Change this circuit to 12 volts

Discussion in 'General Electronics Chat' started by Aurion45, Jul 8, 2015.

  1. Aurion45

    Thread Starter Member

    Aug 7, 2011
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    Hi Everyone,
    I'm thinking of making this circuit for a project for a friend new letter box, to illuminate with Leds over letterbox numbers.
    I want to make this circuit in 12 volts version.
    I will upgrade the following:
    Solar Panel to 12 volts @ 10 Watts http://www.rpc.com.au/catalog/cnpv-12volt-solar-panel-10watt-p-2701.html
    Battery 12 volts @ 15Ah
    I will change the BD140 for TIP32 as the TIP32 is to energised a 12 volt 3 Amp relay, as I'm running more Leds then below.
    Do need to beef up the circuit anywhere else.
    Any help would be greatly appreciated
    [​IMG]
     
  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi A45,
    How many leds more.? and type.?

    When you say 12V, 3A relay, do you mean a 3Amp contact set.?

    E
     
  3. ScottWang

    Moderator

    Aug 23, 2012
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    The TS was shown the LED type in the circuit.
     
  4. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    Hi Scott,
    the TS said:
    as I'm running more Leds then below

    Which suggests more LED's and a possible change of LED type.
    E
     
  5. ScottWang

    Moderator

    Aug 23, 2012
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    Yes, so you just waiting for his reply, and I'm going to give him other suggestion.
     
  6. Aurion45

    Thread Starter Member

    Aug 7, 2011
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    53 @ 3.6 volts 20mA Leds, arranged as below.
    3 amp contacts on relay Yes you're right.

    Solution: 3 x 17 array, 2 extra LEDs =53 leds
    3 x 17 leds
    1 x 2 leds
    [​IMG] [​IMG] [​IMG] [​IMG] [​IMG] [​IMG] R = 68 ohms
    [​IMG] [​IMG] [​IMG] [​IMG] [​IMG] [​IMG] R = 270 ohms
    The wizard says: In solution 0:
    • each 68 ohm resistor dissipates 27.2 mW
    • the wizard thinks 1/4W resistors are fine for your application [​IMG]
    • the 270 ohm resistor dissipates 108 mW
    • the wizard thinks 1/4W resistors are fine for your application [​IMG]
    • together, all resistors dissipate 570.4 mW
    • together, the diodes dissipate 3816 mW
    • total power dissipated by the array is 4386.4 mW
    • the array draws current of 360 mA from the source.
     
  7. ScottWang

    Moderator

    Aug 23, 2012
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    When the current reach up to 100mA then it will be generate the heat, and I guess your application will be get over more, so if you considering the heat then may be using the P mosfet is another options or using a pnp and one n mosfet.
     
  8. Aurion45

    Thread Starter Member

    Aug 7, 2011
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    I don't understand, as I'm not very good at electronics?
    I will draw up a new circuit tomorrow then ask if it is OK.
     
  9. ScottWang

    Moderator

    Aug 23, 2012
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    1. Why you still want to use a relay when the transistor can do the job, but maybe you will need a heat sink for the transistor.

    2. if you want to using the relay then the transistor will no need so big, the 2SA684(1A) or similar transistor can do the job.

    3. If you using the mosfet has low Rds then may be you don't need the heat sink and relay, because it will reducing the power consumption, you just need to care about the power consumption of LED.
     
  10. Aurion45

    Thread Starter Member

    Aug 7, 2011
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    OK if I use a mosfet, how do I replace it for the transistor? what type of Mosfet, eg device number?
    I've no clue how wire this in the circuit.
     
  11. ScottWang

    Moderator

    Aug 23, 2012
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    The circuit is easy, I can do for you, also other members can do, now just waiting for your circuit and total current, and is that just 360 mA?
     
  12. Aurion45

    Thread Starter Member

    Aug 7, 2011
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    Yes as above result solution!
     
  13. Wendy

    Moderator

    Mar 24, 2008
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  14. Aurion45

    Thread Starter Member

    Aug 7, 2011
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    Can I use the Mosfet from terminal 3 of the 555 timer?
    What type of Mosfet is it? as in device name, so I can buy it, please don't forget need to at least 1 amp, so I can add more Leds if I want too?
    [​IMG]
     
  15. ScottWang

    Moderator

    Aug 23, 2012
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    Two kinds of circuit using N/P mosfet, the n mosfet is more easy to get one to suit your need, if you insist to use the relay is ok, just don't forget to using the 2SA684 to replace the TIP32.

    The below are the p mosfet to suit your need, if you choose p mosfet then you should concern the price and which one you can get.
    IRF7204_Pch_20V5.3A_Vgs10V_60mΩ_ICtype.pdf
    APM4435_Pch_30V8A_Vgs4.5V24mΩ_10V16mΩ.pdf
    APM9435_Pch_3B0V4.6A_Vgs4.5V_80mΩ.pdf
    IRF7606_Pch_30V3.6A-Vgs4.5V1.2A_Vgs10V2.4A_90mΩ.pdf

    ChangeThisCircuitTo12V-01_Aurion45_ScottWang.gif

    The below are the n mosfet to suit your need.
    NTMS7N03R2_Nch_30V7A_Vgs10V_23mΩ.pdf
    APM4435_Pch_30V8A_Vgs4.5V24mΩ_10V16mΩ.pdf
    APM9435_Pch_30V4.6A_Vgs4.5V_80mΩ.pdf
    DMN3150L_Nch_30v3.8A_72mΩ.pdf

    ChangeThisCircuitTo12V-02_Aurion45_ScottWang.gif

    You can also go to the online store to choose and buy.
     
  16. Aurion45

    Thread Starter Member

    Aug 7, 2011
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    0
    Last edited: Jul 9, 2015
  17. Wendy

    Moderator

    Mar 24, 2008
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    With a little mod, you could have a joule theif approach, and it will not run out of battery power. If you are interested I will draw a schematic, but I have to be asked.
     
  18. Aurion45

    Thread Starter Member

    Aug 7, 2011
    62
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    Hi Wendy,
    The more the merrier, sound interesting, Yes please, draw a schematic???????????? 12 volts battery
    Andrew
     
  19. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi A45.
    As the 555 is now running on 12V, it will be necessary to modify the P1 and LDR , light/dark sensing.
    What is the resistance of the LDR when and dark and light.?
    E
     
  20. Aurion45

    Thread Starter Member

    Aug 7, 2011
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    0
    Got No idea, as I did not noticed the value missing in the schematic, can we use 10M LDR? also will it work for the existing circuit above "ScottWang" made?
     
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