ripple measured (as well as you can see on an oscilloscope) within 1% of the results of the formula. You can convert p-p into average or you can use some other reference.
If you want to use a page I can't read, you will have to figure out what it means without me.
 

Thread Starter

huntersmile

Joined Jun 10, 2017
34
Read the written instructions in your first image. Max load is 100 ma.

radical 2 is the square root of 2 which is 2 to the 1/2 power which is 1.4142136
That's what I wrote. That's what works to find the p-p ripple of a rectifier into a capacitor.
I found the math in a National Semiconductor book about audio amplifiers. I built the circuit with 1% resistors and capacitors and the p-p output ripple measured (as well as you can see on an oscilloscope) within 1% of the results of the formula. You can convert p-p into average or you can use some other reference.
If you want to use a page I can't read, you will have to figure out what it means without me.

Ok then regarding that. I think I was wrong, I separated the fact that we have C filter on the rectifier output.
As you said, Vpeak at the capacitor is 16.97-Vdiode rectifier (=let 0.7v?) = 16.27V.
Then from here how do I proceed to find the Vripple at Capacitor, please?
 

#12

Joined Nov 30, 2010
18,076
we need the information of Vripple first.
But how to get that?
Read the instructions in the first image of your first post. "The maximum, "ripple factor" is 5%".
Average ripple factor? Peak ripple factor? Peak to peak ripple factor? I don't know. I don't know how to do average ripple because transistors don't use average ripple.

V ripple at the capacitor:
1.4142 C Vrip(p-p)F = I
I said that in post #5.
The easy way to remember the formula? "Radical 2 cerf =I"
You want the capacitance? It's just algebra to pull the C out.

The reason I use the p-p value is because a transistor amplifier drops dead when the lowest possible value of Vcc is below what it needs for the output. The transistor in the next semester doesn't care what the average voltage loss is between rectifier peaks, it runs out of voltage at the maximum voltage loss between rectifier peaks on the filter capacitor. Meanwhile, you can't buy filter capacitors every 5 or 10 percent. They go like, 100uf, 220uf, 470uf, 1000uf. I can't do your homework to 5 digit perfection. If you want an answer like 485.63uf, you're going to have to look it up or wait for some other helper.

I just got reminded why I don't belong in the Homework section.:(

But look carefully at your turns ratio. Did you mean 10:1?
or 10:1 for each half of the secondary?
 

Thread Starter

huntersmile

Joined Jun 10, 2017
34
Hiiiiii!
Btw I dont need to replicate this design into breadboard, so it is just calculation only.

Yes it says maximum ripple factor 5%, but this ripple factor is applicable to the filter output also?
By definition ripple factor= Vripplep-p/Vaverage right? In that case I do not know the average. I only know the peak voltage. And they are not /sqrt(2) or whatsoever since they are already filtered. What is the average filter output voltage?
or, Would you please tell me what is the ripple factor equation for the peak voltage?

Yes, I understood the algebra to get C from the next equation. and I have to use I=101mA (100mA max load and 1mA Izk), I guess.
 

Thread Starter

huntersmile

Joined Jun 10, 2017
34
Read the instructions in the first image of your first post. "The maximum, "ripple factor" is 5%".
Average ripple factor? Peak ripple factor? Peak to peak ripple factor? I don't know. I don't know how to do average ripple because transistors don't use average ripple.

V ripple at the capacitor:
1.4142 C Vrip(p-p)F = I
I said that in post #5.
The easy way to remember the formula? "Radical 2 cerf =I"
You want the capacitance? It's just algebra to pull the C out.

The reason I use the p-p value is because a transistor amplifier drops dead when the lowest possible value of Vcc is below what it needs for the output. The transistor in the next semester doesn't care what the average voltage loss is between rectifier peaks, it runs out of voltage at the maximum voltage loss between rectifier peaks on the filter capacitor. Meanwhile, you can't buy filter capacitors every 5 or 10 percent. They go like, 100uf, 220uf, 470uf, 1000uf. I can't do your homework to 5 digit perfection. If you want an answer like 485.63uf, you're going to have to look it up or wait for some other helper.

I just got reminded why I don't belong in the Homework section.:(

But look carefully at your turns ratio. Did you mean 10:1?
or 10:1 for each half of the secondary?
Btw I dont need to replicate this design into breadboard, so it is just calculation only.

Sorry, I mean, ripple factor= Vrms/Vaverage right?

Then the Vrms is it just peak/square root (3) since it is a sawtooth waveform?

What do you mean by: "Average ripple factor? Peak ripple factor? Peak to peak ripple factor? I don't know. "
I understand that you might want me to get to the answer myself, but I am worried that I make mistake.
Hope you can clarify.

Yes, I understood the algebra to get C from the next equation. and I have to use I=101mA (100mA max load and 1mA Izk), I guess.

Thank youuuuuu!!!
 

MrChips

Joined Oct 2, 2009
19,280
btw, it is standard practice to quote AC voltages in RMS value.

Hence mains voltage @ 120VAC RMS going to a step down 10:1 transformer will produce 12VAC RMS.
Then you convert to 12 x 1.414 = 17V amplitude.

Since these are both multiplication you end up with the same result.

Just wanted to let you know.
 
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