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FreeTrack Forum > FreeTrack : English Forum > Support : Tracking System > How to install HW for 4.5V 800mA DC power supply?

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RSS >  How to install HW for 4.5V 800mA DC power supply?
4ever #1 20/01/2012 - 18h30

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Hi,
I have adapter with output DC 800mA, 4.5V . I want to ask you how to connect 4 IR leds. I would prefer 20mA current for IR led, because I read here, that it should be enough:
http://www.free-track.net/english/hardware/filter_removal/labtec_webcam_pro.php

Actually I'm not sure what leds to buy, but I would prefer 20mA per led. I don't now what max voltage they should have. So should I buy some (and how many Ohms) resistor to give it parallel with leds - to reduce current to 80mA for all leds?

Do I calculate right? 4,5 / 720 / 1000 = 6,25 Ohm (seems little). And what Wattage for the resistor not to be burned? Should I reduce the voltage also?
Blindasabat #2 20/01/2012 - 19h35

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Hi,

Why would you want to make a 4 point model?
If you make a 3 point serial, 15 Ohms will be enough.
4ever #3 20/01/2012 - 21h20

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I thought 4 point model is better.However if it it not, then I will use 3 point model. for 3x20mA. So should I use the resistor in parallel connection to leds?

meaning



+ ------|
        |
 4,5V   |
        +----+----+-----+
        |    |    |     |
       LED  LED   LED   R 15??? Ohm
        |    |    |     |   4,5V
    20  |20  |20  |     |
    mA  |    |mA  | mA  |  760 mA?
        +----+----+-----+
        |
        |
        |  800 mA
        |
        |
- ------



I am not sure if I am right with my assumption: I think maybe I am wrong.
If there are 4 branches, will the 800mA be devided to 200 mA to every branch? Then I should measure resistance of the leds and add resistance of the leds and to add there resistor to every branche. So I calculate 23,5 ohm for every branch that is R1 = 23,5 - LED resistance.



+ ------|
        |
 4,5V   |
        +----+-----+----+
        |    |     |    |
       LED  LED  LED    |
        |    |     |    |  4,5V
       R1    R2   R3    R4
        |    |     |    |
     20 |20  |20   |    |
     mA |mA  |mA   |    |20 mA
        +----+-----+----+
        |
        |
        |  800 mA
        |
        |
- ------



If I would make 3 point serial model, so why should I use 3 resistors? That has no sense. And If I calculate for serial, I have Rx=4,7/(20/1000) = 235 ohm. And Px=235*(20/1000)=4,7W So I don't know, do I calculate it incorrect?  Am I right or not?

The reason why I am not sure to use serial model is that many years ago when I tested to connect 2 leds serial, they did not shine. Only if I did connect them to parallel. So I am not sure if serial connection should work. Did not ever see it but many years ago I tested parallel connections and were OK. Idk if has it something common with properties semi-conductor devices. Maybe somebody could explain.
Edited by 4ever on 20/01/2012 at 21h33.
dewey1 #4 21/01/2012 - 00h23

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You have a few things mixed up.
800mA of the power pack is the maximum current that it can supply.
In your second diagram, only R1,R2 and R3 are required. Remove R4 completely.
This diagram you show is for a parallel setup.

You need to decide what LEDs you plan to use.

If you use Osram SH485P IR LEDs then you need a resistor wired in series with each LED and here is the math:

4.5 VDC is the supply voltage.
1.5 VDC is the LED voltage.
4.5-1.5=3.0 This is the required voltage across each resistor.

3.0/.020=150 Ohm. So for 20mA per LED, R1=R2=R3=150 Ohm.

3.0/.030=100 Ohm. So for 30mA per LED. R1=R2=R3=100 Ohm.

I suggest 5% tolerance 1/2 Watt resistors.

Do you have a voltmeter to check what the voltage of the power pack actually is?
This will determine if you can do a series LED circuit with only one resistor.
Blindasabat #5 21/01/2012 - 00h32

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Build a 3 point model, 4 point is obsolete (but still supported) with the newer Freetrack versions.

I have made two 3 point models myself, both serial with SFH485P leds. I use 4.5V 300mA powersupply and 10-15Ohm resistor. The 800mA is not important, it's just the maximum your psu will put out if it got no resistance.

You can build parallel if you want, but it requires more wiring and you need more resistance, most people use  parallel cause they use batteries with limited voltage (3V). That your serial circuit didn't work had perhaps to do with that you connected a led the wrong way.
4ever #6 21/01/2012 - 09h18

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Thank you for explanatiion.

