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Topic: Power Booster capacitor question (Read 934 times) previous topic - next topic

T_Lex

Pardon the interruption.  I had hoped that my first post would be a bit more dramatic, but I have a question related to this Playground project:
http://www.arduino.cc/playground/Main/RegulatedPositiveVoltageBooster

It states that C1 is a "250v, .1 uF min, 47 uF max" capacitor.

However, new as I am to this hobby, I am unable to find a part number that correlates with those specs - although I could easily be misreading different ways to describe the same thing.

Can someone provide a Radio Shack or other part number for such a capacitor?

(As an aside, there is likely to be a Project Guidance post related to this in the relatively near future, too.)

TIA
T

retrolefty

#1
Mar 01, 2012, 04:28 pm Last Edit: Mar 01, 2012, 04:31 pm by retrolefty Reason: 1
Quote
It states that C1 is a "250v, .1 uF min, 47 uF max" capacitor.

However, new as I am to this hobby, I am unable to find a part number that correlates with those specs - although I could easily be misreading different ways to describe the same thing.


That is because it isn't specifing a specific capacitor but rather the general specifications of any capacitor that would work in that circuit. So:

Capacitor should have a maximum working voltage rating of 250 volts or higher.
Capacitor should have a capacitance rating of any value between .1ufd and 47ufd.

You should have little problem finding a capacitor that would fit those specifications.

Here is a radio shack on-line cap that will fit those specs: http://www.radioshack.com/product/index.jsp?productId=2102515

Lefty

T_Lex

Excellent - thank you!

I did look at that, but what threw me off was the .1uF in the project description and the 1.0 uF in the capacitor description. :)

I see now that it wasn't a "number" issue, so much as a "reading comprehension" issue.  The project writer meant ANY capacitor within that range, but I read that too literally to mean a capacitor with those tolerances. :)

Interestingly, I did pick up on his description of a "small" diode as meaning something like this:
http://www.radioshack.com/product/index.jsp?productId=2036270

Thank you again for your help!
T

DVDdoug

#3
Mar 01, 2012, 09:46 pm Last Edit: Mar 01, 2012, 09:56 pm by DVDdoug Reason: 1
Quote
Interestingly, I did pick up on his description of a "small" diode as meaning something like this:
http://www.radioshack.com/product/index.jsp?productId=2036270
Perfect!

The 1N400x diodes are probably the most popular diodes around.  They are all rated at 1 Amp. (The high resistance values in your circuit will keep the current far below 1A.) The 1N4001 has a PRV (peak reverse voltage) rating of 50V, the 1N4004 is rated for 400V, and the 1N4007 is rated for 1000V.

I usually buy the 1N4007, since they are all very cheap* and about the same price.    But, any of the 1N400x dioes are perfectly acceptable for things that run at the common low-voltages of 5V, 12V, 15V, etc.

"Regular" silicon diodes have two ratings, reverse voltage and forward current.  These are maximum ratings that should not be exceeded.

In the forward direction, the diode "turns on" (starts conducting) at about 1/2 volt.  It's a non-linear device, so unlike a resistor, the voltage does not go over about 0.7V no matter how much current you run through it.

in the reverse direction, (essentially) no current flows (assuming you don't exceed the PRV rating ;) ).   

Since the diode never "sees" high-voltage and high-current at the same time, it doesn't have to dissipate much power.

If you exceed the reverse voltage it can break-down  and start conducting.   In that situation, you can get voltage and current at the same time,and you can fry the diode.  And of course, you can fry the diode if you exceed the forward current rating. 

P.S.
*
  I hope you get 5 or 10 in the package for $1.19 USD at Radio Shack.  They are 6 cents each at  Jameco.

T_Lex

#4
Mar 02, 2012, 03:36 am Last Edit: Mar 02, 2012, 09:03 pm by T_Lex Reason: 1
Ah - thank you for that input, too. :)

In fact, I have another question.  I could not find a 1.5M Ohm resistor, and he puts an emphasis on the 15-1 ratio.  Can I use a 1M Ohm, in line with 5 100k Ohm?  (Probably a silly question, but I'm new at this....)

Edit:
Looked at the Sketch, which says:
Code: [Select]

unsigned int SetVoltage(double voltage) {
 if (voltage>60) {
   return 0; // do not exceed 60V
 }
 
 // This function converts the voltage value (0 to 5V reading; 0 to 60V actual), into a value for an analog input (from 0 to 1024)
 // If your resistor voltage divider is not perfectly 15:1 you'll need to change the formula below
 
 // with ideal resistors and a perfect 5V Analog Reference voltage (AREF), an analog input value of 13.65 would be 1V
 
 return (voltage*12.8);
 // ideal ADC value, per volt     = voltage*(1024/5)/(15/1) = 13.65
 // the real value for my circuit = voltage*(204.8)/(16.5)  = 12.41
 
 // 16.5 was my measured resistor voltage divier ratio.
 // I increased my calculated 12.41 constant to 12.8, by trial and error, in order to get closer to the target voltage
 // My results were +/-0.2V, at a 0.1mA (100Kohm) load when powered by the Arduino.
}


So, since 60v is well beyond what I was going to try for anyway, I think I can do this:
- use a 1M Ohm resistor instead of 1.5M
- change the conditional in the above function to 50 instead of 60 - if the v is greater than 50, return 0
- change the formula to... well.... I'm not sure.  The math doesn't make sense to me.  I get the voltage divider bit (googled it).  But, it seems like a 10:1 ratio would actually make the math easier!

Any help?

Edit2:
Oh... wait.  Let me guess.  The multiplier should be 20.48, which is (1024/5)/(10/1).  Each volt "read" will equal 20.48 volts actual.  It also looks like a bit of trial and error with the multi-meter will be necessary to tweak the multiplier.  Not a big deal, as my voltage does not need to be that precise.

T_Lex

Apologies for the self-reply, I hope that isn't a breach of the forum netiquette.  And, perhaps his is leaking more towards Project Guidance...

Anyway, after manhandling the 2n3904 too much, to the point the base pin broke off :doh: and while I wait to get another one, I noticed that the instructions suggest a MOSFET might also be appropriate in this application.  I have one of those.  But, it isn't clear to me how it would be wired in.

Quote
If you're trying to get more than one watt, then use a darlington pair. With BJT power transistors, the most you'll get is about 4 watts, efficiently. With N-Channel power MOSFETs, it's possible to get 25 watts at about 70% efficiency with a good power inductor (it's very difficult).


Is it possible to replace the 3904 with the 510 MOSFET, or am I misreading the suggestion?

Again, thanks.

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