# How to use voltage regulator?

Discussion in 'Homework Help' started by Michael George, Mar 2, 2016.

1. ### Michael George Thread Starter Member

Feb 8, 2015
52
2
Hello,
I have some questions about using voltage regulator. Would you give me a hand, Please?

If I removed C1, It will become an open circuit. Is that good or bad? I think its better to remove it because it changes the value of C2 and The voltage regulator requires C2 not C1.

Second, Would you tell me how to calculate the values of C2 and C3, Please?

I will use LM7805CV

Thank you very much,

2. ### tracecom AAC Fanatic!

Apr 16, 2010
3,869
1,393
C1 is the main filter capacitor; if you remove it the DC will pulse. The values for C2 and C3 are wrong and the values aren't calculated, but are taken from the datasheet for the LM7805. C2 should be .33uF, C3 should be .1uF, and C4 should be added: 10uF.

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3. ### Jaytronic New Member

Mar 2, 2016
6
1
Hello,

Unless you are doing something different I would stick with the typical application circuits provided in the datasheet. Most linear powers require a cap (C3) for smoothing out the full wave voltage from the diode bridge, C2 is there for additional filtering, C1 is for filtering/smoothing the output voltage.

There is no reason that I can think of that requires any of those caps to be removed. If you want a clean 5VDC then leave them there. As for calculating the values for the caps, it really isn't necessary to do so, just stick with the typical application circuit.

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4. ### AnalogKid Distinguished Member

Aug 1, 2013
4,542
1,251
1. Bad. Do not remove C1.

2. Incorrect. The voltage regulator requires a DC input that *always* is greater than approximately 7 V. This is stated clearly on the datasheet. The value of C1 necessary to make that happens depends on the transformer and the load on the regulator output. Also, the voltage regulator requires an energy source with a very low AC impedance. Both a large bulk filter capacitor and a smaller, high frequency bypass capacitor combine to make that happen.

ak

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5. ### HitEmTrue Member

Jan 25, 2016
32
9
Not meaning to rain on the DIY parade, but you can get a dual supply already built for pretty cheap. You'd still have to mess with connecting the mains to the transformer, but overall there is less that can go wrong (translate: safer).

Like:
https://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_212338_-1

And of course there is the wall-wart to voltage regulator option. Could be put in a box.

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6. ### Michael George Thread Starter Member

Feb 8, 2015
52
2

If I did not remove C1, I think the capacitance at the input of the voltage regulator would be 470.33 uF and datasheet says we need only 0.33 uF. Is that ok?

This website says: A large reservoir capacitor is not necessary, as the regulating action of the I.C. will reduce the amplitude of any AC ripple (within its maximum input voltage range) to just a few millivolts at the output.
This is their schematic:

Is it a good design? or the first one is better?

Thank you,

Last edited: Mar 2, 2016
7. ### Michael George Thread Starter Member

Feb 8, 2015
52
2
Hello HitEm True,
This is a project for my collage. I'm a student at the faculty of engineering (electrical engineering). I can not buy one I have to design and build the project by myself. Also, I would like to learn. Anyway, Thank you for your answer.

8. ### HitEmTrue Member

Jan 25, 2016
32
9
Ok. Well, what else do you see different between the two designs you've posted, besides the removal of C1?

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9. ### Evanguy Member

Dec 21, 2014
81
8
You should use the drawing provided by the chip's maker. its in the data sheet. you wil not improve it by removing parts. they are the minumin parts needed to have it function as intended. why do you so eagerly want to change it with no understanding of why its here and what its doing? And why do you trust that site you linked more then the data sheet. the makers of the regualtor have done more testing the circuit then probably anyone on earth.

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10. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,393
497
The manufacturer provided you with a RECOMMENDED design in the datasheet. You can ignore the manufacturer, like the people from the website you keep quoting. You, the user, get to decide if you will use manufacturer recommended design or design from some people on the internetz whose engineering credentials you don't know.

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11. ### WBahn Moderator

Mar 31, 2012
17,755
4,799
As others have said, how you use a voltage regulator is in conformance with the manufacturer's data sheet. Now, you can go charting your own path (and sometimes that the only way to do it), but you will be sailing in uncharted waters and there be demons off the edge of the map.

The presence of C1 does not change the value of C2. Why there are two capacitors is that each one deals with a different part of the noise. C1 softens the output of the bridge so that you have a mostly-DC signal with some ripple on it. The size of C1 is dictated by the current draw and how much ripple voltage you can tolerate. C2 is intended to feed the higher frequency signals that larger caps can't handle because of their higher effective resistance.

The values of C2 and C3 (and often other recommended components) are contained in the data sheet (usually).

Draw a vertical line between C1 and C2. Everything to the left of the line is in a black box labeled, "Unregulated Power Supply." You can't see C1, so don't worry about it when thinking of the stuff to the right of the line that is in a black box labeled, "Voltage Regulator." Now, having said that, the claim that you "can't see C1" isn't really true, but for this discussion it's close enough.

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12. ### tracecom AAC Fanatic!

Apr 16, 2010
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Asked and answered multiple times. Do it however you want. And it's college, not "collage."

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13. ### AnalogKid Distinguished Member

Aug 1, 2013
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1,251
No.

No. That is a gross overstatement that is misleading. It is correct technically, but only if you read ALL of the words and understand what they mean. You don't. You highlighted the wrong phrase.

The 7805 applications circuits assume that the input is pure DC with no ripple. No ripple. Zero ripple. To get close to that the output of an AC/DC power converter must have a large filter capacitor. The datasheet is describing the chip, not what drives the chip. If you keep asking the same question you will keep getting the same answer.

ak

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14. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
Michael George,

Consider these attachments in light of everything people have told you.

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• ###### compare-02.png
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15. ### Michael George Thread Starter Member

Feb 8, 2015
52
2
Thank you all for answers and giving me your time. I appreciate that and I'm sorry about repeating the question.