So I will use this scheme

paralel A)
+ ---|
       |
4,5V |
       +--+---+
       |    |     |
      LED  LED  LED 1,5V => 0,3W per 1 LED
       |    |     |
      R1    R2   R3   R1,R2,R3=150 Ohm; 3V; 0,46W per Resistor
       |    |     |
   20 |20 |20  |
  mA |mA|mA |
       +---+---+
       |
       |
       |  60 mA
       |
       |
- ------

paralel b)

paralel A)
+ ---|
       |
4,5V |
       +--+---+
       |    |     |
      LED  LED  LED 1,5V => 0,3W per 1 LED
       |    |     |
      R1    R2   R3   R1,R2,R3=100 Ohm; 3V; 0.3W
       |    |     |
   20 |20 |20 |
  mA |mA|mA|
       +--+---+
       |
       |
       |  60 mA
       |
       |
- ------


Or serial

+ ---|
       |
4,5V |
       +
       |
      LED1 = 1,5V
       |
       +
       +
       |
      LED2 = 1,5V
       |
       +
       +
       |
      LED3 = 1,5V
       |
      Rx = ?
       |
       +
       |
       |  20 mA
       |
       |
- ------

Do I really need the resistor for serial connection? I don't think so coy I have 3x1,5 V on Leds.
Edited by 4ever on 21/01/2012 at 09h23.
dewey1 #7 21/01/2012 - 16h45

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Yes you will need a resistor for the series circuit. Use 10 to 15 Ohm as the smallest value.

Remember the Vf of 1.5 VDC is the maximum Voltage Forward (Vf).
You could have a range of Vf from 1.2 VDC to 1.5 VDC. (basically internal resistance)
The LEDs will have tolerance range of Vf usually at test current of 20mA.

There is Vf minimum, Vf typical and Vf maximum. Osram only specifies Vf maximum.

What I have seen is that about 80% of Vf maximum is the Vf typical.
So .80 x 1.5=1.2 VDC.

4.5-3.6= .9 VDC drop required at .060 is 15 Ohm.

This is all assuming that the 4.5 VDC is regulated.
That is why I asked to actually please measure it!!!!!
4ever #8 22/01/2012 - 13h42

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Thank you.

I also did a calculation (serial connection):

   V    I [mA]    R [ohm]        
   1,20    20    60        
   1,20    20    60        
   1,20    20    60        
   0,90    20    45        
SUM    4,50    20    225    Check:    225



   V    I [mA]    R [ohm]        
   1,20    20    =1,2/(20/1000)
   1,20    20    =1,2/(20/1000)
   1,20    20    =1,2/(20/1000)
   0,90    20    =0,9/(20/1000)
SUM    4,50    20    =SUMA(D2:D5)    Check:    =4,5/(20/1000)



But according my calculations, it is 45 ohms. I still don't know why my calculation is not same result. I don't understand when you say "drop required at .060 " - probably "drop" is some elektro-expression?

I will ask the vendor about the voltage because I am not sure if they have exactly these led. What do you mean by "regulated" power supply?
Edited by 4ever on 22/01/2012 at 14h06.
dewey1 #9 22/01/2012 - 14h05

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Some power packs will state for example: 6 VDC at 500 mA.
But in reality the output voltage might be 7 or 8 VDC at no load or very low load.

Regulated means the voltage is relatively constant over a range of different current useage. Example is a 5 VDC USB power pack. The output voltage is typically 5 VDC with a 5% tolerance. (4.75 to 5.25 VDC)

So who is the vendor and where are you located?

Try to get the SH485P IR LEDs. It will save you a lot of time and frustration of "tinkering" or messing around with filing, sanding, and polishing LEDs with the standard bult in lens.

Delete row 5 from your spreadsheet calculations. That was for the 4th LED
Edited by dewey1 on 22/01/2012 at 14h21.
4ever #10 22/01/2012 - 14h13

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I am looking for how to write a personal message to you.

I can buy the devices in local shop, not e-shop.

The power supply is from old Disc-Man that I didn't used for years.

I don't understand when you say "drop required at .060 " - probably "drop" is some elektro-expression?
Edited by 4ever on 22/01/2012 at 14h22.
4ever #11 22/01/2012 - 14h24

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4ever @ 22/01/2012 - 14h13 a dit:

I am looking for how to write a personal message to you.

I can buy the devices in local shop, not e-shop.

The power supply is from old Disc-Man that I didn't used for years.

I don't understand when you say "drop required at .060 " - probably "drop" is some elektro-expression?



4th row is the resistor. 0.9V as you said. In this point our numbers are correct.

You say "drop required at .060" .. Do you mean 0.060 ohm resistance of Led? If so then in this point we are correct.
Edited by 4ever on 22/01/2012 at 14h28.
dewey1 #12 22/01/2012 - 14h31

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Drop is referring voltage drop across a resistor at some current. (.06 A or 60 mA in this example)

Is there a brand name and model or part number on the power pack?
4ever #13 22/01/2012 - 14h38

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Power pack no. RFEA401S
Matsushita Electric Industrial Co ltd
Edited by 4ever on 22/01/2012 at 14h39.
4ever #14 22/01/2012 - 14h44

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4ever @ 22/01/2012 - 14h38 a dit:

Power pack no. RFEA401S
Matsushita Electric Industrial Co ltd



Yep, so I calculated exactly the same number but for the resistor 45 Ohm.
Edited by 4ever on 22/01/2012 at 14h44.
dewey1 #15 22/01/2012 - 15h14

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Well it is useless doing any calculations until you specify a part number and brand of IR LEDs that you can get.

Do they have a selection of IR LEDs to choose from?

This pack?

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