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Topic: Voltage Regulators (LM7805) & Capacitors (Read 7 times) previous topic - next topic

baylorbear

Greetings all --

I've spent the last several hours researching something I don't understand but have failed to figure it out on my own -- so I thought I would ask the experts :)

I've been tinkering with a RFID Door Lock project I found at instructables.com -- but I absolutely do not understand how the values for the capacitors on either side of the voltage regulator in the schematics were selected. 

I've looked at datasheets for both Fairchild and National for the 78XX v-reg's but found that the size of capacitors used in their examples were much smaller than those used in the schematic for the project I've been looking at.

The forums I have sifted through have provided some crazy-intense explanations of how to calculate for the size of a capacitor used with a v-reg, but I'm hoping someone can spell it all out in a much clearer fashion so I can grasp the concepts moving forward.

While I'm on the subject though, ... in the example project I noted above, the author is using a 12vdc Electric Strike Plate for the door lock mechanism.  According to his schematic, he's feeding 15vdc to the electric strike and then to the 7805 v-reg to supply power to the arduino.  Why would you not ALSO throw in a 7812 v-reg before supplying power to the lock mechanism?  And if you were to do so... it begs the question... what size capacitor would you use on the input side of the 7812?

Thanks for your help! :)

CrossRoads

The relay/solenoid can be pretty tolerant of voltage, just have to control the current thru it really to activate it.
If the power in to the 7805 is decent DC, you don't need too big of  a cap.
If its really dirty AC and you are drawing lots of current, then need bigger caps to smooth the AC transitions out or you will see ripple on the 5V side.
If you're just powering digital circuits, you might find that acceptable.
If you have some audio going, then you might not - it will be audible as 60 Hz, 120Hz and maybe 180Hz hum depending on your rectification of the AC.
Designing & building electrical circuits for over 25 years. Check out the ATMega1284P based Bobuino and other '328P & '1284P creations & offerings at  www.crossroadsfencing.com/BobuinoRev17.
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James C4S

Quote
The forums I have sifted through have provided some crazy-intense explanations of how to calculate for the size of a capacitor used with a v-reg, but I'm hoping someone can spell it all out in a much clearer fashion so I can grasp the concepts moving forward.


Without explaining what you don't understand, this thread is probably going to end up in the same direction.

Whenever current is drawn from a regulator, the voltage will start to drop.  There is a certain amount of time it takes before the regulator notices and increases voltage to compensate.  This is known as ripple.  The amount of ripple will be larger if more current is drawn from the regulator.  To compensate, a capacitor provides a temporary bank of energy.  This helps keep the voltage stable resulting in less ripple.

So when picking values of capacitors you need to know the maximum amount of current load and expected amount of ripple. Principle applies to both the input and output side.  The difference is that the voltage ripple will be different.

Keep in mind most designs will calculate the minimum caps need and then select something larger.
Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

baylorbear

Thank you both for your replies!  I will try to be more specific in my questions but I am not beyond doing a little homework, so if you have any specific questions about the hardware I'm dealing with please ask!

With the circuit from the instructables site set up on my breadboard I was able to see that at peak current the circuit was drawing just under half an amp.

Now, judging from the data sheets on the v-regs noticeably smaller selections of smoothing capacitors, I presumed that this was the "default" for the LM7812 under a constant load and at the minimum voltage required to keep the current flowing through the regulator. 

What I don't understand is the relationship between the "default" capacitors in the data sheet with those selected by the original author of the schematics for RFID door lock circuit.  I realize the reason for the larger capacitors has to do with the load placed on the regulator at certain points in time, but it doesn't appear to be a linear relationship (load doubles leads to capacitor size doubled)

What got me thinking about this was the fact that the original schematic doesn't have a 7812 v-reg in circuit... Only a 7805 for the arduino portion of the circuit.  So if I wanted to add that missing 7812, I would need to figure out / calculate the correct smoothing capacitors to the circuit along with the new 7812.

Again, thank you for your response... It is truly appreciated!

James C4S

Well the only capacitors I see specified in the instructable (I can't seem to open the schematic without paying, so I'm not going to bother) are 22pF.  This doesn't sound like a value that was calculated, but just picked.

Typically you'll see 33µF or 47µF on the input and 0.1µF on the output.  Spend some time doing the math and you'll see these are usually large enough, or even overkill, for more applications.
Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

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