Breadboard power supply design

Discussion in 'The Projects Forum' started by Russell_AH, Aug 29, 2013.

  1. Russell_AH

    Thread Starter New Member

    Mar 29, 2013
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    (The title was supposed to be "Breadboard Power Supply Design", but was over 20 characters and causing errors).

    I’d like feedback on the attached design for a breadboard power supply circuit. The idea is for it to work with a variety of transformers/wall warts, although at the moment I plan to mostly use a 16Vac 13VA Hornby transformer. I’d like to use components I already have as much as possible.

    About the design:

    I’ve added two sockets, so it can take 2.1mm or 2.5mm plugs, AC or DC.

    For the diode bridge I would use 1N4001 diodes (these can take about 1A, which is more than any of my supplies are rated at).

    C5 smoothing capacitor: from here: I guess I’d need about 800μF for a 9V regulator or 1.3mF for a 12V regulator (assuming I use the 16Vac transformer mentioned above).

    R1 =2K to limit current to “on-off” LED

    C1 and C2: 0.33 μF and 0.1 μF respectively (as shown in the regulator data sheets)

    78xx: most likely a 9V or 12V regulator. I’ll probably use a female 3-pin header socket so I can easily change them around.

    The Arduino connection will just be a barrel jack to power my Arduino board

    C3 and C4: same as C1 and C2

    The 7805 would provide a 5V signal to use for logic ICs etc

    The two DPDT (centre off) switches toggle which voltage (if any) each power rail is connected to, and the LEDs indicate which one is connected.

    R4, R5: about 100/200Ω

    Can anyone see any fault with the design or values/component choices I’ve made? Any ways to improve it?

    Thanks.
     
  2. #12

    Expert

    Nov 30, 2010
    16,261
    6,768
    Make R4&5 220 ohms and correct the wiring at the bottom right LED and S3.

    The filter capacitor size is not dependent on voltage, they are dependent on current.

    √2 C Er F = I
    C = I/(√2 Er F)
    where:
    Er = the peak to peak voltage drop between power line voltage pulses,
    F is twice the power line frequency
    I is current in amps
    Capacitance is in farads

    Set the capacitor size so that the input voltage does not drop below the voltage required to run the first regulator chip.
     
  3. Russell_AH

    Thread Starter New Member

    Mar 29, 2013
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    Using a peak voltage of 21V (after rectifying) and a minimum voltage of 14.5V (for the 7812) and 11V (for the 7809) supplying 0.8A, I get about 870μF and 570μF respectively. This is the same as I listed above except for the factor of √2 (where does that come from?).

    However, it's unlikely I'll be drawing that much current, so I would probably be able to get away with a smaller capacitor there. Is there any disadvantage to using a larger capacitor here?
     
  4. #12

    Expert

    Nov 30, 2010
    16,261
    6,768
    The√2 comes from the National Semiconductor Voltage Regulator Handbook for 1982, chapter 8, page 6, bottom of the left column.

    Using a larger capacitor than necessary has its own problems, but only if you get ridiculous. Time and a half or double what you calculate as the minimum will work just fine. After all, large electrolytic capacitors are often rated at +20%/-10%
     
  5. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    I would be sure to add current limiting, which will save your breadboard and IC's almost every day. :)

    My breadboard PSU has a switch for 20/50/100mA limit. The switch is just a DPDT centre-off toggle switch; 100-20-50 and the current limiter is as easy as a LM317 and three resistors (and the switch!) which is placed before the 7805 regulator.

    If you use a 78L05 regulator (note the "L") it's a smaller size and limited to about 150mA for its own inbuilt overcurrent limit. That can give you someprotection but 150mA is still enough to blow up a lot of devices if you make a wiring error.

    Mine gets used on 20mA the vast majority of the time, it's perfect for testing LEDs and running most simple circuits like a logic IC or a 555 IC or microcontroller etc.
     
  6. Russell_AH

    Thread Starter New Member

    Mar 29, 2013
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    Thanks you two.

    I'll think about current limiting. I can see how to use the LM317 as a current limiter, but I can't quite see how you would use it with the DPDT switch. Would you be able to elaborate a little please?

    Also, another thought. Would I need to use a heat sink on the 78xx?
     
  7. wayneh

    Expert

    Sep 9, 2010
    12,094
    3,033
    Good advice. My version of "current limiting" is to run most experiments at 5V. Even if I intend the final circuit to use 9V or 12V or whatever, I do as much of the early work as possible at 5V. A lot of mistakes that would be fatal to the ICs at 12V might be briefly survived at 5V.
     
  8. Wendy

    Moderator

    Mar 24, 2008
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  9. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
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    It's very easy, the hardest part is mounting the LM317 and mounting the switch.

    You connect the LM317 before the 7805, with a resistor from Vout to Vadj and then Vadj goes to the Vin pin of the 7805.

    The resistor sets the current limit, at a reference of 1.25v, so a 1 ohm resistor sets current max to 1.25A, a 10 ohm resistor sets current max to 0.125A. (I = 1.25v/R)

    I used a DPDT centre-off switch, so in the centre position the switch does nothing and the current is set to 20mA by the resistor soldered on the LM317.

    In the two outer positions, the switch adds a resistor in parallel to add another 30mA (now is 50mA total) of adds another 80mA (100mA total).

    You need to allow for 5-7mA wasted in the 7805, so the easiest way is to test by trying different resistor values so you get either the short circuit current you want, or the dropout current (current where the 7805 will drop below 5.0v). They will be slightly different.

    From memory, mine delivers about 23mA into a short circuit, and the 7805 drops out at about 20mA.

    It sounds complex but is the simplest thing to add, and is well worth it! :)
     
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